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UNIT I
1
THE MEAN, MEDIAN, MODE, AND
OTHER MEASURES OF CENTRAL
TENDENCY
Unit structure
1.0 Objective s
1.1 Introduction
1.2 Index or Subscript , Notation , Summation Notation
1.3 Averages or Measures of Central Tendency
1.4 Arithmetic Mean
1.4.1 Arithmetic Mean Computed from Grouped Data
1.4.2 Properties of the Arithmetic Mean
1.5 Weighted Arithmetic Mean
1.6 Median
1.7 Mode
1.8 Empirical Relation between the Mean, Median, and Mode
1.9 Geometric Mean
1.10 Harmonic Mean
1.11 Relation between the Arithmetic , Geometric, and Harmonic Means
1.12 Root Mean Square
1.13 Quartiles, Deciles and Percentiles
1.14 Software and Measures of Central Tendency
1.15 Summary
1.16 Exercise
1.17 References
1.0 OBJECTIVES
After going through this chapter, students will able to learn
To present huge data in a summarized form
To calculate and interpret the mean, the median and the mode,
To facilitate comparison
To calculate geometric mean, harmonic mean
To trace precise relationship munotes.in
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To calculate Quartiles, deciles and percentiles
To help in decision -making
1.1 INTRODUCTION
A measure of central tendency is a single value that describes a set of data
by identifying the central position within that set of data. Mean, median
and mode are different measures of central tendency in a numerical data.
The word average is commonly used in day to day conversation like we
often talk about average height of the girls, average student of the class,
and average run rate of the match. When we say average means neither too
good nor b ad. However, in statistics the term average has different
meaning. Average is a single value which representing a group of values
so such a value easy to understand, easy to compute and based on all
observations.
1.2 INDEX OR SUBSC RIPT, NOTATION, SUMMATION NOTATION
Let the symbol X i (read „X subscript i) denote any of the N values X 1, X2,
X3, …….X N assumed by a variable X. The letter i in X i which can stand
for any of the numbers 1, 2, 3, ……, N is called a subscript or index. Any
letter other than i such as j, k, p, q or r could be used also.
Summation Notation:
The symbol ∑
is used to denote the sum of all the ‟s from i = 1 to
N.
∑
= X1 + X 2 + X 3 + ……. + X N
We generally denote this sum simply by ∑ , ∑ .
The symbol ∑ is the Greek capital letter sigma denoting sum.
Ex. ∑
= aX1 + aX 2 + aX 3 + ……. + aX N
= a (X1 + X 2 + X 3 + ……. + X N) = a ∑
, where a
is a constant.
1.3 AVERAGES OR MEASURES OF CENTRAL TENDENCY
There are different ways of measuring the central tendency of a set of
values .
Various authors defined Average differently.
“Average is an attempt to find one single figure to describe whole of
figures.” – Clark munotes.in
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“An average is a single value selected from a group of values to represent
them in some way - a value which is supposed to stand for whole group, of
which it is a part, as typical of all the values in the group.” – A. E. Waugh
“An average is a typical value in the sense that it is sometimes employed
to represent all the individual values in the series or of a variable.” – Ya-
Lun-Chou
Types of Averages:
Arithmetic Mean: a. Simple, b. W eighted
Median
Mode
Geometric Mean
Harmonic Mean
1.4 ARITHMETIC MEAN
The most popular and widely used measure of representing the entire data
by one value is mean or Average.
It simply involves taking the sum of a group of numbers, then dividing
that sum by the total number of values in the group.
Arithmetic mean can be of two types.
a. Simple arithmetic mean
b. Weighted arithmetic mean
A. Simple Arithmetic Mean – Individual Observations:
Calculation of mean in case of individual observations [ i. e. when
frequencies are not given] is very simple. Here, we add all values of the
variable and divide the total by the number of items.
̅ =
= ∑
̅ = Arithmetic Mean; N = number of observations;
∑ = sum of all the values of the variable X i. e.
Ex 1. Find the Arithmeti c mean of following five values 8, 45 , 49, 54, 79.
Sol: We know that, ̅ =
̅ =
=
= 47
Ex 2. Find the Arithmetic mean of following values.
4350, 7200, 6750, 5480, 7940, 3820, 5920, 8450, 4900, 5350.
Sol: We know that, ̅ =
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̅ =
=
=
6416
Short cut method: ̅ = A + ∑
Where A is assumed mean and d is deviation of items from assumed mean
i. e. d = ( ).
Ex 3. Calculate arithmetic mean from following data.
2690, 3670, 4580, 5660, 2750, 2830, 4100, 572 0, 50 40, 4840
Sol: X ( X-A) 2690 -2310 3670 -1330 4580 -420 5660 660 2750 -2250 2830 -2170 4100 -900 5720 720 5040 40 4840 -160 ∑ = -8120
Consider assumed mean, A = 5000
̅ = A + ∑
= 5000 -
=4188
1.4.1 Arithmetic Mean Computed from Grouped Data:
Simple Arithmetic Mean – Discrete series:
Calculation of mean in case of frequencies are given,
̅ = ∑
f = Frequency;
X = the variable
N = Total number of observations i.e. ∑
Here, first multiply the frequency of each row with variable and obtain the
total ∑ and then divide the total by number of observations, i.e. total
frequency.
Ex 4. Following are the marks obtained by 60 students. Calculate
arithmetic mean.
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Marks 15 30 45 60 70 80 No. of students 6 14 15 15 4 6
Sol: Let the marks denoted by X and number of students denoted by f.
Marks X No. of Students f fX 15 6 90 30 14 420 45 15 675 60 15 900 70 4 280 80 6 480 N = 60 ∑ = 2845
̅ = ∑
=
= 47.42
Short cut method: ̅ = A + ∑
Where A is assumed mean and d is deviation of items from assumed mean
i. e. d = ( ),
N = ∑
Ex 5. Calculate arithmetic mean by the short cut method using data from
Ex. 4
Sol: Marks X No. of Students f d = ( ) fd 15 6 -30 -180 30 14 -15 -210 45 15 0 0 60 15 15 225 70 4 25 100 80 6 35 210 N = 60 ∑ = 145
Assumed mean, A = 45
̅ = A + ∑
= 45 +
=47.4166
Simple Arithmetic Mean – Continuous Series:
̅ = ∑
m = mid-point of various classes ; f = the frequency of each class;
N = the total frequency
Here, first obtain the mod -point of each class and denote it by m. munotes.in
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Multiply the se mid -points by the respective frequency of each class and
obtain the total ∑
Divide the total by the sum of the frequency, i.e. N.
Ex 5. From the following data compute arithmetic mean.
Marks 0-10 10-20 20-30 30-40 40-50 50-60 No. of students 5 10 25 30 20 10
Sol:
Marks Mid- point m No. of Students f fm 0-10 5 5 25 10-20 15 10 150 20-30 25 25 625 30-40 35 30 1050 40-50 45 20 900 50-60 55 10 550 N = 100 ∑ = 3300
̅ = ∑
=
= 33
Ex 6 . From the following data compute arithmetic mean.
Class Intervals 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Frequency 4 4 7 10 12 8 5
Sol:
Marks Mid- point m No. of Students f f m 0-10 5 4 20 10-20 15 4 60 20-30 25 7 175 30-40 35 10 350 40-50 45 12 540 50-60 55 8 440 50-60 65 5 325 N = 50 ∑ = 1910
̅ = ∑
=
= 38.2
Short cut method: ̅ = A + ∑
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Where A is assumed mean and d is deviation of items from assumed mean
i. e. d = ( ), m= mid point , N = ∑
Ex 7. Calculate arithmetic mean by the short cut method using data from
Ex. 5.
Sol: Marks Mid- point m No. of Students f d = ( ) fd 0-10 5 5 -30 -150 10-20 15 10 -20 -200 20-30 25 25 -10 -250 30-40 35 30 0 0 40-50 45 20 10 200 50-60 55 10 20 200 N = 100 ∑ = 200
Assumed mean, A = 35
̅ = A + ∑
= 35 -
= 33
1.4.2 Properties of the Arithmetic Mean:
1. The sum of deviation from their arithmetic mean is always equal to
zero.
Symbolically, ∑( ̅ )= 0
Ex 8:
X 10 20 30 40 50 ∑ = 150 X - ̅ -20 -10 0 10 20 ∑ ̅= 0
̅ ∑
=
= 30
When we calculate the deviations of all the items from their arithmetic
mean ( ̅ =30), we find that the sum of the deviations from the arithmetic
mean i. e. ∑( ̅ )= 0
2. The sum of squared deviations of the items from arithmetic mean is
minimum, that is, less than the sum of squared deviations of the items
from any other value.
Ex 9 : X X - ̅ (X - 4)2 2 -2 4 3 -1 1 4 0 0 munotes.in
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5 1 1 6 2 4 ∑ = 20 ∑ ̅= 0 ∑( )̅̅̅ 2= 0
̅ ∑
=
= 4
The sum of the squared deviations is equal to 10 in the above example. If
the deviations are taken from any other value the sum of the squared
deviations are taken from any other value the sum of the squared
deviations would be gre ater than 10.
Let us calculate the squares of the deviations of item from the value less
than the arithmetic mean, say 3
X X - 3 (X - 3)2 2 -1 1 3 0 0 4 1 1 5 2 4 6 3 9 ∑ = 20 ∑( )2= 0
3. Arithmetic mean is NOT independent of change of origin.
If each observation of a series is increased (or decreased) by a constant,
then the mean of these observations is also increased (or decreased) by
that constant.
4. Arithmetic mean is NOT independent of change of scale.
If each observation of a series is multiplied (or divided) by constant, then
the mean of these observations is also multiplied (or divided) by that
constant.
5. If arithmetic mean and number of items of two or more related groups
are given, then we can compute the combined mean using the formula
given below.
̅12 = ̅̅̅̅ ̅̅̅̅
,
Where
̅12 = Combined mean of two groups ;
N1 = Number of items in the first group ; N2 = Number of items in the
second group
̅̅̅̅ = Ari thmetic mean of the first group; ̅̅̅̅ = Arithmetic mean of the
second group
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1.5 WEIGHTED ARITHMETIC MEAN
Arithmetic mean gives equal importance to all the items. When
importance of the items are not same, in these cases we compute weighted
arithmetic mean. The term weighted represents to the relative importance
to the item.
̅w =
= ∑
∑
Where
̅w represent the weighted arithmetic mean; X represent the variable
values i. e. X 1, X2 …… Xn
W represent the weights attached to the variable values i. e. w 1, w2 ……
wn respectively.
To calculate weighted arithmetic mean, multiply the w eight by the
variable X and obtain the total ∑ . Then divide this total by the sum of
the weights, i.e. ∑
In case of frequency distribution, if f1, f2. ….. . fn are the frequencies of the
variable values X 1, X 2,……X n respectively then the weighted arithmetic
mean is given by
̅w = ( ) ( ) ( )
̅w = ∑ ( )
∑
Note: Simple arithmetic mean shall be equal to the arithmetic mean if the
weights are equal.
Ex. 10 Calculate the weighted mean for following data.
X 1 2 5 7 W 2 14 8 32
Sol:
X W WX 1 2 2 2 14 28 5 8 40 7 32 224 ∑ = 56 ∑ = 294 munotes.in
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̅w = ∑
∑ =
= 5.25
Ex. 11 Calculate the weighted mean for following data.
Wages per Day ( X ) 200 150 85 No. of workers ( W ) 25 20 10
Sol:
̅w = ∑
∑ =
= 160.90
Ex. 12. Calculate the weighted mean for following data and compare it
with arithmetic mean
Subject Weight Student X Y Z Physics 2 72 42 52 Chemistry 3 75 52 62 Biology 5 58 88 68
Sol: For Student X,
Arithmetic Mean, ̅X = ∑
=
=
= 67.67
Weighted Arithmetic Mean, ̅wX = ∑
∑ = ( ) ( ) ( )
=
=
= 65.9
For Student Y,
Arithmetic Mean, ̅y = ∑
=
=
= 60.67
Weighted Arithmetic Mean, ̅wY = ∑
∑ = ( ) ( ) ( )
=
=
= 68
For Student Z,
Arithmetic Mean, ̅Z = ∑
=
=
= 60.67 X W WX 200 25 5000 150 20 3000 85 10 850 ∑ = 55 ∑ = 8850 munotes.in
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Weighted Arithmetic Mean, ̅wZ = ∑
∑ = ( ) ( ) ( )
=
=
= 63
1.6 MEDIAN
Median is a middle value in the distribution. Median is a numeric value
that separates the higher half of a set from the lower half . It is the value
that the number of observations above it is equal to the number of
observations below it. The median is thus a positional average.
For example, if the salary of five employees is 6100, 7150, 7250, 7500 and
8500 the median would be 7250.
When odd number of observations are there then the calculations of
median is simple. When an even number of observations are given, there
is no single middle position value and the median is taken to be the
arithmetic mean of two middlemost items.
In the above example we are given the salary of six employees as 6100 ,
7150, 7250, 7500, 8500 and 9000, the median salary would be
Median =
=
= 7375
Hence, in case of even number of observations median may be found by
averaging two middle position values.
Calculations of Median – Individual Observation s:
Arrange the data in ascending or descending order of magnitude.
In a group composed of an odd number of values, add 1 to the total
number of values and divide by 2 gives median value.
Median = size of
th item
Ex. 13 From the fol lowing data, compute the median:
15, 9, 7, 23, 25, 25, 42, 25, 16, 14, 58, 25, 31
Sol: Arrange the numbers in ascending order 7, 9, 14, 15, 16, 23, 25, 25,
25, 25, 31, 42, 58
Median = = size of
th item =
= 7th item = 25
Median = 25
The procedure for calculating median of an even numbered of items is not
as above. The median value for a group composed of an even number of
items is the arithmetic mean of the two middle values – i.e. adding two
values in the middle and dividing by 2
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Ex. 14 From the fol lowing data, compute the median:
451, 502, 523, 512, 622, 612, 754, 732, 701, 721
Sol: Arrange the numbers in ascending order
451, 502, 512, 523, 612, 622, 701, 721, 732, 754
Median = = size of
th item =
= 5.5th item
Size of 5.5th item =
=
Median = 617
Calculations of Median – Discrete Series:
Steps:
1. First arrange the data in ascending or descending order.
2. Find out the cumulative frequencies.
3. Apply formula : Median = size of
4. Find out total in the cumulative frequency column which is equal to
or next higher to that value and determine the value of the
variable corresponding to it. That gives the median value.
Ex. 15 From the following data, find the value of median.
Income ( Rs.) 450 500 630 550 710 580 No. of persons 29 31 21 25 11 35
Sol: Income ( Rs.) Ascending order No. of persons f Cumulative Frequency c.f. 450 29 29 500 31 60 550 25 85 580 35 120 630 21 141 710 11 152 Median = = size of
th item =
= 76.5th item
Size of 76.5th item = Rs. 550 It is median income.
Calculations of Median – Continuous Series:
The following formula is used to calculate median for continuous series.
Median = L + ⁄
x i
L = Lower limit of median class; f = Simple freq. of the median class;
c.f. = Cumulative freq. of the preceding the median class;
i= Class interval of the median class
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Ex. 16 From the following data, find the value of median.
Marks 70-80 60-70 50-60 40-50 30-40 20-30 10-20 No. of students 10 15 26 30 42 31 24
Sol: Arrange the data in ascending order
Marks f c.f. 10-20 24 24 20-30 31 55 30-40 42 97 40-50 30 127 50-60 26 153 60-70 15 168 70-80 10 178
Median = size of
item =
= 89th item
Median lies in the class 30-40 (marked in pink)
Median = L + ⁄
x i
L = 30.
= 89, c.f. = 55, f = 42, i = 10
Median =30 +
x 10
= 30 + 8.09 = 38.09
1.7 MODE
The mode or the modal value is that value in a series of observations
which occurs with the greatest frequency.
For example, the mode of the values 4, 6, 9, 6, 5, 6, 9, 4 would be 6.
Calculations of Mode – Discrete Series:
Ex. 1 7 From the following data, find the value of m ode.
Size of cloth 28 29 30 31 32 33 No. of persons wearing 15 25 45 70 55 20
Sol: The mode or modal size is 31 because the value 31 occurred
maximum number of times.
Calculations of Mode – Continuous Series:
The following formula is used to calculate mode for continuous series.
Mode =
x i ,
L = Lower limit of modal class; f1 = freq. of the modal class; munotes.in
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fo= freq. of the class preceding the m odal class;
f2= freq. of the class succeeding the modal class;
i= Class interval of the modal class
Ex. 1 8 From the following data, find the value of mode. Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No. of students 3 5 7 10 12 15 12 6 2 8
Sol: After observing the table, modal class is 50-60
Mode =
x i ,
= 50 +
x 10
= 50 +
x 10 = 55
1.8 EMPIRICAL RELATION BETWEEN THE MEAN, MEDIAN, AND MODE
Karl Pearson has expressed the relationship between mean, median and
mode as follows:
Mode = Mean – 3 [Mean – Median]
Mode = 3 Median – 2 Mean
If we know any of the two values out of the three, we can compute third
from these relationships.
1.9 GEOMETRIC MEAN
Geometric mean of a set of n observations is the nth root of their product.
G. M. = √( )( )( ) ( ) .
G. M. of 3 values 2, 4, 8 would be
G. M. = √ = √ = 4
For calculation purpose, take the logarithm of both sides
log G. M.=
log G. M.= ∑
G. M. = Antilog [∑
In Discrete series, G. M. = Antilog [∑
In Continuous series, G. M. = Antilog [∑
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Calculations of Geometric Mean – Discrete/ Individual Series:
Ex. 1 9 Daily income of ten families of a particular place is below.
Calculate Geometric Mean . 85 70 15 75 500 8 45 250 40 36
Sol:
X log X 85 1.9294 70 1.8451 15 1.1761 75 1.8751 500 2.6990 8 0.9031 45 1.6532 250 2.3979 40 1.6021 36 1.5563 ∑ log X 17.6373
G. M. = Antilog [∑
= Antilog [∑
= Antilog (1.7637) = 58.03
Calculations of Geometric Mean – Continuous Series:
Ex 20 . Calculate Geometric Mean from following data.
Marks 4-8 8-12 12-16 16-20 20-24 24-28 28-32 32-36 36-40 Frequency 8 12 20 30 15 12 10 6 2
Sol: Marks m.p (m) f log m f log m 4-8 6 8 0.7782 6.2256 8-12 10 12 1.0000 12.0000 12-16 14 20 1.1461 22.922 16-20 18 30 1.2553 37.6590 20-24 22 15 1.3424 20.1360 24-28 26 12 1.4150 16.9800 28-32 30 10 1.4771 14.7710 32-36 34 6 1.5315 9.1890 36-40 38 2 1.5798 3.156 N= 115 ∑ f log m= 143.0386 munotes.in
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G. M. = Antilog [∑
= Antilog [∑
= Antilog (1.2438) = 17. 53
1.10 HARMONIC MEAN
Harmonic mean of a number of observations, none of which is zero, is the
reciprocal of the arithmetic mean of the reciprocals of the given values.
Thus, harmonic mean (H. M.) of n observations x i, i = 1, 2, ….,n is given
by,
H. M. =
∑
=
Calculations of Harmonic Mean – Individual Observations:
Ex. 21 Find the harmonic mean of 4, 36, 45, 50, 75.
Sol: H. M. =
=
=
= 15
Calculations of Harmonic Mean – Discrete Series:
Formula for harmonic mean in Discrete series,
H. M. =
∑
=
∑
Ex. 22 From the following data, Find the harmonic mean.
Marks 10 20 30 40 50 No. of students 20 40 60 30 10
Sol: Marks X f f/X 10 20 2 20 40 2 30 60 2 40 30 0.75 50 10 0.20 N = 160 ∑ ⁄
H. M. =
∑
=
= 23.0215 munotes.in
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Calculations of Harmonic Mean – Continuous Series:
Formula for harmonic mean in continuous series,
H. M. =
∑
Ex. 23 From the following data, compute the value of harmonic mean.
Class interval 10-20 20-30 30-40 40-50 50-60 Frequency 6 8 12 9 5
Sol: Class Interval Mid point (m) f f/m 10 – 20 15 6 0.40 20 – 30 25 8 0.32 30 – 40 35 12 0.3428 40 – 50 45 9 0.2 50 - 60 55 5 0.0909 N = 40 ∑ ⁄
H. M. =
∑
=
= 29.54
1.11 RELATION BETWEEN THE ARITHMET IC, GEOMETRIC AND HARMONIC MEAN
Arithmetic mean is greater than geometric mean and geometric mean is
greater than harmonic mean.
A.M. G. M H. M.
The quality signs hold only if all the numbers X 1, X 2, X 3,…. X n are
identical.
1.12 ROOT MEAN SQUARE
The root mean square (RMS) is defined as the square root of the mean
square (the arithmetic mean of the squares of a set of numbers). It is also
called as the Quadratic average. Sometimes it is denoted by √ ̅ and
given by,
RMS = √ ̅ = √∑
= √∑
It is very useful in fields that study sine waves like electrical engineering.
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Ex. 24 Find RMS of 1, 3, 5, 7 and 9
Sol: RMS = √∑
= √
= √
= √ = 7.28
1.13 QUARTILES, DECILES AND PERCENTILES
From the definition of median that it‟s the middle point which divides the
set of ordered data into two equal parts. I n the same way we can divide the
set into four equal parts and this called quartiles. These values denoted by
Q1, Q 2 and Q 3, are called the first, second and the third quartile
respectively. In the same way the values that divide the data into 10 equal
parts are called deciles and are denoted by D 1, D 2, …., D 9 whereas the
values dividing the data into 100 equal parts are called percentiles and are
denoted by P 1, P 2, …., P 99. The fifth decile and the 50th percentile
corresponds to median.
Formulae:
Quartile :
For individual observations, Qi =(
). No. of observation , i= 1, 2, 3
For discrete series, Qi =(
). N, N= ∑ and i= 1, 2, 3
For continuous series, Q i = L +
. c,
Where, i= 1, 2, 3 , c = size of class interval.
L = Lower limit of the class interval in which lower quartile lies,
f = freq. of the interval in which lower quartile lies,
cf = cumulative freq. of the class preceding the quartile class,
Deciles:
For individual observations, Di =(
). No. of observation, i= 1, 2, …, 9
For discrete series, D i =(
). N, N= ∑ and i= 1, 2,…., 9
For continuous series, D i = L +
. c, i= 1, 2, …, 9
Percentiles:
For individual observations, Pi =(
). value of observation, i= 1,2,…, 99
For discrete series, Pi =(
). N, N= ∑ and i= 1, 2,…., 99
For continuous series, P i = L +
. c, i= 1, 2, …, 99
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Ex. 25 Find the quartiles Q 1, Q3 , D 1, D 5, D 8, P 8, P 50 and P 85 of the
following data 20, 30, 25, 23, 22, 32, 36.
Sol: Arrange data in ascending order, n = 7 i.e. odd number
20, 22, 23, 25, 30, 32, 36
q1 = (
).7 = 1.75 q1 = 2 Q1= 22
q3 = (
).7 = 5 .75 q2 = 6 Q3= 32
d1 = (
).7 = 0.7 d1 = 1 D1= 20
d5 = (
).7 = 3.5 d5 = 4 D5= 25
d8 = (
).7 = 5.6 d8 = 6 D8= 32
p8 = (
).7 = 0.56 p8 = 1 P8= 20
p50 = (
).7 = 3.5 p50 = 4 P50= 25
p85 = (
).7 = 5.95 p85 = 6 P85= 32
Ex. 26 Find Q 1, Q3, D4, P27 for the following data.
X 0 1 2 3 4 5 6 7 8 f 1 9 26 59 72 52 29 7 1 c.f. 1 10 36 95 167 219 248 255 256
Sol. We know that, Qi =(
). N
Q1 = (
).256 = 64 and c.f. just greater than 64 is 95. Hence Q1 = 3
Q3 = (
).256 = 192 and c.f. just greater than 192 is 219. Hence Q3 = 5
D4 =(
).256 = 102.4 and c.f. just greater than 102.4 is 167. Hence D4 = 4
P27 =(
).256 = 69.12 and c.f. just greater than 69.12 is 95. Hence P27 = 3
Ex. 2 7 Find Q 1, Q3, D2, P90 for the following data.
Marks Below 10 10-20 20- 40 40-60 60-80 Above 80 No. of students 8 10 22 25 10 5
Sol: We know that, Qi = L +
. c, Marks Below 10 10-20 20- 40 40-60 60-80 Above 80 f 8 10 22 25 10 5 cf 8 18 40 65 75 80 munotes.in
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Q1 = Size of (N/4)th item = size of (80/4)= 20th item. Q 1 lies in the class
20-40.
L=20, N/4 = 20, cf = 18, f = 22 and c = 20
Q1 = 20 + {(20 – 18)/22}* 20 = 20 + 1.82 = 21.82
Q3 = Size of (3N/4)th item = size of (3*80/4)= 60th item. Q 3 lies in the
class 40 -60.
L=40, 3N/4 = 6 0, cf = 40, f = 25 and c = 20
Q3 = 40 + {( 60 – 40)/25}* 20 = 56
D2 = Size of (2N/10)th item = size of (2*80/10)= 16th item. D 2 lies in the
class 10 -20
L=10, 2N/10 = 16, cf=8, f = 10 and c=10
D2 = 10 + {(16 –8)/10}*10 = 18
P90 = Size of (90N/100)th item = size of (90*80/10)= 72th item. P 90 lies in
the class 60 -80.
L=60, 90N/100 = 72, cf=65, f = 10 and c=20
P90 = 60 + {(72 –65)/10}*20 = 74.
1.14 SOFTWARE AND MEASURES OF CENTRAL TENDENCY
There are many software available to calculate measures of central
tendency. We can use Excel to calculate the standard measures of central
tendency (mean, median and mode). In Microsoft Excel, the mean can
calculated by using one of the functions like AVERAGE, AVERAGEA,
AVERAGEIF, AVERAGEIFS. The mean can be calculated by using th e
MEDIAN function. We can calculate a mode by using the MODE
functio n, GEOMEAN to calculate geometric mean and HARMEAN to
calculate harmonic mean.
We can use SPSS to calculate the standard measures of central tendency
(mean, median and mode). We can get S PSS to compute mean, median
and mode in the command submenu. Go to the Statistics menu, select the
Analyse submenu, and then the Descriptive Statistics submenu and then
the Frequencies option. We can use MINITAB to calculate the standard
measures of centra l tendency using the functions M ean, Median, Mode
ang GMEAN. To compute these go to Stat -Tables -Descriptive statistics.
Using R software one could easily obtain the value of the mean using
summary function.
We could find median value using summary function in R. The
randomForest library can be used to impute the missing values using
Median for numeric variables. Mode is used for missing value imputation
for categorical variables using randomForest library in R. Model can be
easily located graphically. You shouldn‟t be surprised that the R‟s mode munotes.in
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function (mode ()) does not provide a model value. It shows the datatype
of the particular variable which does not comply with our standard
expectation. So how one would find mode using R software? We need to
use table function for finding mode. As you know the table function in R
provides frequency distribution of the variable. Thus the value with
highest frequency is a modal value.
Geometric mean is the only average that is recommended for finding
average growth (decline) rates. It is defined as the nth root of the product
of n terms. Since it is defined in product terms so the observation
shouldn‟t be having zero or negative values. We don‟t have a built-in
function in R for its computation but one could find it by using its formula
directly in R platform.
1.15 SUMMARY
A measure of central tendency is a measure that tells us where the middle
of a group of data lies. Mean, median and mode are the most important
measu res of central tendency. The complete dataset may be represented by
these values. It is not necessary for mean, median and mode to have the
same values. Mean is sensitive to extreme data values. Median is a better
way to understand skewed distribution than mean. It is possible that there
is no mode in the data. Mean and median cannot be zero unless all data
values are zero.
1.16 EXERCISE
1. Find the arithmetic mean of the following distribution:
X 10 30 50 70 89 f 7 8 10 15 10
2. Find the arithmetic mean of the following distribution:
X 3 9 12 14 15 17 f 1 3 4 1 4 2
3. Find the arithmetic mean of the following data.
Class Interval 15-25 25-35 35- 45 45- 55 55-65 65-75 75-85 Frequency 6 11 7 4 4 2 1
4. Find the arithmetic mean of the following data. munotes.in
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Class Interval 10-20 20-30 30-40 40-50 50-60 Frequency 30 27 14 17 2
5. Obtain the median for the following frequency distribution:
X 1 2 3 4 5 6 7 8 9 f 8 10 11 16 20 25 15 9 6
[Ans: Median = 5]
6. Obtain the median from the following data.
X 20-25 25-30 30-35 35-40 40-45 45-50 50-55 55-60 f 35 45 70 105 90 74 51 30
7. Find the mode for the following distribution.
Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 No. of students 5 8 7 12 28 20 10 10 [Ans: Mode = 46.67]
8. Calculate Geometric Mean from following data.
125 1462 38 7 0.22 0.08 12.75 0.5 [ Ans: 6.952]
9. Find the geometric mean, harmonic mean and root mean square of the
numbers 3, 5, 6, 6, 7, 10 and 12.
[Ans: G. M. = 6.43, H. M. = 5.87, RMS = 7.55]
10. Find the arithmetic mean, geometric mean, harmonic mean of
numbers 2, 4 and 8. Check the relation between them.
11. Calculate Quartile 3, Deciles -7 and Percentiles 20 from following
data.
Class 2 - 4 4 – 6 6 – 8 8 - 10 Frequency 3 4 2 1
12. Calculate Q 1, Q2, Q3 D1, D5, D9, P11, P65 from following data.
Wages No. of employees 250.00 – 259.99 8 260.00 – 269.99 10 munotes.in
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270.00 – 279.99 16 280.00 – 289.99 14 290.00 – 299.99 10 300.0 – 309.99 5 310.00 – 319.99 2
1.16 REFERENCES
FUNDAMENTAL OF MATHEMATICAL STATISTICS by S. C.
Gupta and V. K. Kapoor
Statistical Methods by S. P. Gupta
STATISTICS by Murray R. Spiegel, Larry J. Stephens
*****
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2
THE STANDARD DEVIATION AND
OTHER MEASURES OF DISPERSION
Unit structure
2.0 Objectives
2.1 Introduction
2.2 Dispersion, or Variation
2.3 Range
2.4 Semi -Interquartile Range
2.5 Mean Deviation
2.6 10–90 Percentile Range
2.7 Standard Deviation
2.8 Short Methods for C omputing the Standard Deviation
2.9 Propert ies of the Standard Deviation
2.10 Variance
2.11 Charlie r’s Check
2.12 Sheppard’s Correction for Variance
2.13 Empirical Relations between Measures of Dispersion
2.14 Absolute and Relative Dispersion
2.15 Coefficient of Variation
2.16 Standardized Variable and Standard S cores
2.17 Software and Measures of Dispersion
2.18 Summary
2.19 Exercise
2.20 Reference
2.0 OBJECTIVES
After going through this chapter, students will able to learn
To provide the importance of the concept of dispersion
To calculate range, semi -Interquartile range, mean deviation
To explain why measures of dispersion must be reported in addition to
measures of central tendency
To calculate standard devi ation, variance, standard scores
To trace precise relationship
To compare two or more series with regard to their variability munotes.in
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2.1 INTRODUCTION
The measures of central tendency or Averages give us an idea of the
concentration of the observations about the central part of distribution. But
the average alone cannot adequately describe a set of observations. They
must be supported and supplemented by some other measures, called
Dispersion.
2.2 DISPERSION OR VARIATION
Literal meaning of dispersion is ‘sca tteredness’. In two or more
distributions the central value may be the same but still there can be wide
differences in the formation of distribution. Measures of dispersion help us
in studying this important characteristic of a distribution.
Definitions of Dispersion:
1. “Dispersion is the measure of the variation of the items.” – A. L.
Bowley
2. “Dispersion is the measure of extent to which individual item vary.” –
L. R. Connor
3. “The degree to which numerical data tend to spread about an average
value is called variation or dispersion of the data”. – Spiegel
2.3 RANGE
Range is the difference between two extreme observations of the
distribution. Symbolically,
Range = L – S, where L = Largest item, S = smallest item
The relative measure corresponding to range, called the coefficient of
range.
Coefficient of range = ିௌ
ାௌ
Since range is based on two extreme observations, it is not at all a reliable
measure of dispersion.
Ex 1. From the following data, calculate range and coefficient of range. Day Mon Tues Wed Thurs Fri Sat Price 20 21 18 16 22 25
Sol: Range = L – S = 25 – 16 = 9
Coefficient of range = ିௌ
ାௌ munotes.in
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= ଶହିଵ
ଶହାଵ = ଽ
ସଵ = 0.21
For continuous series, find the difference between the upper limit of the
highest class and the lower limit of the lowest class.
Ex 2 . . From the following data coefficient of range.
Marks 10– 20 20 -30 30-40 40-50 50-60 No. of Students 10 12 14 8 6
Sol: Coefficient of range = ିௌ
ାௌ
= ିଵ
ାଵ = ହ
= 0.21
2.4 SEMI -INTERQUARTILE RANGE OR QUARTILE DEVIATION
Semi -Interquartile Range Or Quartile Deviation is given by,
Q. D. =ொయି ொభ
ଶ
Quartile Deviation is a better measure than a range as it makes use of 50%
of the data. But since it ignores the other 50% of the data, it cannot be
considered as a reliable measure.
Q. D. =ொయି ொభ
ଶ
The relative measure corresponding to Q. D., called the coefficient of Q.
D.
Coefficient of Q. D. = ொయష ೂభଶൗ
ொయశ ೂభଶൗ = ொయି ொభ
ொయశ ೂభ
Coefficient of Q. D. can be used to compare the degree of variation in
different distributions.
Computation of Quartile Deviation - Individual Observations:
Ex. 3 Find out Quartile Deviation and Coefficient of Quartile Deviation
from following data.
25 33 45 17 35 20 55
Sol: Arrange the data in ascending order:
17 20 25 33 35 45 55
Q1 = size of [ேାଵ
ସ] th item = size of [ାଵ
ସ] th item = 2nd item munotes.in
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∴ Q1 = 20
Q3 = size of 3 [ேାଵ
ସ] th item = size of 3 [ାଵ
ସ] th item = 6th item
∴ Q3 = 45
Q. D. =ொయି ொభ
ଶ = ସହିଶ
ଶ = 12.5
Coefficient of Q. D. = ொయି ொభ
ொయశ ೂభ = ସହିଶ
ସହାଶ = ଶହ
ହ = 0.455
Computation of Quartile Deviation -Discrete Series:
Ex. 4 Find out Quartile Deviation and Coefficient of Quartile Deviation
from following data.
Marks 10 20 30 40 50 60 No. of Students 7 10 18 12 10 6
Sol: Marks 10 20 30 40 50 60 Frequency f 7 10 18 12 10 6 cf 7 17 35 47 57 63
Q1 = size of [ேାଵ
ସ] th item = size of [ଷାଵ
ସ] th item = 16th item
∴ Q1 = 20
Q3 = size of 3 [ேାଵ
ସ] th item = size of 3 [ଷାଵ
ସ] th item = 48th item
∴ Q3 = 50
Q. D. =ொయି ொభ
ଶ = ହିଶ
ଶ = 15
Coefficient of Q. D. = ொయି ொభ
ொయశ ೂభ
= ହିଶ
ହାଶ = ଷ
= 0.4285
Computation of Quartile Deviation - Continuous Series:
Ex. 5 Find out Quartile Deviation and Coefficient of Quartile Deviation
from following data.
Marks 35-44 45 - 54 55- 64 65 - 74 75 - 84 No. of Students 12 40 33 13 12 munotes.in
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Sol:
Marks 35-44 45 - 54 55- 64 65 - 74 75 - 84 Frequency f 12 40 33 13 12 cf 12 52 75 88 100
Q1 = size of [ே
ସ] th item = size of [ଵ
ସ] th item = 25th item
∴ Q1 lies in the class 45 – 54
Q1 = L + ேସൗି..
* i
L = 45, 𝑁4ൗ= 25, c.f. = 12 [c.f. of previous class], f= 40, i = 9
Q1 = 45 + ଶହ ିଵଶ
ସ* 9 = 47.925
Q3 = size of 3 [ே
ସ] th item = size of 3 [ଵ
ସ] th item = 75th item
∴ Q3 lies in the class 55 -64
Q3 = L + ଷேସൗି..
* i
L = 55, 3𝑁4ൗ= 75, c.f. = 52 [c.f. of previous class], f= 33, i = 9
Q3 = 55 + ହ ିହଶ
ଷଷ* 9 = 61.2727
Q. D. = ொయି ொభ
ଶ
= ଵ.ଶଶିସ.ଽଶହ
ଶ = 6.67
Coefficient of Q. D. = ொయି ொభ
ொయశ ೂభ
= ଵ.ଶଶିସ.ଽଶହ
ଵ.ଶଶାସ.ଽଶହ = .
ଵଽ.ଵଽ = 0.061
2.5 MEAN DEVIATION
Mean deviation is also known as the average deviation.
If xi | fi , i = 1, 2, …, n is the frequency distribution, then mean deviation
from the average A ( usually mean, median or mode).
Since mean deviation is based on all the observations, it is a better
measure of dispersion than range and quartile deviation.
Note: Mea n deviation is least when taken from median munotes.in
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The relative measure corresponding to the mean deviation called the
coefficient of mean deviation and is obtained by,
Coefficient of M. D. = ெ..
ௌ
Computation of Mean deviation – Individual observations
M. D. = ଵ
∑|𝑋−𝐴|
= ଵ
ே ∑|𝐷|, where |𝐷| = |𝑋−𝐴| is the modulus value or absolute value of
the deviation ignoring plus and minus signs.
Ex. 6 Calculate mean deviation and coefficient of mean deviation from
following data:
600, 620, 640, 660, 680
Sol: From above data, Median = 640
Data Deviation from median 640 |𝑫| 600 40 620 20 640 0 660 20 680 40 N= 5 ∑|𝐷| =120
M. D. = ଵ
ே ∑|𝐷| = ଵଶ
ହ = 24
Coefficient of M. D. = ெ..
ௌ
= ଶସ
ସ
= 0.0375
Computation of Mean deviation – Discrete series:
M. D. = ଵ
ே ∑𝑓|𝐷|, where |𝐷| = |𝑋−𝐴|
Ex. 7 Calculate mean deviation from following data.
X 20 21 22 23 24 f 6 15 21 15 6
Sol:
X f c.f |𝐷| f |𝐷| 20 6 6 2 12 munotes.in
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21 15 21 1 15 22 21 42 0 0 23 15 57 1 15 24 6 63 2 12 N = 63 ∑𝑓|𝐷|=54
Median = size of ேାଵ
ଶ th item = size of ଷାଵ
ଶ th item = 32th item
Size of 32th item is 22, hence Median = 22
M. D. = ଵ
ே ∑𝑓|𝐷|
= ହସ
ଷ = 0.857
Computation of Mean deviation – Continuous series:
Here we have to obtain the mid -point of the various classes and take
deviations of these points from median. Formula is same.
M. D. = ଵ
ே ∑𝑓|𝐷|
Ex. 8 Calculate mean and mean deviation from following data.
Size 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Frequency 7 12 18 25 16 14 8
Sol: Size f c.f. m.p (m) |𝑚−35.2| |𝐷| f |𝐷| 0-10 7 7 5 30.2 211.4 10-20 12 19 15 20.2 242.4 20-30 18 37 25 10.2 183.6 30-40 25 62 35 0.2 5.0 40-50 16 78 45 9.8 156.8 50-60 14 92 55 19.8 277.2 60-70 8 100 65 29.8 238.4 N= 100 ∑𝑓|𝐷|=1314.8
Median = size of ே
ଶ th item = size of ଵ
ଶ th item = 50th item
Median lies in the class 30 – 40
Median = L + ேଶൗି..
* i
L= 30, 𝑁2ൗ= 50, c.f. = 37, f = 25, i = 10
Median = 30 + ହିଷ
ଶହ *10 = 35.2 munotes.in
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M. D. = ଵ
ே ∑𝑓|𝐷|
= ଵଷଵସ.଼
ଵ =13.148
2.6 10 –90 PERCENTILE RANGE :
The 10 – 90 percentile range of a set of data is defined by,
10 – 90 percentile range = P 90 – P10
Where P 10 and P 90 are the 10th and 90th for the data.
Semi 10 -90 percentile range = వబି భబ
ଶ
2.7 STANDARD DEVIATION :
Standard deviation is the positive square root of the arithmetic mean of the
squares of the deviations of the given values from their arithmetic mean.
Standard deviation is also known as root mean square deviation as it is the
square root of the mean of the standard deviation from arithmetic mean.
Standard deviation is denoted by the small Greek letter 𝜎 (read as sigma).
Calculation of Standard Deviati on - Individual Observations:
𝜎 =ට∑௫మ
ே , where x = (X -𝑋ത)
Calculation of Standard Deviation - Discrete Series:
𝜎 =ට∑௫మ
ே , where x = (X -𝑋ത)
Calculation of Standard Deviation: Continuous Series:
𝜎 = ට∑ௗమ
ே− ቀ∑ௗ
ேቁଶ
∗𝑖, where d =(ି)
, i = class interval
Ex. 9 Calculate mean and standard deviation from the following data.
Size 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Frequency 7 10 32 43 50 35 23
Sol: Marks m. p. (m) f d = (m-35)/10 d2 fd fd2 0-10 5 7 -3 9 -21 63 10-20 15 10 -2 4 -20 40 20-30 25 32 -1 1 -32 32 30-40 35 43 0 0 0 0 40-50 45 50 1 1 50 50 50-60 55 35 2 4 70 140 munotes.in
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60-70 65 23 3 9 69 207 N = 200 ∑fd =116 ∑fd2= 532
Assumed mean, A = 35
𝑋ത = A + ∑ௗ
ே * i
=35 + ଵଵ
ଶ *10 = 40.8
𝜎 = ට∑ௗమ
ே− ቀ∑ௗ
ேቁଶ
∗𝑖
= ටହଷଶ
ଶ− ቀଵଵ
ଶቁଶ
∗10
= √ 2.66−0.3364 *10 = 1.5243*10 = 15.243
2.8 SHORT METHODS FOR COMPUTING THE STANDARD DEVIATION:
Calculation of Standard Deviation - Individual Observations :
When actual mean is in fractions eg 568.245, it would be too bulky to do
calculations. In such case either the mean may be approximated or the
deviations be taken from assumed mean A. Following is formula if we
take deviations from assumed mean A:
𝜎 = ට∑ௗమ
ே− ቀ∑ௗ
ேቁଶ
, where d = ( X – A)
Ex. 10 Calculate standard deviation with the help of assumed mean.
340, 360, 390, 345, 355, 388, 372, 363, 277, 351
Sol: Consider assumed mean = 364
X d = (X – 364) d2 340 -24 576 360 -4 16 390 26 676 345 -19 361 355 -9 81 388 24 576 372 8 64 363 -1 1 377 13 169 351 -13 169 ∑ d = 1 ∑ d2 = 2689 munotes.in
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𝜎 = ට∑ௗమ
ே− ቀ∑ௗ
ேቁଶ
= ටଶ଼ଽ
ଵ− ቀଵ
ଵቁଶ
= 16.398
Calculation of Standard Deviation - Discrete Series :
Assumed mean method: 𝜎 = ට∑ௗమ
ே− ቀ∑ௗ
ேቁଶ
, where d = ( X – A)
Ex. 11 Calculate standard deviation from the following data.
Size 3.5 4.5 5.5 6.5 7.5 8.5 9.5 Frequency 4 8 21 60 85 30 9
Sol:
Size f d = (X -6.5) d2 fd fd2 3.5 4 -3 9 -12 36 4.5 8 -2 4 -16 32 5.5 21 -1 1 -21 21 6.5 60 0 0 0 0 7.5 85 1 1 85 85 8.5 30 2 4 60 120 9.5 9 3 9 27 81 N = 217 ∑fd = 123 ∑fd2= 375
Assumed mean, A= 6.5
: 𝜎 = ට∑ௗమ
ே− ቀ∑ௗ
ேቁଶ
= ටଷହ
ଶଵ− ቀଵଶଷ
ଶଵቁଶ
= √1.7281−0.3212 = 1.1861
2.9 PROPERTIES OF THE STANDARD DEVIATION
1. Combined standard deviation: We can compute combined standard
deviation of two or more groups. It is denoted by 𝜎12 and given by
𝜎12 = ටேభఙభమ ା ேమఙమమ ାேభௗభమାேమఙమమ
ேభା ேమ
Where 𝜎12 = combined standard deviation;
𝜎ଵ = standard deviation of first group;
𝜎ଶ = standard deviation of second group;
d1 = |𝑋ଵതതത− 𝑋ଵଶതതതതത | ;
d2 = |𝑋ଶതതത− 𝑋ଵଶതതതതത | munotes.in
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2. The standard deviation of the first n natural numbers can obtained by,
𝜎 = ටଵ
ଵଶ (𝑁ଶ−1)
Thus the standard deviation of natural numbers 1 to 20 will be
𝜎 = ටଵ
ଵଶ (20ଶ−1) = ටଵ
ଵଶ 399 = 5.76
3. Standard deviation is always computed from the arithmetic mean
because the sum of the squares of the deviations of items from their
arithmetic mean is minimum.
4. For normal distribution,
Mean ± 1 𝜎 covers 68.27% of the items.
Mean ± 2 𝜎 covers 95.45% of the items.
Mean ± 3 𝜎 covers 99.73% of the items.
2.10 VARIANCE
The square of standard deviation is called the variance and is given by,
Variance = ∑(ି )തതതതమ
ே
i.e. Variance = 𝜎2 or 𝜎 = √𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒
In the frequency distribution where deviations are taken from assumed
mean,
Variance = ൜∑ௗమ
ே− ቀ∑ௗ
ேቁଶ
ൠ* i2, where d =(ି)
and i = class interval
Ex. 12 Calculate standard deviation from the following data.
Marks 10-20 20- 30 30-40 40-50 50-60 60-70 No. of students 2 6 8 12 7 5
Sol: Marks m.p (m) f d=(m-35)/10 d2 fd fd2 10-20 15 2 -2 4 -4 8 20-30 25 6 -1 1 -6 6 30-40 35 8 0 0 0 0 40-50 45 12 1 1 12 12 50-60 55 7 2 4 14 28 60-70 65 5 3 9 15 45 N = 40 ∑fd = 31 ∑fd2= 99 munotes.in
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Variance = ൜∑ௗమ
ே− ቀ∑ௗ
ேቁଶ
ൠ* i2
= ൜ଽଽ
ସ− ቀଷଵ
ସቁଶ
ൠ*102
= (2.475 – 0.6006)*100 = 187.44
2.11 CHARLIE’S CHECK
Some error may be made while calculating the value of mean and standard
deviations using different method. The accuracy of calculations can be
checked by using following formulae.
∑𝑓 (u + 1) = ∑𝑓u + ∑𝑓 = ∑𝑓u + N
∑𝑓 (u + 1)2 = ∑𝑓(u2 + 2u +1) = ∑𝑓u2 + 2 ∑𝑓u + ∑𝑓= ∑𝑓u2 + 2 ∑𝑓u +
N
∑𝑓 (u + 1)3 = ∑𝑓u3 + 3 ∑𝑓u2 + 3∑𝑓𝑢 + N
Ex. 13 Use Charlier’s check to verify mean and the standard deviation. Size 20-30 30-40 40-50 50-60 60-70 70-80 80-90 Freq 9 12 8 10 11 35 15
Sol:
X f m. p. (m) u= (m-55)/i u+1 f(u+1) u2 fu fu2 20-30 9 25 -3 -2 -18 9 -27 81 30-40 12 35 -2 -1 -12 4 -24 48 40-50 8 45 -1 0 0 1 -8 8 50-60 10 55 0 1 10 0 0 0 60-70 11 65 1 2 22 1 11 11 70-80 35 75 2 3 105 4 70 140 80-90 15 85 3 4 60 9 45 135 N=∑f =100 ∑ f(u+1)= 167 ∑fu =67 ∑fu2 =423
∑𝑓 (u + 1) = 167
∑𝑓u + N = 67 +100 =167
∴ ∑𝑓 (u + 1) = ∑𝑓u + N
This provides the required check on the mean.
X f m. p. (m) u= (m-55)/i u+1 f(u+1) f(u+1)2 20-30 9 25 -3 -2 -18 36 30-40 12 35 -2 -1 -12 12 40-50 8 45 -1 0 0 0 munotes.in
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50-60 10 55 0 1 10 10 60-70 11 65 1 2 22 44 70-80 35 75 2 3 105 315 80-90 15 85 3 4 60 240 N=∑f =100 ∑ f(u+1) = 167 ∑ f(u+1)2 ==657
∑𝑓 (u + 1)2 = 657
∑𝑓u2 + 2 ∑𝑓u + N =423 +2*67 +100 = 657
∴ ∑𝑓 (u + 1)2 = ∑𝑓u2 + 2 ∑𝑓u + N
This provides the required check on the standard deviation .
2.12 SHEPPARD’S CORRECTION FOR VARIANCE
The computation of the standard deviation is somewhat in error as a result
of grouping the data into classes (grouping error). To adjust for grouping
error, we use the formula,
Corrected variance = variance from grouped data - మ
ଵଶ
Where i is the class interval size. The correction మ
ଵଶ is called Sheppard’s
correction. It is used for distribution of continuous variables where the
tails tends to zero in both direction.
Ex. 14 Apply Sheppard’s Correction to determine the standard de viation
of the data in Ex. 8
Sol: 𝜎 = 15.243 ∴ 𝜎2 = 232.349 and i= 10.
Corrected variance = variance from grouped data - మ
ଵଶ
= 232.349 - ଵమ
ଵଶ = 224.016
Corrected Standard deviation = √224.016 = 14.9671
2.13 EMPIRICAL RELATIONS BETWEEN MEASURES OF DISPERSION :
There is a fixed relationship between the three measures of dispersion in
normal distribution. munotes.in
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Q. D. = ଶ
ଷ 𝜎 or 𝜎 = ଷ
ଶ Q. D and
M. D. = ସ
ହ 𝜎 or 𝜎 = ସ
ହ M. D
The quartile deviation is smallest, the mean deviation next and the
standard deviation is largest.
2.14 ABSOLUTE AND RELATIVE DISPERSION:
Measures of dispersion may be either absolute or relative. Absolute
measures of dispersion are expressed in the sa me statistical unit in which
the original data are given such as kilograms, tons, rupees etc. These
values may be used to compare the variations in two distributions provided
the variables are expressed in the same units and of the same average size.
In ca se the two sets of data are expressed in different units such as quintals
of sugar versus tons of sugarcane, the absolute measures of dispersion are
not comparable. In such cases measures of relative dispersion is used.
A measure of relative dispersion is the ratio of a measure of absolute
dispersion to an appropriate average. It is sometimes called coefficient of
dispersion.
Relative dispersion = ௦௨௧ ௗ௦௦
௩
2.15 COEFFICIENT OF VARIATION:
Coefficient of is used in problems where we want to compare the
variability of two or more than two series. That series or group for which
the coefficient of variation is greater is said to be more variable or less
consistent, less uniform, less stable or less homogeneous. The series for
which the coefficient of variation is less is said to be less variable or more
consistent, more uniform, more stable or more homogeneous.
If the absolute dispersion is standard deviation 𝜎 and if average is the
mean 𝑋ത, then relative dispersion is called coefficient of variation, it is
denoted by C. V. and is given by,
Coefficient of variation (C.V. ) = ఙ
ത x100
Ex. 15 Calculate arithmetic mean, standard deviation and coefficient of
variation.
Class 23-27 28-32 33-37 38-42 43-47 48-52 53-57 58-62 63-67 68-72 Freq 2 6 7 12 18 13 9 7 4 2
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Sol:
Class m. p. (m) f d = (m-50)/5 d2 fd fd2 23-27 25 2 -5 25 -10 50 28-32 30 6 -4 16 -24 96 33-37 35 7 -3 9 -21 63 38-42 40 12 -2 4 -24 48 43-47 45 18 -1 1 -18 18 48-52 50 13 0 0 0 0 53-57 55 9 1 1 9 9 58-62 60 7 2 4 14 28 63-67 65 4 3 9 12 36 68-72 70 2 4 16 8 32 N = 80 ∑fd = -44 ∑fd2= 380
Mean (𝑋ത) = A + ∑ௗ
ே x i
= 50 + ିସସ
଼ x 5 = 47.25
S. D. (𝜎) = ට∑ௗమ
ே− ቀ∑ௗ
ேቁଶ
x i
= ටଷ଼
଼− ቀିସସ
଼ቁଶ
x 5
= √4.75−0.3025 x 5= 22.23 75
C. V. = ఙ
ത x100
= ଶଶ.ଶଷହ
ସ.ଶହ x100 = 47.06%
2.16 STANDARDIZED VARIABLE AND STANDARD SCORES
The variable that measures the deviation from the mean in units of the
standard deviation is called standardized variable, is independent of the
units used and is given by,
z = ିത
ఙ
If the deviations from the mean are given in units of the standard
deviation, they are said to be expressed in standard units or standard
scores. These are of great value in the com parison of distribution . The
variable z is often used in educational testing, where it is called as a
standard score.
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Ex. 16 Your test score is 160 while the test has a mean of 120 and
standard deviation of 15. If the distribution is normal, what is your z
score? Explain the meaning of the result.
Sol: z = ିത
ఙ = ଵିଵଶ
ଵହ = 2.7
The score is 2. 7 standard deviations above the mean.
Ex. 17 A student received a grade of 84 on a final examination in English
for which mean grade was 76 and the standard deviation was 10. On a
final examination in Science for which mean grade was 82 and the
standard deviation was 16, she received a grade of 90. In which subject
was her relative standing higher?
Sol: Standardized variable z = ିത
ఙ
For English, z = ଼ସି
ଵ = 0.8
For Science, z = ଽି଼ଶ
ଵ = 0.5
Thus, the student had a grade of 0.8 of a standard deviation above the
mean in English but only 0 .5 of a standard deviation above the mean in
science. Thus her relative standing was higher in English.
2.17 SOFTWARE AND MEASURES OF DISPERSION:
The statistical software gives a variety of measures for dispersion. The
dispersion measures are usually given in descriptive statistics. EXCEL and
MINITAB allows for the computation of all the measures discussed above.
The output from MINITAB and STATISTIX has helped clarify some of
the statistical concepts which are hard to understand without some help
from the graphics involved .
Calculating Range In Excel: Excel does not offer a function to compute
range. However, we can easily compute it by subtracting the minimum
value from the maximum value. The formula would be =MAX() -MIN()
where the dataset would be t he referenced in both the parentheses. The
=MAX() and =MIN() functions would find the maximum and the
minimum points in the data. The difference between the two is the range.
Microsoft Excel has two functions to compute quartiles. The inter -quartile
range has to be calculated as the difference between the quartile 3 and
quartile 1 values Quartiles can be calculated using =QUARTILE.INC() or
=QUARTILE.EXC(). Both functions calculate the quartiles by calculatin g
the percentiles on the data. Excel offers two fu nctions, =STDEV.S() for
sample standard deviation, and =STDEV.P() for population standard
deviation.Excel with two different functions: =VAR.P() for population
variance, and =VAR.S() for sample variance. Minitab may be used to munotes.in
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compute descriptive statistic s for numeric variables, including the mean,
median, mode, standard deviation, variance and coefficient of variance. To
compute these go to Stat -Tables -Descriptive statistics.
You can use SPSS to calculate the measures of dispersion such as range,
semi -interquartile range, standard deviation and variance. We can get
SPSS to compute these in the command submenu. Go to the Statistics
menu, select the Analyse submenu, and then the Descriptive Statistics
submenu and then the Frequencies option. We can use MINITAB to
calculate the measures of dispersion the functions Q1, Q3, Range StDev,
Variance and CorfVar
2.18 SUMMARY
A measure of dispersion indicates the scattering of data. Di spersion is the
extent to which values in a distribution differ from the average of the
distribution. The measure of dispersion displays and gives us an idea about
the variation and the central value of an individual item. The range and
interquartile range are generally ineffective to measure the dispersion of
set of data. The useful measure that describes the dispersion of all the
values is standard deviation or variance. Dispersion can prove very
effective in association with central tendency in making an y statistical
decision.
2.19 EXERCISE
1. Calculate Quartile deviation (Q. D.), Mean Deviation (M. D. ) from
mean for the following data.
Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 No. of Students 6 5 8 15 7 6 3
[Ans: Q.D. = 11.23, Mean = 33.4, M.D from mean = 13.184 ]
2. Calculate Mean Deviation (M. D.) from mean for the following data
Size 2 4 6 8 10 12 14 16 f 2 2 4 5 3 2 1 1
[Ans: Mean = 8, M.D from mean = 2.8 ]
3 Calculate Mean Deviation and its coefficient from mean for the
following data .
Size 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70 -80 Freq 5 8 12 15 20 14 12 6
[Ans: Median = 43, M.D = 15.37, Coe. Of M. D. = 0.357] munotes.in
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4. Find the standard deviation of the following data.
i. 12, 6, 7,3,15, 10, 18, 5
ii. 9, 3, 8, 8, 9, 8, 9,18
[Ans: i. St. dev. 𝜎= 4.87, ii. St. dev. 𝜎= 3.87]
5. Find the standard deviation of the following data. Age 20-25 25-30 30-35 35-40 40-45 45-50 No. of persons 170 110 80 45 40 35
Take assumed average = 32.5
[Ans: Standard deviation 𝜎= 7.936]
6. Calculate the standard deviation from the following data by short
method.
240.12, 240.13, 240.15, 240.12, 240.17, 240.15, 240.17, 240.16, 240.22,
240.21
7. Calculate standard deviation from the following data by short method. Salary 45 50 55 60 65 70 75 80 No. of persons 3 5 8 7 9 7 4 7 [Ans: Standard deviation = 10.35]
8. Calculate arithmetic mean, standard deviation and coefficient of
variation. Class 20-25 25-30 30-35 35-40 40-45 Frequency 1 22 64 10 3
[Ans: 𝑋ത = 32.1, S. D. (𝜎) =3.441, C. V. = 10.72]
2.20 REFERENCE
FUNDAMENTAL OF MATHEMATICAL STATISTICS by S. C
Gupta and V. K. Kapoor
Statistical Methods by S. P. Gupta
STATISTICS by Murray R. Spiegel, Larry J. Stephens
***** munotes.in
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3
INTRODUCTION TO R
Unit structure
3.0 Objectives
3.1 Introduction
3.2 Basic syntax
3.3 Data types
3.4 Variables
3.5 Operators
3.6 Control statements
3.7 R-functions
3.8 R –Vectors
3.9 R – lists
3.10 R Arrays
3.11 Summary
3.12 Exercise
3.13 References
3.0 OBJECTIVES
After going through this chapter, students will able to learn
1. Understand the different data types, variables in R.
2. Understand the basics in R prog ramming in terms of operators,
control statements
3. Use of built-in and user defined function
4. Understand the different data structures in R.
3.1 INTRODUCTION
R is programming language and software environment for statistical
computing and graphics. It is an open source programming language. It
was designed by Ross Ihaka and Robert Gentleman at the University of
Auckland, New Zealand. in 1993 . It was released on 31 -Oct-2014 by the R
Development Core Team . It is widely used by researchers from diverse
disciplines to estimate and display results and by teachers of statistics and
research methods. Today, millions of analysts, researchers, and brands
such as Facebook, Google, Bing, Accenture, and Wipro are using R to
solve complex issues. The applications of R are not limited to just one
sector, we can see the use of R in banking, e -commerce, finance, and
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It is f reely available on www.r -project.org or can also download from
CRAN (Comprehensive R Archive Network) website http://CRAN.R -
project.org .
R Command Prompt:
We will be using RStudio . Once we have R environment setup, then it’s
easy to start R command prompt by just typing the following command at
command prompt −
$ R
This will launch R interpreter and you will get a prompt > where you can
start typing your program
> “Hello, World!”
[1] “Hello, world!”
Usually, we will write our code inside scripts which are
called RScripts in R.
Write the below given code in a file
1 print (“Hello, World!”)
and save it as myfirstprogrm.R and then run it in console by writing:
Rscript my firstprogram .R
It will produce following output
[1] "Hello, World!"
3.2 BASIC SYNTAX
Any program in R is made up of three things: Variables, Comments, and
Keywords . Varia bles are used to store the data.
Comments are us ed to improve code readability. They are like helping text
in your R program. Single comment is written using # in the beginning of
the statement.
Eg. # This is my first R program
Keywords are reserved words that hold a spe cific meaning to the compiler.
Keyword cannot be used as a variable name, function name.
Following are the Reserved words in R: if, else, while, repeat, for,
function, in, next, break, TRUE, FALSE, NU LL, Inf, NaN, NA,
NA_integer_, NA_real_, NA_complex_, NA_character etc
We can view these keywords by using either help (reserved) or ?reserved
R is case sensitive language.
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3.3 DATA TYPES
Variables can store data of different types and different types can do
different things. Variables are the reserved memory location to store
values. As we create a variable in our program some space is reserved in
memory.
Following are the data types use d in R programming.
Data type Example Description Numeric 50, 25.65, 999 Decimal values Logical True, False Data with only two possible values
which can be constructed as true/false Character ‘A’, “Excellent”,
’50.50’ A character is used represent string
values. Integer 5L, 70L, 9876L L tells R to store the value as an
integer. Complex X= 5+4i A complex value in R defined as the
pure imaginary value i. Raw A raw data type is used to holds raw
bytes.
We can use the class ( ) function to check the data type of a variable.
# numeric
a < - 25.5
class (a)
# complex
a < - 10+5i
class (a)
# integer
a < - 100L
class (a)
# logical/boolean
a < - TRUE
class (a)
# character/string
a < - “I am doing R programming”
class (a)
output:
[1] “numeric” munotes.in
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[1] “complex”
[1] “integer”
[1] “logical”
[1] “character”
3.4 VARIABLES
Variables are used to store the information to be manipulated in the R
program. A variable in R can store an atomic vector, group of atomic
vectors or a combination of many R -objects. A valid variable name
consists of letters, numbers and the dot or underline characters. The
variable name must start with a letter or the dot not followed by a number.
Ex – valid - a , a_b, a.b , a 1 , a1. , a.c
Invalid - 2a, _a
R does not have a command for declaring a variable. A variable is created
the moment you first assign a value to it.
In R, the assignment can be denoted in three ways:
1. = (Simple Assignment)
2. <- (Leftward Assignment)
3. -> (Rightward Assignment)
name = “Ajay”
gender < - “Male”
age < - 25
Here, name, gender and age are variables an d “Ajay”, “Male”, 25 are
values.
To print/output variable, you do not need any function. You can just type
the name of the variable.
name = “Ajay”
O/P:
[1] “Ajay” #auto print the value of name variable
However, R have a print() and cat() function s which are used to print the
value of the variable . The cat( ) function combines multiple values into a
continuous print output.
Cat (“My name is” , name , “\n”)
Cat (“my age is”, age, “ \n”)
O/P: My name is Ajay
My age is 25
ls() function: To know all the variables currently available in the
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# using equal to operator
a = “Good morning”
# using leftward operator
b < - “Good morning”
# using leftward operator
“Good morning - > c
print(ls())
O/P: “a” “b” “c”
# List the variables starting with the pattern "var".
> print(ls (pattern="var"))
The variables starting with dot (.) are hidden, they can be listed using
"all.names=TRUE" argument to ls() function.
> print(ls(all.name=TRUE))
rm() function: This is a built in function used to delete an wanted
variables .
> rm( variable )
# using equal to operator
a = “Good morning”
# using leftward operator
b < - “Good morning”
# using leftward operator
“Good morning - > c
# Removing variable
rm(a)
print(a)
O/P: Error in print(a) : object ‘a’ not found
All the variables can be deleted by using the rm() and ls() function
together.
> rm(list=ls())
> print(ls())
3.5 OPERATORS
Operators are the symbols directing the compiler to perform various
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operator, and each operator performs a different task. Operators simulate
the various mathematical, logical, and decision operations performed on
a set of Complex Numbers, Integers, and Numericals as input operands .
Types of Op erators used in R programming:
• Arithmetic Operators
• Relational Operators
• Logical Operators
• Assignment Operators
• Miscellaneous Operators
Arithmetic Operators:
Arithmetic operators are used with numeric values to perform common
mathematical operations
< - , = Assignment A < - 5 ; b=10 + Addition x <- c( 2,5.5,6) ; y < - c(8, 3, 4);
print(x+ y)
# O/P [1] 10.0 8.5 10.0 - Subtraction x <- c( 2, 5.5,6); y <- c(8, 3, 4);
print(x - y) * Multiplication x<- c( 2,5.5,6); y <- c(8, 3, 4); print(x*y) / division x <- c( 2,5.5,6); y <- c(8, 3, 4); print(x/y) %% remainder x<- c( 2,5.5,6); y <- c(8, 3, 4); print(x%%y) #O/P [1] 2.0 2.5 2.0 %/% gives quotient x <- c( 2,5.5,6); y <- c(8, 3, 4) ;print(x%/%y) # O/P 0 1 1 ^ ** exponent x <- c( 2,5.5,6) ; y <- c(8, 3, 4); print(x^y) #O/P 256.000 166.375 1296.000
Relational Operators: Relational/ Comparison operators are used to
compare two values
> Greater than x <- c(2,5.5,6,9) ; y <- c(8,2.5,14,9); print(x>y) # O/P [1] FALSE TRUE FALSE FALSE < Less than x <- c(2,5.5,6,9) ; y <- c(8,2.5,14,9) ; print(x < y) ; #O/P [1] TRUE FALSE TRUE FALSE munotes.in
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<= Less than equal to x <- c(2,5.5,6,9) ;y<- c(8,2.5,14,9) print(x<=y) #O/P [1] TRUE FALSE TRUE TRUE >= Greater than equal to x <- c(2,5.5,6,9) ;y <- c(8,2.5,14,9) print(x>=y) #O/P [1] FALSE TRUE FALSE TRUE == Equal x < - c(2,5.5,6,9); y < - c(8,2.5,14,9) print(x==y) #O/P [1] FALSE FALSE FALSE TRUE != Not equal x <- c(2,5.5,6,9) ; y <- c(8,2.5,14,9) print(x!=y) # O/P [1] TRUE TRUE TRUE FALSE
Logical Operators: Logical operators are used to combine conditional
statements.
& Element wise Logical AND x <- c(3, 1, TRUE, 2+3i); y <- c(4, 1,
FALSE, 2+3i)
print(x&y );
# O/P [1] TRUE TRUE FALSE TRUE | Element wise Logical OR x <- c(3, 0, TRUE, 2+2i); y <- c(4, 5, FALSE, 2+3i) print(x|y) # O/P [1] TRUE TRUE TRUE TRUE ! Element wise Logical NOT x<- c(8, 0, FALSE, 4+4i); print(!x) # O/P [1] FALSE TRUE TRUE FALSE && Takes first element of both the vectors and gives the TRUE only if both are TRUE. x <- c(3,0,TRUE, 8+9i); y<- c(1,3,TRUE, 3+4i) print(x&&y) # O/P [1] TRUE || Logical OR operator. It returns TRUE if one of the statement is TRUE. x <- c(4, 0,TRUE, 8+9i); y<- c(3, 5, TRUE, 2+3i) print(x||y) # O/P [1] TRUE
Miscellaneous Operators: Miscellaneous operators are used to anipulate
data:
: Create a series of numbers in sequence x <- 2:9 print(x) # [1] 2 3 4 5 6 7 8 9 %in% Find out if an element belongs to a vector x <- 8 ; y <- 12 ; z <- 1:10 print(x %in% z) ; print(y %in% z) # O/P [1] TRUE [1] FALSE munotes.in
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%*% It is used to multiply a matrix with its transpose
3.6 CONTROL STATEMENTS
Control statements are expressions used to control the execution and flow
of the program based on the conditions provided in the statements.
if condition: if statement checks the expression provided in the
parenthesis is true or not true. The block of code insi de if statement will be
executed only when the expression evaluates to be true.
Syntax :
if (expression ) {
// statements will execute if expression is true.
}
a < - 500
if (a > 100) {
print((x, “is greater than 100”))
}
O/P: [1] “500 is greater than 100”
If ….. else condition : If expression evaluates to be true, then the if block
of code will be executed, otherwise else block of code will be executed.
Syntax:
if(expression){
// statements will execute if expression is true.
}
else{
// statements will execute if expression is false.
}
a < - 500
if (a > 100) {
print(a, “is greater than 100”)
} else {
print(a, “is smaller than 100”)
}
O/P: [1] “500 is greater than 100” munotes.in
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Repeat loop: Repeat loop executes the same code again and again until
stop condition met.
Syntax:
repeat { commands
if (condition){ break
}
}
a <- 1
repeat {
print(a)
a =a+1
if (a>5){ break
}
}
O/P:
[1] 1
[1] 2
[1] 3
[1] 4
[1] 5
return statement: return statement is used to return the result of an
executed function and returns control to the calling function .
Syntax:
retur n(expression)
Example: func < - function(a) {
if(a > 0){
return (“POSITIVE”)
}else if (a < 0){
return(“NEGATIVE”)
}else{
return( “ZERO”)
}
}
fun(1)
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fun(-1)
O/P :
“POSITIVE”
“NEGATIVE”
“ZERO”
next statement: next statement is useful when we want to skip the current
iteration of a loop without terminating it.
Syntax:
next
Example:
a < - 1:8
#Print even numbers
for( i in a){
if(i%%2 !=0){
next
}
print(i)
}
O/P:
[1] 2
[1] 4
[1] 6
[1] 8
break statement: The break keyword is a jump statement that is used to
terminate the loop at a particular iteration
Syntax:
if (test_ expression) {
break
}
switch Statement: A switch statement is a selection control mechanism.
Switch case is a multiway branch statement. It allows a variable to be
tested for equality against a list of value s. If there is more than one match
for a specific value, then the switch statement will return the first match
found of the value matched with the expression.
Syntax:
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Example:
a < - switch( 2, “Nagpur”, “Mumbai”, “Delhi”,“Raipur”)
print(a)
O/P:
[1] “Mumbai”
while loop: The while loop executes the same code again and again until
stop condition is met.
Syntax:
While (test_expression) {
Statement
}
Example:
a < - c(“Hello”, “World”)
count < - 1
while (count < 5) {
print(a)
count = count + 1
}
O/P
[1] “Hello” “World”
[1] “Hello” “World”
[1] “Hello” “World”
[1] “Hello” “World”
for loop: The for loop can be used to execute a group of statements
repeatedly depending upon the number of elements in the object. It is an
entry controlled loop, in this loop the test condition is tested first, then the
body of the loop executed, the loop body would not be executed if the test
condition is false.
Syntax:
for (value in vector) {
statements
}
Example:
v < - LETTERS[1:5]
for ( x in v
}) { munotes.in
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print(x)
O/P:
[1] “A”
[1] “B”
[1] “C”
[1] “D”
[1] “E”
Example:
for (x in c(-5, 8, 9, 11))
{ print(x)
}
O/P:
[1] -5
[1] 8
[1] 9
[1] 11
Nested for-loop: Nested loops are similar to simple loops. Nested means
loops inside loop. R programming allows using one loop inside another
loop. In loop nesting, we can put any type of loop inside of any other type
of loop. For example, a if loop can be inside a for loop or vice versa.
Moreover, nested loops are used to manipulate the matrix .
for ( i in 1:3)
{
for ( j in 1:i)
{
print( i * j)
}
}
O/P:
[1] 1
[1] 2
[1] 4
[1] 3
[1] 6
[1] 9
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3.7 R -FUNCTIONS A set of statements which are organized together to perform a specific task
is known as a function. A function is a set of statements organized
together to perform a specific task. R has a large number of in -built
functions and the user can create their own functions. An R function is
created by using the keyword function .
Syntax :
Function_name < - function (arg1, arg2, …..)
{ function body
}
The different components of function are -
Function name is the actual name of the function.
An argument is placeholder. In function, argument are optional means
a function may or may not contain arguments, and these arguments
have default values also.
The function body contains a set of statements which defines what the
function does.
Return value is the last expression in the function body which is to be
evaluated.
R also has two types of function, i.e. Built in function and user
defined function.
Built -in function: The functions which are already defined in the
programming framework are known as built in functions. Simple
examples of built -in functions are seq (), mean(), amx(), sum(), paste(…)
etc. They are directly called by user written programs.
print(seq(50, 60))
O/P: [1] 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60
print(mean(30, 40))
O/P: [1] 35
User defined Function: R allows us to create our own function in our
program. They are specific to what a user wants and once created they can
be used like built-in functions.
Example:
areaofCircle < - function (radius){
area = pi*radius^2
return(area)
}
print(areaofCircle(2))
O/P: [1] 12.56637 munotes.in
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Example :
# create a function without an argument.
a.function < - function (){
for(i in 1:5) {
b < - i^2
print( b)
}
}
# call the function a.function without supplying an argument
a.function()
O/P
[1] 1
[1] 4
[1] 9
[1] 16
[1] 25
# create a function with an argument.
a.function < - function (a){
for(i in 1:a) {
b < - i^2
print(b)
}
}
# call the function a.function without supplying 5 as an argument
a.function(5)
O/P
[1] 1
[1] 4
[1] 9
[1] 16
[1] 25
Calling a function with argument values :
#Create a function with argument
a.function < - function(x,y,z) {
esult < - x * y + z
print( result)
}
# call the function by position of arguments
a.function(4, 2, 10)
# call the function by names of the arguments
a.function(x=10, y=4, z=2)
O/P:
[1] 18 munotes.in
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[1] 42
Calling a function with default argument:
#Create a function with argument
a.function < - function(x = 5, y= 7) {
result < - x * y
print( result)
}
# call the function without giving any arguments
a.function()
# call the function with giving new values of the argument.
a.function(10, 6)
O/P:
[1] 35
[1] 60
3.8 R –VECTORS
A vector is a basic data structure. In R, a sequence of elements which
share the same data type is known as vector. A vector supports logical,
integer, double, character, complex, or raw data type. A vector length is
basically the number of elements in the vector, and it is calculated with the
help of the length() function.
Vector is classified into two parts, i.e., Atomic vectors and Lists . There is
only one difference between atomic vectors and lists. In an atomic vector,
all the elements are of the same type, b ut in the list, the elements are of
different data types. The elements which are contained in vector known
as components of the vector. We can check the type of vector with the
help of the typeof() function.
Creation of atomic vector
Single Element Vector: when you write just one value, it becomes a
vector of length 1.
print (“xyz”)
print(25.5)
print(TRUE)
O/P
[1] “xyz”
[1] 25.5
[1] TRUE
Multiple Elements vector:
1. Using the colon ( : ) operator:
# Creating a sequence from 1 to 8
v< - 1:8 munotes.in
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print (v)
# Creating a sequence from 1.5 to 8.5
v< - 1.5:8.5
print (v)
O/P:
[1] 1 2 3 4 5 6 7 8
[1] 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5
2. Using sequence (seq) operator:
# Create a vector from 1 to 5 incrementing by 0.6
print (seq( 1, 5, by = 0.6 ))
O/P [1] 1.0 1.6 2.2 2.8 3.4 4.0 4.6
3. Using the c () function : The non character values are converted to
character type if one of the elements is a character.
x < - c(‘mango’, ‘yellow’, 10, TRUE)
print(x)
O/P
[1] “mango” “yellow”, “10”, “TRUE”
Accessing Vector Elements: Elements of a Vector are accessed using
indexing. The [ ] brackets are used for indexing. Indexing starts with
position 1. Giving a negative value in the index drops that element from
result. TRUE , FALSE or 0 and 1 can also be used for indexing.
# Accessing vector elem ents using position
x < - c(“Jan”, “Feb”, “Mar”, “Ap ril”, “May”, “Jun”, “July”, “Aug”,
“Sept”, “Oct”, “Nov”, “Dec”)
a < - x[c(2,4,8)]
print(a)
# Accessing vector elements using logical indexing.
b < - x[c(TRUE, TRUE, FALSE, FALSE, TRUE, FALSE, FALSE,
FALSE, TRUE, TRUE, FALSE, FALSE)]
print (b)
# Accessing vector elements using negative indexing.
c < - x[c(-1, -3, -4, -8, -9, -10, -11)]
print(c )
# Accessing vector elements using 0/1indexing.
d < - x[c(0,1,0,0,1,0,0,0,0,0,0,1)]
print(d )
O/P
[1] “Feb” “April” “Aug”
[1] “Jan” “Feb” “May” “Sept” “Oct”
[1] “Feb” “May” “Jun” “July” “Dec”
[1] “Feb” “May” “Dec”
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Vector Manipulation:
Vector arithmetic : Two vectors of same length can be added, subtracted,
multiplied or divided giving the result as a vector output.
#create two vectors.
x < - c(2,7,3,4,0,10)
y < - c(3,10,0,7,1,1)
add.result < - x + y
print(add.result)
multi.result < - x * y
print(multi.result)
O/P:
[1] 5 17 3 11 1 11
[1] 6 70 0 28 0 10
Vector Element Recycling: If we apply arithmetic op erations to two
vectors of unequal length, then the elements of the shorter vector are
recycled to complete the operations.
x < - c(2,7,3,4,0,10)
y < - c(3,10)
# y becomes c(3,10,3,10,3,10)
add.result < - x + y
print(add.result)
O/P: [1] 5 17 6 14 3 20
Vector Element Sorting: Elements in a vector can be sorted using
the sort() function.
x < - c(2,7,3, -11, 4,0,210)
sort.result < - sort(x)
print(sort.result)
resort.result < - sort(x, decreasing - TRUE)
print(resort.result)
O/P:
[1] -11 0 2 3 4 7 210
[1] 210 7 4 3 2 0 -11
3.9 R – LISTS
Lists are heterogeneous data structures. Lists are the R objects which
contain elements of different types. These are also one -dimensional data
structures. A list can be a list of vectors, list of matrices, a list of
characters and a list of functions and so on. List is created
using list() function.
Creating a List:
#Create a list containing strings, numbers, vectors)
list_1 < - list(“Apple”, “Mango”, 25.25, 60.5, c(16,25,36))
print(list_1) munotes.in
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O/P:
[[1]]
[1] “Apple”
[[2]]
[1] “Mango”
[[3]]
[1] 25.25
[[4]]
[1] 60.5
[[5]]
[1] 16 25 36
Naming List Element: The list elements can be given and they can be
accessed using these names.
list_1 < - list(c(“Mon”, “Tues”, “Wed”), matrix(c(1,2,3,4,5,6), nrow = 2)
#Give names to the elements in the list.
names(list_1) < - c(“Days of Week”, “Matrix”)
print(list_1)
O/P
$Days of Week
[1] “Mon” “Tues” “Wed”
$Matrix
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 4 5 6
Accessing List Elements:
Elements of the list can be accessed by the index of the element in the list.
In case of named lists it can also be accessed using the names.
list_1 < - list(c(“Mon”, “Tues”, “Wed”), matrix(c(1,2,3,4,5,6), nrow = 2)
names(list_1) < - c(“Days of Week”, “Matrix”)
print(list_1[1])
print(list_1 $Matrix)
O/P:
$Days of Week
[1] “Mon” “Tues” “Wed”
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 4 5 6
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Manipulating List Elements:
We can add, delete and update list elements as shown below. We can add
and delete elements only at the end of a list. But we can update any
element.
list_1 < - list(c( “Mon”, “Tues”, “Wed”), matrix(c(1,2,3,4,5,6), nrow = 2)
names(list_1) < - c(“Days of Week”, “Matrix”)
#add element at the end of the list
list_1[3] < - “Add Element”
print (list_1[4])
O/P
[[1]]
[1] “Add Element”
Merging Lists:
You can merge many lists into one list by placing all the lists inside one
list() function.
list_a < - list(1,2)
list_b < - list(“Ankit”, “Pooja”)
#merge tow lists
merged.list < - c(list_a, list_b)
print(merged.list)
[[1]]
[1] 1
[[2]]
[1] 2
[[3]]
[1] Ankit
[4]]
[1] Pooja
Converting List to vector:
A list can be converted to a vector so that the elements of the vector can be
used for further manipulation. All the arithmetic operations on vectors can
be applied after the list is converted into vectors.
list_a < - list(10:13)
print( list_a)
list_b < - list(20:23)
print( list_b)
#Convert the lists to vectors
x1 < - unlist (list_a)
x2 < - unlist (list_b)
print(x1 )
print(x2 )
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print( add)
O/P
[[1]]
[1] 10 11 12 13
[[2]]
[1] 20 21 22 23
[1] 10 11 12 13
[1] 20 21 22 23
[1] 30 32 34 36
3.10 R ARRAYS
Arrays are the R data objects which can store data in more than two
dimensions. In R, an array is created with the help of the array() function.
This array() function takes a vector as an input and to create an array it
uses vectors values in the dim parameter.
#create two vectors of different lengths
v1 < - c(1,2,3)
v2 < - c(4,5,6,7,8,9)
#Take these vectors as input to the array
array1 < - (c(v1,v2), dim = c(3,3,2))
print(array1)
O/P
, , 1
[,1] [,2] [,3]
[1,] 1 4 7
[2,] 2 5 8
[3,] 3 6 9
, , 2
[,1] [,2] [,3]
[1,] 1 4 7
[2,] 2 5 8
[3,] 3 6 9
Naming Columns and Rows: We can give names to the rows, columns
and matrices in the array by using the dimnames parameter.
#create two vectors of different lengths
v1 < - c(1,2,3)
v2 < - c(4,5,6,7,8,9)
column.names < - c(“Col1”, “Col2”, “Col3”)
row.names < - c(“Row1”, “Row2”, “Row3”)
matrix.names < - c(“Matrix1”, “Matrix2”)
array1 < - array(c(v1,v2),dim = c(3,3,2), dimnames = list(row.names,
column.names, matrix.names))
print(array1)
O/P: munotes.in
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, , Matrx1
Col1 Col2 Col3
Row1 1 4 7
Row2 2 5 8
Row3 3 6 9
, , Matr ix2
Col1 Col2 Col3
Row1 1 4 7
Row2 2 5 8
Row3 3 6 9
Accessing Array Elements:
#create two vectors of different lengths
v1 < - c(1,2,3)
v2 < - c(4,5,6,7,8,9)
column.names < - c(“Col1”, “Col2”, “Col3”)
row.names < - c(“Row1”, “Row2”, “Row3”)
matrix.names < - c(“Matrix1”, “Matrix2”)
array1 < - array(c(v1,v2),dim = c(3,3,2), dimnames = list(row.names,
column.names, matrix.names))
#Print the second row of the second matrix of the array.
print(array1[2,2])
#Print the element in the first row and 3rd column of the first matrix.
print(array1[1,3,1])
#Print the first Matrix
print(array1[, ,1])
O/P
Col1 Col2 Col3
2 5 8
[1] 7
Col1 Col2 Col3
Row1 1 4 7
Row2 2 5 8
Row3 3 6 9
Manipulating Array Element: As array is made up matrices in multiple
dimensions, the operations on elements of array are carried out by
accessing elements of the matrices.
#create two vectors of different lengths
v1 < - c(5,9,3 )
v2 < - c(10,11,12,13,14,15 )
#Take these vectors as input to the array
array1 < - array(c(v1,v2),dim = c(3,3,2))
#create two vectors of different lengths
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V4< - c(6,0,11,3,14,1,2,6,9 )
array2 < - array(c( v3,v4 ),dim = c(3,3,2))
#create matrices from these arrays
matrix1 < - array1[, , 2]
matrix2 < - array2[, , 2]
#add the matrices
add1 < - matrix1+matrix2
print(add1)
O/P:
[,1] [,2] [,3]
[1,] 10 20 26
[2,] 18 22 28
[3,] 6 24 30
3.11 SUMMARY
R is world's most widely used statistics programming language . It is the 1
choice of data scientists R is taught to solve critical business applications.
In addition, R is a full -fledged programming language, with a rich
complement of mathematical functions, matrix operations and control
structures. It is very easy to write your own functions. In this chapter we
covered basic programming to different types of data objects of R with
suitable examples in simple and easy steps.
3.12 EXERCISE
1. Find the output of following code.
1) b= "15"
a = switch ( b,
"5"="Hello A",
"10"="Hello B",
"15"="Hello C",
"20"="Hello D" )
print (a)
2) a= 1
b = 2
y = switch (a+b, "Hello, A", "Hello B", "Hello C", "Hello D" )
print (y)
3) # Create vegetable vector
vegetable <- c('Potato' , 'Onion' ,’Brinjal’ , 'Pumpkin' )
for ( x in vegetable) {
print(x)
}
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4) for ( i in c (5, 10, 15, 20, 0, 25)
{
if (i == 0)
{
break
}
print (i)
}
print(“outside loop”)
5) for ( i in c (5, 10, 15, 20, 0, 25)
{
if (i == 0)
{
next
}
print (i)
}
print(“outside loop”)
6) a < - 10
b<- 14
count=0
if(a{
cat(a,"is a smaller number \n")
count=1
}
if(count==1){
cat("Block is successfully execute")
}
7) a <-1
b<-24
count=0
while (a cat(a,"is a smaller number \n")
a=a+2
if(x==15)
break
}
8) a < -24
if(a%%2==0){
cat(a," is an even number")
}
if(a%%2!=0){
cat(a," is an odd number")
}
9) x <- c("Hardwork","is","the","key","of","success") munotes.in
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if("key" %in% x) {
print("key is found")
} else {
print("key is not found")
}
10) Rectangle = function(l=6, w=5){
area = l * w
return(area)
}
print(Rectangle(3, 4))
print(Rectangle(w = 9, l = 3))
print(Rectangle())
3.13 REFERENCES
The Art of R Programming: A Tour of Statistical Software Design by
Norman Matloff
Beginning R – The Statistical Programming Language by Mark
Gardener
https://www.javatpoint.com/
https://www.ict.gnu.ac.in/content/r -programming
https://www.geeksforgeeks.org/
*****
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UNIT I I
4
MOMENTS, SKEWNESS,
AND KURTOSIS
Unit Structure
4.1 Objective
4.2 Introduction
4.3 Moments
4.3.1 Moments for Grouped Data.
4.3.2 Relations between Moments.
4.3.3 Computati on of Moments for Grouped Data.
4.4 Charlie’s C heck and Sheppard’s Corrections.
4.5 Moments in Dimensionless Form
4.6 Skewness
4.7 Kurtosis
4.8 Software Computation of Skewness and Kurtosis .
4.9 Summary
4.10 Exercise
4.11 List of References
4.1 OBJECTIVE
After going through this unit, you will able to :
Define Moments and calculate for ungroup and group data.
Explain types of moments.
Find relation between raw and central moment.
Used Charlier’s check method in computing moments by coding
method.
Define Sheppard’s correction for moments.
Define moments in dimens ional form.
Define Skewness and Kurtosis.
Calculate moments, Skewness and Kurtosis using software.
4.2 INTRODUCTION
The measure of central tendency (location) and measure of dispersion
(variation) both are useful to describe a data set but both of them fail to tell
anything about the shape of the distribution. We need some other certain
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known as Skewness and Kurtosis .Moments are statistical measures that
give certain characteristics of the distribution . Moments provide su fficient
information to reconstruct a frequency distribution function . Moments are
a set of statistical parameters to measure a distribution. Four moments are
commonly used: 1st moments for Average, 2nd for Variance, 3rd for
Skewness and 4th moment for Kurtosis.
4.3 MOMENTS
The arithmetic mean of the rth power of deviations taken either from mean,
zero or from any arbitrary origin is called moments. Assume there is
sequence of random variables 𝑥ଵ,𝑥ଶ,𝑥ଷ,………𝑥. The first sample
moment, usually called the average is defined b y first moments. Three
types of moments are defined as follow:
When the deviations are computed from the arithmetic mean, then such
moments are called moments about mean (mean moments) or sometimes
calle d central moments, denoted by 𝜇 and given as follows: Hence for
ungroup data,
i) The first moment about A, as 𝜇ଵ=∑(௫ି௫̅)
.
ii) The second moment about A, as 𝜇ଶ=∑(௫ି௫̅)మ
.
iii) The third moment about A, as 𝜇ଷ=∑(௫ି௫̅)య
.
iv) The fourth moments about A, as 𝜇ସ=∑(௫ି௫̅)ర
When the deviations of the values are computed from any arbitrary value
(provisional mean) say A, then such moments are called moments about
arbitrary or provisional mean denoted by 𝜇(𝑎).Hence for ungroup data,
i) The first moment about A, as 𝜇ଵ(𝑎)=∑(௫ି)
.
ii) The second moment about A, as 𝜇ଶ(𝑎)=∑(௫ି)మ
.
iii) The third moment about A, as 𝜇ଷ(𝑎)=∑(௫ି)య
.
iv) The fourth moments about A, as 𝜇ସ(𝑎)=∑(௫ି)ర
When the deviations of the values are computed from the origin or zero,
then such moments are called momen ts abo ut origin or raw moments
denoted by 𝜇(𝑎)
i) The first moment about origin , as 𝜇′ଵ=∑(௫)
.
ii) The second moment about origin , as𝜇′ଶ=∑(௫)మ
.
iii) The third moment abou t origin , as𝜇′ଷ=∑(௫)య
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iv) The fourth moments about origin , as 𝜇′ସ=∑(௫)ర
Example 1: Find raw moments for the following data: 5, 8, 12, 4, 6.
Solution: 𝑥 𝑥ଶ 𝑥ଷ 𝑥ସ 5 8 2 4 6 25 64 4 16 36 125 512 8 64 216 625 4096 16 256 1296 𝑥=25 𝑥ଶ=145 𝑥ଷ=925 ∑𝑥ସ=6289
i) The first moment about origin , as 𝜇′ଵ=∑(௫)
=ଶହ
ହ=5 .
ii) The second moment about origin , as𝜇′ଶ=∑(௫)మ
=ଵସହ
ହ=29.
iii) The third moment abou t origin , as𝜇′ଷ=∑(௫)య
=ଽଶହ
ହ=185 .
iv) The fourth moments about origin , as 𝜇′ସ=∑(௫)ర
=ଶ଼ଽ
ହ=1257.8
4.3.1 Moments for Grouped Data :
1. Moments about arbitrary point :Let 𝑥 represents a variable occurring
with frequency 𝑓, in a given distribution, then the 𝑖௬ moment 𝜇(𝑎) about
𝐴 is defined as
𝜇(𝑎)=∑(௫ି)
ே, where 𝑁=∑𝑓.
We generally find moments upto 𝑖=4.
∴we can write :
i) The first moment about A, as 𝜇ଵ(𝑎)=∑(௫ି)
ே .
ii) The second moment about A, as 𝜇ଶ(𝑎)=∑(௫ି)మ
ே .
iii) The third moment about A, as 𝜇ଷ(𝑎)=∑(௫ି)య
ே .
iv) The fourth moments about A, as 𝜇ସ(𝑎)=∑(௫ି)ర
ே.
Example 2: For the following distribution find all four moments about 5. X 2 4 6 8 10 F 4 6 12 5 3
Solution: let prepared table first, 𝑥 𝑓 (𝑥−5) 𝑓(𝑥−5) 𝑓(𝑥−5)ଶ 𝑓(𝑥−5)ଷ 𝑓(𝑥−5)ହ 2 4 -3 -12 36 -108 324 4 6 -1 -6 6 -6 6 6 12 1 12 12 12 12 8 5 3 15 45 135 405 10 3 5 15 75 375 1875 Total 30 24 174 408 2622 munotes.in
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Moments about arbitrary A = 5 is given by
The first moment about A, as 𝜇ଵ(𝑎)=∑(௫ି)
ே=ଶସ
ଷ=0.8 .
The second moment about A, as 𝜇ଶ(𝑎)=∑(௫ି)మ
ே=ଵସ
ଷ=5.8 .
The third moment about A, as 𝜇ଷ(𝑎)=∑(௫ି)య
ே=ସ଼
ଷ=13.6 .
The fourth moments about A, as 𝜇ସ(𝑎)=∑(௫ି)ర
ே=ଶଶଶ
ଷ=87.4.
2. Moments about mean (Central moments):
These are moments about the Arithmetic Mean 𝑥̅. Hence when A is taken
as 𝑥̅, we obtain these moments. Thus it is given by
i) The first moment about 𝑥̅, as
𝜇ଵ=∑(௫ି௫̅)
ே .
ii) The second moment about 𝑥̅, as
𝜇ଶ=∑(௫ି௫̅)మ
ே .
iii) The third moment about 𝑥̅, as
𝜇ଷ=∑(௫ି௫̅)య
ே .
iv) The fourth moments about 𝑥̅, as
𝜇ସ=∑(௫ି௫̅)ర
ே.
From the definition of the mean 𝑥̅ and the standard deviation 𝜎, it
immediately follows that 𝜇ଵ=0 , 𝜇ଶ=𝜎ଶ and 𝜇ଷ measure the asymmetry
of the curve. These moments are important study the nature of the
distribution.
Example 3: Find the central moments for the following distribution: X 1 2 3 4 5 F 2 5 6 5 2
Solution:
𝑥 𝑓 𝑓𝑥 (𝑥−𝑥̅) 𝑓 (𝑥−𝑥̅ ) 𝑓 (𝑥−−𝑥 ଶ) 𝑓(𝑥−𝑥̅)ଷ 𝑓(𝑥−𝑥̅)ସ 1 2 2 -2 -4 8 -16 32 2 5 10 -1 -5 5 -5 5 3 6 18 0 0 0 0 0 4 5 20 1 5 5 5 5 5 2 10 2 4 8 16 32 Total 20 60 0 26 0 74 munotes.in
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Here , 𝑥̅=∑௫
ே=
ଶ=3.
Therefore, the central moments are given by
i) The first moment about 𝑥̅, as
𝜇ଵ=∑(௫ି௫̅)
ே=
ଶ=0 .
ii) The second moment about 𝑥̅, as
𝜇ଶ=∑(௫ି௫̅)మ
ே=ଶ
ଶ=1.3 .
iii) The third moment about 𝑥̅, as
𝜇ଷ=∑(௫ି௫̅)య
ே=
ଶ=0 .
iv) The fourth moments about 𝑥̅, as
𝜇ସ=∑(௫ି௫̅)ర
ே=ସ
ଶ=3.7.
3. Moments about origin(Raw moments):
As the name suggests, taking A as the origin ( A = 0), we get these
moments. Thus it is given by
i) The first moment about Origin , as
𝜇ଵ′=∑௫
ே .
ii) The second moment about Origin, as
𝜇ଶ′=∑௫మ
ே .
iii) The third moment about Origin , as
𝜇ଷ′=∑௫య
ே .
iv) The fourth moments about Origin , as
𝜇ସ′=∑௫ర
ே.
Note that for first moment about origin is mean of the data.
Example 4: Find the raw moments for the following data: X -1 0 1 2 3 4 F 2 4 3 7 3 1
Solution : lets prepared table 𝑥 𝑓 𝑓𝑥 𝑓𝑥ଶ 𝑓𝑥ଷ 𝑓𝑥ସ -1 2 -2 2 -2 2 0 4 0 0 0 0 1 3 3 3 3 3 2 7 14 28 56 112 3 3 9 27 81 243 4 1 4 16 64 256 Total 20 28 76 202 616
Therefore, the raw moments are given by
i) The first moment about Origin , as munotes.in
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𝜇ଵᇱ=∑௫
ே=ଶ଼
ଶ=1.4 .
ii) The second moment about Origin, as
𝜇ଶᇱ=∑௫మ
ே=
ଶ=3.8 .
iii) The third moment about Origin , as
𝜇ଷᇱ=∑௫య
ே=ଶଶ
ଶ=10.1 .
iv) The fourth moments about Origin , as
𝜇ସᇱ=∑௫ర
ே=ଵ
ଶ=30.8.
4.3.2 Relations between Moments:
We studied three different types of moments. Now it is very useful to
simplifying relation between them. We will now give inter -relation
between various moments and solve example using these relations.
Relation between moments about arbitrary point and the central moment:
i) 𝜇ଵ=𝜇ଵ(𝑎)−𝜇ଵ(𝑎)=0
ii) 𝜇ଶ=𝜇ଶ(𝑎)−𝜇ଵ(𝑎)ଶ
iii) 𝜇ଷ=𝜇ଷ(𝑎)−3𝜇ଵ(𝑎)𝜇ଶ(𝑎)+2𝜇ଵ(𝑎)ଷ
iv) 𝜇ସ=𝜇ସ(𝑎)−4𝜇ଵ(𝑎)𝜇ଷ(𝑎)+6𝜇ଵ(𝑎)ଶ𝜇ଶ(𝑎)−3𝜇ଵ(𝑎)ସ
Conversely the moments 𝜇(𝑎)′𝑠 about A in term of 𝜇′𝑠are given as
follow s:
i) 𝜇ଵ(𝑎)=𝑥̅−𝐴
ii) 𝜇ଶ(𝑎)=𝜇ଶ+𝜇ଵ(𝑎)ଶ
iii) 𝜇ଷ(𝑎)=𝜇ଷ+3𝜇ଶ𝜇ଵ(𝑎)+𝜇ଵ(𝑎)ଷ
iv) 𝜇ସ(𝑎)=𝜇ସ+4𝜇ଷ𝜇ଵ(𝑎)+6𝜇ଶ𝜇ଵ(𝑎)ଶ+𝜇ଵ(𝑎)ସ
Relation between Raw moments and central moments:
Recall that, the raw moments 𝜇′ are obtained from the general moments
𝜇(𝑎) when A is taken as ‘0’.
Hence taking A as ‘0’and replacing 𝜇(𝑎) by corresponding 𝜇′ in the
formula, we get
i) 𝜇ଵ=𝜇′ଵ−𝜇′ଵ=0
ii) 𝜇ଶ=𝜇ଶ′−𝜇ଵ′ଶ
iii) 𝜇ଷ=𝜇ଷ′−3𝜇ଵ′𝜇ଶ′+2𝜇ଵ′ଷ
iv) 𝜇ସ=𝜇ସ′−4𝜇ଵ′𝜇ଷ′+6𝜇ଵ′ଶ𝜇ଶ′−3𝜇ଵ′ସ
Conversely the moments 𝜇′ in term of 𝜇 are given as follows:
i) 𝜇ଵ′=𝑥̅
ii) 𝜇ଶ′=𝜇ଶ+𝜇ଵ′ଶ munotes.in
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iii) 𝜇ଷ′=𝜇ଷ+3𝜇ଶ𝜇ଵ′+𝜇ଵ′ଷ
iv) 𝜇ସ′=𝜇ସ+4𝜇ଷ𝜇ଵ′+6𝜇ଶ𝜇ଵ′ଶ+𝜇ଵ′ସ
Example 5: The first four central moments of a distribution are 0, 3, 5, 10.
If the mean of the distribution is 2, find the moments about 3.
Solution: We have 𝐴=3,𝑥̅=2,𝜇ଵ=0,𝜇ଶ=3,𝜇ଷ=5,𝜇ସ=10.
Using the relation between central moments and arbitrary moments,
i) 𝜇ଵ(𝑎)=𝑥̅−𝐴=2−3=−1.
ii) 𝜇ଶ(𝑎)=𝜇ଶ+𝜇ଵ(𝑎)ଶ=3+(−1)ଶ=4
iii) 𝜇ଷ(𝑎)=𝜇ଷ+3𝜇ଶ𝜇ଵ(𝑎)+𝜇ଵ(𝑎)ଷ=5+3(3)(−1)+
(−1)ଷ=−5.
iv) 𝜇ସ(𝑎)=𝜇ସ+4𝜇ଷ𝜇ଵ(𝑎)+6𝜇ଶ𝜇ଵ(𝑎)ଶ+𝜇ଵ(𝑎)ସ
=10+4(5)+6(3)(−1)ଶ+(−1)ସ=49.
Example 6: The first four raw moments about the origin are 2, 12, 74 and
384. Find the mean 𝑥̅ and the first four central moments.
Solution: We already define the raw moments about the origin i.e.
𝜇(𝑎)′s with 𝐴=0. Given that 𝜇ଵ(𝑎)=2,𝜇ଶ(𝑎)=12,𝜇ଷ(𝑎)=
74,𝑎𝑛𝑑 𝜇ସ(𝑎)=384,𝑤𝑖𝑡ℎ 𝐴=0.
Therefore, Mean =𝑥̅=𝜇ଵ(𝑎)+𝐴=2+0=2.
Using the relation between raw moments and central moments
i) 𝜇ଵ=𝜇ଵ(𝑎)−𝜇ଵ(𝑎)=2−2=0
ii) 𝜇ଶ=𝜇ଶ(𝑎)−𝜇ଵ(𝑎)ଶ=12−2ଶ=8
iii) 𝜇ଷ=𝜇ଷ(𝑎)−3𝜇ଵ(𝑎)𝜇ଶ(𝑎)+2𝜇ଵ(𝑎)ଷ=74−
3(2)(12)+2(2ଶ)=10
iv) 𝜇ସ=𝜇ସ(𝑎)−4𝜇ଵ(𝑎)𝜇ଷ(𝑎)+6𝜇ଵ(𝑎)ଶ𝜇ଶ(𝑎)−3𝜇ଵ(𝑎)ସ
=284−4(2)(74)+6(2ଶ)(12)−3(2ସ)
=384−592+288−48=128.
Example 7: The first four central moments for a distribution are 0,3,0 and
7. If the mean 𝑥̅ of the distribution is 4, find the first four raw moments.
Solution: The raw moments are the moments about origin. Given that
𝜇ଵ=0,𝜇ଶ=3,𝜇ଷ=0 𝑎𝑛𝑑 𝜇ସ=7 𝑤𝑖𝑡ℎ 𝑥̅=4.
Using the relation between central moments and raw moments.
i) 𝜇ଵᇱ=𝑥̅=4
ii) 𝜇ଶᇱ=𝜇ଶ+𝜇ଵ′ଶ=3+4ଶ=17.
iii) 𝜇ଷᇱ=𝜇ଷ+3𝜇ଶ𝜇ଵᇱ+𝜇ଵ′ଷ=0+3(3)(4)+4ଷ=100.
𝜇ସᇱ=𝜇ସ+4𝜇ଷ𝜇ଵᇱ+6𝜇ଶ𝜇ଵ′ଶ+𝜇ଵ′ସ munotes.in
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4.3.3 Computat ion of Moments for Grouped Data:
We have already found mean and standard deviation for the continuous
data (grouped data). Now to calculate moments for the continuous data we
used coding method (Short method).
When the values of 𝑥 are not consecutive, but equally spaced at an interval
of length ′𝑐′. We need to divide the expression by ′𝑐′. It is called change of
scale by ′𝑐′.
Where we take 𝑥=𝑎+𝑐𝑢 𝑜𝑟 𝑢=௫ି
.
We give below the effect of change of origin and scale on mo ments.
Let 𝑥=𝑎+𝑐𝑢 ,
∴ 𝑥̅=𝑎+𝑐𝑢ത.
i)The moments of 𝑥 about A are given by
𝜇(𝑎)=∑𝑓𝑢
𝑁×𝑐
ii) The central moments of 𝑥 are given by
𝜇=∑𝑓(𝑢−𝑢ത)
𝑁×𝑐
Note: When A = 0, we get the raw moment.
Example 8: Find the central moments for the following data: Class interval 0-20 20-40 40-60 60-80 Frequency 4 7 6 3
Solution: first find mean by coding method, taking 𝑎=30
Here, 𝑢ത=∑௨
ே=଼
ଶ=0.4.
The central moments of 𝑥 are given by C.I F Cla
ss
Mar
ks
(x) 𝑢
=𝑥−𝑎
𝑐 F
u (𝑢
−0.4) 𝑓(𝑢
−0.4) 𝑓(𝑢
−0.4)ଶ 𝑓(𝑢
−0.4)ଷ 𝑓(𝑢
−0.4)ସ 0-
20 4 10 -1 -4 -1.4 -5.6 7.84 -10.976 15.3664 20-
40 7 30 0 0 -0.4 -2.8 3.92 -0.448 0.1792 40-
60 6 50 1 6 0.6 3.6 2.16 1.296 0.7776 60-
80 3 70 2 6 1.6 4.8 7.68 12.288 19.6608 Tot
al 2
0 8 0 21.6 2.16 35.984 munotes.in
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𝜇=∑𝑓(𝑢−𝑢ത)
𝑁×𝑐
i) 𝜇ଵ=∑(௨ି௨ഥ)భ
ே×𝑐ଵ=
ଶ×20=0.
ii) 𝜇ଶ=∑(௨ି௨ഥ)మ
ே×𝑐ଶ=ଶଵ.
ଶ×20ଶ=432.
iii) 𝜇ଷ=∑(௨ି௨ഥ)య
ே×𝑐ଷ=ଶ.ଵ
ଶ×20ଷ=864.
iv) 𝜇ସ=∑(௨ି௨ഥ)ర
ே×𝑐ସ=ଷହ.ଽ଼ସ
ଶ×20ସ=2,87,872.
4.4 CHARLIE’S CHECK AND SHEPPARD’S CORRECTIONS
A check which can be used to verify correct computations in a table of
grouped classes. For example, consider the following table with specified
class limits and frequencies 𝑓. The class marks 𝑥 are then computed as
well as the rescaled fr equencies 𝑢, which are given by
𝒖𝒊=𝒇𝒊−𝒙𝟎
𝒄
Where the class mark is taken as 𝑥=44.5 and the class interval is 𝑐=
10. The remaining quantities are then computed as follows. Class interval 𝑥 𝑓 𝑢 𝑓𝑢 𝑓𝑢ଶ 𝑓(𝑢+1)ଶ 0-9 4.5 2 -4 -8 32 18 10-19 14.5 3 -3 -9 27 12 20-29 24.5 11 -2 -22 44 11 30-39 34.5 20 -1 -20 20 0 40-49 44.5 32 0 0 0 32 50-59 54.5 25 1 25 25 100 60-69 64.5 7 2 14 28 63 Total 100 -20 176 236
In order to compute the variance , note that
𝑉(𝑢)=∑𝒇𝒊𝒖𝒊𝟐
∑𝒇𝒊−ቆ∑𝒇𝒊𝒖𝒊
∑𝒇𝒊ቇଶ
=176
100−൬−20
100൰ଶ
=1.72
So the variance of the original data is
𝑉(𝑥)=𝑐ଶ𝑉(𝑢)=100×1.72=172. munotes.in
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Charlier's check makes use of the additional column 𝑓(𝑢+1)ଶ added to
the right side of the table. By noting that the identity
𝒇𝒊(𝒖𝒊+𝟏)𝟐= 𝒇𝒊(𝒖𝒊𝟐+𝟐𝒖𝒊+𝟏)
=𝒇𝒊𝒖𝒊𝟐+𝟐𝒖𝒊+𝒇𝒊
connects columns five through seven, it can be checked that the
computations have been done correctly. In the example above, 236 = 176 +2 (- 20) +100 (8) Hence, the computations pass Charlier's check.
Charlier's check in computing moments by the coding method uses the
following identities:
𝑓(𝑢+1)=𝑓𝑢+𝑓
𝑓(𝑢+1)ଶ=𝑓𝑢ଶ+2𝑓𝑢+𝑓
𝑓(𝑢+1)ଷ=𝑓𝑢ଷ+3𝑓𝑢ଶ+3𝑓𝑢+𝑓
𝑓(𝑢+1)ସ
=𝑓𝑢ସ+4𝑓𝑢ଷ+6𝑓𝑢ଶ+4𝑓𝑢
+𝑓
Sheppard’s Corrections:
When the frequency distribution, consists of interval, we take 𝑥 as the
class mark of the interval and use this 𝑥 in all the formulae.
While doing this, it is assumed that all the values in the interval
concentrate at the class mark. But this assumption may not be always true
and we are likely to get some errors in this calculation.
The well -known statistician Sheppa rd gave the corrected values of the
moments as follows:
𝜇ଵ(𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑 )=𝜇ଵ
𝜇ଶ(𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑 )=𝜇ଶ−𝑐ଶ
12
𝜇ଷ(𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑 )=𝜇ଷ
𝜇ସ=𝜇ସ−1
2𝑐ଶ𝜇ଶ+7
240𝑐ସ munotes.in
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Where ‘c’ is the length of the class -interval, which is the same as the
spacing between the mid values.
Note that even though this correction has great mathematical significance,
we need not use these corrections in practice because the error is too small
hence negligible and also in statistic, we look for estimates, which are
approximate values .
Example 9:Apply Sheppard’s corrections to determine the moments abo ut
the mean for the data Class Interval 0-10 10-20 20-30 30-40 40-50 Frequency 1 2 9 2 6
Solution: Lets prepared table, taking 𝐴=25. Class Interval Class Mark (X) f 𝑢
=𝑥−𝑎
𝑐 𝑓𝑢 (𝑢−0.5) 𝑓(𝑢−0.5) 𝑓(𝑢−0.5)ଶ 𝑓(𝑢−0.5)ଷ 𝑓(𝑢−0.5)ସ 0-10 5 1 -2 -2 -2.5 -2.5 12.5 -15.625 39.0625 10-20 15 2 -1 -2 -1.5 -3 4.5 -6.75 10.125 20-30 25 9 0 0 -0.5 -4.5 2.25 -1.125 0.5625 30-40 35 2 1 2 0.5 1 0.5 0.25 0.125 40-50 45 6 2 12 1.5 9 13.5 20.25 30.375 Total 20 10 0 33.25 -3 80.25
Here, 𝑢ത=∑௨
ே=ଵ
ଶ=0.5.
The central moments of 𝑥 are given by
𝜇=∑𝑓(𝑢−𝑢ത)
𝑁×𝑐
i) 𝜇ଵ=∑(௨ି௨ഥ)భ
ே×𝑐ଵ=
ଶ×20=0.
ii) 𝜇ଶ=∑(௨ି௨ഥ)మ
ே×𝑐ଶ=ଷଷ.ଶହ
ଶ×20ଶ=665.
iii) 𝜇ଷ=∑(௨ି௨ഥ)య
ே×𝑐ଷ=ିଷ
ଶ×20ଷ=−1200.
iv) 𝜇ସ=∑(௨ି௨ഥ)ర
ே×𝑐ସ=଼.ଶହ
ଶ×20ସ=6,42,000.
Sheppard gave the corrected values of the moments as follows: munotes.in
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𝜇ଵ(𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑 )=𝜇ଵ=0, 𝜇ଶ(𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑 )=𝜇ଶ−మ
ଵଶ=665−ଶమ
ଵଶ=
631.67
𝜇ଷ(𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑 )=𝜇ଷ=−1200
𝜇ସ=𝜇ସ−ଵ
ଶ𝑐ଶ𝜇ଶ+
ଶସ𝑐ସ=6,42,000−ସ
ଶ×665+
ଶସ×1,60,000=
5,13,666.67.
4.5 MOMENTS IN DIMENSIONLESS FORM
To avoid particular units, we can define the dimensionless central
moments as
𝑎=𝜇
𝜎
Where 𝜎 is the standard deviation, so, as we have 𝜎=√𝜇ଶ ,
We already know that for central moments, 𝜇ଵ=0,𝜇ଶ=𝜎ଶ.
So, we get 𝑎=0 𝑎𝑛𝑑 𝑎ଵ=1.
4.6 SKEWNESS
Skewness is one more concept which deals with the symmetry or rather
asymmetry of the values of distribution around its central value. When a
frequency distribution is plotted on a chart, an ideal distribution by a nice,
symmetric, bell -shaped curve around the central value. Such a distribution
is called symmetric distribution or a normal distribution. However in
practice every distribution that we across need not be normal. Their graph
will be asym metric or skew. Such distributions are called skewed
distribution.
Definition: Skewness defined by famous statistician Garrett “ A
distribution is said to be skewed when the mean and median fall at
different points of the distribution and balance is shift ed to one side or the
other to left or right. ”
Types of Skewness: In order to unde rstand this concept we draw the
following graphs, where 𝑥̅= Mean, 𝑀= Median and 𝑀= Mode of the
distributions.
Figure 4.1
munotes.in
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It is clear from the diagram that
i) Represents a symmetric distribution, for which Mean = Median =
Mode.
ii) Represents a positive skewed distribution for which Mode < Median
iii) Represents a negative skewed distribution for which Mean < Median
Measure of Skewness:
Since mean, median a nd mode are different for a skewed distribution, the
simplest measure would be the difference between two of these in pairs.
Though such measures are simple to calculate, their main drawback is the
following: these measures are expressed with respect to th e corresponding
units of the distribution. Therefore two distributions with different units
cannot be compared. In order to overcome this difficulty, relative
measures are defined. These are called Coefficients of Skewness.
Karl Pearson’s Coefficient of S kewness: it is defined as
𝑆=𝑀𝑒𝑎𝑛−𝑀𝑜𝑑𝑒
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 =𝑥̅−𝑀
𝜎
Using the relation between mean, median and mode:
Mean – Mode = 3 (Mean – Median), we can write
𝑆=3(𝑀𝑒𝑎𝑛−𝑀𝑒𝑑𝑖𝑎𝑛)
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛=3(𝑥̅−𝑀)
𝜎
Interpretation on 𝑆
i) If 𝑆 is positive then the distribution is positively skewed.
ii) If 𝑆 is negative then the distribution is negatively skewed.
iii) If 𝑆=0 then the distribution is symmetric.
iv) Theoretica lly the limits of 𝑆 are from -3 to +3.
Example 10: For the following ungrouped data find the Karl Pearson’s
Coefficient of Skewness.
12,18,25,15, 16, 10, 8 15, 27,14
Solution: For the Karl Pearson’s Coefficient of Skewness we need to find
mean, mode and standard deviation of the data.
Mean = 𝑥̅=∑௫
=ଵଶାଵ଼ାଶହାଵହାଵାଵା଼ାଵହାଶାଵସ
ଵ=ଵ
ଵ=16.
Mode = 15 ( number which repeated maximum time)
𝑥ଶ=144+324+625+225+256+100+64+225+729
+196=2,888 munotes.in
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𝜎=ඨ∑𝑥ଶ
𝑛−(𝑥̅)ଶ=ඨ2,888
10−(15)ଶ=√288.8−225=√63.8=7.99
Therefore, the Karl Pearson’s Coefficient of Skewness is
𝑆=𝑀𝑒𝑎𝑛−𝑀𝑜𝑑𝑒
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 =𝑥̅−𝑀
𝜎=16−15
7.99=1
7.99=0.125
Example 11: For the following grouped data find th e Karl Pearson’s
Coefficient of Skewness. Also interpret the type of distribution. C.I 0-4 4-8 8-12 12-16 16-20 F 1 3 10 4 2
Solution: First we find mean, mode and standard deviation. C.I F x fx 𝑓𝑥ଶ 0-4 1 2 2 4 4-8 3 6 18 108 8-12 10 10 100 1000 12-16 4 14 56 784 16-20 2 18 36 648 Total 20 212 2,544
Mean = 𝑥̅=∑௫
ே=ଶଵଶ
ଶ=10.6.
Standard deviation =
𝜎=ඨ∑𝑓𝑥ଶ
𝑁−(𝑥̅)ଶ=ඨ2,544
20−(10.6)ଶ=√127.2−112.36=√14.84
=3.85.
𝑀𝑜𝑑𝑒=𝑙ଵ+𝑓ଵ−𝑓
2𝑓ଵ−𝑓−𝑓ଶ×ℎ
𝑀𝑜𝑑𝑒=8+10−3
2(10)−3−4×4=10.15.
Therefore, the Karl Pearson’s Coefficient of Skewness is
𝑆=𝑀𝑒𝑎𝑛−𝑀𝑜𝑑𝑒
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 =𝑥̅−𝑀
𝜎=10.6−10.15
3.85=0.45
3.85=0.12.
Bowley’s Coefficient of Skewness:
This measure is based on Quartiles, hence it is also known as Quartile
Coefficient of Skewness. It is given by
𝑺𝑩=𝑸𝟑+𝑸𝟏−𝟐𝑸𝟐
𝑸𝟑−𝑸𝟏
The limits of Bowley’s Coefficient of Skewness are between -1 to +1.
Example 12: Find the Bow ley’scoefficient of Skewness for the following
information are given: 𝑄ଵ=12.5,𝑄ଶ=17.2,𝑄ଷ=24.7
munotes.in
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Solution: Given that
𝑄ଵ=12.5,𝑄ଶ=17.2,𝑄ଷ=24.7
The Bowley’s coefficient of Skewness is given by
𝑆=𝑄ଷ+𝑄ଵ−2𝑄ଶ
𝑄ଷ−𝑄ଵ=24.7+12.5−2(17.2)
24.7−12.5=2.8
12.2=0.23
Example 13:Find the Bowley’s coefficient of Skewness for the following
distribution: X 1 3 5 7 9 11 F 3 8 14 20 18 7
Solution: Let find all three quartiles for the distribution: X F cf(Cummulative frequency) 1 3 3 3 8 11 5 14 25 7 20 45 9 19 64 11 7 71 Total N =71
Therefore,
𝑄ଵ=𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 1൬𝑁+1
4൰൨௧
𝑖𝑡𝑒𝑚
=𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 1൬71+1
4൰൨௧
𝑖𝑡𝑒=𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 18𝑡ℎ 𝑖𝑡𝑒𝑚=5
𝑄ଶ=𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 2൬𝑁+1
4൰൨௧
𝑖𝑡𝑒𝑚
=𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 2൬71+1
4൰൨௧
𝑖𝑡𝑒=𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 36𝑡ℎ 𝑖𝑡𝑒𝑚=7
𝑄ଷ=𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 3൬𝑁+1
4൰൨௧
𝑖𝑡𝑒𝑚
=𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 3൬71+1
4൰൨௧
𝑖𝑡𝑒=𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 54𝑡ℎ 𝑖𝑡𝑒𝑚=9
The Bowley’s coefficient of Skewness is given by
𝑆=𝑄ଷ+𝑄ଵ−2𝑄ଶ
𝑄ଷ−𝑄ଵ=9+5−2(7)
9−5=0
.4=0.
Therefore, the distribution is symmetric.
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4.7 KURTOSIS
Kurtosis in Greek means ‘Bulginess’. In statistics, Kurtosis refers to the
degree of flatness or peakedness around the mode of a frequency curve.
The measure of kurtosis is with respect to a normal curve, which is
accepted as a yardstick to decide the nature of oth er curves.
In other words measures of kurtosis tell us to what extent the given
distribution is flat or peaked with respect to the standard normal curve.
Figure 4.2
i) Normal curve is called Mesokurtic (M).
ii) Flat one is called Platykurtic (P).
iii) Peaked is c alled Leptokurtic (L).
Measures of Kurtosis:
The most prominent measure of kurtosis is the coefficient 𝛽ଶ, given by
𝛽ଶ=𝜇ସ
𝜇ଶଶ
Where 𝜇′𝑠are the moment about mean 𝑥̅.
Bigger value of 𝛽ଶ gives more peak to the distributions. For normal
distribution 𝛽ଶ= 3.
Hence the given distribution is :
i) Leptokurtic, if 𝛽ଶ> 3.
ii) Mesokurtic, if 𝛽ଶ= 3.
iii) Platykurtic, if 𝛽ଶ< 3.
Example 14: for the following distribution find 𝛽ଵ and 𝛽ଶ and comment
on the Skewness and Kurtosis of the distribution.
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X 2 3 4 5 f 4 3 2 1 Solution: First calculate moments about mean for the given distribution.
𝑥̅=∑𝑓𝑥
∑𝑓=2(4)+3(3)+4(2)+5(1)
4+3+2+1=30
10=3. 𝑥 𝑓 (𝑥−3) 𝑓(𝑥−3) 𝑓(𝑥−3)ଶ 𝑓(𝑥−3)ଷ 𝑓(𝑥−3)ସ 2 4 -1 -4 4 -4 4 3 3 0 0 0 0 0 4 2 1 2 2 2 2 5 1 2 2 4 8 16 Total 10 0 10 6 22
The central moments are given by
i) The first moment about 𝑥̅, as
𝜇ଵ=∑(௫ି௫̅)
ே=
ଵ=0 .
ii) The second moment about 𝑥̅, as
𝜇ଶ=∑(௫ି௫̅)మ
ே=ଵ
ଵ=1 .
iii) The third moment about 𝑥̅, as
𝜇ଷ=∑(௫ି௫̅)య
ே=
ଵ=0.6 .
iv) The fourth moments about 𝑥̅, as
𝜇ସ=∑(௫ି௫̅)ర
ே=ଶଶ
ଵ=2.2.
𝛽ଵ=𝜇ଷଶ
𝜇ଶଷ=(0.6)ଶ
(1)ଷ=0.36.
𝛽ଶ=𝜇ସ
𝜇ଶଶ=2.2
(1)ଶ=2.2
Since 𝛽ଵ≠0, the curve is not symmetric Also 𝜇ଷ=0.6>0.
Therefore, the curve is positively skewed.
Since 𝛽ଶ=2.2<3.
Therefore, the curve is flat as compared to the normal curve.
Hence the distribution is platykurtic.
4.8 SOFTWARE COMPUTATION OF SKEWNESS AND KURTOSIS
To compute Skewness and Kurtosis by using different software are given
below:
Sigma Magic : Using the Sigma Magic software, calculating the Skewness
and Kurtosis is relatively straightforward. Just add a new Basic Statistics
template to Excel by clicking on Stat > Basic Statistics. Copy and paste munotes.in
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the data for which you want to Skewness and Kurto sis into the input area
and then click on Compute Outputs. The analysis results will include the
Skewness and Kurtosis values.
Excel : You could also calculate these values in Excel by using the formula
=SKEW(…) for the Skewness value, =KURT(…) for the Kurt osis value.
Minitab : If you use the Minitab software, you can copy and paste the data
into Minitab and then click on Stat > Basic Statistics > Display
Descriptive Statistics. Then select the data column and then click on OK.
This will print out the quarti les for the sample values. If you want the
Skewness and Kurtosis values, you have to go back to the menu and click
on Statistics and select the checkbox next to Skewness and Kurtosis in the
statistics options. Note that the values provided by Minitab may b e slightly
different from Excel and Sigma Magic software.
4.10 SUMMARY
In this unit, we have discussed:
Moments and its types for ungroup and grouped data.
The relation between raw, arbitrary and central moments.
The effect of change of origin and scale on moments.
Charlie’s check, and Shepha rd’s Correction for Moments.
Skewness and about symmetry of distribution.
Kurtosis .
4.11 EXERCISE
1.The first four moments of a distribution are 1, 4, 10 and 46 respectively.
Compute the moment coefficients of skewness and kurtosis and comment
upon the nature of the distribution.
2. Compute the first four central moments from the following data. Also
find the two beta coefficients.
X 5 10 15 20 25 30 35 f 8 15 20 32 23 17 5
3. The first four central moments of a distribution are 0, 2.5, 0.7 and 18.75.
Examine the skewness and kurtosis of the distribution.
4. Calculate first four central moments for the following distribution: Class interval 0-4 4-8 8-12 12-16 16-20 Frequency 5 8 13 9 5
5. Find the first four arbitrary moments about A = 7 for the following: munotes.in
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10, 5, 8, 7, 2, 3, 12, 14
6. Find raw moment for the following data: C.I 5-10 10-15 15-20 20-25 25-30 f 3 4 7 4 2
7. The first four central moments of a distribution are 0, 15, 36, 78. If the
mean of the distribution is 8, find the moments about A = 5.
8. The first four raw moments about origin are 4, 16, 33, 89. Find mean
and the first four central moments.
9. For the following data verify Charlie’s check : C.I 2-8 8-14 14-20 20-26 f 1 4 3 2
10. Find the first four central moments using coding method, also find
Sheppard’s correction for moments. C.I 0-5 5-10 10-15 15-20 20-25 25-30 f 3 8 12 13 7 2
11. For the following data , find Karl Pearson’s coefficient of Skewness
and also find the type of distribution:
i) 12,15,17, 12,8,25,16,6,7,41
ii) 3,7,8, 12, 15 X 1 2 3 4 5 6 7 f 2 8 12 15 18 9 6 C.I 0-2 2-4 4-6 6-8 8-10 frequency 4 7 13 10 6
12. Find the Bowely’s coefficient of Skewness for each of the following:
i) 𝑄ଵ=165.5,𝑄ଶ=184.3,𝑄ଷ=196.7.
ii) 2,8,7,12,14,17,20.
X 1 2 3 4 5 6 7 F 2 8 12 15 18 9 6
4.12 LIST OF REFERENCES
Statistics by Murry R. Spiegel, Larry J. Stephens. Publication
McGRAWHILL INTERNATIONAL.
Fundamental Statistic by S.C. Gupta
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5
ELEMENTARY PROBABILITY
THEORY
Unit Structure
5.1 Objective
5.2 Introduction
5.3 Definitions of Probability
5.4 Conditional Probability
5.4.1 Independent and Dependent Events, Mutually Exclusive
Events
5.5 Probability Distributions
5.6 Mathematical Expectation
5.7 Combinatorial Analysis
5.8 Combinations, Stirling’s Approximation to n!
5.9 Relation of Probability to Point Set Theory, Euler or Venn Diagrams
and Probability
5.10 Summary
5.11 Exercise
5.12 List of References
5.1 OBJECTIVE
After going through this unit, you will able to :
Determine the probability of different experimental results.
Explain the concept of probability.
Calculate probability for simple, compound and complimentary
events.
Conditional probability and i ts examples.
Independent events and multiplication theorem of probability.
Probability distribution and its Expected value of probability
distribution.
Combination and Stirling’s number approximation.
Relations between probability and set theory with help of Venn
diagram.
5.2 INTRODUCTION Some time s in daily life certain things come to mind like “I will be success
today’, I will complete this work in hour, I will be selected for job and so munotes.in
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on. There are many possible results for these things but we are happy
when we get required result. Probability theory deals with experiments
whose outcome is not predictable with certainty. Probability is very useful
concept. These days many field in comp uter science such as machine
learning, computational linguistics, cryptography, computer vision,
robotics other also like science, engineering, medicine and management.
Probability is mathematical calculation to calculate the chance of
occurrence of some happening , we need some bas ic concept on random
experiment , sample space, and events.
Basic concept of probability:
Random experiment: When experiment can be repeated any number of
times under the similar conditions but we get different results on same
experiment, also result is not predictable such experiment is called random
experiment. For.e.g. A coin is tossed, A die is rolled and so on.
Outcomes: The result which we get from random experiment is called
outcomes of random experiment.
Sample space: The set of all possible outcomes of random experiment is
called sample space. The set of sample space is denoted by S and number
of elements of sample space can be written as 𝑛(𝑆). For e.g. A die is
rolled, we get ={1,2,3,4,5,6} , 𝑛(𝑆)=6.
Events: Any subset of the sample space is called an event. Or a set of
sample point which satisfies the required condition is called an events.
Number of elements in event set is denoted by 𝑛(𝐸).For example in the
experiment of throwing of a dia. The sample space is
S = {1, 2, 3, 4, 5, 6 } each of the following can be an event :
i) A: even number i.e. A = { 2, 4, 6} ii) B: multiple of 3 i.e. B = { 3, 6}
iii) C: prime numbers i.e. C = { 2, 3, 5}.
Types of events:
Impossible event: An event which does not occurre d in random
experiment is called impossible event. It is denoted by ∅ set. i. e. 𝑛(∅)=
0. For example getting number 7 when die is rolled. The probability
measure assigned to impossible event is Zero.
Equally likely events : when all events get equal chance of occurrences is
called equally likely events. For e.g. Events of occurrence of head or tail
in tossing a coin are equally likely events.
Certain event: An event which contains all sample space elements is
called certain events. i.e. 𝑛(𝐴)=𝑛(𝑆).
Mutually e xclusive events: Two events A and B of sample space S, it
does not have any common elements are called mutually exclusive events.
In the experiment of throwing of a die A: number less than 2 , B: multiple
of 3. There fore 𝑛(𝐴∩𝐵)=0
Exhaustive events: Two events A and B of sample space S, elements of
event A and B occurred together are called exhaustive events. For e.g. In a munotes.in
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thrown of fair die occurrence of even number and occurrence of odd
number are exhaustive events. There fore 𝑛(𝐴∪𝐵)=1.
Complement event: Let S be sample space and A be any event than
complement of A is denoted by 𝐴̅ is set of elements from sample space S,
which does not belong to A. For e.g. if a die is thrown, S = {1, 2, 3, 4, 5,
6} and A: odd numbers, A = {1, 3, 5}, then 𝐴̅={2,4,6}.
5.3 DEFINITIONS OF PROBABILITY
Probability: For any random experiment, sample space S with required
chance of happing event E than the probability of event E is define as
𝑃(𝐸)=𝑛(𝐸)
𝑛(𝑆)
Basic properties of probability:
1) The probability of an event E lies between 0 and 1. i.e. 0≤𝑃(𝐸)≤1.
2) The probability of impossible event is zero. i.e. 𝑃(∅)=0.
3) The probability of certain event is unity. i.e. 𝑃(𝐸)=1.
4) If A and B are exhaustive events than probability of 𝑃(𝐴∪𝐵)=1.
5) If A and B are mutually exclusive events than probability of 𝑃(𝐴∩
𝐵)=0.
6) If A be any event of sample space than probability of complement of
A is given by 𝑃(𝐴)+𝑃(𝐴̅)=1⇒∴𝑃(𝐴̅)=1−𝑃(𝐴).
Probability Axioms:
Let S be a sample space. A probability function P from the set of all event
in S to the set of real numbers satisfies the following three axioms for all
events A and B in S.
i) 𝑃(𝐴)≥0 .
ii) 𝑃(∅)=0 and𝑃(𝑆)=1.
iii) If A and B are two disjoint sets i.e. 𝐴∩𝐵=∅) than the probability of
the union of A and B is 𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵).
Theorem : Prove that for every event A of sample space S, 0≤𝑃(𝐴)≤
1.
Proof: 𝑆=𝐴∪𝐴̅ , ∅=𝐴∩𝐴̅.
∴1=𝑃(𝑆)=𝑃(𝐴∪𝐴̅)=𝑃(𝐴)+𝑃(𝐴̅)
∴1=𝑃(𝐴)+𝑃(𝐴̅)
⇒𝑃(𝐴)=1−𝑃(𝐴̅)or𝑃(𝐴̅)=1−𝑃(𝐴).
If 𝑃(𝐴)≥0. than P( 𝐴̅)≤1.
∴for every event 𝐴; 0≤𝑃(𝐴)≤1.
Addition theorem of probability:
Theorem: If A and B are two events of sample space S, then probability of
union of A and B is given by 𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∩𝐵). munotes.in
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Proof: A and B are two events of sample space S.
Now from diagram probability of union of two events A and B is given by,
𝑃(𝐴∪𝐵)=𝑃(𝐴∩𝐵ത)+𝑃(𝐴∩𝐵)+𝑃(𝐵∩𝐴̅)
But 𝑃(𝐴∩𝐵ത)=𝑃(𝐴)−𝑃(𝐴∩𝐵) and 𝑃(𝐵∩𝐴̅)=𝑃(𝐵)−𝑃(𝐴∩𝐵).
∴𝑃(𝐴∪𝐵)=𝑃(𝐴)−𝑃(𝐴∩𝐵)+ 𝑃(𝐴∩𝐵)+𝑃(𝐵)−𝑃(𝐴∩𝐵)
∴𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∩𝐵).
Note: The above theorem can be extended to three events A, B and C as
shown below:
𝑃(𝐴∪𝐵∪𝐶)=𝑃(𝐴)+𝑃(𝐵)+𝑃(𝐶)−𝑃(𝐴∩𝐵)−𝑃(𝐵∩𝐶)
−𝑃(𝐶∩𝐴)+𝑃(𝐴∩𝐵∩𝐶)
Example 1: A bag contains 4 black and 6 white balls; two balls are
selected at random. Find the probability that balls are i) both are different
color s. ii) both are of same colors.
Solution: Total number of balls in bag = 4 blacks + 6 white = 10 balls
To select two balls at random, we get
𝑛(𝑆)=𝐶(10,2)=45.
i) A be the event to select both are different colors.
∴𝑛(𝐴)=𝐶(4,1)×𝐶(6,1)=4×6=24.
𝑃(𝐴)=𝑛(𝐴)
𝑛(𝑆)=24
45=0.53.
ii) To select both are same colors.
Let Abe the event to select both are black balls
𝑛(𝐴)=𝐶(4,2)=6
𝑃(𝐴)=𝑛(𝐴)
𝑛(𝑆)=6
45
Let B be the event to select both are white balls.
𝑛(𝐵)=𝐶(6,2)=15 S 𝐴∩𝐵ത 𝐵∩𝐴̅ 𝐴∩ B munotes.in
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𝑃(𝐵)=()
(ௌ)=ଵହ
ସହ .
A and B are disjoint event.
∴ The required probability is
𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)=
ସହ+ଵହ
ସହ=ଶଵ
ସହ=0.467.
Example 2: From 40 tickets marked from 1 to 40, one ticket is drawn at
random. Find the probability that it is marked with a multiple of 3 or 4.
Solution: From 40 tickets marked with 1 to 40, one ticket is drawn at
random
𝑛(𝑆)=𝐶(40,1)=40
it is marked with a multiple of 3 or 4, we need to select in two parts.
Let A be the event to select multiple of 3,
i.e. A = { 3,6,9,….,39}
𝑛(𝐴)=𝐶(13,1)=13
𝑃(𝐴)=𝑛(𝐴)
𝑛(𝑆)=13
40
Let B be the event to select multiple of 4.
i.e. B = {4,8,12, …., 40}
𝑛(𝐵)=𝐶(10,1)=10
𝑃(𝐵)=𝑛(𝐵)
𝑛(𝑠)=10
40.
Here A and B are not disjoint.
𝐴∩𝐵be the event to select multiple of 3 and 4.
i.e. 𝐴∩𝐵 = {12,24,36}
𝑛(𝐴∩𝐵)=𝐶(3,1)=3
𝑃(𝐴∩𝐵)=𝑛(𝐴∩𝐵)
𝑛(𝑆)=3
40
∴ The required probability is
𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∩𝐵)=13
40+10
40−3
40=20
40=0.5.
Example 3: If the probability is 0.45 that a program development job; 0.8
that a networking job applicant has a graduate degree and 0.35 that applied
for both. Find the probability that applied for atleast one of jobs. If number
of graduate are 500 then how many are not applied for jobs?
Solution: Let Probability of program development job= 𝑃(𝐴)=0.45.
Probability of networking job = 𝑃(𝐵)=0.8.
Probability of both jobs = 𝑃(𝐴∩𝐵)=0.35. munotes.in
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Probability of atleast one i.e. to find 𝑃(𝐴∪𝐵).
𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∩𝐵)
𝑃(𝐴∪𝐵)=0.45+0.8−0.35=0.9
Now there are 500 application, first to find probability that not applied for
job.
𝑃(𝐴∪𝐵തതതതതതത)=1−𝑃(𝐴∪𝐵)=1−0.9=0.1
Number of graduate not applied for job = 0.1×500=50 .
Check your Progress:
1. A card is drawn from pack of 52 cards at random. Find the probability
that it is a face card or a diamond card.
2. If 𝑃(𝐴)=ଷ
଼and (𝐵)=ହ
଼ , 𝑃(𝐴∪𝐵)=
଼ than find i) 𝑃(𝐴∪𝐵തതതതതതത) ii)
𝑃(𝐴∩𝐵).
3. In a class of 60 students, 50 passed in computers, 40 passed in
mathematics and 35 passed in both. What is the probability that a
student selected at random has i) Passed in atleast one subject, ii)
failed in both the subjects, iii) passed in only one subject.
5.4 CONDITIONAL PROBABILITY
In many case we come across occurrence of an event A and for the same
are required to find out the probability of occurrence an event B which
depend on event A . This kind of problem is called conditional probability
problems.
Definition: Let A and B be two events. The conditional probability of
event B, if an event A has occurred is defined by the relation,
𝑃(𝐵|𝐴)=(∩)
()if and only if 𝑃(𝐴)>0.
In case when 𝑃(𝐴)=0,𝑃(𝐵|𝐴) is not define because 𝑃(𝐵∩𝐴)=0 and
𝑃(𝐵|𝐴)=
which is an indeterminate quantity.
Similarly, Let A and B be two events. The conditional probability of event
A, if an event B has occurred is defined by the relation,
𝑃(𝐴|𝐵)=(∩)
() If and only if 𝑃(𝐵)>0.
Example 4: A pair of fair dice is rolled. What is the probability that the
sum of upper most face is 6, given that both of the nu mbers are odd?
Solution: A pair of fair dice is rolled, therefore 𝑛(𝑆)=36.
A to select both are odd number, i.e. A = {(1,1), (1,3), (1,5), (3,1), (3,3),
(3,5), (5,1),(5,3), (5,5)}.
𝑃(𝐴)=𝑛(𝐴)
𝑛(𝑆)=9
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B is event that the sum is 6, i.e. B = { ((1,5),(2,4), (3,3),(4,2), (5,1)}.
𝑃(𝐵)=𝑛(𝐵)
𝑛(𝑆)=5
36
𝐴∩𝐵 = { (1,5), (3,3), (5,1)}
𝑃(𝐴∩𝐵)=𝑛(𝐴∩𝐵)
𝑛(𝑆)=3
36
By the definition of conditional probability,
𝑃(𝐵|𝐴)=𝑃(𝐴∩𝐵)
𝑃(𝐴)=336ൗ
936ൗ=1
3.
Example 5: If A and B are two events of sample space S, such th at
𝑃(𝐴)=0.85,𝑃(𝐵)=0.7and 𝑃(𝐴∪𝐵)=0.95. Find i) 𝑃(𝐴∩𝐵), ii)
𝑃(𝐴|𝐵), iii) 𝑃(𝐵|𝐴).
Solution: Given that 𝑃(𝐴)=0.85,𝑃(𝐵)=0.7and 𝑃(𝐴∪𝐵)=0.95.
i) By Addition theorem,
𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∩𝐵)
0.95=0.85+0.7−𝑃(𝐴∩𝐵)
𝑃(𝐴∩𝐵)=1.55−0.95=0.6.
ii) By the definition of conditional probability ,
𝑃(𝐴|𝐵)=(∩)
()=.
.=0.857.
iii) 𝑃(𝐵|𝐴)=(∩)
()=.
.଼ହ=0.706
Example 6: An urn A contains 4 Red and 5 Green balls. Another urn B
contains 5 Red and 6 Green balls. A ball is transferred from the ur n A to
the urn B, then a ball is drawn from urn B. find the probability that it is
Red.
Solution: Here there are two cases of transferring a ball from urn A to B.
Case I: When Red ball is transferred from urn A to B.
There for probability of Red ball from urn A is 𝑃(𝑅)=ସ
ଽ
After transfer of red ball, urn B contains 6 Red and 6 Green balls.
Now probability of red ball from urn B = 𝑃(𝑅|𝑅)×𝑃(𝑅)=
ଵଶ×ସ
ଽ=
ଶସ
ଵ଼.
Case II: When Green ball is transferred from urn A to B.
There for probability of Green ball from urn A is 𝑃(𝐺)=ହ
ଽ
After transfer of red ball, urn B contains 5 Red and 7 Green balls. munotes.in
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Now probability of red ball from urn B = 𝑃(𝑅|𝐺)×𝑃(𝐺)=ହ
ଵଶ×ହ
ଽ=
ଶହ
ଵ଼.
Therefore required probability =ଶସ
ଵ଼+ଶହ
ଵ଼=ସଽ
ଵ଼=0.4537.
Check your progress:
1. A family has two children. What is the probability that both are boys,
given at least one is boy?
2. Two dice are rolled. What is the condition probability that the sum of
the numbers on the dice exceeds 8, given that the first shows 4 ?
3. Consider a medical test that screens for a COVID -19 in 10 people in
1000. Suppose that the false positive rate is 4% and the false negative
rate is 1%. Then 99% of the time a person who has the condition tests
positive for it, and 96% of the time a person who does not have the
condition tests negative for it. a) What is the probability that a
randomly chosen person who tests positive for the COVID -19 actually
has the disease? b) What is the probability that a randomly chosen
person who tests negative for th e COVID -19 does not indeed have the
disease?
5.4.1 Independent and Dependent Eve nts, Mutually Exclusive Events:
Independent events:
Two events are said to be independent if the occurrence of one of them
does not affect and is not affected by the occurrence or non -occurrence of
other.
i.e. 𝑃൫𝐵𝐴ൗ൯=𝑃(𝐵) or 𝑃൫𝐴𝐵ൗ൯=𝑃(𝐴).
Multiplication theorem of probability: If A and B are any two events
associated with an experiment, then the probability of simultaneous
occurrence of events A and B is given by
𝑃(𝐴∩𝐵)=𝑃(𝐴)𝑃൫𝐵𝐴ൗ൯
Where 𝑃൫𝐵𝐴ൗ൯ denotes the conditional probability of event B given that
event A has already occur red.
OR
𝑃(𝐴∩𝐵)=𝑃(𝐵)𝑃൫𝐴𝐵ൗ൯
Where 𝑃൫𝐴𝐵ൗ൯ denotes the conditional probability of event A given that
event B has already occurred.
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11.5.1 For Independent events multiplication theorem:
If A and B are independent events then multiplication theorem can be
written as,
𝑷(𝑨∩𝑩)=𝑷(𝑨)𝑷(𝑩)
Proof. Multiplication theorem can be given by,
If A and B are any two events associated with an experiment, then the
probability of simultaneous occurrenc e of events A and B is given by
𝑃(𝐴∩𝐵)=𝑃(𝐴)𝑃൫𝐵𝐴ൗ൯
By definition of independent events, 𝑃൫𝐵𝐴ൗ൯=𝑃(𝐵) or 𝑃൫𝐴𝐵ൗ൯=𝑃(𝐴).
∴𝑷(𝑨∩𝑩)=𝑷(𝑨)𝑷(𝑩).
Note:
1) If A and B are independent event then, 𝐴̅ and 𝐵ത are independent event.
2) If A and B are independent event then, 𝐴̅ and B are independent event.
3) If A and B are independent event then, A and 𝐵ത are independent event.
Example 7: Manish and Mandar are trying to make Software for
company. Probability that Manish can be succes s is ଵ
ହ and Mandar can be
success is ଷ
ହ, both are doing independently. Find the probability that i)
both are success. ii) Atleast one will get success. iii) None of them will
success. iv) Only Mandar will success but Manish will not success.
Solution: Let probability that Manish will success is 𝑃(𝐴)=ଵ
ହ=0.2.
Therefore probability that Manish will not success is 𝑃(𝐴̅)=1−𝑃(𝐴)=
1−0.2=0.8.
Probability that Mandar will success is 𝑃(𝐵)=ଷ
ହ=0.6.
Therefore probability that Mandar will not success is 𝑃(𝐵ത)=1−𝑃(𝐵)=
1−0.6=0.4.
i) Both are success i.e. 𝑃(𝐴∩𝐵).
𝑃(𝐴∩𝐵)=𝑃(𝐴)×𝑃(𝐵)=0.2×0.6=0.12∵A and B are
independent events.
ii) Atleast one will get success. i.e. 𝑃(𝐴∪𝐵)
By addition theorem,
𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∩𝐵)=0.2+0.6−0.12=0.68.
iii) None of them will success. 𝑃(𝐴∪𝐵തതതതതതത) or 𝑃(𝐴̅∩𝐵ത)
[ ByDeMorgan’s law both are same]
𝑃(𝐴∪𝐵തതതതതതത)=1−𝑃(𝐴∪𝐵)=1−0.68=0.32.
Or
If A and B are independent than 𝐴̅ and 𝐵ത are also independent. munotes.in
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𝑃(𝐴̅∩𝐵ത)=𝑃(𝐴̅)×𝑃(𝐵ത)=0.8×0.4=0.32.
iv) Only Mandar will success but Manish will not success. i.e. 𝑃(𝐴̅∩𝐵).
𝑃(𝐴̅∩𝐵)=𝑃(𝐴̅)×𝑃(𝐵)=0.8×0.6=0.48
Example 8: 50 coding done by two students A and B, both are trying
independently. Number of correct coding by student A is 35 and
student B is 40. Find the probability of only one of them will do
correct coding.
Solution: Let probability of student A get correct coding is 𝑃(𝐴)=ଷହ
ହ=
0.7
Probability of student A get wrong coding is 𝑃(𝐴̅)=1−0.7=0.3
Probability of student B get correct coding is 𝑃(𝐵)=ସ
ହ=0.8
Probability of student B get wrong coding is 𝑃(𝐵ത)=1−0.8=0.2.
The probabi lity of only one of them will do correct coding.
i.e. A will correct than B will not or B will correct than A will not.
𝑃(𝐴∩𝐵ത)+𝑃(𝐵∩𝐴̅)=𝑃(𝐴)×𝑃(𝐵ത)+𝑃(𝐵)×𝑃(𝐴̅).
=0.7×0.2+0.8×0.3=0.14+0.24
=0.38
Example 9: Given that 𝑃(𝐴)=ଷ
,𝑃(𝐵)=ଶ
, if A and B are independent
events than find i) 𝑃(𝐴∩𝐵), ii) 𝑃(𝐵ത), iii) 𝑃(𝐴∪𝐵), iv) 𝑃(𝐴̅∩𝐵ത).
Solution: Given that 𝑃(𝐴)=ଷ
,𝑃(𝐵)=ଶ
.
i) A and B are independent events,
∴𝑃(𝐴∩𝐵)=𝑃(𝐴)×𝑃(𝐵)=3
7×2
7=6
49=0.122
𝑖𝑖) 𝑃(𝐵ത)=1−𝑃(𝐵)=1−ଶ
=ହ
=0.714.
iii) By addition theorem,
𝑃(𝐴∪𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∩𝐵)=3
7+2
7−6
49=29
49=0.592.
iv) 𝑃(𝐴̅∩𝐵ത)=𝑃(𝐴∪𝐵തതതതതതത)=1−𝑃(𝐴∪𝐵)=1−0.592=0.408.
Check your progress:
1. If 𝑃(𝐴)=ଶ
ହ ,𝑃(𝐵)=ଵ
ଷ and if A and B are independent events, find
(𝑖)𝑃(𝐴∩𝐵),(𝑖𝑖)𝑃(𝐴∪𝐵),(𝑖𝑖𝑖)𝑃(𝐴̅∩𝐵ത).
2. The probability that A , B and C can solve the same problem
independently are ଵ
ଷ,ଶ
ହ𝑎𝑛𝑑ଷ
ସ respectively. Find the probability that i)
the problem remain unsolved, ii) the problem is solved, iii) only one
of them solve the pro blem.
3. The probability that Ram can shoot a target is ଶ
ହ and probability of
Laxman can shoot at the same target is ସ
ହ. A and B shot independently.
Find the probability that (i) the target is not shot at all, (ii) the target is
shot by at least one of them. (iii) the target shot by only one of them.
iv) target shot by both. munotes.in
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5.5 PROBABILITY DISTRIBU TIONS
In order to under stand the behavior of a random variable, we may want to
look at its average value. For probability we need to find Average is
called expected value of random variable X. for that first we have to learn
some basic concept of random variable.
Random Variable: A probability measurable real valued functions, say
X, defined over the sample space of a random experiment with respective
probability is called a random variable.
Types of random variables: There are two type of random variable.
Discrete Rando m Variable: A random variable is said to be discrete
random variable if it takes finite or countably infinite number of values.
Thus discrete random variable takes only isolated values.
Continuous Random variable: A random variable is continuous if its set
of possible values consists of an entire interval on the number line.
Probability Distribution of a random variable: All possible values of
the random variable, along with its corresponding probabilities, so
that∑ =1
ୀ1, is called a probability distribution of a random variable.
The probability function always follows the following properties:
i) 𝑃(𝑥)≥0 for all value of 𝑖.
ii) ∑𝑃=1
ୀଵ.
The set of values 𝑥 with their probability 𝑃 constitute a discrete
probability distribution of th e discrete variable X.
For e.g. Three coins are tossed, the probability distribution of the discrete
variable X is getting head. X= 𝑥 0 1 2 3 𝑃(𝑥) 18 38 38 18
5.6 MATHEMATICAL EXPECTA TION
All the probability information of a random variable is contained in
probability mass function for random variable, it is often useful to
consider various numerical characteristics of that random variable. One
such number is the expectation of a random variable.
If random variable X takes values 𝑥ଵ,𝑥ଶ,…..𝑥 with corresponding
probabilities 𝑃ଵ,𝑃ଶ,…….𝑃 respectively, then expectation of random
variable X is
𝐸(𝑋)=∑𝑝𝑥
ୀଵwhere ∑𝑃=1
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Example 10 : In Vijay sales every day sale of number of laptops with his
past experience the probability per d ay are given below: No. of laptop 0 1 2 3 4 5 Probability 0.05 0.15 0.25 0.2 0.15 0.2
Find his expected number of laptops can be sale?
Solution: Let X be the random variable that denote number of laptop sale
per day.
To calculate expected value, 𝐸(𝑋)=∑𝑝𝑥
ୀଵ
𝐸(𝑋)=(0×0.05)+(1×0.15)+(2×0.25)+(3×0.2)+(4×0.15)
+(5×0.2)
𝐸(𝑋)= 2.85 ~3
Therefore expected number of laptops sale per day is 3.
Example 11 : A random variable X has probability mass function as
follow: X=𝑥 -1 0 1 2 3 P(𝑥) K 0.2 0.3 2k 2k
Find the value of k, and expected value.
Solution: A random variable X has probability mass function,
∑𝑃=1
ୀଵ.
⇒ k + 0.2 + 0.3 + 2k +2k = 1
⇒5k = 0.5
⇒ k = 0.1
Therefore the probability distribution of random variable X is X=𝑥 -1 0 1 2 3 P(𝑥) 0.1 0.2 0.3 0.2 0.2
To calculate expected value, 𝐸(𝑋)=∑𝑝𝑥
ୀଵ
𝐸(𝑋)=(−1×0.1)+(0×0.2)+(1×0.3)+(2×0.2)+(3×0.2)=
1.2 .
Example 12 : A box contains 5 white and 7 black balls. A person draws 3
balls at random. He gets Rs. 50 for every white ball and losses Rs. 10
every black ball. Find the expectation of him.
Solution: Total number of balls in box = 5 white + 7 black = 12 balls.
To sele ct 3 balls at random, 𝑛(𝑠)=𝐶(12,3)=ଵଶ×ଵଵ×ଵ
ଷ×ଶ×ଵ=220.
Let A be the event getting white ball.
A takes value of 0, 1, 2 and 3 white ball.
Case I : no white ball. i.e. A = 0, munotes.in
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𝑃(𝐴=0)=𝐶(7,3)
220=35
220
Case II: one white ball i.e. A = 1,
𝑃(𝐴=1)=𝐶(5,1)×𝐶(7,2)
220=105
220
Case III: two white balls i.e. A = 2,
𝑃(𝐴=2)=𝐶(5,2)×𝐶(7,1)
220=70
220
Case IV: three white balls i.e. A = 3,
𝑃(𝐴=3)=𝐶(5,3)
220=10
220
Now let X be amount he get from the game.
Therefore the probability distribution of X is as follows: X= 𝑥 -30 30 90 150 P(𝑥) 35
220 105220 70220 10220
To calculate expected value, 𝐸(𝑋)=∑𝑝𝑥
ୀଵ
𝐸(𝑋)=ቀ−30×ଷହ
ଶଶቁ+ቀ30×ଵହ
ଶଶቁ+ቀ90×
ଶଶቁ+ቀ150×ଵ
ଶଶቁ=Rs.
45.
5.7 COMBINATORIAL ANALYSIS
Multiplication Rule:
If the procedure can be broken into first and second stages, and if there are
m possible outcomes for the first stage and for each of these outcomes,
there are n possible outcomes for second stage, then the total procedure
can be carried out in the designa te order, in 𝑚×𝑛 ways.
This principle can be extended to a general form as follows:
Theorem : If a process consists of n steps, and
i) The first step can be performed by 𝑛ଵ ways.
ii) The second step can be performed by 𝑛ଶ ways.
iii) The 𝑖௧ step can be performed by 𝑛 ways.
Then the whole process can be completed by 𝑛ଵ×𝑛ଶ×……×𝑛
different ways.
Example 13 : There are 8 men and 7 women in a drama company. How
many way the director has to choose a couple to play lead roles in a stage
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Solution: The director can choose a man (task 1) in 8 ways and then a
woman (task 2) in 7 ways. Then by multiplication rul e he can choose a
couple from 8×7=56 ways.
Example 14 : How many four digits numbers can be formed contains each
of the digits 7, 8, and 9 exactly once?
Solution: To construct four digits number we have four places.
_______ _______ _______ _________
Thousand place Hundred place Ten place Unit place
First for ‘7’ there are 4 places, for ‘8’ there are 3 places and for ‘9’ there
are 2 places. For last digit, we can choose any of 0,1,2,3,4,5,6 so there will
be 7 digits.
Thus these can be done by 4×3×2×7=168 ways.
Example 15 : To generate typical personal identification number (PIN) is
a sequence of any four symbols chosen from the letters in the alphabet and
the digits , How many different PIN’s are generated?
i) repetition is not allowed.
ii) repetition is allowed.
Solution: There are 26 letters of alphabets and 10 digits. Total different
symbols are 36.
i) Repetition is not allowed:
There are four place to generate PIN with four symbols,
First place can be filled by 36 ways, second place can be filled by 35
ways, third place can be filled by 34 ways and last fourth place can be
filled by 33 ways.
By the multiplication rule,
Therefore these can be done by 36×35×34×33=1,413,720 ways.
ii) Repetition is allowed:
Since repetition is allowed, so each place can be filled by 36 ways,
By multiplication rule,
These can be done by 36×36×36×36=1,679,616 ways.
Check your progress:
1. A license plate can be made by 2 letters followed by 3 digits. How
many different license plates can be made if i) repetition is not allowed.
ii) Repetition is allowed.
2. Mr. Modi buying a personal computer system is offered a choice of 4
models of basic units, 2 models keyboard, and 3 models of printer.
How many distinct systems can be purchased?
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Counting elements of disjoint sets with Addition Rule:
In above section we have discussed counting problem that can be solved
using possibility tree. Here we discuss counting problem that can be
solved using the operation sets like union , intersection and the difference
between two sets.
The addition rule:
If a task can be performed in m ways and another task in n ways assuming
that these two tasks cannot perform simultaneously, then the performing
either task can be accomplished in any one of the 𝑚+𝑛 ways.
In general from as follows:
If there are 𝑛ଵ,𝑛ଶ,𝑛ଷ,……,𝑛 different objects in m different sets
respectively and the sets are disjoint, then the number of ways to select an
object from one of the m sets is 𝑛ଵ+𝑛ଶ+ 𝑛ଷ+ ……+𝑛
Example 16 : How many different number of signals that can be sent by 5
flags of different colours taking one or more at a time ?
Solution: Let number of signal made by one colour flag =5 ways.
Number of signal made by two colours flag =5×4=20 ways.
Number of signal made by three flag colours =5×4×3=60 ways.
Number of signal made by four flag colours =5×4×3×2=120 ways.
Number of signal made by five flag colours =5×4×3×2×1=120
ways.
Using Addition rule we get,
Therefore total number of signals =5+20+60+120+120=325
ways.
Example 17 : There are 4 different English books, 5 different Hindi books
and 7 different Marathi books. How many ways are there to pick up an
pair of two books not both with the same subjects?
Solution: One English and one Hindi book is chosen, that selection can be
done by =4×5=20 ways.
One Engli sh and one Marathi book is chosen, that selection can be done
by =4×7=28 ways.
One Hindi and one Marathi book is chosen, that selection can be done by
=5×7=35 ways.
These three types of selection are disjoint, therefore by addition rule,
Total selection ca n be done by =20+28+35=83 ways.
Additive Principle with Disjoint sets:
Given two sets A and B, both sets are disjoint i.e. if 𝐴∩𝐵=∅ , than
|𝐴∪𝐵|=|𝐴|+|𝐵|. munotes.in
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Example 18 : In college 200 students visit to canteen every day of which
80 likes coffee and 70 likes tea. If no one student like both than find i)
number of students like atleast one of them? ii) number of students like
none of them?
Solution: Total number of stud ents = 200
Total number of students who like coffee = |𝐴|=80.
Total number of students who like tea = |𝐵|=70.
Total number of students like atleast one |𝐴∪𝐵|=|𝐴|+|𝐵|=80+
70=150.
Total number of students like none of them =200−150=50.
Definition: An r -combination of n distinct objects is an unordered
selection, or subset, of r out of the n objects. We use 𝐶(𝑛,𝑟)𝑜𝑟𝐶 to
denote the number of r -combinations. This number is called as binomial
number.
If 𝑥ଵ,𝑥ଶ,𝑥ଷ,…..,𝑥 are n distinct o bjects, and r is any integer, with
1≤𝑟≤𝑛. Therefore selecting r -objects from n objects is given by
𝐶(𝑛,𝑟)=𝑛!
𝑟!(𝑛−𝑟)!
Example 19 : How many elements of set 3 -bit string with weight 2?
Solution : there are 3 -bit with weight 2, i.e. 𝑛=3,𝑟=2.
These can be done by = 𝐶(𝑛,𝑟)=𝐶(3,2)=3.
Therefore the bit string is 011, 101, 110.
Example 20 : A bag contains 4 red marbles and 5 green marbles. Find the
number of ways that 4 marbles can be selected from the bag, if selection
contain i) No restriction of colors. ii) all are of same colors.
Solution: Total number of marbles: 4 Red + 5 Green = 9 marbles
To select 4 marbles from the bag with condition,
i) No restriction of colors:
These can be done by : 𝐶(9,4)=126 ways.
ii) All are of same colors:
First select th e colors by :𝐶(2,1)=2.
If all is Red in colors than these can be done by =𝐶(4,4)=1.
If all is Green in colors then these can be done by =𝐶(5,4)=5.
Therefore total number of ways =2×1×5=10 ways.
Example 21 : There are 10 members in a societ y who are eligible to attend
annual meeting. Find the number of ways a 4 members can be selected
that
i) No restriction munotes.in
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ii) If 2 of them will not attend meeting together.
iii) If 2 of them will always attend meeting together.
Solution:
i) To select 4 members from 10 members, it can be done by =
𝐶(10,4)=210 ways.
ii) If 2 of them will not attend meeting together,
Let A and B denote the 2 members who will not attend meeting
together.
i.e. A or B but not both are together , these can be done by =2×
𝐶(8,3)=112 ways.
It possible that both will not attend meeting, i.e. Neither A nor B
will attend meeting, these can be done by =𝐶( 8,2)=28 ways.
Therefore total number of ways = 112 + 28 =140 ways.
iii) If 2 of them will attend meeting together,
Let A an d B denote the 2 members who will attend meeting together.
i.e. A or B =𝐶(8,2)=28 ways.
It possible that both will not attend meeting, i.e. Neither A nor B will
attend meeting, these can be done by =𝐶( 8,4)=70 ways.
Therefore total number of ways = 28 +70 = 88 ways.
Example 21 : How many diagonal has a regular polygon with n sides?
Solution : The regular polygon with n sides has n vertices. Any two
vertices determine either a side or diagonal. Therefore these can be done
by =𝐶 (𝑛 ,2)= (ିଵ)
ଶ . But there are n sides which are not diagonal.
Therefore total number of diagonals are = (ିଵ)
ଶ− 𝑛= మି
ଶ−ଶ
ଶ=
మି ଷ
ଶ= (ିଷ)
ଶ diagonals.
r-combinations with Repetition Allowed:
Till now, we have seen the formula for the number of combinations when
r objects are chosen from the collection of n distinct objects. The
following results is very important to find the number of selection of n
objects when not all n are distinct.
The num ber of selection with repetition of r objects chosen from n types of
objects is
𝐶(𝑛+𝑟−1,𝑟)
Example 22 : How many ways are there to fill a box with a dozen marbles
chosen five different colors of marbles with the requirement that at least
one fruit of each colors is picked?
Solution: One can pick one marble of each colors and then the remaining
seven marbles in any way. There is no choice in picking one marble of munotes.in
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each type. The choice occurs in picking the remaining 7 marbles from 5
colors. By the resul t of r -combination with repetition allowed,
These can be done by =𝐶(5+7−1,7)=𝐶(11,7)=330 ways.
Example 23 : How many solution does the following equation 𝑥ଵ+𝑥ଶ+
𝑥ଷ+𝑥ସ=15 have 𝑥ଵ,𝑥ଶ,𝑥ଷ, and 𝑥ସ are non -negative integers?
Solution: Assume we have four types of unknown 𝑥ଵ,𝑥ଶ,𝑥ଷ, and 𝑥ସ .
There are 15 items or units (since we are looking for an integer solution).
Every time an item is selected it adds one to the type it picked it up.
Observe that a solution corresponds to a w ay of selecting 15 items from
set of four elements. Therefore, it is equal to r -combinations with
repetition allowed from set with four elements, we have
𝐶(4+15−1,15)=𝐶(18,15)=𝐶(18,3)=18×17×16
3×2×1=816
Example 24 : In how many ways can a teacher choose one or more
students from 5 students?
Solution: Let set of student are 5, therefore total number of subsets are
2ହ=32 .
To select one or more students, we must deleted empty set,.
Therefore total number of selection = 32 – 1= 31 ways.
5.8 COMBINATIONS, STIRLING’S APPROXIMATION TO N!
A helpful and commonly used approximate relationship for the evaluation
of the fac torials of large numbers is Stirling’s approximation. It is a good
approximation, leading to accurate results even for small values of n. it is
given by
𝑛!≈
√2𝜋𝑛
Where e = 2:71828 is the natural base of logarithms .
5.9 RELATION OF PROBABILITY TO POINT SET THEORY, EULER OR VENN DIAGRAMS AND
PROBABILITY
a) Relation of Probability to Point Set Theory :
In discrete probability we assume well defined experiment such as
flipping a coin or rolling a die. Each individual result which could
occur is called an outcome. The set of all outcomes is called sample
space, and any subset of the sample space is called an event.
The union of two or more sets is the set that contains all the elements of
the two or more sets. Union is denoted by the symbol ∪.The general munotes.in
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probability addition rule for the union events states that 𝑃(𝐴∪𝐵)=
𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∩𝐵). Where 𝐴∩𝐵 is the intersection of the two
sets.
Euler or Venn Diagrams and Probability:
In probability, a Venn diagram is a figure with one or more circles inside a
rectangle that describes logical relation between events. The rectangle in a
Venn diagram represents the sample space or the universal set, that is , the
set of all possible outco mes. A circle inside the rectangle represents an
event, that is, a subset of the sample space. We consider the following
Venn diagram involving two events, 𝐴 and 𝐵.
In the above diagram, we have two events 𝐴 and 𝐵 within the sample
space 𝑆(or Universal set)
White and Red region represent event A, Black and Red region represent
event B, only Red region represent 𝐴∩𝐵, and White, Black and Red
region together represents 𝐴∪𝐵. Also Blue region represent 𝐴∪𝐵തതതതതതത.
If the circle s do not overlap than 𝐴 and 𝐵 are mutually exclusive
events.
i.e𝐴∩𝐵=∅.
If the circles are overlap than 𝐴 and 𝐵 are intersecting each other .
i.e𝐴∩𝐵≠∅.
The region outside both the circles but within the rectangle represents
complement of union of both the events. i.e. 𝐴∪𝐵തതതതതതത.
Within each divided region of a Venn diagram, we can add data in one of
the following ways:
The outcomes of the event,
The number of outcomes in the event,
The probability of the event.
,
𝐴
a
n
d
𝐵
,
𝐴
a
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d
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5.10 LET US SUM UP
In this unit we have learn :
Basic concept and Definitions of Probability.
Conditional Probability.
Independent and Dependent Events, Mutually Exclusive Events.
Probability Distributions for discrete distribution.
Mathematical Expectation for probability distribution.
Relation between Population, Sample Mean, and Variance.
Combinations, Stirling’s Approximation to n!.
Relations of Probability to Point Set Theory with represent Euler or
Venn Diagrams.
5.11 UNIT END EXERCISES
1. A card is drawn at random from well shuffled pack of card find the
probability that it is red or king card.
2. There are 30 tickets bearing numbers from 1 to 15 in a bag. One ticket
is drawn from the bag at random. Find the probability that the ticket
bears a number , which is even , or a multiple of 3.
3. In a group of 200 persons, 1 00 like sweet food items, 120 like salty
food items and 50 like both. A person is selected at random find the
probability that the person (i). Like sweet food items but not salty
food items (ii). Likes neither.
4. A bag contains 7 white balls & 5 red balls. One ball is drawn from
bag and it is rep laced after noting its color. In the second draw again
one ball is drawn and its color is noted. The probability of the event
that both the balls drawn are of different colors.
5. The probability of A winning a race is ଵ
ଷ& that B wins a race is ଷ
ହ. Find
the probability that (a). either of the two wins a race.b), no one wins
the race.
6. Three machines A, B & C manufact ure respectively 0.3, 0.5 & 0.2 of
the total production. The percentage of defective items produced by
A, B & C is 4 , 3 & 2 percent respectively. for an item chosen at
random , what is the probability it is defective.
7. An urn A contains 3 white & 5 black balls. Another urn B contains 5
white & 7 black balls. A ball is transferred from the urn A to the urn
B, then a ball is drawn from urn B. find the probability that it is white.
8. A husband & wife appear in an interview for two vacancies in the
same post. The probability of husband selection is ଵ
& that of wife’s
selection is ଵ
ହ. What is the probability that, a). both of them will be munotes.in
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selected.b). only one of them will be selected.c). none of them will be
selected?
9. A problem statistics is given to 3 students A,B & C whose chances of
solving if are ଵ
ଶ, ଷ
ସ&ଵ
ସ respectively. What is the probability that the
problem will be solved?
10. A bag contains 8 white & 6 red balls. Find the probability of drawing
2 balls of the same color.
11. Find the probability of drawing an ace or a spade or both from a deck
of cards?
12. A can hit a target 3 times in a 5 shots, B 2 times in 5 shots & C 3 times
in a 4 shots. they fire a volley. What is the probability that a).2 shots
hit? b). at least 2 shots hit?
13. A purse contains 2 silver & 4 cooper coins & a second purse contains
4 silver & 4 cooper coins. If a coin is selected at random from one of
the two purses, what is the probability that it is a silver coin?
14. The contain of a three urns are : 1 white, 2 red, 3 green balls; 2 white,
1 red, 1 green b alls & 4 white, 5 red, 3 green balls. Two balls are
drawn from an urn chosen at random. This are found to be 1 white &
1 green. Find the probability that the balls so drawn come from the
second urn.
15. Three machines A,B& C produced identical items. Of there respective
output 2%, 4% & 5 % of items are faulty. On a certain day A has
produced 30 % of the total output, B has produced 25% & C the
remainder. An item selected at random is found to be faulty. What are
the chances that it was produced by the machin e with the highest
output?
16. A person speaks truth 3 times out of 7. When a die is thrown, he says
that the result is a 1. What is the probability that it is actually a 1?
17. There are three radio stations A, B and C which can be received in a
city of 1000 families. The following information is available on the
basis of a survey:
(a) 1200 families listen to radio station A
(b) 1100 families listen to radio station B.
(c) 800 families listen to radio station C.
(d) 865 families listen to radio station A & B.
(e) 450 families listen to radio station A & C.
(f) 400 families listen to radio station B & C.
(g) 100 families listen to radio station A,B & C.
The probability that a family selected at random listens at least to one
radio station.
18. The probability distribution of a random variable x is as follows. X 1 3 5 7 9 P(x) K 2k 3k 3k K munotes.in
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Find value of (i). K (ii). E(x)
19. A player tossed 3 coins. He wins Rs. 200 if all 3 coins show tail, Rs.
100 if 2 coins show tail, Rs. 50 if one tail appears and loses Rs. 40 if
no tail appears. Find his mathematical expectation.
20. The probability distribution of daily demand of cell ph ones in a mobile
gallery is given below. Find the expected mean .
Demand 5 10 15 20 Probability 0.4 0.22 0.28 0.10
21. If 𝑃(𝐴)=ସ
ଵହ ,𝑃(𝐵)=
ଵହ and if A and B are independent events, find
(𝑖)𝑃(𝐴∩𝐵),(𝑖𝑖)𝑃(𝐴∪𝐵),(𝑖𝑖𝑖)𝑃(𝐴̅∩𝐵ത).
22. If 𝑃(𝐴)=ହ
ଽ ,𝑃(𝐵ത)=ଶ
ଽ and if A and B are independent events, find
(𝑖)𝑃(𝐴∩𝐵),(𝑖𝑖)𝑃(𝐴∪𝐵),(𝑖𝑖𝑖)𝑃(𝐴̅∩𝐵ത).
23. If 𝑃(𝐴)=0.65 ,𝑃(𝐵)=0.75 and 𝑃(𝐴∩𝐵)=0.45, where A and B
are events of sample space S , find (𝑖)𝑃(𝐴|𝐵) ,(𝑖𝑖)𝑃(𝐴∪𝐵),
(𝑖𝑖𝑖)𝑃(𝐴̅∩𝐵ത).
24. A box containing 5 red and 3 black balls, 3 balls are drawn at random
from box. Find the expected number of red balls drawn.
25. Two fair dice are rolled. X denotes the sum of the numbers appearing
on the uppermost faces of the dice. Find the expected value.
26. A bag contains 5 black marbles and 6 white marbles. Find the number
of ways that five marbles can be drawn from the bag such that it
contains i) No restriction ii) no black marbles, iii) 3 black and 2 white,
iv) at least 4 black, v) All are of same colors .
27. A student is to answer 8 out of 10 questions on an exam. Find the
number of ways that the student can chose the 8 questions if i) No
restriction, ii) student must answer the first 4 questions, iii) student
must answer atleast 4 out of the five questions.
28. There are 12 points in a given plane, no three on the same line. i) How
many triangle are determine by the points? ii) How many of these
triangle contain a particular point as a vertex?
29. How many committees of two or more can be selected from 8 people?
30. Find the number of combinations if the letters of the letters of the
word EXAMINATION taken out at a time.
5.12 LIST OF REFERENCES
Statistics byMurry R. Spiegel, Larry J. Stephens. Publication
McGRAWHILL INTERNATIONAL.
Fundamental Mathematics and Statistics by S.C. Gupta and V.K
kapoor Mathematical Statistics by J.N. Kapur and H.C. Saxena.
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6
ELEMENTARY SAMPLING
THEORY
Unit Structure
6.1 Objective
6.2 Introduction
6.3 Sampling Theory
6.3.1 Random Samples and Random Numbers,
6.3.2 Sampling With and Without Replacement,
6.4 Sampling Distributions,
6.4.1 Sampling Distribution of Means,
6.4.2 Sampling Distribution of Proportions,
6.4.3 Sampling Distributions of Differences and Sums,
6.5 Standard Errors,
6.6 Summary
6.7 Exercise
6.8 List of References
6.1 OBJECTIVE
After going through this chapter you will able to know:
Sampling and its requirements in statistics.
Random sampling with and without replacement.
Sampling distribution of Mean, Proportions, difference and sum.
Standard errors in sampling distribution.
Some software to use for sampling.
6.2 INTRODUCTION
In the previous chapters, we have discussed probability theory. In this
chapter, we will introduce some basic concepts in statistics. The basic idea
of statistical inference is to assume that the observed data is generated
from some unknown probability distribution, which is often assumed to
have a known functional form up to some unknown parameters. The
purpose of statistical inference is to develop theory and methods to make
inference on the unknown parameters based on observed data.
Sampli ng theory provides the tools and techniques for data collection
keeping in mind the objectives to be fulfilled and nature of population .
Sample surveys collect information on a fraction of total population
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The concept of sampling has a huge implementation and its application is
seen in many vital fields. The importance of sampling theory is when it
comes into play while making statistical analysis with different efficiency
levels, ther e are three different methods of sampling. We have adequately
thrown light on the process and methods of sampling.
6.3 SAMPLING THEORY
Often we are interested in drawing some valid inferences about a large
group of individuals or objects called population in statistics. Instead of
studying the entire population, which may be difficult or even impossible
to study, we may study only a small portion of the population. Our
objective is to draw valid inferences about certain facts for the population
from results found in the sample; a process known as statistical inferences.
The process of obtaining samples is called sampling and theory
concerning the sampling is called sampling theory .
The sampling theory definition of the statistic is the creation of a sample
set. This is recognized as one of the major processes. It retains the
accuracy in bringing out the correct statistical information. The population
tree is huge set and it turns out to be exhausting for the actual study and
estimation process. Bot h money and time get exhausting in the process.
The creation of the sample set saves time and effort and is a vital theory in
the process of statistical data analysis.
Process of Sampling :
In this part of the chapter, we will discuss a few details regarding the
process of sampling. So the steps are mentioned in the steps below:
The first step is a wise choice of the population set.
The second step is focusing on the sample set and the size of it.
Then, one needs to choose an identifiable property ba sed on which the
samples will be created out of the population set.
Then, the samples can be chosen using any of the types of sampling
theory – Simple random, systematic, or stratified. Each of them is
thoroughly discussed in the article ahead.
Checking th e inaccuracy, if there is any.
Hence, the set is achieved in the result.
Sampling can be done in their different method and they are given below:
1. Simple random type.
2. Systematic Sampling.
3. Stratified sampling.
6.3.1 Random Samples and Random Numbers:
Definition: Simple random sampling is defined as a sampling technique
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being selected in the sample. Here the selection of items entirely depends
on luck or probability, and therefore t his sampling technique is also
sometimes known as a method of chances. For e.g. Using the lottery
method is one of the oldest ways and is a mechanical example of random
sampling. In this method, the researcher gives each member of the
population a number. Researchers draw numbers from the box randomly
to choose samples. The use of random numbers is an alternative method
that also involves numbering the population. The use of a number table
similar to the one below can help with this sampling technique.
Simp le random sampling (SRS) is a method of selection of a sample
comprising of n a number of sampling units out of the population having N
number of sampling units such that every sampling unit has an equal
chance of being chosen.
Simple random sampling metho ds:
Researchers follow these methods to select a simple random sample:
1. They prepare a list of all the population members initially, and then
each member is marked with a specific number ( for example, there
are nth members, then they will be numbered from 1 to N).
2. From this population, researchers choose random samples using two
ways: random number tables and random number generator software.
Researchers prefer a random number generator software, as no human
interference is necessary to generate samples.
Advantages of simple random sampling :
1. It is a fair method of sampling, and if applied appropriately, it helps to
reduce any bias involved compared to any other sampling method
involved.
2. Since it involves a large sample frame, it is usually easy to pick a
smaller sample size from the existing larger population.
3. The person conducting the research doesn’t need to have prior
knowledge of the data he/ she is collecting. Once can ask a question to
gather the researcher need not be subject expert.
4. This sampling method is a fundamental method of collecting the data.
You don’t need any technical knowledge. You only require essential
listening and recording skills.
5. Since the population size is vast in this type of sampling method, there
is no restri ction on the sample size that the researcher needs to create.
From a larger population, you can get a small sample quite quickly.
6. The data collected through this sampling method is well informed;
more the sample better is the quality of the data.
Disadvantages:
1. Sampling is not feasible where knowledge about each element or unit
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2. The sampling procedures must be correctly desi gned and followed
otherwise, what we call as wild sample would crop up with mis -
leading results.
3. Each type of sampling has got its own limitations.
4. There are numerous situations in which units, to be measured, are
highly variable. Here a very large sample is required in order to yield
enough cases for achieving statistically r eliable information.
5. To know certain population characteristics like population growth
rate, population density etc. census of population at regular intervals
is more appropriate than studying by sampling.
6.3.2 Samplin g With and Without Replacement:
Selection with Replacement (SWR): In this case, a unit is selected from
a population with a known probability and the unit is returned to the
population before the next selection is made (after recording its
characteristic). Thus, in this method at each s election, the population size
remains constant and the probability at each selection or draw remains the
same. Under this sampling plan, a unit has chances of being selected more
than once. For example a card is randomly drawn from a pack of cards and
placed back in the pack, after noting its face value before the next card is
drawn. Such a sampling method is known as sampling with replacement.
There are 𝑁possible samples of size n from a population of N units in
case of sampling with replacement.
Sampling with out replacement (SWOR) : In this selection procedure, if
a unit from a population of size N selected, it is not returned to the
population. Thus, for any subsequent selection, the population size is
reduced by one. Obviously, at the time of the first selection, the population
size is N and the probability of a unit being selected randomly is 1𝑁ൗ; for
the second unit to be randomly selected, the population size is (𝑁−1)
and the probability of selection of any one of the remaining sampling unit
is 1(𝑁−1)ൗ , similarly at the third draw, the probability of selection is
1(𝑁−2)ൗ and so on.
6.4 SAMPLING DISTRIBUTIONS
Sampling distribution is a statistic that determines the probability of an
event based on data from a small group within a large population. Its
primary purpose is to establish representative results of small samples of a
comparatively larger population. Since the population is too large to
analyze, the smaller group is selected and repeatedly sampled, or analyzed.
The gathered data, or statistic, is used to calculate the likely occurrence, or
probability, of an event. Using a sampling distribution simplifies the
process of making inferences, or conclusions, abo ut large amounts of data.
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The idea behind a sampling distribution is that when you have a large
amount of data (gathered from a large group, the value of a statistic from
random samples of a small group will inform you of that statistic’s value
for the en tire group. Once the data is plotted on a graph, the values of any
given statistic in random samples will make a normal distribution from
which you can draw inferences.
Each random sample selected will have a different value assigned to the
statistic being studied. For example, if you randomly sample data three
times and determine the mean, or the average, of each sample, all three
means are likely to be different and fall somewhere along the graph. That's
variability. You do that many times, and event ually the data you plot
should look like a bell curve . That process is a sampling distribution.
Factors that influence sampling distribution:
The sampling distribution’s variability can be measured either by standard
deviation, also called “ standard error of the mean ,” or population variance ,
depending on the context and inferences you are trying to draw. They both
are mathematical formulas that measure the spread of data points in
relation to the mean.
There are three primary fac tors that influence the variability of a sampling
distribution. They are:
The number observed in a population: This variable is represented
by "N." It is the measure of observed activity in a given group of data.
The number observed in the sample: This variable is represented by
"n." It is the measure of observed activity in a random sample of data
that i s part of the larger grouping.
The method of choosing the sample: How the samples were chosen
can account for variability in some cases.
Types of distributions :
There are three standard types of sampling distributions in statistics.
1. Sampling Distribution of Means.
2. Sampling Distribution of Proportions.
3. Sampling Distributions of Differences and Sums.
6.4.1 Sampling Distribution of Means:
The most common type of sampling distribution is of the mean. It focuses on
calculating the mean of every sample group chosen from the population and
plotting the data points. The graph shows a normal distribution where the center
is the mean of the sampling distri bution, which represents the mean of the entire
population.
The mean of the sampling distribution of the mean is the mean of the
population from which the scores were sampled. Therefore, if a population
has a mean μ, then the mean of the sampling distribut ion of the mean is
also 𝜇. The symbol 𝜇ത is used to refer to the mean of the sampling munotes.in
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distribution of the mean. Therefore, the formula for the mean of the
sampling distribution of the mean can be written as:
𝜇ത=𝜇
The standard deviation of the sampling distribution of the mean is
computed as follows:
𝜎ത=𝜎
√𝑁
That is, the standard deviation of the sampling distribution of the mean is
the population Standard deviation divided by √𝑁, the sample size (the
number of scores used to compute a mean). Thus, the larger the sa mple
size, the smaller the Standard deviation of the sampling distribution of the
mean.
For sampling is drawn without replacement,
The mean of the sampling distribution of means given by
𝜇ത=𝜇
The standard deviation of the sam pling distribution of means is given by
𝜎ത=ఙ
√ேටேି
ேିଵ.
Example 1: A population consist s of the five numbers 5, 6, 7, 12, and 15 .
Conside r all possible samples of size 2 that can be drawn with replacem ent
from this population. Find
a) the mean of the population,
b) the standard devi ation of the population,
c) the mean of the sampling distribution of means,
d) the standard deviation of the sampling d istribution of means.
Solution: Here 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑁=5,𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 𝑛=2.
a) The mean of the population is given by
𝜇=5+6+7+12+15
5=45
5=9
b) The standard deviation of the population is given by
𝜎=ඨ(𝑥−𝑥̅)ଶ
𝑁
𝜎=ඨ(5−9)ଶ+(6−9)ଶ+(7−9)ଶ+(12−9)ଶ+(15−9)ଶ
5
𝜎=ඨ16+9+4+9+36
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𝜎=ඨ74
5=3.85
c) The mean of the sampling distribution of means:
Hare 5(5)=25 samples of size 2, that can be drawn with
replacement, i.e. Samples are
(5,5),(5,6),(5,7),(5,12),(5,15),(6,5),(6,6),(6,7),(6,12),(6,15),
(7,5),(7,6),(7,7),(7,12),(7,15),
(12,5),(12,6),(12,7),(12,12),(12,15),(15,5),(15,6),(15,7),
(15,12),(15,15)
Therefore , the corresponding sample means are
5,5.5,6,8.5,10,5.5,6,6.5,9,
10.5,6,6.5,7,9.5,11,
8.5,9,9.5,12,13.5,10,
10.5,11,13.5,15.
The mean of the sampling distribution of means is given by
𝜇ത=𝑆𝑢𝑚 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 𝑚𝑒𝑎𝑛𝑠
25=225
25=9.
i.e.𝜇ത=𝜇
d) The standard deviation of the sampling distribution of means.
𝜎ത=ඩ(5−9)ଶ+(5.5−9)ଶ+(6−9)ଶ+⋯………+(11−9)ଶ+
(13.5−9)ଶ+(15−9)ଶ
25
= ඨ185
25=√7.4=2.72
Therefore, 𝜎ത=ఙ
√ே=ଷ.଼ହ
√ଶ=2.72.
Example 2: A population consists of the five numbers 7, 9, 10, 14, and 20.
Consider all possible samples of size 2 that can be drawn without
replacement from this population. Find
a) the mean of the population,
b) the standard deviation of the population,
c) the mean of the sampling distribution of means, munotes.in
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d) the standard deviation of the sampling distribution of means.
Solution : Here 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑁=5,𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 𝑛=2.
a) The mean of the population is given by
𝜇=7+9+10+14+20
5=60
5=12
b) The standard deviation of the population is given by
𝜎=ඨ(𝑥−𝑥̅)ଶ
𝑁
𝜎=ඨ(7−12)ଶ+(9−12)ଶ+(10−12)ଶ+(14−12)ଶ+(20−12)ଶ
5
𝜎=ඨ25+9+4+4+64
5
𝜎=ඨ106
5=4.60
c) The mean of the sampling distribution of means:
Here 5C2 = 10 samples of size 2, that can be drawn with out
replacement, i.e. (all sample are distinct selection). Samples are
(7,9),(7,10),(7,14),(7,20),(9,10),(9,14),(9,20),
(10,14),(10,20),(14,20)
Therefore, the corresponding sample means are
8,8.5,10.5,13.5,9.5,11.5,14.5,
12,15,17.
The mean of the sampl ing distribution of means is given by
𝜇ത=𝑆𝑢𝑚 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 𝑚𝑒𝑎𝑛𝑠
10=120
10=12.
i.e.𝜇ത=𝜇
d) The standard deviation of the sampling distribution of means. munotes.in
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𝜎ത
=ඩ(8−12)ଶ+(8.5−12)ଶ+(10.5−12)ଶ+⋯………+(12−12)ଶ
+(15−12)ଶ+(17−12)ଶ
10
= ඨ79.5
10=√7.95=2.81
Therefore, 𝜎ത=ఙ
√ேටேି
ேିଵ=ସ.
√ଶටହିଶ
ହିଵ=2.81.
6.4.2 Sampli ng Distribution of Proportions:
This sampling distribution focuses on proportions in a population.
Samples are selected and their proportions are calculated. The mean of the
sample proportions from each group represent the proportion of the entire
population.
Suppose random samples of size 𝑛 are drawn from a population in which
the proportion with a characteristic of interest is 𝑝.
The Sampling Distribution of Proportion measures the proportion of
success, i .e. a chance of occurrence of certain events, by dividing the
number of successes i.e. chances by the sample size ’n’. Thus, the sample
proportion is defined as
𝑝=𝑥
𝑛
Therefore the mean 𝜇 and standard deviation 𝜎 are given by
𝜇=𝑝,
𝜎=ට𝑝𝑞
𝑛
Where 𝑞 is probability of non -occurrence of event, which is given
by𝑞=1−𝑝.
The following formula is used when population is finite, and the sampling
is made without the replacement:
𝜎=ඨ𝑁−𝑛
𝑁−1ට𝑝𝑞
𝑛
If 𝑛 is large, and 𝑝 is not too close to 0 or 1, the binomial distribution can
be approximated by the normal distribution. Practically, the Normal
approximation can be used when both 𝑛𝑝≥10.
Once we have the mean and standard deviation of the survey data, we can
find out the probability of a sample proportion . Here, the Z score
conversion formula will be used to find out the required probability, i.e. munotes.in
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𝑍=ିఓ
ఙ.
Example 3:A random sample of 100 students is taken from the population
of all part -time students in the Maharashtra, for which the overall
proportion of females is 70%. Find sample mean and sample standard
deviation.
Solution: Here 𝑛=100,𝑝=70%=
ଵ=0.7
∴𝑞=1−𝑝=1−0.7=0.3
the mean 𝜇 is given by
𝜇=𝑝=0.7
and standard deviation 𝜎 is given by
𝜎=ට𝑝𝑞
𝑛=ඨ0.7×0.3
100=ඨ0.21
100=0.0458.
Example 4: Suppose it is known that 47% of Indian own smart phone. If a
random sample of 50 Indians were surveyed, what is the probability that
the proportion of the sample who owned smart p hone is between 50% and
54%?
Solution: Here we have, 𝑛=50,𝑝=47%=0.47,
∴𝑞=1−𝑝=1−0.47=0.53.
Now, we should check our conditions for the sampling distribution of the
sample proportion.
𝑛𝑝=50×0.47=23.5 ≥10,𝑛𝑞=50×0.53=26.5≥10.
Since both the condition satisfy,
∴ The sampling distribution that is approximately normal with mean and
standard deviation,
𝜇=𝜇=0.47,
𝜎=ට𝑝𝑞
𝑛=ඨ0.47×0.53
50=0.07
Now, T he probability that the proportion of the sample who owned smart
phone is between 50% and 54 % is given by
𝑃(0.50<𝑍<0.54)=𝑃൬0.50−0.47
0.07<𝑍<0.54−0.47
0.07൰
=𝑃(0.429<𝑍<1)
=𝑃(𝑍<1)−𝑃(𝑍<0.429)
=0.8413−0.6627
=0.1786 munotes.in
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∴ If the true proportion of Indians who own smart phone is 47%, then
there would be a 17.86% chance that we would see a sample proportion
between 50% and 54% when the sample size is 50.
6.4.3 Sampling Distrib utions of Differences and Sums:
Statistical analyses are very often concerned with the difference between
means. A typical example is an experiment designed to compare the mean
of a control group with the mean of an experimental group. Inferential
statistics used in the analysis of this type of experiment depend on the
sampling distribution of the difference between means.
The sampling distribution of the difference between means can be thought
of as the distribution that would result if we repeated the following three
steps over and over again:
1. Sample 𝑛ଵ scores from Population 1 and 𝑛ଶ scores from Population 2.
2. compute t he means of the two samples 𝑀ଵ and 𝑀ଶ.
3. compute the difference between means, M 1 - M2. The distribution of
the differences between means is the sampling distribution of the
difference between means.
As you might expect, the mean of the sampling distribut ion of the
difference between means is:
𝜇ெభିெమ=𝜇ெభ−𝜇ெమ
Which says that the mean of the distribution of differences between
sample means is equal to the difference between population means.
From the variance sum law , we know that:
𝜎ଶ
ெభିெమ=𝜎ଶ
ெభ+𝜎ଶ
ெమ
We can write the formula for the standard deviation of the sampling
distribution of the difference between means as
∴𝜎ெభିெమ=ට𝜎ଶெభ+𝜎ଶெమ
Similarly we say about the sampling distribution of the sum between
means is given by:
𝜇ெభାெమ=𝜇ெభ+𝜇ெమ
𝜎ெభାெమ=ට𝜎ଶெభ+𝜎ଶெమ
Example 5: Let 𝑆ଵ be a variable that stands for any of the elements of the
population 4, 6, 8 and 𝑆ଶ be a variable that stands for any of the elements
of the population 3, 5. Compute a) 𝜇ௌభ, b) 𝜇ௌమ, c) 𝜇ௌభିௌమ, d)𝜎ௌభ, e) 𝜎ௌమ,
and f) 𝜎ௌభିௌమ.
Solution:
a) Here 𝑆ଵ has sample of population 4, 6, 8. munotes.in
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𝜇ௌభ=4+6+8
3=18
3=6.
b) Here 𝑆ଶ has sample of population 3, 5.
𝜇ௌమ=3+5
2=4
c) The population consisting of the differences of any member of 𝑆ଵ
and any member of 𝑆ଶ mean is given by
𝜇ௌభିௌమ
=(4−3)+(6−3)+(8−3)+(4−5)+(6−5)+(8−5)
6
=12
6=2.
Therefore, we can verify that
𝜇ௌభିௌమ=𝜇ௌభ−𝜇ௌమ=6−4=2.
d) 𝜎ௌభ=ට(ସି)మା(ି)మା(଼ି)మ
ଷ=ට଼
ଷ=√2.67=1.63
e) 𝜎ௌమ=ට(ଷିସ)మା(ହିସ)మ
ଶ=ටଶ
ଶ=1.
f) The population consisting of the differences of any member of 𝑆ଵ
and any member of 𝑆ଶStandard deviation is given by
𝜎ௌభିௌమ=ඩ(1−2)ଶ+(3−2)ଶ+(5−2)ଶ+(−1−2)ଶ+
(1−2)ଶ+(3−2)ଶ
6
=ඨ22
6=ඨ11
3=1.91
Therefore, we can verify that,
∴𝜎ௌభିௌమ=ට𝜎ଶௌభ+𝜎ଶௌమ=√2.67+1=√3.67=1.91
Example 6: The battery life of smart phone manufacture r A have a mean
lifetime of 1050 days with a sta ndard deviation of 150 days, while those
of manufacture r B have a mean lifetime of 800 days with a standard
deviation of 120 days. If random samples of 100 batteries of each brand
are tested, what is the probabi lity that the brand A batteries will have a
mean lifetime that is at least (a) 2 00 days and (b) 280 days more than the
brand B batteries?
Solution: Let 𝑋തതത and 𝑋തതതത denote the mean lifetimes of samples A and B,
respectively. Then munotes.in
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𝜇ಲതതതതିಳതതതത=𝜇ಲതതതത− 𝜇ಳതതതത=1050−800=250 𝑑𝑎𝑦𝑠.
𝜎ಲതതതതିಳതതതത=ඨ𝜎ಲതതതതଶ
𝑁+ 𝜎ಳതതതതଶ
𝑁 =ඨ(150)ଶ
100+(120)ଶ
100=ඨ36900
100
=√369=19.21
Therefore, the standardized variable for the difference in means is
𝒁=(𝑋തതത−𝑋തതതത)−(𝜇ಲതതതതିಳതതതത)
𝜎ಲതതതതିಳതതതത
and is very closely normally distributed .
a) the probability that the brand A batteries will have a mean lifetim e
that is at least 1200 days,
𝑃(𝑍>200)=0.5+𝑃൬𝑍=200−250
19.21൰=0.5+𝑃(𝑍>−2.6)
=0.5+0.4953=0.9953.
b) the probability batteries will have a mean lifetime that is at least
960 days more than the brand B batteries ,
𝑃(𝑍>280)=0.5−𝑃൬𝑍=280−250
19.21൰=0.5−𝑃(𝑍=1.56)
=0.5 −0.4406=0.0594.
6.5 STANDARD ERRORS
Another measure is standard error, which is the standard deviation of the
sampling distribution of an estimator. The idea is that if we draw a number
of repeated samples of fixed size 𝑛 from a population having a mean 𝜇 and
variance 𝜎ଶ , each simple mean, say 𝑥̅, will have a different value. Here
𝑥̅it isa random variable and hence it has a distribution. The standard
deviation of 𝑥̅is called standard error . It has been proved that the standard
error ′𝜎௫̅′of the mean 𝑥̅based on a sample of size 𝑛 is,
𝜎௫̅= 𝜎
√𝑛
From above formula, it is obvious that the larger the sample size, the
smaller the standard error and vice -versa. The advantage of considering
standard error instead of a standard deviation is that this measure is not
influenced by the extreme values present in a population under
consideration.
In reality neither we use 𝜎 to calculate the standard error of 𝑥̅nor we take
more than one sample. As a matter of fact, what we do is, that we select
only one sample, find its standard deviation 𝑠 and use the following
formula to find out the standard error of 𝑥̅ .i.e. munotes.in
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𝑆.𝐸.(𝑥̅)=𝑠
√𝑛
Standard error is commonly used in testing of hypothesis and interval
estimation. Many distributions, which are originally not normally
distributed, have been taken as normal by considering the distribution of
mean 𝑥̅ for a large 𝑛.
6.6 SUMMARY In this unit we have learn:
In sampling theory we have Random Samples, Sampling With and
Without Replacement of sample.
Sampling Distributions and its types.
Standard Errors of sampling distribution.
6.7 EXERCISE 1. A population consists of the four numbers 8, 11, 12, and 19 . Consider
all possible samples of size 2 that can be drawn with replacement from
this population. Find
a) the mean of the population,
b) the standard deviation of the population,
c) the mean of the sampling distribution of means,
d) the standard deviation of the sampling distribution of means.
2. A population consists of the seven numbers 3, 5, 7, 9, 11, 13, and 15.
Consider all possible samples of size 2 that can be drawn without
replacement from this population. Find
a) the mean of the populat ion,
b) the standard deviation of the population,
c) the mean of the sampling distribution of means,
d) the standard deviation of the sampling distribution of means.
3. Let 𝑆ଵ be a variable that stands for any of the elements of the
population 5, 8, 12 and 𝑆ଶ be a variable that stands for any of the
elements of the population 2, 6. Compute a) 𝜇ௌభ, b) 𝜇ௌమ, c) 𝜇ௌభିௌమ,
d)𝜎ௌభ, e) 𝜎ௌమ, and f) 𝜎ௌభିௌమ.
4. A certain t ype of electric light bulb has a mean lifetime of 1500 h and a
standard deviation of 150 h. Three bulbs are connected so that when
one burns out, another will go on. Assuming that the lifetimes are
normally distributed, what is the probability that lightin g will take place
for (a) at least 5000 h and (b) at most 4200 h?
5. Two distances are measured as 27.3 centimeters (cm) and 15.6 cm with
standard deviations (standard errors) of 0.16 cm and 0.08 cm, munotes.in
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respectively. Determine the mean and standard deviation of (a) the sum
and (b) the difference of the distances.
6. A and B play a game of ‘‘heads and tails,’’ each tossing 50 coins. A
will win the game if she tosses 5 or more heads than B; otherwise, B
wins. Determine the odds against A winning any particular game.
7. The average income of men in a city is Rs. 20,000 with standard
deviation Rs. 10,500 and the average income of women is Rs. 16,000
and standard deviation Rs. 8,000. There are 100 sample selected from
population. Find the probability of income betw een Rs. 15,000 to Rs.
18000.
8. A sample of 300 items selected at random had 32 defective items. Find
mean and standard deviation of sampling distribution of proportion.
9. Ball bearings of a given brand weigh 0.50 g with a standard deviation
of 0.02 g. What is the probability that two lots of 1000 ball bearings
each will differ in weight by more than 2 g?
10. Find the probability that in 120 tosses of a fair coin (a) less than 40%
or more than 60% will be heads and (b) 5/8 or more will be heads.
6.8 LIST OF REFERENCES
Statistics by Murry R. Spiegel, Larry J. Stephens. Publication
McGRAWHILL INTERNATIONAL.
Fundamental Mathematics and Statistics by S.C. Gupta and V.K
kapoor
Mathematical Statistics by J.N. Kapur and H.C. Saxena.
*****
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UNIT III
7
STATISTICAL ESTIMATION
THEORY
Unit Structure
7.0 Objectives
7.1 Basic definitions
7.1.1 Population
7.1.2 Sample
7.1.3 Parameter
7.1.4 Statistic
7.1.5 Sampling distribution
7.1.6 Parameter Space
7.1.7 Estimator
7.1.8 E stimate
7.2 Point estimation
7.2.1 Unbiasedness
7.2.2 Consistency
7.2.3 Efficiency
7.2.4 Minimum variance unbiased estimator
7.2.5 Uniformly minimum variance unbiased estimator
7.2.6 Likelihood function
7.2.7 Sufficiency
7.3 Interval estimation
7.3.1 Probable error
7.4 Summary
7.5 Exercise
7.6 Refer ences
7.0 OBJECTIVES
To understand basic definitions related to point estimation.
To find the best p oint estimators to represent population
characteristics .
In this chapter, students can learn the requirements of a good
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To find an appropriate confidence interval for the population
parameters .
7.1 BASIC DEFINITIONS
7.1.1 Population :
A collection of all well-defined objects under study is called
population.
Example : Suppose we want to study the economic conditions of
primary teachers in Maharashtra, then the group of all primary
teachers in the state of Maharashtra is a population.
7.1.2 Sample :
A well defined finite subset of the population is called a sample.
Example : Suppose we want to study the economic conditions of
primary teachers in the state of Maharashtra, then the few primary
teachers (set of few teachers ) in the state of Maharashtra forms a
sample.
7.1.3 Parameter :
An unknown constant of a population that summarises or describes an
aspect of the population (such as a mean or a standard deviation ) is
called parameter . Let f (x, θ) be the pdf of a random variable ‘ X’
having an unknown constant θ.
7.1.4 Statistic :
Any function of a sample value (observed value) is called a statistic.
The sample statistic is constants but it differ from sample to sample.
7.1.5 Sampling distribution :
The probability distribution of the sample statistic is called a sampling
distribution.
7.1.6 Parameter space :
The set of all admissible value s of a parameter of the distribution is
called parameter space. It is denoted by Θ.
Example : X ~ Normal ( μ, σ2)
Θ = {(μ, σ2) / -∞<μ<∞, σ>0}
7.1.7 Estimator :
Let 𝑥ଵ,𝑥ଶ,…,𝑥 be a sample of size n taken from a distribution
having pdf f (x, θ) where θ ∈ Θ is an unknown parameter. A function
T = T ( 𝑥ଵ,𝑥ଶ,…,𝑥) which maps sample space (S) to parameter space
Θ is called an estimator. In other words, If a statistic T =
T(𝑥ଵ,𝑥ଶ,…,𝑥) is used to estimate θ, and its value belongs to
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7.1.8 Estimate :
A particular value of an estimator corresponding to the given sample
values is called an estimate of the population parameter.
In the theory of estimation , there are two parts, 1) Point Estimation, 2)
Interval Estimation.
7.2 POINT ESTIMATION
Let 𝑥ଵ,𝑥ଶ,…,𝑥 be a sample of size n taken from a distribution having
pdf (probability density function) f (x, θ) where θ ∈ Θ is an unknown
parameter. The method of using sample statistic ‘T’ to estimate the
value of parameter θ, which is a po int on real line R, is called “Point
Estimation”.
Requirements of good and reliable e stimators:
1. Unbiasedness
2. Consistency
3. Efficiency
4. Sufficiency
7.2.1 Unbiasedness :
An estimator T = T ( 𝑥ଵ,𝑥ଶ,…,𝑥) is said to be an unbiased estimator of θ
iff;
𝐸(𝑇)= 𝜃, ∀ 𝜃∈Θ,
a parameter ic function 𝜙 (𝜃) is said to be estimable if there exists a
statistic h (T) such that, 𝐸൫ℎ(𝑇)൯= 𝜙 (𝜃) ∀ 𝜃∈Θ.
7.2.1.1 The b ias of an estimator :
An estimator T = T (x 1, x2, …, x n) is said to be a biased estimator with a
bias (ఏ)
if;
𝐸(𝑇)= 𝜃+𝑏 (𝜃)
𝑛 ∀ 𝜃∈Θ
Or
𝐸(𝑇− 𝜃)= 𝑏 (𝜃)
𝑛 ∀ 𝜃∈Θ
If 𝑏 (𝜃)>0, then estimator T is called biased estimator with an
upward (positive) bias (ఏ)
.
If 𝑏 (𝜃)<0, then estimator T is called biased estimator with a
downward (negative) bias (ఏ)
.
If 𝑏 (𝜃)=0, then estimator T is called an unbiased estimato r. munotes.in
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Example : Let x 1, x2, x3 be independent observation from Poisson ( 𝜆),
then show that, 𝑇= ௫భା ௫మା ௫య
ଷ is an unbiased estimator of 𝜆.
Since 𝑥ଵ, 𝑥ଶ,𝑥ଷ are i.i.d. Poisson ( 𝜆).
𝐸[𝑋ଵ]= 𝐸[𝑋ଶ]= 𝐸[𝑋ଷ]= 𝜆
Consider;
𝐸[𝑇]=𝐸ቂ௫భା ௫మା ௫య
ଷቃ= ଵ
ଷ 𝐸[𝑥ଵ+ 𝑥ଶ+ 𝑥ଷ]= 𝜆
Hence, T is an unbiased estimator of 𝜆.
7.1.1.1 MSE of an estimator :
Mean square error (MSE) of an estimator T = T (x 1, x2, …, xn) is,
𝑀𝑆𝐸=𝐸(𝑇− 𝜃)ଶ
7.1.1 Consistency :
A sequence of the estimator 𝑇=𝑇(𝑥ଵ,𝑥ଶ,…,𝑥) is said to be a
consistent estimator for parameter 𝜃 if any given 𝜖> 0,
𝑃 [|𝑇− 𝜃|< 𝜖] →1 𝑎𝑠 𝑛 →∞ ∀ 𝜖>0
Difference between 𝑇 & 𝜃 becomes smaller and smaller as n goes to a
large number (infinity) .
A sequence of an estimator 𝑇=𝑇(𝑥ଵ,𝑥ଶ,…,𝑥) is said to be a
consistent estimator for parameter 𝜃 if;
𝐸(𝑇)= 𝜃 ∀ 𝜃∈Θ
𝑉[𝑇] →0 𝑎𝑠 𝑛 → ∞
Example : Show that the sample mean is a consistent estimator of the
population mean.
Let 𝑥ଵ,𝑥ଶ,…,𝑥 be a random sample of size n from a population with
mean 𝜇 and variance 𝜎ଶ.
First, we have to define the sample mean,
𝑥 ഥ= ∑௫
సభ
𝐸 [𝑥]= 𝜇 & 𝑉 [𝑥]= 𝜎ଶ ∀ 𝑖=1,2,…,𝑛
𝐸 [𝑥̅]=𝐸 ቂ∑௫
సభ
ቃ
= ∑ா [௫]
= ఓ
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𝐸 [𝑥̅]= 𝜇
Consider,
𝑉 [𝑥̅]=𝑉ቂ∑௫
సభ
ቃ
= ଵ
మ ∑𝑉 (𝑥 )
= ଵ
మ 𝑛𝜎ଶ
𝑉 [𝑥̅]= ఙమ
Lim → ஶ𝐸 [𝑥̅]= 𝜇
& 𝑉 [𝑥̅]= ఙమ
→0.
Therefore, the sample mean is a consistent estimator of the population
mean.
Remark : If 𝑇 is a consistent estimator for 𝜃, then 𝜙 (𝑇) is a consistent
estimator for 𝜙 (𝜃), where 𝜙 is a continuous function.
7.1.2 Efficiency :
An estimator T 1 is said to be more efficient than estimator T 2 of
parameter 𝜃 if V (T1) < V (T2). The relative efficiency of T 2 with
respect to T 1 is defined as,
𝑒= [்భ]
[்మ].
Example : Let 𝑥ଵ,𝑥ଶ,…,𝑥 be a random sample of size n from a Normal
(𝜇, 𝜎ଶ). Then select the most efficient estimator between 𝑥̅ (Mean) &
𝑥 (Median).
We know that,
𝐸 [𝑥̅]= 𝜇
𝑉 [𝑥̅]= ఙమ
→0 𝑎𝑠 𝑛 → ∞
& 𝐸 [𝑥]= 𝜇
𝑉 [𝑥]= ቀగ
ଶቁ .ቀఙమ
ቁ .
Now we have to calculate efficiency as,
𝑒= [௫̅]
[௫]= ଶ
గ<1.
Therefore, the sample mean is a more efficient estimator than the sample
median.
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7.1.3 Minimum Variance Unbiased Estimator ( MVUE ):
An estimator T = T ( 𝑥ଵ,𝑥ଶ,…,𝑥) is said to be minimum variance
unbiased estimator of parameter 𝜃 if,
1) 𝐸 [𝑇]= 𝜃 ∀ 𝜃∈Θ and,
2) 𝑉 [𝑇]<𝑉 [𝑇ᇱ] ; where, 𝑇ᇱis any other unbiased estimator of 𝜃.
7.1.4 Uniformly minimum variance unbiased estimator (UMVUE) :
Let 𝜃 be the unknown parameter & Θ be the parameter space of 𝜃. Let
𝑈 (𝜃) be the set of a class of unbiased estimator of 𝑇 (𝜃) such that,
𝐸 [𝑇ଶ]< ∞ ∀ 𝜃∈Θ
i.e.,
𝑈 (𝜃)= {𝑇: 𝐸 [𝑇]= 𝜃,𝐸 [𝑇ଶ]< ∞ ∀ 𝜃∈Θ}
Then, 𝑇 ∈𝑈 (𝜃) is UMVUE of 𝜃 if,
𝐸 [𝑇− 𝜃]ଶ ≤𝐸 [𝑇− 𝜃]ଶ ∀ 𝜃∈Θ & 𝑇∈𝑈 (𝜃).
7.1.5 Likelihood function :
Let 𝑥ଵ,𝑥ଶ,…,𝑥 be a sample of size n taken from a distribution
having pdf f (x, θ) where θ ∈ Θ is an unknown parameter. Then the
likelihood function of 𝑥ଵ,𝑥ଶ,…,𝑥 is defined as
𝐿 (𝑋ଵ,𝑋ଶ,…,𝑋,𝜃)=𝑓൫𝑋ଵ,𝜃൯ .𝑓൫𝑋ଶ,𝜃൯ .𝑓൫𝑋ଷ,𝜃൯… 𝑓൫𝑋,𝜃൯
𝐿 ൫𝑋,𝜃൯= ∏𝑓(𝑥,𝜃)
ୀଵ .
T is said to be the maximum likelihood estimator of 𝜃, which
maximizes the likelihood function 𝐿 ൫𝑋,𝜃൯.
7.1.6 Sufficiency :
Definition I A stati stic 𝑇=𝑇(𝑥ଵ,𝑥ଶ,…,𝑥) based on a sample of size
n having pmf /pdf 𝑓(𝑋,𝜃) 𝜃∈Θ is said to sufficient statistic for 𝜃 if
and only if the information contains in T about 𝜃 is same as the
information contains in 𝑥ଵ,𝑥ଶ,…,𝑥 about 𝜃.
Definition II A statistic 𝑇=𝑇(𝑥ଵ,𝑥ଶ,…,𝑥) based on a sample of size
n having pmf /pdf 𝑓(𝑋,𝜃) 𝜃∈Θ is said sufficient statistic for 𝜃, if and
only if the conditional distribution of 𝑥ଵ,𝑥ଶ,…,𝑥 given T is
independent of 𝜃.
Definition III (Neyman factorization criteri on) If 𝑥ଵ,𝑥ଶ,…,𝑥 is a
sample of size n having pmf /pdf 𝑓(𝑋,𝜃) 𝜃∈Θ and
𝑇=𝑇(𝑥ଵ,𝑥ଶ,…,𝑥) be a statistic which is said to be sufficient for 𝜃 if
and only if the joint probability distribution function of 𝑥ଵ,𝑥ଶ,…,𝑥
can be expressed as a product of a function of T and 𝜃, and function of
𝑥ଵ,𝑥ଶ,…,𝑥 only.
i.e., 𝐿 ൫𝑋,𝜃൯=𝑔(𝑇,𝜃) .ℎ൫𝑋൯,
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then T is said to be sufficient statistic for 𝜃, where 𝑔(𝑇,𝜃) is a function of
𝑇 & 𝜃 only and ℎ൫𝑋൯ is a function of 𝑥ଵ,𝑥ଶ,…,𝑥 only.
Example : Let 𝑥ଵ,𝑥ଶ,…,𝑥 be a random sample from a population
having pdf,
𝑓(𝑥)=൜𝜃𝑥ఏ ି ଵ,0 ≤𝑥≤1
0,𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
The likelihood function of 𝑥ଵ,𝑥ଶ,…,𝑥 is,
𝐿 ൫𝑋,𝜃൯= ∏𝜃 𝑥ఏିଵ
– ଵ
= 𝜃∏𝑥ఏିଵ
ିଵ
= ൣ𝜃 ∏𝑥ఏ൧ ቂଵ
∏௫ቃ
𝐿 ൫𝑋,𝜃൯= 𝑔(𝑇,𝜃) .ℎ൫𝑋൯
Where, (𝑇,𝜃)= 𝜃 ∏𝑥ఏ , ℎ൫𝑋൯= ଵ
∏௫
Therefore, from the Neyman factorization criteri on ∏𝑥 is sufficient
statistic for 𝜃.
7.2 INTERVAL ESTIMATION
A confidence interval for an unknown parameter is an interval of
possible values for the parameter. It is constructed so that, with a
chosen degree of confidence, the actual value of the parameter lies
within the lower and upper bounds of the interval.
Let T1 and T2 be two statistics such that,
𝑃 (𝑇ଵ> 𝜃)= 𝛼ଵ … (1)
and 𝑃 (𝑇ଶ< 𝜃)= 𝛼ଶ … (2)
where 𝛼ଵ and 𝛼ଶ are constants independent of 𝜃. Equation s (1) & (2) can
be combined to give
𝑃 (𝑇ଵ < 𝜃 < 𝑇ଶ)=1− 𝛼 … (3)
where 𝛼= 𝛼ଵ+ 𝛼ଶ
Example : If we take a large sample from a normal population with
mean 𝜇 and standard deviation 𝜎, then 100(1−𝛼)% confidence
interval for a population mean at 𝛼=0.05 𝑖𝑠
𝑍= 𝑥̅− 𝜇
𝜎
√𝑛ൗ ~ 𝑁 (0,1)
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In general, confidence interval for 𝜇 can be constructed by using the
following normal probability approach,
𝑃 ൫𝑍ఈ/ଶ ≤𝑍 ≤𝑍ଵିఈ/ଶ൯=1−𝛼,
Where 𝑃(𝑍<𝑍ఈ/ଶ)=𝛼/2.
In particular case, 𝛼=0.05 then 𝑍ఈ/ଶ=−𝑍ଵିഀ
మ=−1.96 implies,
𝑃 (−1.96 ≤𝑍 ≤1.96)=0.95 … (From Normal Probability Tables)
⟹𝑃 ቆ−1.96 ≤ ௫̅ି ఓ
ఙ
√ൗ ≤1.96ቇ=0.95
⟹𝑃 ቀ𝑥̅−1.96 ఙ
√ ≤ 𝜇 ≤ 𝑥̅ +1.96 ఙ
√ ቁ=0.95
Thus, 𝑥̅ ±1.96 ఙ
√ are 95% confidence limits for the unknown parameter
𝜇, the population mean and interval ቀ𝑥̅−1.96 ఙ
√,𝑥̅ +1.96 ఙ
√ ቁ is
called the 95% confidence interval.
Remark For a small sample case with unknown variance, to construct
a confidence interval for a population mean, use students -t
distribution.
7.2.1 Probable error :
Probable error defines the half-range of an interval about a mean for
the distribution, such that half of the values from the distribution will
lie within the interval and half outside.
Thus for a symmetric distribution it is equivalent to half
the interquartile range, or the median absolute deviation.
For a normal distribution probable error is 0.6745𝜎
7.3 EXERCISE
1. What do you understand by point estimation?
2. Explain the terms statistic and its sampling distribution.
3. Define estimator and estimate.
4. Write a note on unbiased ness, consiste ncy, the efficien cy of an
estimator. Also , define sufficient statistics.
5. Define MVUE and UMVUE.
6. Define Neyman factorization criteria.
7. Construct a confidence interval for a population mean where
population variance is known. munotes.in
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7.4 SUMMARY
In this chapter we studied basic definitions related to point estimation
of population parameters.
The good estimator must satisfy conditions of Unbiasedness,
Consistency, Efficiency and Sufficiency.
Uniformly Minimum variance unbiased estimators popularly used
techniques to estimate population parameters.
In this chapter we studied confidence interval estimation of the
population parameter.
7.5 REFERENCES
1. Gupta S. C. and Kapoor V. K., 2011, Fundamentals of Mathematical
Statistics, 11th Ed, Sultan and Chand.
2. Rohatgi V. K, 1939, Introduction to Probability and Statis tics, Wiley
3. Murray R. Spiegel, Larry J. Stephens, STATISTICS, 4th Ed,
McGRAW – HILL I NTERNATIONAL.
4. J.N. KAPUR and H.C. SAXENA, 2005, MATHEMATICAL
STATISTICS, 12th rev, S.Chand.
5. Agrawal B. L, 2003, Programmed Statistics , 2nd Ed, New Age
International.
*****
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8
STATISTICAL DECISION THEORY
Unit Structure
8.0 Objectives
8.1 Basic definitions
8.1.1 Statistical decision
8.1.2 Hypothesis
8.1.3 Null hypothesis
8.1.4 Alternative hypothesis
8.1.5 Simple and composite hypothesis
8.1.6 Errors in the test of significance
8.1.7 Critical region
8.1.8 Level of significance
8.1.9 Test statistic
8.1.10 One-tailed and two -tailed tests
8.1.11 Central limit theorem
8.1.12 Critical value
8.1.13 P-value
8.1.14 Power of the test
8.1.15 Procedure for test a hypothesis
8.2 Large sample test
8.2.1 One sample Z -test for a mean
8.2.2 Two sample Z -test for the difference of two mean
8.2.3 Test for a testing population proportion
8.2.4 Test for testing equality of population proportion
8.3 Small sample test
8.3.1 A t-test for testing a population mean
8.3.2 Two sample t -test for the difference of two mean
8.3.3 Paired t -test
8.4 Control chart
8.4.1 A lot acceptance sampling plan
8.4.2 OC curve
8.4.3 Control chart
8.4.4 Variable control chart
8.4.5 Attribute control chart
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8.6 Exercise
8.7 References
8.0 OBJECTIVES
In this chapter , student s can learn,
Basic definitions related to the testing of hypothesis/decision making
Decision making through Testing of hypothesis
Large and small sample tests
Decision Making through statistical c ontrol chart
8.1 BASIC DEFINITIONS
8.1.1 Statistical decision :
The decisions are made based on observations of a phenomenon that
carry out probabilistic laws that are not completely known.
8.1.2 Hypothesis :
A definite statement about the population parameter is called a
hypothesis. A hypothesis is a claim to be tested.
Example : A particular scooter gives an average of 50 km per litre.
8.1.3 Null hypothesis :
A hypothesis having no difference is called the null hypothesis.
Example : The population mean is 𝜇 the hypothesis will be 𝐻: 𝜇=
𝜇.
8.1.4 Alternative hypothesis :
A hypothesis that is accepted in the case 𝐻 is rejected is called the
alternative hypothesis and usually denoted by 𝐻ଵ. It is exactly
opposite to H 0.
Example : If 𝐻: 𝜇= 𝜇 i.e., the population has a specified mean 𝜇,
then the alternative hypothesis could be;
i. 𝐻ଵ: 𝜇≠ 𝜇 (𝜇 > 𝜇 𝑜𝑟 𝜇 < 𝜇) … Two-tailed alternative
ii. 𝐻ଵ: 𝜇 > 𝜇 … Right tailed alternative
iii. 𝐻ଵ: 𝜇 < 𝜇 … Left tailed alternative
8.1.5 Simple and composite hypothesis :
A statistical hypothesis that completely specifies the population
parameter is called a simple hypothesis, and the hypothesis that does
not specify the population parameter is called a composite hypothesis.
Example : If 𝑥ଵ,𝑥ଶ,…,𝑥 is a random sample from normal with mean
𝜇 and variance 𝜎ଶ then 𝐻: 𝜇= 𝜇 and 𝜎ଶ= 𝜎ଵଶ is a simple munotes.in
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hypothesi s. The following fully not specified hypotheses is called a
composite hypothes is.
1) 𝐻ଵ: 𝜇≠ 𝜇 2) 𝐻: 𝜎ଶ≠𝜎ଶ 3) 𝐻: 𝜇= 𝜇 𝑎𝑛𝑑 𝜎ଶ>𝜎ଶ etc.
8.1.6 Errors in the test of significance :
The main objective in the sampling theory is to draw a valid
inference about the population parameters based on sample results.
In practice, we decide to accept or reject the lot after examining a
sample drawn from it. In sampling theory, we are liable to commit
two types of errors: a rejection of a good lot and acceptance of a bad
lot.
i. Type -I error: Rejecting 𝐻 when 𝐻 is true.
ii. Type -II error: Accepting 𝐻 when it is false (Accepting 𝐻 when
𝐻ଵis true).
iii. Size of Type -I and Type -II errors
𝑃 [𝑅𝑒𝑗𝑒𝑐𝑡 𝐻 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑖𝑠 𝑡𝑟𝑢𝑒]=𝑃 [𝑅𝑒𝑗𝑒𝑐𝑡 𝐻 | 𝐻]=
𝑃 [𝑅𝑒𝑗𝑒𝑐𝑡 𝑎 𝑙𝑜𝑡 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑖𝑠 𝑔𝑜𝑜𝑑]=𝛼
𝑃 [𝐴𝑐𝑐𝑒𝑝𝑡 𝐻 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑖𝑠 𝑤𝑟𝑜𝑛𝑔]=𝑃 [𝐴𝑐𝑐𝑒𝑝𝑡 𝐻 | 𝐻ଵ]=
𝑃 [𝐴𝑐𝑐𝑒𝑝𝑡 𝑎 𝑙𝑜𝑡 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑖𝑠 𝑏𝑎𝑑]= 𝛽
In the above probabilities, 𝛼 & 𝛽 are called the Type -I & Type -II errors,
respectively.
The four types of decisions are shown in the table as follows. Actual Situation Decision Reject 𝐻 Accept 𝐻 𝐻 is true Type-I Error Correct Decision 𝐻 is false Correct Decision Type-II Error
8.1.7 Critical region:
A region in sample space S which amounts to a rejection of 𝐻 is
called a critical region or rejection region of 𝐻.
8.1.8 Level of significance :
The probability ‘ 𝛼’ is that the value of the test statistic belongs to the
critical region, known as ‘level of significance’. That is the
probability of the occurrence of the type I error is the level of
significance. Usually, we use the level of significance of 5% or 1%.
The le vel of significance is always fixed in advance before
collecting the sample information.
8.1.9 Test statistic :
A function of sample observations is used to test 𝐻 is called test
statistic.
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8.1.10 One-tailed and two-tailed tests:
A function of any statistical hypothesis where the alternative
hypothesis is one -tailed (right -tailed or left -tailed) is called a one -
tailed test.
Example : A test for testing the mean of a population
𝐻: 𝜇= 𝜇 Versus 𝐻ଵ: 𝜇 > 𝜇 (Right Tailed) or 𝐻ଵ: 𝜇 < 𝜇 (Left
Tailed), is a one -tailed test. In the right -tailed test, the critical region
lies entirely in the right tail of the sampling distribution of 𝑋ത, while
for the left tailed test, the critical region lies entirely in the left tail of
the sam pling distribution of 𝑋ത.
A test of any statistical hypothesis where the alternative hypothesis
is two -tailed such as; 𝐻: 𝜇= 𝜇 Versus 𝐻ଵ: 𝜇≠ 𝜇 (𝜇 >
𝜇 𝑜𝑟 𝜇 < 𝜇) is known as a two -tailed test, and in such a case, the
critical region is giv en by the portion of the area lying in both the
tails (sides) of the probability curve of the test statistic 𝑋ത.
8.1.11 Central limit theorem :
In many cases, the exact probability distribution of the test statistics
T cannot be obtained. The difficulty is overcome using the normal
approximation. The probability distribution of standardized T is
assumed to be N (0, 1) as the sample size 𝑛 → ∞ (i.e., n is
sufficiently large). The corresponding theorem in support of the
normal approximation is known as the central limit theor em.
Case -I: Parent population is Normal :
Let the random sample drawn from 𝑁 (𝜇,𝜎ଶ). By the definition of a
random sample, vari ates values 𝑥ଵ,𝑥ଶ,…,𝑥 of the sample are
independent and identically distributed as 𝑁 (𝜇,𝜎ଶ) then the sample mean
(𝑋ത) is distributed normally with 𝜇 and variance 𝜎ଶ
𝑛ൗ i.e.,
𝑋ത ~ 𝑁 (𝜇,𝜎ଶ
𝑛ൗ). The result shows how the precision of a sample mean
increases as the sample size increases and 𝑍= ௫̅ି ఓ
ఙ√⁄ ~ 𝑁 (0,1), standard
normal variate.
Case -II: Parent population is Non -Normal :
If the population from which the random sample is drawn has a non-
normal distribution with finite mean 𝜇 and finite standard deviation 𝜎 then
the variate, by owing to central limit theorem,
𝑍= 𝑥̅− 𝜇
𝜎√𝑛⁄ ~ 𝑁 (0,1) 𝑎𝑠 𝑛 → ∞
8.1.12 Critical value:
The value of the test statistic, which separates the critical (rejection)
region and acceptance region, is called the ‘critical value’ . It
depends upon (1) The level of significance 𝛼 used and (2) The
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8.1.13 P-value :
Another approach for testing is to find out the ‘p’ value at which 𝐻 is
significant. That is, to find the smallest level of significance, 𝛼 at
which 𝐻 is rejected. About the acceptance or rejection of 𝐻, the
experimenter can himself decide the level 𝛼 by comparing it with the
p-value. The criterion for this is that if the p -value is less than or equal
to 𝛼, reject 𝐻 otherwise, accept 𝐻.
Example :
Fig.4: P -value for One -tailed Z -Test
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8.1.14 Power of test:
𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝑇𝑒𝑠𝑡=𝑃 [𝑅𝑒𝑗𝑒𝑐𝑡𝑖𝑛𝑔 𝐻 𝑔𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 𝐻ଵ 𝑖𝑠 𝑡𝑟𝑢𝑒]
= 𝑃 [𝑅𝑒𝑗𝑒𝑐𝑡𝑖𝑛𝑔 𝐻 | 𝐻ଵ 𝑖𝑠 𝑡𝑟𝑢𝑒]
=1− 𝑃 [𝐴𝑐𝑐𝑒𝑝𝑡𝑖𝑛𝑔 𝐻 | 𝐻ଵ 𝑖𝑠 𝑡𝑟𝑢𝑒]
=1− 𝑃 [𝑇𝑦𝑝𝑒 𝐼𝐼 𝐸𝑟𝑟𝑜𝑟]
=1− 𝛽
8.1.15 The p rocedure of testing of hypothesis :
In any testing of a hypothesis, it is a stepwise procedure that leads to
rejection or acceptance of the null hypothesis based on samples drawn
from the population. The various steps in testing a statistical hypothesis
are as follows;
i. Set up the null hypothesis 𝐻.
ii. Set up the alternative hypothesis 𝐻ଵ, this will enable us to decid e
whether we have to use a one -tailed (right or left) test or a two -tailed
test.
iii. Choose an appropriate level of significance ( 𝛼), i.e., 𝛼 is fixed in
advance.
iv. Choose the appropriate test statistic Z or T and find its value under the
null hypothesis 𝐻.
v. Determine the critical values and critical region corresponding to the
level of significance and the alternative hypothesis ( 𝑍ఈ for one -tailed
𝐻ଵor 𝑍ఈଶ⁄ for two -tailed 𝐻ଵ).
vi. Decision rule: We c ompare the calculated value of Z, with the tabulated
value of 𝑍ఈ if 𝐻ଵ is one -tailed & compare | Z| with 𝑍ఈଶ⁄ (For
symmetrical distribution). If the calculated value is greater than the
tabulated value, then we reject 𝐻 at 𝛼 % level of significance and
conclude that there is a significant difference at 𝛼 % level of
significance otherwise accept 𝐻 at 𝛼 % significance level and
conclude that there is no significant difference at 𝛼 % level of
significance.
8.2 LARGE SAMPLE TESTS
8.2.1 Test for testing of a population mean :
Let consider 𝑥={𝑥ଵ,𝑥ଶ,…,𝑥} be a random sample of size n taken
from a norma lly distributed population having population mean 𝜇 and
population variances 𝜎ଶ respectively.
We have to test the hypothesis,
𝐻: 𝜇= 𝜇
Against
𝐻ଵ: 𝜇≠ 𝜇 OR
𝐻ଵ: 𝜇 > 𝜇 OR munotes.in
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𝐻ଵ: 𝜇 < 𝜇
Under 𝐻, the test statistic is,
𝑍= ௫̅ିఓబ
ఙ√⁄ ~ 𝑁 (0,1),
where n is the sample size, 𝑥̅ is the sample mean, 𝜎 population standard
deviation.
Let 𝑍ఈ be the critical value at 𝛼 level of significance. We compare the
calculated value of Z, with the tabulated value of 𝑍ఈ. Where 𝑃(𝑍≤𝑍ఈ)=
𝛼.
𝐻ଵ is two -tailed ( 𝜇≠ 𝜇 )
|𝑍|≥𝑍ఈଶ⁄, then reject H 0.
𝐻ଵ is one -tailed ( 𝜇 > 𝜇)
𝑍≥𝑍ଵିఈ, then reject H 0.
𝐻ଵ is one -tailed (𝜇 <𝜇 )
𝑍≤𝑍ఈ, then reject H 0.
8.2.2 Test for the testing difference of the two population mean :
Let consider 𝑥ଵ={𝑥ଵଵ,𝑥ଵଶ,…,𝑥ଵభ} and 𝑥ଶ={𝑥ଶଵ,𝑥ଶଶ,…,𝑥ଶమ} be
two independent normally distributed samples having population
means 𝜇ଵ and 𝜇ଶ and population variances 𝜎ଵଶ and 𝜎ଶଶ respectively.
𝑥̅ଵ and 𝑥̅ଶ are the sample means.
We have to test the hypothesis,
𝐻: 𝜇ଵ−𝜇ଶ=𝜇
Against
𝐻ଵ: 𝜇ଵ≠ 𝜇ଶ OR
𝐻ଵ: 𝜇ଵ−𝜇ଶ>𝜇 OR
𝐻ଵ: 𝜇ଵ−𝜇ଶ<𝜇
Under 𝐻, the test statistic is,
𝑍= (௫̅భି ௫̅మ)ିఓబ
ඨቈభమ
భାቈమమ
మ ~ 𝑁 (0,1).
Let 𝑍ఈ be the critical value at 𝛼 level of significance . We compare the
calculated value of Z, with the tabulated value of 𝑍ఈ. Where 𝑃(𝑍≤𝑍ఈ)=
𝛼.
𝐻ଵ is two-tailed (𝜇ଵ≠ 𝜇ଶ )
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𝐻ଵ is one -tailed ( 𝜇ଵ−𝜇ଶ>𝜇 )
𝑍≥𝑍ଵିఈ, then reject H 0.
𝐻ଵ is one -tailed ( 𝜇ଵ−𝜇ଶ<𝜇 )
𝑍≤𝑍ఈ, then reject H 0.
8.2.4 Test for testing of the population proportion :
We have to test the hypothesis,
𝐻: 𝑃= 𝑃
Against
𝐻ଵ: 𝑃≠ 𝑃 OR
𝐻ଵ: 𝑃 > 𝑃 OR
𝐻ଵ: 𝑃 < 𝑃
Under 𝐻, the test statistic is,
𝑍= ି బ
ටುబೂబ
~ 𝑁 (0,1),
where n is the sample size, p is the sample proportion.
Let 𝑍ఈ be the critical value at 𝛼 level of significance. We compare the
calculated value of Z, with the tabulated value of 𝑍ఈ. Where 𝑃(𝑍≤𝑍ఈ)=
𝛼.
𝐻ଵ is two -tailed ( 𝑃≠ 𝑃 )
|𝑍|≥𝑍ఈଶ⁄, then reject H 0.
𝐻ଵ is one -tailed ( 𝑃 > 𝑃)
𝑍≥𝑍ଵିఈ, then reject H 0.
𝐻ଵ is one -tailed ( 𝑃 < 𝑃 )
𝑍≤𝑍ఈ, then reject H 0.
8.2.4 Test for testing equality of two population Proportion :
We have to test the hypothesis,
𝐻: 𝑃ଵ= 𝑃ଶ
Against
𝐻ଵ: 𝑃ଵ≠ 𝑃ଶ OR
𝐻ଵ: 𝑃ଵ> 𝑃ଶ OR
𝐻ଵ: 𝑃ଵ< 𝑃ଶ
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Under 𝐻, the test statistic is,
𝑍= (𝑝ଵ− 𝑝ଶ)
ට𝑃𝑄ቀ1
𝑛ଵ+ 1
𝑛ଶቁ ~ 𝑁 (0,1)
𝑃= 𝑛ଵ𝑝ଵ+ 𝑛ଶ𝑝ଶ
𝑛ଵ+ 𝑛ଶ 𝑄=1− 𝑃
Let 𝑍ఈ be the critical value at 𝛼 level of significance. We compare the
calculated value of Z, with the tabulated value of 𝑍ఈ. Where 𝑃(𝑍≤𝑍ఈ)=
𝛼.
𝐻ଵ is two -tailed ( 𝑃ଵ≠ 𝑃ଶ )
|𝑍|≥𝑍ఈଶ⁄, then reject H 0.
𝐻ଵ is one -tailed ( 𝑃ଵ> 𝑃ଶ)
𝑍≥𝑍ଵିఈ, then reject H 0.
𝐻ଵ is one -tailed ( 𝑃ଵ< 𝑃ଶ )
𝑍≤𝑍ఈ, then reject H 0.
8.3 SMALL SAMPLE TESTS
8.3.1 Test for the population mean:
Let consider 𝑥={𝑥ଵ,𝑥ଶ,…,𝑥} be a random sample of size n (n small)
taken from a normally distributed population having population mean
𝜇 and population variances 𝜎ଶ (mean and variances are unknown)
respectively.
We have to test the hypothesis,
𝐻: 𝜇= 𝜇
Against
𝐻ଵ: 𝜇≠ 𝜇 OR
𝐻ଵ: 𝜇 > 𝜇 OR
𝐻ଵ: 𝜇 < 𝜇
Under 𝐻, the test statistic is,
𝑡= 𝑥̅− 𝜇
𝑠√𝑛⁄ ~ 𝑡ିଵ𝑑.𝑓.
Where n is the sample size, 𝑥̅ is the sample mean, 𝑠 sample standard
deviation.
Let 𝑡(ିଵ,ఈ) be the critical value based on students t -distribution at n -1
degrees of freedom and 𝛼 level of significance. We compare the munotes.in
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calculated value of t, with the tabulated value of 𝑡(ିଵ,ఈ). Where 𝑃(𝑇≤
𝑡(ିଵ,ఈ))=𝛼.
𝐻ଵ is two -tailed ( 𝜇≠ 𝜇 )
|𝑡|≥𝑡(ିଵ,ఈଶ⁄), then reject H 0.
𝐻ଵ is one -tailed ( 𝜇 > 𝜇)
𝑡≥𝑡(ିଵ,ଵିఈ), then reject H 0.
𝐻ଵ is one -tailed (𝜇 <𝜇 )
𝑡≤𝑡(ିଵ,ఈ), then reject H 0.
8.3.2 Test for the difference of two population means (two sample s):
Let consider 𝑥={𝑥ଵ,𝑥ଶ,…,𝑥భ} and 𝑦={𝑦ଵ,𝑦ଶ,…,𝑦మ} be two
independent normally distributed samples having unknown population
means 𝜇ଵ and 𝜇ଶ and unknown population variances 𝜎ଵଶ and 𝜎ଶଶ
respectively.
𝑥̅ and 𝑦ത are the sample arithmetic means.
We have to test the hypothesis,
𝐻: 𝜇ଵ−𝜇ଶ=𝜇
Against
𝐻ଵ: 𝜇ଵ≠ 𝜇ଶ OR
𝐻ଵ: 𝜇ଵ−𝜇ଶ>𝜇 OR
𝐻ଵ: 𝜇ଵ−𝜇ଶ<𝜇
Under 𝐻, the test statistic is,
𝑡= തି ത
௦ටభ
భାభ
మൗ ~ 𝑡భାమିଶ𝑑.𝑓. where 𝑠ଶ= (భିଵ)௦ೣమା (మିଵ)௦మ
భା మିଶ.
Let 𝑡(భାమିଶ,ఈ) be the critical value based on students t -distribution at
𝑛ଵ+𝑛ଶ−2 degrees of freedom and 𝛼 level of significance. We compare
the calculated value of t, with the tabulated value of 𝑡(భାమିଶ,ఈ). Where
𝑃(𝑇≤𝑡(భାమିଶ,ఈ))=𝛼.
𝐻ଵ is two -tailed ( 𝜇≠ 𝜇 )
|𝑡|≥𝑡(భାమିଶ,ఈଶ⁄), then reject H 0.
𝐻ଵ is one -tailed ( 𝜇 > 𝜇)
𝑡≥𝑡(భାమିଶ,ଵିఈ), then reject H 0.
𝐻ଵ is one -tailed (𝜇 <𝜇 )
𝑡≤𝑡(భାమିଶ,ఈ), then reject H 0. munotes.in
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8.3.3 Paired t -test for difference of m ean:
Let consider {𝑥,𝑦};𝑖=1,2,…,𝑛 be n -pair dependent normally
distributed samples having unknown population means 𝜇ଵ and 𝜇ଶ and
unknown population variances 𝜎ଵଶ and 𝜎ଶଶ respectively.
We have to test the hypothesis,
𝐻: 𝜇ଵ− 𝜇ଶ=0
Against
𝐻ଵ:𝜇ଵ− 𝜇ଶ≠ 0 OR
𝐻ଵ: 𝜇ଵ− 𝜇ଶ >0 OR
𝐻ଵ: 𝜇ଵ − 𝜇ଶ< 0
Under 𝐻, the test statistic is,
𝑡= തି ത
௦√⁄ ~ 𝑡ିଵ𝑑.𝑓. where 𝑑= 𝑋ത− 𝑌ത & 𝑠ଶ= ଵ
ିଵ∑൫𝑑−𝑑̅൯ଶ
Let 𝑡(ିଵ,ఈ) be the critical value based on students t -distribution at n -1
degrees of freedom and 𝛼 level of significance. We compare the
calculated value of t, with the tabulated value of 𝑡(ିଵ,ఈ). Where 𝑃(𝑇≤
𝑡(ିଵ,ఈ))=𝛼.
𝐻ଵ is two -tailed ( 𝜇≠ 𝜇 )
|𝑡|≥𝑡(ିଵ,ఈଶ⁄), then reject H 0.
𝐻ଵ is one -tailed ( 𝜇 > 𝜇)
𝑡≥𝑡(ିଵ,ଵିఈ), then reject H 0.
𝐻ଵ is one -tailed (𝜇 <𝜇 )
𝑡≤𝑡(ିଵ,ఈ), then reject H 0.
8.4 STATISTICAL CONTROL CARTS
8.4.1 A lot acceptance sampling plan (LASP) :
A lot acceptance sampling plan (LASP) is a sampling scheme and a set
of rules for making decisions. The decision, based on counting the
number of defectives in a sample, can be to accept the lot, reject the lot,
or even, for multiple or sequential sampling schem es, to take another
sample and then repeat the decision process.
8.4.2 OC curve :
Operating Characteristic (OC) Curve plots the probability of accepting
the lot (Y -axis) versus the lot fraction or percent defectives (X -
axis). The OC curve is the primary tool for displaying and
investigating the properties of a LASP.
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8.4.3 Control charts :
Control charts are a statistical process control tool used to determine if
a manufacturing or business process is in a state of control . It is more
appropriate to say that the control charts are the graphical device for
Statistical Process Monitoring (SPM). Traditional control charts are
mostly de signed to monitor process parameters when an underlying
form of the process distributions are known.
8.4.4 Variable control chart ( 𝑿ഥ Chart) :
Dr Walter A. Shewhart proposed a general model for control charts in
1920. Let 𝑤 be a sample statistic that measures some continuously
varying quality characteristic of interest (e.g., thickness), and suppose
that the mean of 𝑤 is 𝜇௪, with a standard deviation of 𝜎௪. Then the
centre line, the Upper Control Limit (UCL), and the Lower Control
Limit (LCL) are:
𝑈𝐶𝐿=𝜇௪+𝑘𝜎௪
Center Line =𝜇௪
𝐿𝐶𝐿=𝜇௪+𝑘𝜎௪
where k is the distance of the control limits from the cent re line, expressed
in terms of standard deviation units. When 𝑘 is set to 3, we speak of
3−𝑠𝑖𝑔𝑚𝑎 control charts. Historically, 𝑘=𝟑 has become an accepted
standard in the industry. The centerline is the process mean, which in
general is unknown. We replace it with a target or the average of all the
data. The quantity that we plot is the sample average, 𝑋ത The chart is called
the 𝑋ത chart.
We also have to deal with the fact that σ is, in general, unknown. Here we
replace σw with a given standard value, or we estimate it by a function of
the average standard deviation.
8.4.5 Attributes control charts :
The Shewhart control chart plots quality characteristics that can be
measured and expressed numerically. We measure weight, height,
position, thickness, etc. If we cannot represent a particular quality
characteristic numerically, or if it is impractical to do so, we then often
resort to u sing a quality characteristic to sort or classify an item that is
inspected into one of two "buckets".
An example of a common quality characteristic classification would be
designating units as "conforming units" or "nonconforming units".
Another quality characteristic criteria would be sorting units into "non -
defective" and "defective" categories. Quality characteristics of that type
are called attributes.
Control charts dealing with the number
of defects or nonconformities are called c charts (for the count). munotes.in
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Control charts dealing with the proportion or fraction of defective
product s are called p charts (for proportion).
There is another chart that handles defects per unit, called the u chart
(for the unit). This applies when we wish to work with the a verage
number of nonconformities per unit of product.
8.6 SUMMARY
Students can get an idea about the testing of a hypothesis and make
decisions about parameters of interest. In this chapter, we briefly studied
various large/small One -sample and two -samp le tests for mean and
proportion along with variable and attribute control charts are also
discussed.
8.5 EXERCISE
1. Explain the term hypothesis and its types.
2. Define Type -I and Type -II errors.
3. Explain the terms level of significance, p -value and power of a test.
4. Write down the stepwise procedure of testing of hypothesis.
5. Write down the procedure for testing the equality of two population
proportion s.
6. Write down the procedure to test the specified population mean, in the
case of a small sample.
7. Explain in detail paired t -Test for difference Mean.
8.7 REFERENCES
Gupta S. C. and Kapoor V. K., 2011, Fundamentals of Mathematical
Statistics, 11th Ed, Sultan and Chand.
Rohatgi V. K, 1939, Introduction to Probability and Statis tics, Wiley
Murray R. Spiegel, Larry J. Stephens, STATISTICS, 4th Ed,
McGRAW – HILL I NTERNATIONAL.
J.N. KAPUR and H.C. SAXENA, 2005, MATHEMATICAL
STATISTICS, 12th rev, S.Chand.
Agrawal B. L, 2003, Programmed Statistics , 2nd Ed, New Age
International.
Kanji G. K., 20 06, 100 Statistical tests, 3rd Ed, SAGE Publication.
***** munotes.in
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9
STATISTICS IN R
Unit Structure
9.0 Objectives
9.1 Descriptive statistics in R
9.2 Normal distribution
9.3 Binomial distribution
9.4 Frequency distribution
9.5 Data i mport and export
9.6 Summary
9.7 Exercise
9.8 References
9.0 OBJECTIVES
Use of R -software to find basic statistical measures
Use of R -software to simulate Normal Distributions
Use of R -software to simulate Binomial Distributions
Data import and Export in R.
9.1 DESCRIPTIVE STATISTICS IN R -SOFTWARE Sr. No. Descriptive Statistics Syntax in R Output 1. Observation Vector (c-function) x <- c(1,4,7,12,19,15,21,20) x [1] 1 4 7 12 19 15 21 20 2. Arithmetic mean AM<-mean(x) AM [1] 12.375 3. Mode # Create the function. getmode <- function(x) { uniqv <- unique(x) uniqv[which.max(tabulate(match(x, uniqv)))] } v <-c(2,1,2,3,1,2,3,4,1,5,5,3,2,3); mode<-getmode(v) Mode [1] 2
4. median Med=median(x) Med [1] 13.5 munotes.in
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9.2 NORMAL DISTRIBUTION
R has four inbuilt functions to generate normal distribution. They are
described below.
Sr. no. Function Name Syntax Description 1. Density function dnorm(x, mean, sd) This function gives the height of the probability distribution at each point for a given mean and standard deviation. 2. Cumulative Probability pnorm(x, mean, sd) This function gives the probability of a normally distributed random number less than the value of a given number. It is also called the "Cumulative Distribution Function”. 3. Inverse function qnorm(p, mean, sd) This function takes the probability value and gives a number whose cumulative value matches the probability value. 4. Random Number rnorm(n, mean, sd) This function is used to generate random numbers whose distribution is normal. It takes the sample size as input and generates that many random numbers.
Following is the description of the parameters used in the above
functions :
x: is a vector of numbers.
p: is a vector of probabilities.
n: is a number of observations(sample size).
mean: is the mean value of the sample data. Its default value is zero.
sd: is the standard deviation. Its default value is 1.
9.3 BINOMIAL DISTRIBUTION
R has four in -built functions to generate binomial distribution. They are
described below.
Sr. no. Function Name Syntax Description 1. Probability dbinom(x, size, prob) This function gives the munotes.in
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probability density distribution at each point. 2. Cumulative Probability pbinom(x, size, prob) This function gives the cumulative probability of an event. It is a single value representing the probability. 3. Inverse function qbinom(p, size, prob) This function takes the
probability value and
gives a number whose
cumulative value
matches the probability
value. 4. Random Number rbinom(n, size, prob) This function generates
required number of
random values of given
probability from a given
sample.
Following is the description of the parameters used −
x: is a vector of numbers.
p: is a vector of probabilities.
n: is a number of observations.
Size: is the number of trials.
prob: is the probability of success of each trial.
9.4 FREQUENCY DISTRIBUTION IN R
Table function in R -table(), performs categorical tabulation of data
with the variable and its frequency. table() function is also helpful in
creating Frequency tables with the condition and cross -tabulation
Syntax
x <- c(1,2,3,2,4,2,5,4,6,7,8,9)
freq <- data.frame(table(x))
Output
> freq x Freq
1 1 1
2 2 3
3 3 1
4 4 2
5 5 1
6 6 1
7 7 1
8 8 1
9 9 1 munotes.in
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9.5 DATA IMPORT AND EXPORT
i. Data Importing :
The sample data is frequently observed in Excel format and needs to be
imported into R before use. For this, we can use the function read.xls from
the gdata package. It reads from an Excel spreadsheet and returns a data
frame.
library(gdata) # load gdata package
> help(read.xls) # documentation
> mydata = read.xls("mydata.xls") # read from the first sheet
ii. Data export :
There are numerous methods for exporting R objects into other formats.
A tab delimited tex file
write.table(mydata, "c:/mydata.txt", sep=" \t")
MS-Excel Spread sheet
library(xlsx)
write.xlsx(mydata, "c:/mydata.xlsx")
9.7 SUMMARY
R is a programming language and free software environment for statistical
computing and graphics supported by the R Core Team and the R
Foundation for Statistical Computing. It is widely used for data analysis
purposes in analytical industries. In this chapter, we studied various
descriptive measures in R, probabilities, quanti les and random number
generation syntax of the normal and binomial distribution.
9.6 EXERCISE
1. Write down the R -command for the arithmetic mean and compute for
the given data.
1, 4, 7, 12, 19, 15, 21 and 20.
2. Generate 10 random numbers using R -command from a normal
distribution with mean zero and standard deviation one.
3. Calculate cumulative distribution function at point zero for normal
distribution with mean zero and standard deviation one.
4. Compute probability density function at 1, 2, 3, 4, 5 for normal
distribution with mean zero and standard deviation one.
5. Compute quantiles at 0.1, 0.2, 0.3, 0.4, 0.5 for binomial distribution
with n=5 and p=0.7.
6. Generate 10 random numbers using R -command from a binomial
distribution with n=5 and p=0.5. munotes.in
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7. Explain fr equency distribution in R.
9.8 REFERENCES
Gupta S. C. and Kapoor V. K., 2011, Fundamentals of Mathematical
Statistics, 11th Ed, Sultan and Chand.
R.B. Patil, H.J. Dand and R. Bhavsar, 2017, A Practical Approach
using R, 1st ed, SPD.
https://www.tutorialspoint.com/r/r_normal_distribution.htm
https://cran.r -project.org/bin/windows/base/
*****
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149 UNIT IV
10
SMALL SAMPLING THEORY
Unit Structure
10.0 Objectives
10.1 Introduction
10.2 Student’s t distribution
10.3 Graph of t-distribution
10.4 Critical values of t
10.5 Application of t -distribution
10.6 Test of Hypothesis and Significance
10.7 Confidence Interval
10.8 t-Test for Difference of Means
10.9 Degrees of Freedom
10.10 The F -Distribution
10.11 Summary
10.12 Reference for further reading
10.13 Exercises
10.14 Solution to Ex ercises
10.15 Tables of t -distribution an d F-distribution
10.0 OBJECTIVES
In this chapter we will study about the test suitable for small samples i.e
sample size less than or equal to 30 for which the tests studied in previous
chapters are not applicable. We will also study the test for equality of
variance.
10.1 INTRODUCTION
The entire large sample theory was based on the application of “Normal
Test”. However if the sample size n is small, the distribution of the various
statistics, e.g. 𝑍=௫̅ିఓ
(
√)𝑜𝑟𝑍=ି
ඥொetc., are far from normality and as such
‘normal test’ cannot be applied if ‘n’ is small. In such cases exact sample
tests, pioneered by W.S. Gosset (1908) who wrote under the pen name of
Student, and later on developed and extended by Prof. R. A. Fisher (1926), munotes.in
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150 are used. In the following sections we shall discuss i) t – test and ii) F -
test.
The exact sample tests can, however, be applied to large samples though
the converse is not true. In all exact sample tests, the basic assumption is
that “the population(s) from which the sample(s) is(are) drawn is(are)
normal, i.e., the parent population( s) is(are) normally distributed.”
10.2 STUDENT’S T DISTRIBUTION
Let x i(i = 1,2,….,n) be a random sample of size n from a normal
population with mean 𝜇 and variance 𝜎ଶ. Then Student’s t is defined by
the statistic:
𝑡=𝑡=𝑥̅−𝜇
௦
√ ,𝑠ଶ=1
𝑛−1(𝑥−𝑥̅)ଶୀ
ୀଵ,𝑥̅=∑𝑥
𝑛
where 𝑥̅ is the sample mean and 𝑠ଶis an unbiased estimate of the
population variance 𝜎ଶ, and it follows Student’s distribution with 𝑣=
(𝑛−1) degree of freedom with probability density function:
𝑓(𝑡)=1
√𝑣𝐵ቀଵ
ଶ,௩
ଶቁ⋅1
ቀ1+௧మ
௩ቁೡశభ
మ ,−∞<𝑡∞
10.3 GRAPH OF t-DISTRIBUTION
The probability density function of t -distribution with n degrees of
freedom is:
𝑓(𝑡)=𝐶.ቆ1+𝑡ଶ
𝑛ቇି(శభ)
మ
,−∞<𝑡<∞
Since f(t) is an even function, the probability curve is symmetric about the
line t =0. As t increases, f(t) decreases rapidly and tends to zero as 𝑡→∞,
so that t -axis is an asymptote to the curve. We know that
𝜇ଶ=𝑛
𝑛−2 ,𝑛>2 ; 𝛽ଶ=3(𝑛−2)
𝑛−4,𝑛>4
Hence for n >2, 𝜇ଶ>1 i.e., the variance of t -distribution is greater than
that of standard normal distribution and for n >4, 𝛽ଶ>3 and thus t -
distribution is more flat on the top than the normal curve. In fact, for small
n, we have
𝑃(|𝑡|≥𝑡)>−𝑃(|𝑍|≥𝑡),𝑍 ~ 𝑁(0,1)
i.e, the tails of the t -distribution have a greater probability (area) than the
tails of standard normal distribution. Moreover we can check that for large munotes.in
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151 n, t-distribution tends to standard normal distribution. Graph of t -
distribution is given by the following diagram
10.4 CRITICAL VALUES OF t
The critical values of t at level of significance 𝛼 and degree of freedom v
for two tailed test are given by the equation:
𝑝{|𝑡|>𝑡௩(𝛼)}=𝛼
𝑝{|𝑡|≤𝑡௩(𝛼)}=1−𝛼
The values 𝑡௩(𝛼) have been tabulated in table, for different values of
𝛼𝑎𝑛𝑑𝑣 are given at the end of the chapter.
Since t -distribution is symmetric about t= 0, we get
𝑃൫𝑡<𝑡௩(𝛼)൯+𝑃൫𝑡<−𝑡௩(𝛼)൯=𝛼⇒2𝑃(൫𝑡>𝑡௩(𝛼)൯=𝛼
⇒𝑃൫𝑡>𝑡௩(𝛼)൯=𝛼
2∴𝑃൫𝑡>𝑡௩(2𝛼)൯=𝛼
𝑡௩(2𝛼) (from the tables at the end of the chapter) gives the significant
value of t for a single tail test(Right tail or L eft tail since the distribution is
symmetrical), at level of significance 𝛼&𝑣 degree of freedom.
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152 Hence the significant values of t at level of significance ′𝛼′ for a single
tailed test can be obtained from those of two tailed test by looking the
values at level of significance 2𝛼.For example
𝑡଼(0.05) 𝑓𝑜𝑟 𝑠𝑖𝑛𝑔𝑙𝑒 𝑡𝑎𝑖𝑙 𝑡𝑒𝑠𝑡=𝑡଼(0.1) 𝑓𝑜𝑟 𝑡𝑤𝑜 𝑡𝑎𝑖𝑙 𝑡𝑒𝑠𝑡=1.86
𝑡ଵହ(0.01) 𝑓𝑜𝑟 𝑠𝑖𝑛𝑔𝑙𝑒 𝑡𝑎𝑖𝑙 𝑡𝑒𝑠𝑡=𝑡ଵହ(0.02) 𝑓𝑜𝑟 𝑡𝑤𝑜 𝑡𝑎𝑖𝑙 𝑡𝑒𝑠𝑡=2.602
10.5 APPLICATION OF t-DISTRIBUTION
The t -distribution has a whole number of applications in Statistics, some
of which are given here below:
i) to test if the sample mean (𝑥̅) differs significantly from the
hypothetical value 𝜇 of the population mean.
ii) to test the significance of the difference between two sample means.
iii) to test the significance of an observed sample correlation coefficient
and sample regression co efficient.
iv) to test the significance of observed partial correlation coefficient.
10.6 TEST OF HYPOTHESIS AND SIGNIFICANCE
Suppose we want to test :
i) if a random sample x i(i=1,2,….,n) of size n has been drawn from a
normal population with a specified mean say 𝜇 orif the sample mean
differs significantly from the hypothetical value of 𝜇 of the
population mean.
i.e 𝐻∶𝜇= 𝜇 ,𝐻ଵ∶𝜇≠𝜇𝑜𝑟𝜇>𝜇𝑜𝑟𝜇<𝜇where 𝐻 is the null
hypothesis and 𝐻ଵ is the alternative hypothesis
ii) Calculate
𝑡=𝑥̅−𝜇
ௌ
√,𝑆ଶ =1
𝑛−1(𝑥−𝑥̅)ଶୀ
ୀଵ
iii) degree of freedom df = v = n -1
iv) from the table calculate, 𝑡௩(𝛼)
v) Conclusion: Reject H 0if calculated |t| > tabulated t and Do not reject
H0 if calculated |t| ≤ tabulated t .
Remark: We know, the sample variance:
𝑠ଶ=1
𝑛∑(𝑥−𝑥̅)ଶ⇒𝑠ଶ=1
𝑛(𝑛−1)𝑆ଶ⇒𝑠ଶ
𝑛−1=𝑆ଶ
𝑛 munotes.in
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153 Hence for numerical problems, the test statistic t stated above will become
𝑡=𝑥̅−𝜇
ௌ
√=𝑥̅−𝜇
௦
√ିଵ
Eg: 1) A machinist is making engine parts with axle diameter of 0.7 inch.
A random sample of 10 parts show a mean diameter of 0.742 inch with a
standard deviation of 0.04 inch. Compute the statistic you would use to
test whether the work is meeting the specifications and state the
conclusion .
Solution:
i) 𝐻∶𝜇=0.7,𝐻ଵ∶𝜇≠0.7
ii) 𝑥̅=0.742,𝑠=0.04,𝑛=10,
𝑡=𝑥̅−𝜇
௦
√ିଵ=0.742−0.7
.ସ
√ଽ= 0.042×√9
0.04=3.15
iii) degree of freedom i.e v = n -1, v = 9
iv) 𝑡௩(𝛼)=𝑡ଽ(0.05)=𝑡ଽ(0.025)=2.262
v) conclusion 𝑡=3.15>𝑡ଽ(0.05) =2.262
Since calculated ‘t’ is greater than the tabulated ‘t’ we reject the null
hypothesis H0 i.e the product is not conforming to specifications.
Eg: 2) The mean weakly sales of soap bars in departmental stores was
146.3 bars per store. After an advertising campaign the mean weekly sales
in 22 stores for a typical week increased to 153.7 and showed a standard
deviation of 17.2. Was the advertising campaign successful?
Solution:
i) 𝐻∶𝜇=146.3,𝐻ଵ∶𝜇>146.3
ii) 𝑥̅=153.7,𝑠=17.2,𝑛=22
𝑡=𝑥̅−𝜇
௦
√ିଵ=153.7−146.3
ଵ.ଶ
√ଶଵ= 7.4×√21
17.2=1.97
iii) degree of freedo m f i.e v = n -1, v = 21
iv) iv) 𝑡௩(𝛼)=𝑡ଶଵ(0.05)=𝑡ଶଵ(0.025)=1.72
v) conclusion 𝑡=1.97>𝑡ଶଵ(0.05) =1.72
Since calculated ‘t’ is greater than the tabulated ‘t’ we reject the null
hypothesis H 0 i.e the advertising campaign was successful in
promoting the sales.
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154 Eg: 3) A random sample of 10 boys had the following I.Q.’s is 70, 120,
110, 101, 88, 83, 95, 98, 107, 10 0. Do these data support the assumption
of a population mean I.Q of 100?
Solution: First we calculate find the value of 𝑥̅&𝑆ଶ.
Here n = 10, ∑𝑥=972∴𝑥̅=∑௫
=ଽଶ
ଵ= 97.2
𝑆ଶ=1
𝑛−1∑(𝑥−𝑥̅)ଶ=1
9(1833.6)=203.73⇒𝑆= √203.73
=14.273
i) 𝐻∶𝜇=100,𝐻ଵ∶𝜇≠100
ii) 𝑥̅=97.2,𝑆=203.73,𝑛=10
𝑡=|𝑥̅−𝜇|
ௌ
√=|97.2−100|
ଵସ.ଶଷ
√ଵ= 2.8×√10
14.273=0.6203
iii) degree of freedo m d f i.e v = n -1, v = 9
iv) 𝑡௩(𝛼)=𝑡ଽ(0.05)=𝑡ଽ(0.025)=2.262
v) conclusion 𝑡=0.6203<𝑡ଽ(0.05) =2.262
Since calculated ‘t’ is less than the tabulated ‘t’ we accept the null
hypothesis H 0 i.e the data are consistent with the assumption of mean I.Q
of 100 in the population.
10.7 CONFIDENCE INTERVAL
As done with normal distribution in earlier chapter, we can define 95%,
99 % or other confidence intervals by using the table given at the end of
this chapter. We can estimate within specified limits of confidence the
population mean 𝜇. In general confidence limits are given by the formula
𝑥̅±𝑡௩(𝛼)𝑆
√𝑛
In specific the 95 % c onfidence interval is given by
ቆ𝑥̅−𝑡௩(0.05)𝑆
√𝑛,𝑥̅+𝑡௩(0.05)𝑆
√𝑛ቇ
and 99% confidence interval is given by
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155 ቆ𝑥̅−𝑡௩(0.01)𝑆
√𝑛,𝑥̅+𝑡௩(0.01)𝑆
√𝑛ቇ
Eg: 1) A random sample of 16 values from a normal population showed a
mean of 41.5 inches and the sum of squares of deviations from this mean
= 135 sq. inches. Obtain the 95 % and 99 % confidence limits for
population mean .
Solution: n = 16, 𝑥̅=41.5,∑(𝑥−𝑥̅)ଶ=135
∴𝑆ଶ=1
𝑛−1∑(𝑥−𝑥̅)ଶ=1
15(135)=9⇒𝑆=3
from the table of t -distribution, we get 𝑡ଵହ(0.05)=2.131&𝑡௩(0.01)=
2.94795% confidence limits for population mean are given by
𝑥̅±𝑡ଵହ(0.05)𝑆
√𝑛 =41.5±2.131×3
√16=41.5±2.131×0.75
⇒39.902<𝜇<43.098
99% confidence limits for population mean are given by
𝑥̅±𝑡ଵହ(0.01)𝑆
√𝑛 =41.5±2.947×3
√16=41.5±2.947×0.75
⇒39.29<𝜇<43.71
10.8 t-TEST FOR DIFFERENCE OF MEANS
Suppose we want to test if two independent samples
𝑥(𝑖=1,2,3,⋯⋯,𝑛ଵ)and𝑦(𝑗=1,2,3,⋯⋯,𝑛ଶ)of size n 1 and n 2 have
been drawn from two normal populations with means 𝜇௫(𝜇ଵ) &𝜇௬(𝜇ଶ)
respectively.
Under the null hypothesis (H 0) that the samples have been drawn from the
normal populations with mean 𝜇௫(𝜇ଵ)&𝜇௬(𝜇ଶ) and the under the
assum ption that the population variance are equal
i.e 𝜎௫ଶ=𝜎௬ଶ =𝜎ଶ, the statistic
𝑡=ቀ𝑥̅−𝑦ത−൫𝜇௫−𝜇௬൯ቁ
𝑆൬ටଵ
భ+ଵ
మ൰ =𝑥̅−𝑦ത−𝑑
𝑆ටଵ
భ+ଵ
మ ,𝑑 =𝜇௫−𝜇௬
Where 𝑥̅=∑௫
భ ,𝑦ത=∑௬ೕ
మ & 𝑆ଶ=ଵ
భାమିଶቂ∑(𝑥−𝑥̅)ଶ+∑൫𝑦−𝑦ത൯ଶቃ is
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156 Students t -distribution with 𝑣=(𝑛ଵ−1)+(𝑛ଶ−1)=(𝑛ଵ+𝑛ଶ−2)
degree s of freedom .
Paired t-test for Difference of means:
Let us now consider the case when (i) the sample sizes are equal i.e. n 1 =
n2 = n and (ii) the two samples are not independent but the sample
observations are paired together, i.e., the pair of observations (𝑥,𝑦),1≤
𝑖≤𝑛 corresponds to the same ith sample unit. The pro blem is to test if the
sample means differ significantly or not.
For example, suppose we want to test the efficacy of a particular drug, say,
for inducting sleep. Let x i and y i (i =1,2,……,n) be the readings, in hours
of sleep, on the ith individual, before and after the drug is given
respectively. Here instead of applying the difference of the means test
discussed above in the same section we apply the paired t -test given
below:
Here we consider the increments, 𝑑=𝑥−𝑦 ,1≤𝑖≤𝑛
Under the null Hypothesis, H 0 that increments are due to fluctuations of
sampling, i.e., the drug is not responsible for these increments, the statistic
is
𝑡=𝑑̅
ௌ
√ =𝑑̅×√𝑛
𝑆
where
𝑑̅=1
𝑛𝑑
ୀଵ&𝑆ଶ=1
𝑛−1൫𝑑−𝑑̅൯ଶ=1
𝑛−1൝𝑑ଶ
ୀଵ−(∑𝑑
ୀଵ)ଶ
𝑛ൡ
ୀଵ
follows Student’s t -distribution with (n - 1) degree of freedom.
Eg: 1) For a random sample of 10 pigs fed on diet A, the increases in
weight in pounds in a certain period were: 10, 6, 16, 17, 13, 12, 8, 14, 15,
9. For another sample of 12 pigs, fed on Diet B, the increase in the same
period were: 7, 13, 22, 15, 12, 14, 18, 8, 21, 23, 10, 17.Test whether diets
A and B differ significantly as regards to their effect on increase in weight.
Solution:
i) Null Hypothesis , 𝐻: 𝜇௫=𝜇௬, i.e., there is no significant difference
between t he mean increase in weight due to diets A and B.
Alternative hypothesis, 𝐻ଵ:𝜇௫≠𝜇௬(two tailed)
ii) xi 𝑥−𝑥̅ (𝑥−𝑥̅)ଶ yj 𝑦−𝑦ത ൫𝑦−𝑦ത൯ଶ 10 -2 4 7 -8 64 6 -6 36 13 -2 4 16 4 16 22 7 49 munotes.in
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157 17 5 25 15 0 0 13 1 1 12 -3 9 12 0 0 14 -1 1 8 -4 16 18 3 9 14 2 4 8 -7 49 15 3 9 21 6 36 9 -3 9 23 8 64 10 -5 25 17 2 4 Total 120 0 120 180 0 314
𝑥̅=∑𝑥
𝑛ଵ =120
10=12 ,𝑦ത=∑𝑦
𝑛ଶ=180
12=15
𝑆ଶ=1
𝑛ଵ+𝑛ଶ−2ቂ∑(𝑥−𝑥̅)ଶ+൫𝑦−𝑦ത൯ଶቃ=1
20[120+314]
𝑆ଶ=21.7⇒𝑆=√21.7=4.6583
𝑡=𝑥̅−𝑦ത−𝑑
𝑆ටଵ
భ+ଵ
మ =12−15−0
4.6583ටଵ
ଵ+ଵ
ଵଶ =−3×√120
4.6583×√22= −32.8634
21.8494
⇒𝑡= −1.5041
iii) df = v = 10 + 12 - 2 = 20
iv) 𝑡ଶ(0.05)=2.086
v) Conclusion: |𝑡|=1.5041<𝑡ଶ(0.05)=2.086
Since calculated value of t is less tha n the tabulated value, the null
hypothesis H 0 is accepted at 5% level of significance and we may
conclude that the two diets do not differ significantly as regards their
effect on increase in weight.
Eg: 2) The yields if two types ‘Type A’ and ‘Type B’ of grains in pounds
per acre in 6 replications are given below. What comments would you
make on the difference in the mean yield.
Replication 1 2 3 4 5 6 Yield of Type A 20.5 24.6 2306 29.98 30.37 23.83 Yield of Type B 24.86 26.39 28.19 30.75 29.98 22.04
Solution: i) 𝐻:𝜇௫=𝜇௬,𝐻ଵ: 𝜇௫≠𝜇௬
ii)𝑡=ௗത
ቀೄ
√ቁ ,𝑑̅=∑𝑑
ୀଵ,𝑑 =𝑥−𝑦,𝑣=𝑛−1
𝑆ଶ=1
𝑛−1ቊ∑𝑑ଶ−(∑𝑑)ଶ
𝑛ቋ
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158 replication type A type B d d2 1 20.5 24.86 -4.36 2 24.6 26.39 -1.79 3 23.06 28.19 -5.13 4 29.98 30.75 -0.77 5 30.37 29.97 0.4 6 23.83 22.04 1.79 total -9.86 52.4876
(𝑖𝑖𝑖)𝑆ଶ=1
5ቈ52.4876−(−9.86)ଶ
6=1
552.4876−97.2196
6൨
=36.2843
5=7.2569
∴𝑆=√7.2569 =2.6939
𝑑̅= −ଽ.଼
= −1.6433,𝑡=ௗത
ೄ
√ =ିଵ.ସଷଷ×√
ଶ.ଽଷଽ =−1.4942
iv) s𝑡ହ(0.05)=2.571
v) Conclusion: |𝑡|=1.4942<𝑡ହ(0.05)=2.571
Since the calculated value of t is less than the tabulated value of t we
accept H 0 at 5% level of significance i.e., there is no major difference
between the mean yield of two types diets.
10.9 DEGREES OF FREEDOM
In order to discuss the statistic t as discussed above, it is necessary to use
observations obtained from a sample as well as certain population
parameters. If these parameters are unknown, they must be estimated from
the sample.
The number of degrees of freedom of a statistic, generally denoted by v, is
defined as the number N of independent observation in the sample (i.e. the
sample size) minus the number k of population parameters, which must be
estimated from sample observations. In symbols,
𝑣=𝑁−𝑘.
In case of the statistic t, the number of independent observations in the
sample is N, from which we compute 𝑥̅&𝑆. However, since we must
estimate 𝜇,𝑘=1and 𝑣=𝑁−1.
10.10 THE f-DISTRIBUTION
If X and Y are two independent chi -square variate s with v 1 and v 2 degree
of freedom respectively, then F -statistic is defined by
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159 𝐹=𝑋/𝑣ଵ
𝑌/𝑣ଶ
In other words, F is defined as the ration of two independent chi-square
variates divided by their respective degrees of freedom and it follows
Snedecor’s F-distribution with (𝑣ଵ,𝑣ଶ) degree of freedom with probability
function given by:
𝑓(𝐹)=ቀ௩భ
௩మቁೡభ
మ
𝐵ቀ௩భ
ଶ,௩మ
ଶቁ⋅𝐹ೡభ
మି ଵ
ቀ1 + ௩భ
௩మ𝐹ቁ(ೡభ శ ೡమ)
మ
Remark: The sampling distribution of F -statistic does not involve any
population parameters and depends only on the degrees of freedom 𝑣ଵ&𝑣ଶ.
F-test for Equality of Two population Variances.
Suppose we want to test (i) whether two independent samples x i,
(i=1,2,…n 1) and y j, (j =1,2,…n 2) have been drawn from the normal
population with the same variance 𝜎ଶ or (ii) whether the two independent
estimates of the population variance are homogenous or not.
Under the Null hypothesis 𝐻: 𝜎௫ଶ=𝜎௬ଶ=𝜎ଶ i.e., the population variances
are equal, or two independent estimates of the population variance are
homogenou s, the statistics F is given by
𝐹=𝑆ଶ
𝑆ଶ
where
𝑆ଶ=1
𝑛ଵ−1∑(𝑥−𝑥̅)ଶ, 𝑆ଶ=1
𝑛ଶ−1∑(𝑦−𝑦ത)ଶ
are unbiased estimates of the common population variance 𝜎ଶ obtained
from two independent samples and it follows Snedecor’s F -distribution
with (𝑣ଵ,𝑣ଶ)=(𝑛ଵ−1,𝑛ଶ−1) degree of freedom.
The shaded in the diagram indicates the acceptance region (1−𝛼) and the
unshaded region indicates the rejection region 𝛼.
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160 Eg: 1) In one sample of 8 observations, the sum of the squares of
deviations of the sample values from the sample mean was 84.4 and in the
other sample of 10 observations it was 102.6. Test whether this difference
is significant at 5% LOS given that 5% point of F for v 1 =7 and v 2 = 9
degree of freedom is 3.29.
Solution : 𝐻:𝜎௫ଶ =𝜎௬ଶ i.e., t he estimate of variance given by the samples
are homogenous,
𝐻ଵ∶ 𝜎௫ଶ≠𝜎௬ଶ
𝑛ଵ=8,𝑛ଶ=10,∑(𝑥−𝑥̅)ଶ=84.4,∑(𝑦−𝑦ത)ଶ=102.6
𝑆௫ଶ=1
𝑛ଵ−1∑(𝑥−𝑥̅)ଶ =1
7×84.4=12.0571
𝑆௬ଶ=1
𝑛ଶ−1∑(𝑦−𝑦ത)ଶ=1
9×102.6= 11.4
𝐹=𝑆௫ଶ
𝑆௬ଶ=12.0571
11.4=1.0576
tabulated 𝐹,ଽ(0.05)=3.29
𝐹=𝐹=1.0576<𝐹.ହ(7,9)=3.29
∴𝐴𝑐𝑐𝑒𝑝𝑡 𝐻 𝑎𝑡 5% 𝐿𝑂𝑆
Eg : 2) Two random samples gave the following results: Sample no Size sum of squares of deviation from the mean 1 10 90 2 12 108 Test at 5% LOS whether there is a difference in variance.
{Given 𝐹.ହ(9,11)=2.9 ,𝐹.ହ(11,9)=3.1}
Solution:𝐻:𝜎௫ଶ=𝜎௬ଶ ,𝐻ଵ:𝜎௫ଶ≠𝜎௬ଶ
𝑛ଵ=10,𝑛ଶ=12,∑(𝑥−𝑥̅)ଶ=90,∑(𝑦−𝑦ത)ଶ = 108
𝑆௫ଶ=1
𝑛ଵ−1∑(𝑥−𝑥̅)ଶ =1
9×90=10
𝑆௬ଶ=1
𝑛ଶ−1∑(𝑦−𝑦ത)ଶ=1
11×108=9.8182
𝐹=𝑆௫ଶ
𝑆௬ଶ=10
9.8182=1.0185
𝐹=1.0185 <𝐹.ହ(9,11)=2.9
∴𝑎𝑐𝑐𝑒𝑝𝑡 𝐻
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161 10.11 SUMMARY
In this chapter we had learnt about to apply t -test for the sample size less
than or equal to 30. We had also learnt to test if the sample mean (𝑥̅)
differs significantly from the hypothetical value 𝜇 of the population mean
and to test the significance of the difference between two sample means.
WE had also learnt about the confidence interval and the F -test to whether
the two independent estimates o f the population variance are homogenous
or not.
10.13 EXERCISES
1) A researcher is interested in determing whether or not review sessions
affect exam performance. The independent variable, a review session,
is administered to a sample of students (n=9) in an attempt to
determine if this has an effect on the dependent variable, exam
performance. Based on the information gathered in previous
semesters, the researcher knows that the population mean for a given
exam is 24. The sample mean is 25. with a S.D of 4 , LOS = 5%.
2) You conduct a survey of a sample of 25 members of this year’s
graduating marketing students and find that average GPA is 3.2. The
standard deviation of the sample is 4. Over the last year the average
GPA has been 3.0. Is the GPA of this year’s students significantly
different from long run average?
3) The heights of 10 males of given locality are found to be 70, 67, 62,
68, 61, 68, 70, 64, 64, 66 inches. Is it reasonable to believe that the
average height is greater than 64 inches?
4) A random sample of 16 values from a normal population showed a
mean of 41.5 inches and the sum of squares of deviations from this
mean = 135 sq. inches. Show that the assumptions of mean 43.5
inches for the population is not reasonable. Obtain the 95 % and 99 %
confidence limits.
5) Below are given the gain in weights (in kgs) of pigs fed on two Diets
A and B
Test if the two diets differ significantly as regards to their effect on
increase in weight.
6) Samples of two types of electric light bulbs were tested for length of
life and following data were obtained:
Type I Type II
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162 Sample Size 8 7 Sample mean 1234 1036 Sample S.D. 36 40
Is the difference in the means sufficient to warrant that Type I is
superior to Type II regarding length of life?
7) Two laboratories carry out independent estimates of a particular
chemicals in a medicine produced by a certain firm. A sample is taken
from each batch, halved and the separate halves sent to the two
laboratories. The following data is obtained.
no. of samples 10 mean value of the diff. of estimates 0.6 sum of the squares of their diff. from their mean 20
Is the diffe rence significant at 5% LOS.
8) A certain stimulus administered to each of the following 12 patients
resulted in the following increase of blood pressure: 5, 2, 8, -1, 0,-2, 1,
5, 0,4 and 6. Can it be obtained that the stimulus will, in general, be
accompanied by an increase in blood pre ssure?
9) Two independent samples of 8 and 7 items respectively had the
following values of the variables:
Sample I 9 11 13 11 15 9 12 14 Sample II 10 12 10 14 9 8 10
Do the estimates of population variance differ significantly?
10.14 SOLUTION TO EXERCISES
10.15 TABLES OF t-DISTRIBUTION AND f-DISTRIBUTION t DISTRIBUTION : CRITICAL VALUES
OF t
Significance level
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163 Degrees of Two-tailed test: 10% 5% 2% 1% 0.2% 0.1% freedom One-tailed test: 5% 2.5% 1% 0.5% 0.1% 0.05% 1 6.314 12.706 31.821 63.657 318.309 636.619 2 2.920 4.303 6.965 9.925 22.327 31.599 3 2.353 3.182 4.541 5.841 10.215 12.924 4 2.132 2.776 3.747 4.604 7.173 8.610 5 2.015 2.571 3.365 4.032 5.893 6.869 6 1.943 2.447 3.143 3.707 5.208 5.959 7 1.894 2.365 2.998 3.499 4.785 5.408 8 1.860 2.306 2.896 3.355 4.501 5.041 9 1.833 2.262 2.821 3.250 4.297 4.781 10 1.812 2.228 2.764 3.169 4.144 4.587 11 1.796 2.201 2.718 3.106 4.025 4.437 12 1.782 2.179 2.681 3.055 3.930 4.318 13 1.771 2.160 2.650 3.012 3.852 4.221 14 1.761 2.145 2.624 2.977 3.787 4.140 15 1.753 2.131 2.602 2.947 3.733 4.073 16 1.746 2.120 2.583 2.921 3.686 4.015 17 1.740 2.110 2.567 2.898 3.646 3.965 18 1.734 2.101 2.552 2.878 3.610 3.922 19 1.729 2.093 2.539 2.861 3.579 3.883 20 1.725 2.086 2.528 2.845 3.552 3.850 21 1.721 2.080 2.518 2.831 3.527 3.819 22 1.717 2.074 2.508 2.819 3.505 3.792 23 1.714 2.069 2.500 2.807 3.485 3.768 24 1.711 2.064 2.492 2.797 3.467 3.745 25 1.708 2.060 2.485 2.787 3.450 3.725 26 1.706 2.056 2.479 2.779 3.435 3.707 27 1.703 2.052 2.473 2.771 3.421 3.690 28 1.701 2.048 2.467 2.763 3.408 3.674 29 1.699 2.045 2.462 2.756 3.396 3.659 30 1.697 2.042 2.457 2.750 3.385 3.646 32 1.694 2.037 2.449 2.738 3.365 3.622 34 1.691 2.032 2.441 2.728 3.348 3.601 36 1.688 2.028 2.434 2.719 3.333 3.582 38 1.686 2.024 2.429 2.712 3.319 3.566 40 1.684 2.021 2.423 2.704 3.307 3.551 42 1.682 2.018 2.418 2.698 3.296 3.538 44 1.680 2.015 2.414 2.692 3.286 3.526 46 1.679 2.013 2.410 2.687 3.277 3.515 48 1.677 2.011 2.407 2.682 3.269 3.505 50 1.676 2.009 2.403 2.678 3.261 3.496 60 1.671 2.000 2.390 2.660 3.232 3.460 70 1.667 1.994 2.381 2.648 3.211 3.435 80 1.664 1.990 2.374 2.639 3.195 3.416 munotes.in
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164 90 1.662 1.987 2.368 2.632 3.183 3.402 100 1.660 1.984 2.364 2.626 3.174 3.390 120 1.658 1.980 2.358 2.617 3.160 3.373 150 1.655 1.976 2.351 2.609 3.145 3.357 200 1.653 1.972 2.345 2.601 3.131 3.340 300 1.650 1.968 2.339 2.592 3.118 3.323 400 1.649 1.966 2.336 2.588 3.111 3.315 500 1.648 1.965 2.334 2.586 3.107 3.310 600 1.647 1.964 2.333 2.584 3.104 3.307 1.645 1.960 2.326 2.576 3.090 3.291
F Distribution : Critical Values of F (5%significancelevel)
v1 1 2 3 4 5 6 7 8 9
10 12 14 16 18 20
v2
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165
(continued)
f Distribution : Critical Values of f (5%significancelevel)
v1 25 30 35 40 50 60 75 100 150
200
v2
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166 750 1.52 1.47 1.44 1.41 1.37 1.34 1.30 1.26 1.22 1.20 1000 1.52 1.47 1.43 1.41 1.36 1.33 1.30 1.26 1.22 1.19
f Distribution : Critical Values of f(1%significancelevel)
v11 2 3 4 5 6 7 8 9 10
12 14 16 18 20
v2
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167
continued)
f Distribution : Critical Values of f (1%significancelevel)
v1 25 30 35 40 50 60 75 100 150
200
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168
10.12 REFERENCE FOR FURTHER READING
Following books are recommended for further reading:
Statistics by Murray R, Spiegel, Larry J. Stephens, Mcgraw Hill
International Publisher, 4th edition
Fundamental of Mathematical Statistics by S. C. Gupta and V. K.
Kapoor, Sultan Chand and Sons publis her, 11th edition
*****
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169 11
THE CHI -SQUARE TEST
Unit Structure
11.0 Objectives
11.1 Introduction
11.2 Properties of Chi Square variate
11.3 The Chi Square test for Goodness of fit
11.3.1 Decision Criterion
11.4 Test for Independence of Attributes
11.5 Yate’s Correction for continuity
11.6 Test in r c Contingency Table
11.7 Coefficient of Contingency
11.8 Correlation of Attributes
11.9 Additive Property of Chi Square variate
11.10 Summary
11.11 Exercises
11.12 Solution to Exercises
11.13 Table of Chi -Square distribution
11.14 Reference for further reading
11.0 OBJECTIVES
The chi -square test is a non -parametric test that compares two or more
variables from randomly selected data. It helps find the relationship
between two or more variables. The chi square distribution is a theoretical
or mathematical distribution which has wide applicability in statistical
work. The term ‘chi square’ (pronounced with a hard ‘ch’) is used because
the G reek letter χ is used to define this distribution. It will be seen that the
elements on which this distribution is based are squared, so that the
symbol χ 2 is used to denote the distribution.
11.1 INTRODUCTION
We know that if the probability distribution of the discrete random
variable X is known we can find the probability distribution of the random
variable 𝑌=𝑋ଶ. One may be interested in knowing whether we can find
the probability distribution of random varia ble 𝑌=𝑋ଶ if the probability
distribution of a continuous random variable X is known. The answer is
affirmative and in particular, if X is a standard normal variate (a
continuous random variable) so that its probability density function is
given by munotes.in
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170 𝑓(𝑥)= 1
√2𝜋𝑒ቀିభ
మቁ௫మ ,−∞<𝑥<∞ =0 , 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
then 𝑌=𝑋ଶis also a continuous random variable whose probability
density function is given by
𝑔(𝑦)=1
√2𝜋1
ඥ𝑦𝑒ି
మ , 𝑦≥0
=0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Here, t he distribution of Y is known as Chi -square distribution with one
degree of freedom .
More generally, if 𝑋ଵ,𝑋ଶ,⋯⋯𝑋 are n independent standard normal
variates, then the distribution of random variable 𝑈=𝑋ଵଶ+𝑋ଶଶ+⋯⋯+
𝑋ଶis given by the probability density function
ℎ(𝑢)=1
2
మΓቀ
ଶቁ𝑒ିೠ
మ𝑢
మିଵ,𝑢≥0,𝑛∈ℕ
=0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
where Γቀ
ଶቁ is called gamma
ଶand is given by Γቀ
ଶቁ=∫𝑒ି௧ஶ
𝑡
మିଵ 𝑑𝑡
Here, the distribution of random variable U is called Chi -sqaure
distribution with n degrees of freedom n degrees of freedom and U is
called a Chi -square variate. Generally, a chi square variate is denoted by
the square of greek letter chi, i.e 𝜒ଶ. Thus if 𝜒ଶ denotes a Chi -square
variate with n degrees of freedom, th en its probability density function is
given by
𝑓(𝜒ଶ)=1
2
మΓቀ
ଶቁ𝑒ିഖమ
మ(𝜒ଶ)
మିଵ,𝜒ଶ≥0,𝑛∈ℕ
=0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
11.2 PROPERTIES OF CHI -SQUARE VARIATE WITH n DEGREES OF FREEDOM
Let 𝜒ଶ denote a Chi -square variate with n degrees of freedom. Then w e
have the following properties:
1) The probability density function of 𝜒ଶ is given by
𝑓(𝜒ଶ)=1
2
మΓቀ
ଶቁ𝑒ିഖమ
మ(𝜒ଶ)
మିଵ,𝜒ଶ≥0,𝑛∈ℕ
=0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
2) The mean of 𝜒ଶ is E(𝜒ଶ) = n
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171 3) The variance of 𝜒ଶ is 𝑉(𝜒ଶ) = 2n
4) The mode of 𝜒ଶ = n – 2
5) The frequency curve is given by 𝑦=𝑓(𝜒ଶ) lies in the first quadrant
and it is positively skewed curve, its tail on the right extends upto
infinity , as given in the below diagram
𝑝=Pr[𝑋≥ 𝜒ଶ] is denoted by 𝛼 and 𝜒ଶ= 𝜒,ఈଶ
6) The total area under the Chi -square curve is 1.
7) 𝑝(𝜒ଶ>𝑐)=𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 𝑦=𝑓(𝜒ଶ) 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑜𝑓 𝜒ଶ=𝑐
8) For a chi square variate with n degrees of freedom, if 𝒑(𝝌𝟐>𝑐)=
𝜶,𝒕𝒉𝒆𝒏 𝒄 𝒊𝒔 𝒅𝒆𝒏𝒐𝒕𝒆𝒅 𝒃𝒚 𝝌𝒏,𝜶𝟐 𝑖.𝑒 𝑝൫𝜒ଶ>𝜒,ఈଶ൯=𝛼
9) 𝜒,ఈଶ is called 𝛼 probability point of chi square distribution with n
degrees of freedom.
Eg: 1) If a random variable X follows chi square distribution with 10
degrees of freedom find i) x 0 ii) x 1& iii) 𝛼 such that 𝑝(𝑋>𝑥)=
0.95,𝑝(𝑋≤𝑥ଵ)=0.01 & 𝑝(𝑋>18.3)=𝛼
Solution : n = 10 – degree of freedom
i) to find x 0 such that p(X> x 0) = 0.95 ⇒𝑥=𝜒ଵ,.ଽହଶ=3.9403
ii) to find x 1 such that p(X ≤ x 1) = 0.01
𝑝(𝑋≤𝑥ଵ)=1−𝑝(𝑋>𝑥ଵ) ⇒𝑝(𝑋>𝑥ଵ)=1−0.01 =0.99
⇒𝑝(𝑋>𝑥ଵ)=0.99⇒𝑥ଵ=𝜒ଵ,.ଽଽଶ =2.5582
iii) 𝑝(𝑋>18.3)=𝛼⇒𝜒ଵ,ఈଶ=18.3⇒𝛼=0.05
Eg: 2) A random variable Y follows chi square distribution with S.D 4,
Find y 0 if 𝑝(𝑌≤𝑦)=0.05.
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172 Solution: 𝑆.𝐷=4⇒𝑣𝑎𝑟=16⇒2𝑛=16⇒𝑛=8
𝑝(𝑌≤𝑦)=0.05⇒𝑝(𝑌≤𝑦)=1−𝑝(𝑌>𝑦)⇒𝑝(𝑌>𝑦)
=1−0.05=0.95
𝑝(𝑌>𝑦)=0.95⇒𝑦=𝜒଼,.ଽହଶ⇒𝑦 =2.7326
Eg: 3) If a random variable X follows chi square distribution with S.D 4.
Find mean and mode.
Solution: 𝑆.𝐷=4⇒𝑉𝑎𝑟=16,𝑏𝑢𝑡 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑜𝑓 𝜒ଶ=2𝑛⇒𝑛=
8𝑚𝑒𝑎𝑛 𝑜𝑓 𝜒ଶ=𝑛=8 & 𝑚𝑜𝑑𝑒 𝑜𝑓 𝜒ଶ=𝑛−2=6
11.3 THE CHI SQUARE TEST FOR GOODNESS OF FIT
When we come across some observations on a random variable, our
curiosity may tempt us to investigate whether it can be considered to be a
random variable following a certain specified probability law. A technique
to test whether a given frequency distribution of a random var iable follows
a certain specified distribution (known as theoretical distribution) was
proposed by Karl Pearson. It involves a test statistic that can be shown to
follow Chi -square distribution under certain assumptions and the test is
known as Chi-square test for goodness of fit.
Suppose we have an observed frequency distribution with n classes having
observed frequencies
𝑂ଵ,𝑂ଶ,𝑂ଷ,⋯⋯,𝑂 𝑤𝑖𝑡ℎ ∑𝑂ୀ
ୀଵ=𝑁. (Here the classes may correspond
to the discrete values of a variable or groups of va lues of a variable or
even to the groups corresponding to an attribute).
Further, suppose that according to our assumption to be called the null
hypothesis H 0, the expected frequencies are
𝐸ଵ,𝐸ଶ,𝐸ଷ,⋯⋯,𝐸 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 ∑𝐸=𝑁ୀ
ୀଵ.
Then the test statistic proposed by Karl Pearson is given by 𝜒ଶ=
∑(ை ି ா)మ
ாୀ
ୀଵ
It can be shown that under the assumptions
1) observations are drawn independently and at random.
2) null hypothesis H 0 is true i.e the expected frequencies are
𝐸ଵ,𝐸ଶ,𝐸ଷ,⋯⋯,𝐸 respectively.
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173 3) total number of observations made = N is large
4) observed frequencies O i’s are large
5) expected frequencies E i’s are large so that the termsை ି ா
ா can be
considered to be negligible.
The test statistic 𝜒ଶ=∑(ை ି ா)మ
ாୀ
ୀଵ follow Chi -square distribution with
(n-1) degrees of freedom.
For all practical purposes, the assumptions regarding the distribution of
test statistic 𝜒ଶ can be considered to be reasonable if
a) the number of observations = N ≥ 50
b) the expected frequencies E i’s ≥ 5.
In case when the expected frequencies is less than 5 we require to combine
more than one neighboring classes so that the expected frequency for such
class is not less than 5.
11.3.1 DECISION CRITERION
While performing a test for goodness of fit, we shall use the following
decision criteria:
𝑅𝑒𝑗𝑒𝑐𝑡 𝐻 𝑖𝑓 𝜒ଶ=(𝑂 − 𝐸)ଶ
𝐸ୀ
ୀଵ>𝜒(ିଵ),ఈଶ
𝐷𝑜 𝑛𝑜𝑡 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻 𝑖.𝑒 𝑎𝑐𝑐𝑒𝑝𝑡 𝐻 𝑖𝑓 𝜒ଶ≤ 𝜒(ିଵ),ఈଶ
Eg: 1) The following data repres ents the last digit of the cars passing at a
certain traffic signal observed during last 30 minutes for 180 cars.
last digit 0 1 2 3 4 5 6 7 8 9 frequency 12 20 14 12 21 18 17 26 19 21
Can we retain at 5% level of significance that all the digits are equally
likely to occur?
Solution: we want to test whether all digits are equally likely to occur. If
all digits are equally likely, 𝑝=ଵ
ଵ.𝐸=𝑁𝑝=180×ଵ
ଵ=18
𝐻:𝐴𝑙𝑙 𝑑𝑖𝑔𝑖𝑡𝑠 𝑎𝑟𝑒 𝑒𝑞𝑢𝑎𝑙𝑙𝑦 𝑙𝑖𝑘𝑒𝑙𝑦 𝑡𝑜 𝑜𝑐𝑐𝑢𝑟
𝐻ଵ:𝑛𝑜𝑡 𝐻 𝑖.𝑒 𝑙𝑜𝑔𝑖𝑐𝑎𝑙 𝑛𝑒𝑔𝑎𝑡𝑖𝑜𝑛 𝑡𝑜 𝐻(𝑖.𝑒 𝑎𝑙𝑙 𝑑𝑖𝑔𝑖𝑡𝑠 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑒𝑞𝑢𝑎𝑙𝑙𝑦 𝑙𝑖𝑘𝑒𝑙𝑦) munotes.in
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174 𝐿𝑂𝑆=5% 𝑖.𝑒 𝛼=0.05,𝑁=180≥50 𝑎𝑛𝑑 𝐸=18≤5,𝑛=10𝜒ଶ
=(𝑂−𝐸)ଶ
𝐸
ୀ𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑐ℎ𝑖𝑞 𝑠𝑞𝑢𝑎𝑟𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜 𝑛 𝑤𝑖𝑡ℎ 9 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚
Decision criteria is given by
𝑟𝑒𝑗𝑒𝑐𝑡 𝐻 𝑖𝑓 𝜒ଶ>𝜒ଽ,.ହଶwhere 𝜒ଽ,.ହଶ=16.9190
𝑑𝑜 𝑛𝑜𝑡 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻 𝑖𝑓 𝜒ଶ≤𝜒ଽ,.ହଶ,𝜒ଶ=(𝑂−𝐸)ଶ
𝐸
ୀ
Digit observed freq. (Oi) Exp. freq.
(Ei) Oi - Ei (𝑶𝒊−𝑬𝒊)𝟐
𝑬𝒊
0 12 18 -6 36/18 1 20 18 2 4/18 2 14 18 -4 16/18 3 12 18 -6 36/18 4 21 18 3 9/18 5 18 18 0 0 6 17 18 -1 1/18 7 26 18 8 64/18 8 19 18 1 1/18 9 21 18 3 9/18 total 176/18 = 9.7778
𝜒ଶ= (𝑂 − 𝐸)ଶ
𝐸
ୀ=9.7778<16.9190=𝜒ଽ,.ହଶ
do not reject H 0 i.e accept H 0.
Eg: 2) As per Mendel’s theory according to the shape and color, certain
variety of pea that can be classified into four categories Round and yellow,
Round and green, Angular and yellow, Angular and green occur in the
proportion of 9:3:3:1. To test this a sample of N = 128 peas was taken and
the following were the observed frequencies
RY – 66 RG – 28 AY – 29 AG – 5
Perform the chi square test for goodness of fit.
Solution : N = 128, n =4
The probability of occurrence and Expected frequencies are given by
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175 category pi Ei = Npi RY 9/16 128 X 9/16 = 72 RG 3/16 128 X 3/16 = 24 AY 3/16 128 X 3/16 = 24 AG 1/16 128 X1/16 = 8
∴𝐻:𝑇ℎ𝑒 𝑓𝑜𝑢𝑟 𝑐𝑎𝑡𝑒𝑔𝑜𝑟𝑖𝑒𝑠 𝑜𝑓 𝑝𝑒𝑎𝑠 𝑖.𝑒 𝑅𝑌,𝑅𝐺,𝐴𝑌,𝐴𝐺 have expected
frequencies 72, 24, 24, 8 resp.
H1: not H 0
LOS = 5%, 𝛼=0.05
Decision criteria is given by 𝑅𝑒𝑗𝑒𝑐𝑡 𝐻 𝑖𝑓 𝜒ଶ>𝜒ଷ,.ହଶ=7.8147
Do not reject H 0 if 𝜒ଶ≤7.8147,𝑤ℎ𝑒𝑟𝑒 𝜒ଶ=∑൫ை –ா൯మ
ா
ୀଵ
category of peas obs. freq.(Oi) Exp. freq. (Ei) 𝑂−𝐸 (𝑂−𝐸)ଶ𝐸 RY 66 72 -6 0.5 RG 28 24 4 0.6667 AY 29 24 5 1.0417 AG 5 8 -3 1.125 total 3.3334
𝜒ଶ=3.3334<𝜒ଷ,.ହଶ=7.8147
⇒𝐴𝑐𝑐𝑒𝑝𝑡 𝐻 𝑎𝑡 5% 𝐿𝑂𝑆.
Eg: 3) Four identical coins are tossed 100 times and the following results
are obtained.
no. of heads (x) 0 1 2 3 4 frequency 8 29 40 19 4
Are there sufficient evidences to conclude that the coins are biased at 5%
LOS.
Solution: let p denote the probability of getting a head with each of the
four coins,
X: no. of heads follows binomial distribution with n= 4, p
X ~ B(n, p) i.e X ~ B(4,p)
H0: the coins are unbiased i.e p =1/2
H1 : the coins are baised i.e p ≠ ½
LOS = 5 %, 𝛼=0.05
𝑋 ~ 𝐵(4,𝑝)⇒𝑝(𝑥)= ൬4
𝑥൰𝑝௫𝑞ସି௫
𝑝(0)=𝑝𝑞ସ=1
16 ,𝑝(1)=4𝑝𝑞ଷ=4
16,𝑝(2)=6𝑝ଶ𝑞ଶ=6
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176 𝑝(3)=4𝑝ଷ𝑞=4
16,𝑝(4)=𝑝ସ𝑞=1
16
Decision criteria Reject 𝐻 𝑖𝑓 𝜒ଶ>𝜒ସ,.ହଶ=9.4877
Expected frequencies 𝐸=𝑁 ⋅𝑝(𝑥)
x obs. freq. Oi Exp. freq. Ei Oi - Ei (𝑂−𝐸)ଶ𝐸 0 8 100p(0) = 6.25 1.75 0.49 1 29 100p(1) = 25 4 0.64 2 40 100p(2) = 37.5 2.5 0.1667 3 19 100p(3) = 25 -6 1.44 4 4 100p(4) = 6.25 -2.25 0.81 total 3.5467
𝜒ଶ=3.5467<𝜒ସ,.ହଶ=9.4877
𝐴𝑐𝑐𝑒𝑝𝑡 𝐻𝑎𝑡 5% 𝐿𝑂𝑆⇒𝑐𝑜𝑖𝑛𝑠 𝑎𝑟𝑒 𝑢𝑛𝑏𝑖𝑎𝑠𝑒𝑑
Eg: 4) The random variable X denotes the number of street accidents per
week.
X 0 1 2 3 4 5 6 7 obs. freq. 15 30 28 14 8 4 0 1 exp. freq. 14 27 27 18 9 4 1 0
Test whether the random variable X follows Poisson distribution with
parameter m =2 at 1% l evel of significance.
Solution: As the expected frequency is less than 5 for X = 5, 6 & 7 we
combine them into one class
X 0 1 2 3 4 5-7 obs. freq. 15 30 28 14 8 5 exp. freq. 14 27 27 18 9 5
H0: X follows Poisson distributions with parameter m =2
H1: not H 0
LOS = 1%, 𝛼=0.01
Decision criteria is given by 𝑅𝑒𝑗𝑒𝑐𝑡 𝐻 𝑖𝑓 𝜒ଶ>𝜒ହ,.ଵଶ=15.086
Do not reject H 0 if 𝜒ଶ≤𝜒ହ,.ଵଶ,𝑤ℎ𝑒𝑟𝑒 𝜒ଶ=∑൫ை – ா൯మ
ா
ୀଵ
i X obs. freq. Oi exp. freq. Ei (Oi – Ei) (𝑂−𝐸)ଶ𝐸 1 0 15 14 1 0.0714 2 1 30 27 3 0.3333 munotes.in
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177 3 2 28 27 1 0.0370 4 3 14 18 -4 0.8889 5 4 8 9 -1 0.1111 6 5-7 5 5 0 0 total 1.4417
𝜒ଶ=1.4417<𝜒ହ,.ଵଶ=15.086
Accept H 0 i.e at 1% level of significance the hypothesis that the variable X
follows Poisson distribution with parameter m =2 is retainable.
11.4 TEST FOR INDEPENDENCE OF ATTRIBUTES
In this section, we consider the presence of two attributes among units
from single population and the interest is centered around the possible
dependence or independence of the attributes. In other words, on the basis
of data regarding two attributes for some units from the population, we
shall investigate whether the observed data provide sufficient reasons to
reject the claim that the two attributes are independent of each other for
the population under consideration. Such a test is called test for
indep endence of attributes.
To understand the mechanism of the test, we consider the following table
representing data for two attributes known as 2 2 contingency table. It
is called a contingency table as it represents the information which can be
attributed to chance as the information is regarding randomly selected
persons.
2 2 contingency table is given as below:
Attribute B total B1 B2 Attribute A A1 a b a+b A2 c d c+d a+c b+d a+b+c+d= N
With the help of the test statistic 𝜒ଶ=ே(ௗି)మ
(ା)(ା)(ାௗ)(ାௗ) we can perform
a test for testing the hypothesis H0: the attributes A and B under the
consideration are independent against the logical alternative H 1: not H 0i.e
the attributes A and B are dependent subject to the condition s N ≥ 50 and
each of the observed frequ encies a, b, c, d ≥ 5.
The decision criterion at level of significance = α is given by
𝑅𝑒𝑗𝑒𝑐𝑡 𝐻:𝑖𝑓 𝜒ଶ>𝜒ଵ,ఈଶ,𝑑𝑜 𝑛𝑜𝑡 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻∶𝑖𝑓 𝜒ଶ≤𝜒ଵ,ఈଶ
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178 where 𝜒ଶ=ே(ௗି)మ
(ା)(ା)(ାௗ)(ାௗ) where a,b,c,d are the observed
frequencies with a+b+c+d = N.
Eg: 1) The following results are obtained at the end of six months of a
kind of psychotherapy given to a group of 120 patients and also for
another group of 120 patients who were not given the psychotherapy.
psychotherapy given not given condition improved 71 42 condition did not improved 49 78
Can we conclude at 5% LOS that the psychotherapy is effective ?
Solution: H0: Psychotherapy is not effective
H1: psychotherapy is effective
LOS = 5% ,𝛼=0.05
Decision criteria Reject 𝐻 𝑖𝑓𝑓 𝜒ଶ>𝜒ଵ,.ହଶ=3.8415
N = 240, a =71, b = 42, c = 49, d = 78
𝜒ଶ=𝑁(𝑎𝑑−𝑏𝑐)ଶ
(𝑎+𝑏)(𝑎+𝑐)(𝑏+𝑑)(𝑐+𝑑) =240(71×78−49×42)ଶ
113⋅120⋅120⋅127
𝜒ଶ=240×12110400
206654400 = 14.0645
⇒𝜒ଶ=14.0645>𝜒ଵ,.ହଶ= 3.8415
⇒ Reject H 0 at 5% level of significance, we may say that the
Psychotherapy is effective at 5% LOS.
11.5 YATE’S CORRECTION FOR CONTINUITY
When the cell frequencies a, b, c, d as observed in case of the four classes
corresponding to two attributes are small, we cannot use the test statistic
𝜒ଶ as defined in previous section i.e for when the assumption a, b, c, d are
greater than or equal to 5 does not hold, the distribution of
𝜒ଶ=ே(ௗି)మ
(ା)(ା)(ାௗ)(ାௗ) cannot be considered to be Chi -square with one
degree of freedom.
In such a case i.e if the cell frequencies a, b, c, d are not all greater than or
equal to 5, we make the followin g adjustment called Yate’s correction.
if ad < bc add ½ to a and b and subtract ½ from both b and c both.
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179 With this adjustments, we get the test statistic as 𝜒ଶ=ேቀ|ௗି|ିಿ
మቁమ
(ା)(ା)(ାௗ)(ାௗ)
The decision criterion at level of significance = 𝛼 is given by
𝑅𝑒𝑗𝑒𝑐𝑡 𝐻:𝑖𝑓 𝜒ଶ>𝜒ଵ,ఈଶ,𝑑𝑜 𝑛𝑜𝑡 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻∶𝑖𝑓 𝜒ଶ≤𝜒ଵ,ఈଶ
where 𝜒ଶ=ேቀ|ௗି|ିಿ
మቁమ
(ା)(ା)(ାௗ)(ା)
Eg: 1) In an experiment on immunization of cattle from tuberculosis the
following results were obtained
affected unaffected Inoculated 11 31 Not inoculated 14 4
Examine the effect of vaccine in controlling the incidence of the disease at
1% LOS.
Solution: H0: the attributes are independent
H1: the attributes are not independent
LOS = 1%, 𝛼=0.01
Decision criteria Rej ect 𝐻 𝑖𝑓𝑓 𝜒ଶ>𝜒ଵ,.ଵଶ=6.6349
N = 60, a = 11, b = 31, c = 14, d = 4
𝜒ଶ =𝑁ቀ|𝑎𝑑−𝑏𝑐|−ே
ଶቁଶ
(𝑎+𝑏)(𝑎+𝑐)(𝑏+𝑑)(𝑐+𝑑) =60(|11×4−31×14|−30)ଶ
42⋅18⋅25⋅35
𝜒ଶ=60(390−30)ଶ
661500=60×360×360
661500=11.7551
𝜒ଶ=11.7551>𝜒ଵ,.ଵଶ=6.6349
⇒𝑟𝑒𝑗𝑒𝑐𝑡 𝐻 𝑎𝑡 1% 𝐿𝑂𝑠
We can say at 1% level of significance that the Inoculation and affection
due to disease are dependent.
11.6 TEST IN r c CONTINGENCY TABLE
If we have two attributes A and B classified into r and c classes
respectively denoted by 𝐴ଵ,𝐴ଶ,⋯⋯𝐴 & 𝐵ଵ,𝐵ଶ,⋯⋯𝐵 then the observed
frequency can be put in a tabular form with r rows and c columns called
r c contingency table.
If we use Oij (i = 1,2,3,……,r , j = 1,2,3,……..,c)to denote the observed
frequency for attribute class A iBj, then r c contingency table can be
represented as shown below: munotes.in
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180 Attribute B1 B2 B3 BC TOTAL A1 O11 O12 O13 O1C a1 A2 O21 O22 O23 O2C a2 A3 O31 O32 O33 O3C a3 AR OR1 OR2 OR3 ORC ar TOTAL b1 b2 b3 bc N
Here a i’s represent total observed frequencies for attribute classes A i’s and
bj’s represent the same for classes B j’s, N being the overall total frequency.
Then with these notations, we can write down the test statistic as
𝜒ଶ= ∑∑൫ைೕିாೕ൯మ
ாೕ ,𝑤ℎ𝑒𝑟𝑒 𝐸=ೕ
ேare the expected frequencies.
With the help of the test statistic 𝜒ଶ= ∑∑൫ைೕିாೕ൯మ
ாೕ we can perform a
test for testing the hypothesis H 0: the attributes A and B under the
consideration are independent against the logical alternative H 1: not H 0i.e
the attributes A and B are dependent subject to the conditions N ≥ 50 and
each of the observed frequencies ≥ 5 (Oij ≥ 5).
The d ecision criterion at level of significance = 𝛼 is given by
𝑅𝑒𝑗𝑒𝑐𝑡 𝐻:𝑖𝑓 𝜒ଶ>𝜒(ିଵ)(ିଵ),ఈଶ
𝑑𝑜 𝑛𝑜𝑡 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻∶𝑖𝑓 𝜒ଶ≤𝜒(ିଵ)(ିଵ),ఈଶ
Eg: 1) Using the data given in the following table decide whether we can
conclude that standard of clothing of a salesman has significant effect on
his performance in field selling at 5% LOS.
Performance in Field Selling Disappointing Satisfactory Excellent Total Poorly dressed 21 15 6 42 Well dressed 24 35 26 85 Very well dressed 35 80 58 173 Total 80 130 90 300
Solution: H0: attributes are independent
LOS = 5%, 𝛼=0.05
Reject H 0 iff 𝜒ଶ>𝜒ସ,.ହଶ=9.4877,𝑤ℎ𝑒𝑟𝑒 𝜒ଶ = ∑∑൫ைೕ – ாೕ൯మ
ாೕ
𝐸ଵଵ=𝑎ଵ𝑏ଵ
𝑁=42⋅80
300=11.2,𝐸ଵଶ=𝑎ଵ𝑏ଶ
𝑁=42⋅130
300=18.2
𝐸ଵଷ=𝑎ଵ𝑏ଷ
𝑁 =42⋅90
300=12.6,𝐸ଶଵ =𝑎ଶ𝑏ଵ
𝑁=85⋅80
300=22.67 munotes.in
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181 𝐸ଷଵ=𝑎ଷ𝑏ଵ
𝑁=173⋅80
300=46.13,𝐸ଷଶ=𝑎ଷ𝑏ଶ
𝑁=173⋅130
300=74.97
𝐸ଷଷ=𝑎ଷ𝑏ଷ
𝑁 =173⋅90
300=51.9
Oij Eij Oij - Eij ൫𝑂−𝐸൯ଶ𝐸 21 11.2 9.8 8.575 15 18.2 -3.2 0.5626 6 12.6 -6.6 3.4571 24 22.67 1.33 0.078 35 36.83 -1.83 0.0909 26 25.5 0.5 0.0098 35 46.13 -11.13 2.6854 80 74.97 5.03 0.3375 58 51.9 6.1 0.7170 total 16.5133 𝜒ଶ=16.5133>𝜒ସ,.ହଶ=9.488
⇒𝑟𝑒𝑗𝑒𝑐𝑡 𝐻 𝑎𝑡 5% 𝐿𝑂𝑆
We decide to reject H 0 at 5% level of significance and conclude that the
standard of clothing of a salesman has significant effect on his
performance in field selling.
11.7 COEFFICIENT OF CONTINGENCY
A measure of the degree of relationship, association, or dependence of the
classifications in a contingency table is given by
𝐶= ඨ𝜒ଶ
𝜒ଶ+𝑁
which is called the coefficient of contingency. The larger the C, the greater
is the degree of association. The number of rows and columns in the
contingency table determines the maximum val ue of C, which is never
greater than 1. If the number of rows and columns of a contingency table
is equal to k, the maximum value of C is given by ටିଵ
.
Eg: 1) Using the data given in the following table decide whether we can
conclude that standard of clothing of a salesman has significant effect on
his performance in field selling at 5% LOS.
munotes.in
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182 Performance in Field Selling Disappointing Satisfactory Excellent Total Poorly dressed 21 15 6 42 Well dressed 24 35 26 85 Very well dressed 35 80 58 173 Total 80 130 90 300 Also find the coefficient of contingency.
Solution: H0: attributes are independent
LOS = 5%, 𝛼=0.05
Reject H 0 iff 𝜒ଶ>𝜒ସ,.ହଶ=9.4877,𝑤ℎ𝑒𝑟𝑒 𝜒ଶ = ∑∑൫ைೕ – ாೕ൯మ
ாೕ
𝐸ଵଵ=𝑎ଵ𝑏ଵ
𝑁=42⋅80
300=11.2,𝐸ଵଶ=𝑎ଵ𝑏ଶ
𝑁=42⋅130
300=18.2
𝐸ଵଷ=𝑎ଵ𝑏ଷ
𝑁 =42⋅90
300=12.6,𝐸ଶଵ =𝑎ଶ𝑏ଵ
𝑁=85⋅80
300=22.67
𝐸ଷଵ=𝑎ଷ𝑏ଵ
𝑁=173⋅80
300=46.13,𝐸ଷଶ=𝑎ଷ𝑏ଶ
𝑁=173⋅130
300=74.97
𝐸ଷଷ=𝑎ଷ𝑏ଷ
𝑁 =173⋅90
300=51.9 Oij Eij Oij - Eij ൫𝑂−𝐸൯ଶ𝐸 21 11.2 9.8 8.575 15 18.2 -3.2 0.5626 6 12.6 -6.6 3.4571 24 22.67 1.33 0.078 35 36.83 -1.83 0.0909 26 25.5 0.5 0.0098 35 46.13 -11.13 2.6854 80 74.97 5.03 0.3375 58 51.9 6.1 0.7170 total 16.5133
𝜒ଶ=16.5133>𝜒ସ,.ହଶ=9.488 ⇒𝑟𝑒𝑗𝑒𝑐𝑡 𝐻 𝑎𝑡 5% 𝐿𝑂𝑆
We decide to reject H 0 at 5% level of significance and conclude that the
standard of clothing of a salesman has significant effect on his
performance in field selling.
𝐶= ඨ𝜒ଶ
𝜒ଶ+𝑁= ඨ16.5133
16.5133+300=0.2284 munotes.in
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183 11.8 CORRELATION OF ATTRIBUTES
Because classifications in a contingency table often describe
characteristics of individuals or objects, they are often referred to as
attributes , and the degree of dependence, association, or relationship is
called the correlation of attributes . For k X k tables, we define
𝑟= ඨ𝜒ଶ
𝑁(𝑘 − 1)
as the correlation coefficient between attributes(or classification). This
coefficient lies between 0 and 1. For 2 2 tables in which k = 2, the
correlation is often called tetrachoric correlation.
11.9 ADDITIVE PROPERTY OF CHI SQUARE VARIATE
Suppose that the results of repeated experiments yield sample values of 𝜒ଶ
given by
𝜒ଵଶ,𝜒ଶଶ,𝜒ଷଶ,⋯⋯,𝜒ଶ with 𝑣ଵ,𝑣ଶ,𝑣ଷ,⋯⋯,𝑣 degrees of freedom,
respectively. Then the result of all these experiment can be considered
equivalent to 𝜒ଶ value given by 𝜒ଵଶ+𝜒ଶଶ+ 𝜒ଷଶ+ ⋯⋯+𝜒ଶ
with 𝑣ଵ+𝑣ଶ+ 𝑣ଷ+ ⋯⋯+𝑣 degrees of freedom.
11.10 SUMMARY
In this chapter we discussed about the chi-square test which is a non -
parametric test that compares two or more variables from randomly
selected data. It helps to find the relationship between two or more
variables. The chi square distribution is a theoretical or mathematical
distribution whic h has wide applicability in statistical work. We had also
seen the various properties of Chi -square variate along with different tests
to check whether the variable follows Chi -square distribution, along with
Yate’s correction.
11.11 EXERCISES
1) If X is a chi square variate with 17 degreed of freedom
find𝑥 ,𝑥ଵ 𝑎𝑛𝑑 𝛼 such that
𝑝(𝑋>𝑥)=0.01,𝑝(𝑋≤𝑥ଵ)=0.95 & 𝑝(𝑋≤8.67)=𝛼.
2) The following table gives the result of investigation in the association
of eye color and hair color. Can we deduce that the two attributes are
independent? Hair color munotes.in
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184 Brown Black Eye color Blue 75 25 Brown 65 35
3) The following data show the classification of individuals with respect
to gender and literacy in a random sample of 200. Test the data for
independence of attributes using chi square at 1% LOS. literate illiterate male 95 5 female 75 25
4) The eyesight of 100 randomly selected people from a town were tested
with the following results: poor eyesight good eyesight male 200 350 female 200 250 Can we conclude at 5% level of significance that gender has no bearing
on the quality of eyesight?
5) The following table shows results of inoculation against cholera not attacked attacked Inoculated 446 4 Not inoculated 291 9 Can we say at 5% LOS that inoculation is effective in controlling
susceptibility of cholera?
6) The following are the results of the tests performed on two brands of
tyres manufactured by a manufacturer. Brand A Brand B Lasted more than 30000km 27 38 Failed to last 30000 km 18 27
Use chi square at 5% LOS to test whether we can say that the two
brands of tyres differ significantly or not as regards their lifespan.
7) Determine at 1% LOS whether vaccination can be regarded as a
preventive measure for small pox on the basis of following report:
Out of 1482 persons in a locality exposed to small pox, 368 in all were
attacked. Out of 1482 persons, 343 had been vaccinated and of these
only 35 were attacked.
8) During a market research survey organized by ABC Ltd. the
households were asked whether they used “Beauty Soap”(the price of
which is Rs. 7.25 a piece) and whether their per capita monthly
expenditure exceeded Rs. 800 w ith the following results. munotes.in
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185 Monthly per capita expenditure Exceeded Rs 800 Did not exceed Rs. 800 Whether they used yes 6 20 “Beauty Soap” no 4 30
Can we say the use of “Beauty Soap” depends on monthly per capita
expenditure at 5% LOS.
9) In a household survey conducted in certain locality, the following
information is collected. Whether exclusive Indoor toilet facility is available yes no owned house 9 4 rented house 21 16
Use chi square test for independence at 1% LOS and give your
conclusions.
10) The following data refer to an investigation carried out to examine the
effect of T.V as a medium of advertisement on the turnover of the
company’s manufacturing and selling consumer products. A random
sample of 40 companies was selected. An alyze the data and comment
on your findings. Annual turnover exceeding one crore rupees Yes No T.V advertisement 7 3 No T.V advertisement 10 20
11) A random sample of students of Mumbai University was selected and
asked their opinion about autonomous college. The results are given
below. The same number of each gender was included within each
class group. Test the hypothesis at 5% LOS that opinions are
independent of the class groupings. favoring autonomous colleges opposed to autonomous college First Year 120 80 Second Year 130 70 Third year 70 30 Post Graduation 80 20
12) Test for independence between health and working capacity from one
following data: Health very good good fair working good 20 25 15 Capacity bad 10 15 15 munotes.in
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186 13) ABC Ltd. employ a large number of handicapped persons. The
following is an account of the performance of 200 randomly chosen
employees of the company: performance above average average below average handicapped 29 31 20 non handicapped 55 30 35 Can we retain at 5% LOS that handicapped employees are equally
efficient as the non -handicapped employees of the company.
14) In a survey 100 couples are interviewed and they were asked to give
an opinion on the importance of amiable nature of partner in selection
of a bride or a groom. The ranks were given independently by them as
I or II or III. Rankings by wives I II III Ranking I 25 12 8 by II 10 23 8 Husbands III 5 5 4 Use chi square test for independence and comment at 5% LOS.
15) A socio economic survey conducted in 1981 in Mumbai revealed the
following results: Monthly family income Below 1200 1200 to 1800 1800 and above No child 18 15 12 one child 31 34 25 Two or more children 81 51 63 Can we regard at 1% LOS that the number of children in the family
has no association with monthly income?
16) The data in the table were collected on how individuals prepared their
taxes and their education level. The null hypothesis is that the way
people prepare their taxes (computer software or pen and paper) is
independent of their education level. The following table is the
contingency table.
Education Tax Prepare High School Bachelors Masters Computer Software 23 35 42 Pen and Paper 45 30 25 Find the co efficient of contingency.
11.12 SOLUTION TO EXERCISE Q. No. Solution Q. No. Solution munotes.in
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187 1 𝑥=33.4087,𝑥ଵ =27.5871, 𝛼=0.05 2 2.38, accept H0 3 15.686, reject H0 4 3.84, donot reject H0 5 3.552, accept H0 6 0.026, accept 7 51.157 reject 8 0.6652 accept 9 0.2122, accept 10 2.762, accept 11 12.046 , reject 12 1.9097, accept 13 4.904, accept 14 10.8125, accept 15 4.305, accept 16 0.236
11.13 TABLE OF CHI -SQUARE DISTRIBUTION
11.14 REFERENCE FOR FURTHER READING
Following books are recommended for further reading:
Statistics by Murray R, Spiegel, Larry J. Stephens, Mcgraw Hill
International Publisher, 4th edition
Fundamental of Mathematical Statistics by S. C. Gupta and V. K.
Kapoor, Sultan Chand and Sons publisher, 11th edition
***** chi suaredf\area0.9950.990.9750.950.90.750.50.250.10.050.0250.010.00510.000040.000160.000980.003930.015790.101530.454941.32332.705543.841465.023896.63497.8794420.010030.02010.050640.102590.210720.575361.386292.772594.605175.991467.377769.2103410.5966330.071720.114830.21580.351850.584371.212532.365974.108346.251397.814739.348411.3448712.8381640.206990.297110.484420.710721.063621.922563.356695.385277.779449.4877311.1432913.276714.8602650.411740.55430.831211.145481.610312.67464.351466.625689.2363611.070512.832515.0862716.7496 60.675730.872091.237341.635382.204133.45465.348127.840810.6446412.5915914.4493816.8118918.5475870.989261.239041.689872.167352.833114.254856.345819.0371512.0170414.0671416.0127618.4753120.2777481.344411.64652.179732.732643.489545.070647.3441210.2188513.3615715.5073117.5345520.0902421.9549591.734932.08792.700393.325114.168165.898838.3428311.3887514.6836616.9189819.0227721.6659923.58935102.155862.558213.246973.94034.865186.73729.3418212.5488615.9871818.3070420.4831823.2092525.18818 112.603223.053483.815754.574815.577787.5841410.34113.7006917.2750119.6751421.9200524.7249726.75685123.073823.570574.403795.226036.30388.4384211.3403214.845418.5493521.0260723.3366626.2169728.29952133.565034.106925.008755.891867.04159.2990712.3397615.9839119.8119322.3620324.735627.6882529.81947144.074674.660435.628736.570637.7895310.1653113.3392717.1169321.0641423.6847926.1189529.1412431.31935154.600925.229356.262147.260948.5467611.0365414.3388618.2450922.3071324.9957927.4883930.5779132.80132 165.142215.812216.907667.961659.3122411.9122215.338519.3688623.5418326.2962328.8453531.9999334.26719175.697226.407767.564198.6717610.0851912.7919316.3381820.4886824.7690427.5871130.1910133.4086635.71847186.26487.014918.230759.3904610.8649413.6752917.337921.6048925.9894228.869331.5263834.8053137.15645196.843977.632738.9065210.1170111.6509114.56218.3376522.7178127.2035730.1435332.8523336.1908738.58226207.433848.26049.5907810.8508112.4426115.4517719.3374323.8276928.4119831.4104334.1696137.5662339.99685 218.033658.897210.282911.5913113.239616.3443820.3372324.9347829.6150932.6705735.4788838.9321741.40106228.642729.5424910.9823212.3380114.0414917.2396221.3370426.0392730.8132833.9244436.7807140.2893642.79565239.2604210.1957211.6885513.0905114.8479618.137322.3368827.1413432.006935.1724638.0756341.638444.18128249.8862310.8563612.4011513.8484315.6586819.0372523.3367328.2411533.1962436.4150339.3640842.9798245.558512510.5196511.5239813.1197214.6114116.4734119.9393424.3365929.3388534.3815937.6524840.6464744.314146.92789 2611.1602412.1981513.843915.3791617.2918820.8434325.3364630.4345735.5631738.8851441.9231745.6416848.289882711.8075912.878514.5733816.151418.113921.749426.3363431.5284136.7412240.1132743.1945146.9629449.644922812.4613413.5647115.3078616.9278818.9392422.6571627.3362332.6204937.9159241.3371444.4607948.2782450.993382913.1211514.2564516.0470717.7083719.7677423.5665928.3361333.7109139.0874742.5569745.7222949.5878852.335623013.7867214.9534616.7907718.4926620.5992324.4776129.3360334.7997440.2560243.7729746.9792450.8921853.67196
munotes.in
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188 UNIT V
12
CURVE FITTING AND THE METHOD OF
LEAST SQUARES
Unit Structure
12.0 Objectives
12.1 Introduction
12.2 Relationship between variables
12.3 Curve fitting
12.4 Equations of Approximating Curves
12.5 Freehand Method o f Curve Fitting
12.6 The Straight line Method
12.7 Least Square Curve fitting
12.7.1 Straight Line
12.7.2 Parabola
12.7.3 Non-Linear relationship
12.8 Regression
12.9 Applications to Time Series
12.10 Problems involving more than two variables
12.11 Summary
12.12 Exercises
12.13 Solution to Exercises
12.14 Logarithm tables
12.15 Reference for further reading
12.0 OBJECTIVES
The main objective is to study the fitting of curves using the method of
least square which minimizes the sum of the square of the errors.
12.1 INTRODUCTION
When we come across data for two variables and think about the relation
between each other, two objects come to our mind:
1) To translate the possible relationship into a mathematical
equation(called equation of curve)
2) To exploit the relationship between the variables for estimating the
value of one variable corresponding to a given value of the other
variable.
munotes.in
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189 Curve fitting helps us in serving the first object directly, i.e., it leads to a
mathematical equation describing the relationship between the variables.
Further, using this equation one can estimate the value of the other
variable.
12.2 RELATIONSHIP BETWEEN VARIABLES
Very often in practice a relationship is found to exist between two (or
more) variables. For example, weights of adult males depend to some
degree on their heights, the circumferences of circles depend on their radii,
and the pressure of a given mass of g as depends on its temperature and
volume.
It is frequently desirable to express this relationship in mathematical form
by determining an equation that connects the variables.
12.3 CURVE FITTING
To determine an equation that connects variables, a first step is to collect
data that show corresponding values of the variables under consideration.
For example, suppose X and Y denote respectively, the height and weight
of adult males; then a sample of N individuals would reveal the heights
𝑋ଵ,𝑋ଶ,𝑋ଷ,⋯⋯,𝑋ே and the corresponding weights 𝑌ଵ,𝑌ଶ,𝑌ଷ,⋯⋯𝑌ே.
A next step is plot the points (𝑋ଵ,𝑌ଵ),(𝑋ଶ,𝑌ଶ),(𝑋ଷ,𝑌ଷ),⋯⋯,(𝑋ே,𝑌ே) on a
rectangular coordinate system. The resulting set of points is some times
called a scatter diagram . From the scatter diagram it is often possible to
visualize a smooth curve that approximates the data. Such a curve is called
an approximating curve .
For example, in the following diagram the data appear to be approximated
by a straight line, and so we call that a linear relationship exists between
the variables.
00.511.522.53
024681012linear relationship munotes.in
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190 In the following diagram, however the relationship exists between the
variables, it is not a linear relationship, and so we call it a non -linear
relationship.
The general problem of finding equations of approximating curves that fit
the given sets of data is called curve fitting.
12.4 EQUATIONS OF APPROXIMATING CURVES
Various common types of approximating curves and their equations are
given below for reference. All letters other than x and y shall be treated as
constants. The variables x and y are termed as independent variable and
dependent variable respectively, role s of the variables can be interchanged
as per the requirement.
Straight Line 𝑦=𝑎+𝑏𝑥
Parabola or quadratic curve 𝑦=𝑎+𝑏𝑥+𝑐𝑥ଶ
Cubic curve 𝑦=𝑎+𝑏𝑥+𝑐𝑥ଶ+𝑑𝑥ଷ
Quadratic curve 𝑦=𝑎+𝑏𝑥+𝑐𝑥ଶ+𝑑𝑥ଷ+𝑒𝑥ସ
nth degree curve 𝑦=𝑎+𝑎ଵ𝑥+𝑎ଶ𝑥ଶ+⋯⋯+𝑎𝑥
The above equations refer to the polynomial equations of the degree one,
two, three, four and n respectively. The following are some examples of
other equations most commonly used.
Hyperbola 𝑦=ଵ
బ ା భ௫
Exponential curve 𝑦=𝑎⋅𝑏௫
Geometric curve 𝑦=𝑎⋅𝑥
Gompertz Curve 𝑦=𝑝𝑞ೣ
Logistic Curve 𝑦=ଵ
ೣ ା
00.511.522.5
024681012Non linear relationship munotes.in
Page 191
191 To decide which curve to be used, it is helpful to obtain scatter diagrams
of the transformed variables. For example, if a scatter diagram of
log𝑦versus x shows a linear relationship, then the equation is of the 𝑦=
𝑎⋅𝑏௫ , while if log𝑦 versus log𝑥 shows a linear relationship, the equation
is of the form 𝑦=𝑎⋅𝑥.
12.5 FREEHAND METHOD OF CURVE FITTING
Individual judgment can often be used to d raw an approximating curve to
fit a set of data. This is called a freehand method of curve fitting. If the
type of equation of this curve is known, it is possible to obtain the
constants in the equation by choosing as many points on the curve as there
are constants in the equation. For example, if the curve is a straight line,
two points are necessary; if it is a parabola, three points are necessary. The
method has the disadvantage that different observers will obtain different
curves and equations.
12.6 THE STRAIGHT LINE METHOD
The simplest type of the approxi mating curve is a straight line , whose
equation can be written
𝑦=𝑎+𝑏𝑥
Given any two points (𝑥ଵ,𝑦ଵ)&(𝑥ଶ,𝑦ଶ) on the line, the constants a and b
can be determined. The resulting equation of the straight line can be
written as
𝑦−𝑦ଵ=𝑚(𝑥−𝑥ଵ),𝑚 =𝑦ଶ−𝑦ଵ
𝑥ଶ−𝑥ଵ
where m is known as the slope of the line.
When the equation is written in the form of 𝑦=𝑎+𝑏𝑥, the constant ‘b’
denotes the slope m. The constant ‘a’ is the value of y when x = 0 is called
the y -intercept.
12.7 LEAST SQUARE CURVE FITTING
We can find the trend curve by fitting a mathematical equation. The
method is more precise and can be used even for forecasting. We can fit
either a straight line or a curve to the given data. We fit a straight line
𝑦=𝑎+𝑏𝑥, where a and b are constants. We determine the constants a
and b so that the following conditions are fulfilled:
i) The sum of the deviations of all the values of y from their trend values
is zero, when, deviations above the line are given positive sign and
deviations below, negative i.e if 𝑦is the trend value obtained from the
trend line, and y is the ac tual value in the data.
∑(𝑦−𝑦)=0
ii) The sum of the squares of the deviations is the least, i.e., munotes.in
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192 ∑(𝑦−𝑦)ଶ is minimum.
The method gets the name ‘least square method’ because of this second
property. This is also called line of best fit . In a sense this line is like
arithmetic mean since arithmetic mean is a single value possessing the
above two properties.
Remark: When the value of x is given in terms of years or value of x is
big then we apply the following technique to get the values of x:
1) If the value of n is odd then we take the value of central most
observation i.e value of ାଵ
ଶth observation as 0 and we add 1 as we
move downward and we subtract 1 as we move upwards.
2) If the value if n is even then we take the value of
ଶ th observation as -
1 and we add 2 as we move downward and subtract 2 as we move
upwards.
12.7.1 Straight Line :
Suppose o two variables x and y, n pairs of observations, say
(𝑥ଵ,𝑦ଵ),(𝑥ଶ,𝑦ଶ),⋯⋯,(𝑥,𝑦) are available and we want to fit a linear
curve of the form 𝑦=𝑎+𝑏𝑥 to the data. Then according to the least
squares method we have to find a and b such that ∑(𝑦−𝑦)ଶ is minimum
where 𝑦=𝑎+𝑏𝑥.
Thus, we have to minimize 𝐷=∑(𝑦−𝑎−𝑏𝑥). The first order
conditions are ఋ
ఋ=0 𝑎𝑛𝑑 ఋ
ఋ=0
which are known as normal equations and they are given by
∑𝑦=𝑎𝑛+𝑏∑𝑥 𝑎𝑛𝑑 ∑𝑥𝑦 =𝑎∑𝑥+𝑏 ∑𝑥ଶ
It can be easily verified that the second order condition for minima are
satisfied and therefore the best choice of a and b can be made by finding
their values satisfying the normal equations. This equations will be used to
fit a linear curve to the given data.
Eg: 1) Fit a straight line of the form y = a + bx using least square method x 0 1 2 3 4 y 1 2.9 4.8 6.7 8.6
Solution: n = 5 x y xy x2 0 1 0 0 1 2.9 2.9 1 2 4.8 9.6 4 3 6.7 20.1 9 4 8.6 34.4 16 total 10 24.0 67.0 30 munotes.in
Page 193
193 The straight line equation is given by 𝑦=𝑎+𝑏x. To find values of a and
b we solve the following equations
∑𝑦=𝑎𝑛 +𝑏∑𝑥,∑𝑥𝑦=𝑎∑𝑥+𝑏∑𝑥ଶ --- (1)
by substituting the values in eq (1) , we get
24=5𝑎+10𝑏−−−−(2)
67=10𝑎+30𝑏−−−−(3)
multiplying eq (2) by 2 we get 48=10𝑎+20𝑏−−−−(4)
subtracting eq (4) by eq (3), we get 19=10𝑏⇒𝑏=1.9
substituting 𝑏=1.9 in eq (2) gives 𝑎=1
∴𝑦=1+1.9𝑥
is the required straight line.
Eg: 2) Fit a straight line of the form 𝑦=𝑎+𝑎ଵ𝑥 x 1 2 3 4 6 8 y 2.4 3.1 3.5 4.2 5 6
Solution: n = 6 x y xy x2 1 2.4 2.4 1 2 3.1 6.2 4 3 3.5 10.5 9 4 4.2 16.8 16 6 5 30 36 8 6 48 64 total 24 24.2 113.9 130
The straight line equation is given by 𝑦=𝑎+𝑎ଵx. To find values of 𝑎
and 𝑎ଵ we solve the following equations
∑𝑦=𝑎𝑛 +𝑎ଵ∑𝑥,∑𝑥𝑦=𝑎∑𝑥+𝑎ଵ∑𝑥ଶ - - - (1)
by substituting the values in eq (1) , we get
24.2=6𝑎+24𝑎ଵ −−−(2)
113.9=24𝑎+130𝑎ଵ −−−(3)
multiplying eq (2) by 4, we get 96.8=24𝑎+96𝑎ଵ−−−(4)
subtracting eq (4) from eq (3), we get 17.1=34𝑎ଵ⇒𝑎ଵ=0.5029
subs 𝑎ଵ=0.5029 in (2) gives 𝑎=2.0217
∴ 𝑦=2.0217+0.5029𝑥
is the required straight line.
Eg: 3) Fit a straight line using least square method and estimate the
exchange rate for the year 1993 -94 and 1984 -85.
munotes.in
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194 year 1985 -
86 1986 -
87 1987-88 1988-89 1989-90 1990-91 1991-92 Exchange rate 12.24 12.78 12.97 14.48 16.65 17.94 24.47
Solution: n = 7
The straight line equation we take is 𝑦=𝑎+𝑏x. To find values of a and b
we solve the following equations
∑𝑦=𝑎𝑛 +𝑏∑𝑥,∑𝑥𝑦=𝑎∑𝑥+𝑏∑𝑥ଶ --- (1)
year ex. rate(y) x xy x2 1985-86 12.24 -3 -36.72 9 1986-87 12.78 -2 -25.56 4 1987-88 12.97 -1 -12.97 1 1988 -89 14.48 0 0 0 1989 -90 16.65 1 16.65 1 1990 -91 17.94 2 35.88 4 1991 -92 24.47 3 73.41 9 total 111.53 0 50.69 28
by su bstituting the values in eq (1) , we get
111.53=7𝑎−−−−(2)
50.69=28𝑏−−−−(3)
from eq (2), we get 𝑎=ଵଵଵ.ହଷ
=15.9328
from eq (3), we get 𝑏=ହ.ଽ
ଶ଼=1.8104
∴𝑦=15.9328+1.8104𝑥 −−−−(4)
is the required straight line.
To the get exchange rate for the year 1993 -94 , we subs x = 5 in eq (4)
∴𝑦=15.9328+1.8104(5)=24.9848
To the get the exchange rate for the year 1984 -85, we subs x = -4 in eq (4)
∴𝑦=15.9328+1.8104(−4)=8.6912
12.7.2 Parabola :
Suppose we want to fit a second degree curve called parabola of the form
𝑦=𝑎+𝑏𝑥+𝑐𝑥ଶ to n pairs of observations
(𝑥ଵ,𝑦ଵ),(𝑥ଶ,𝑦ଶ),⋯⋯,(𝑥,𝑦). Then according to the least squares
method we have to minimize 𝐷=∑(𝑦−𝑎−𝑏𝑥−𝑐𝑥ଶ)ଶ with respect to
a, b and c. The first order conditions are given by
𝛿𝐷
𝛿𝑎=0,𝛿𝐷
𝛿𝑏=0 𝑎𝑛𝑑𝛿𝐷
𝛿𝑐=0
Applying these we get,
∑𝑦=𝑛𝑎+𝑏∑𝑥+𝑐∑𝑥ଶ
∑𝑥𝑦=𝑎∑𝑥+𝑏∑𝑥ଶ+𝑐∑𝑥ଷ
∑𝑥ଶ𝑦=𝑎∑𝑥ଶ+𝑏∑𝑥ଷ+𝑐∑𝑥ସ munotes.in
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195 Eg: 1) Fit a parabola of the form 𝑦=𝑎+𝑏𝑥+𝑐𝑥ଶ using least square
method x 0 1 2 y 1 6 17
Solution: n=3 x y x2 x3 x4 xy x2y 0 1 0 0 0 0 0 1 6 1 1 1 6 6 2 17 4 8 16 34 68 total 3 24 5 9 17 40 74
The equation of the parabola is given by 𝑦=𝑎+𝑏𝑥+𝑐𝑥ଶ. To find
values of 𝑎,𝑏 𝑎𝑛𝑑 𝑐 we solve the following equations
∑𝑦=𝑛𝑎+𝑏∑𝑥+𝑐∑𝑥ଶ,∑𝑥𝑦=𝑎∑𝑥+𝑏∑𝑥ଶ+𝑐∑𝑥ଷ,
∑𝑥ଶ𝑦=𝑎∑𝑥ଶ+𝑏∑𝑥ଷ+𝑐∑𝑥ସ −−−−(∗)
subs values in eq (* ), we get
24=3𝑎+3𝑏+5𝑐 −−−−(1)
40=3𝑎+5𝑏+9𝑐 −−−−(2)
74=5𝑎+9𝑏+17𝑐 −−−−(3)
Subtracting eq (1) by (2), we get 16 = 2b +4c ⇒8=𝑏+2𝑐−−−(4)
𝑚𝑢𝑙 𝑒𝑞 (2) 𝑏𝑦 5 ,𝑤𝑒 𝑔𝑒𝑡 200=15𝑎+25𝑏+45𝑐−−−(5)
𝑚𝑢𝑙 𝑒𝑞 (3)𝑏𝑦 3,𝑤𝑒 𝑔𝑒𝑡 222=15𝑎+27𝑏+51𝑐−−−(6)
𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑡𝑖𝑛𝑔 𝑒𝑞 (5) 𝑏𝑦 (6),𝑤𝑒 𝑔𝑒𝑡 22=2𝑏+6𝑐
⇒11=𝑏+3𝑐−−−(7)
𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑛𝑔 𝑒𝑞 (4) 𝑏𝑦 (7),𝑤𝑒 𝑔𝑒𝑡 3=𝑐
𝑠𝑢𝑏 𝑐=3 𝑖𝑛 𝑒𝑞 (4) 𝑤𝑒 𝑔𝑒𝑡 𝑏=2
𝑠𝑢𝑏 𝑏=2,𝑐=3 𝑖𝑛 (1),𝑤𝑒 𝑔𝑒𝑡 𝑎=1
∴𝑦=1+2𝑥+3𝑥ଶis the required equation of the parabola.
Eg: 2) Fit a parabola of the form 𝑦=𝑎+𝑏𝑥 +𝑐𝑥ଶ year 1989 1990 1991 1992 1993 1994 1995 1996 no.of students 15 17 20 25 30 31 30 32 Also find the no. of students for the year 2000 & 1987.
Solution: n = 8 year no. of students(y) x x2 x3 x4 xy x2y 1989 15 -7 49 -343 2401 -105 735 1990 17 -5 25 -125 625 -85 425 1991 20 -3 9 -27 81 -60 180 1992 25 -1 1 -1 1 -25 25 munotes.in
Page 196
196 1993 30 1 1 1 1 30 30 1994 31 3 9 27 81 93 279 1995 30 5 25 125 625 150 750 1996 32 7 49 343 2401 224 1568 total 200 0 168 0 6216 222 3992
The equation of the parabola is given by 𝑦=𝑎+𝑏𝑥+𝑐𝑥ଶ. To find
values of 𝑎,𝑏 𝑎𝑛𝑑 𝑐 we solve the following equations
∑𝑦=𝑛𝑎+𝑏∑𝑥+𝑐∑𝑥ଶ,∑𝑥𝑦=𝑎∑𝑥+𝑏∑𝑥ଶ+𝑐∑𝑥ଷ,
∑𝑥ଶ𝑦=𝑎∑𝑥ଶ+𝑏∑𝑥ଷ+𝑐∑𝑥ସ −−−−(∗)
subs values in eq ( *), we get
200 =8𝑎+168𝑐 −−−−(1)
222=168𝑏 −−−−(2)
3992=168𝑎+6216𝑐 −−−−(3)
𝑓𝑟𝑜𝑚 (2),𝑤𝑒 𝑔𝑒𝑡 𝑏=222
168=1.3214
𝑚𝑢𝑙 (1)𝑏𝑦 21 ,𝑤𝑒 𝑔𝑒𝑡 4200=168𝑎+3528𝑐 −−−(4)
𝑠𝑢𝑏 (4)𝑏𝑦 (3),𝑤𝑒 𝑔𝑒𝑡 208= −2688𝑐 ⇒𝑐=−208
2688=−0.0774
𝑠𝑢𝑏 𝑐= −0.0774 𝑖𝑛 (1),𝑤𝑒 𝑔𝑒𝑡 𝑎=200−168(−0.0774)
8=26.6254
∴𝑦=26.6254+1.3214𝑥−0.0774𝑥ଶ−−−−(4)
is the required equation of the parabola.
To get the no. of students for the year 2000 we substitute x = 15 in (4)
∴𝑦=26.6254+1.3214(15)−0.0774(15)ଶ=29.0314
To get the no. of students for the year 1987 we substitute x = -11 in (4)
∴𝑦=26.6254+1.3214(−11)−0.0774(−11)ଶ=2.7246
Eg: 3) Fit a parabola of the form 𝑦=𝑎+𝑏𝑥+𝑐𝑥ଶ
year 1985 1986 1987 1988 1989 milk (in 100 litres) 20 25 27 35 38 Find the production of milk for the year 1982 and 1995.
`Solution: n =5 year milk (y) x x2 x3 x4 xy x2y 1985 20 -2 4 -8 16 -40 80 1986 25 -1 1 -1 1 -25 25 1987 27 0 0 0 0 0 0 munotes.in
Page 197
197 1988 35 1 1 1 1 35 35 1989 38 2 4 8 16 76 152 total 145 0 10 0 34 46 292
The equation of the parabola is given by 𝑦=𝑎+𝑏𝑥+𝑐𝑥ଶ. To find
values of 𝑎,𝑏 𝑎𝑛𝑑 𝑐 we solve the following equations
∑𝑦=𝑛𝑎+𝑏∑𝑥+𝑐∑𝑥ଶ,∑𝑥𝑦=𝑎∑𝑥+𝑏∑𝑥ଶ+𝑐∑𝑥ଷ,
∑𝑥ଶ𝑦=𝑎∑𝑥ଶ+𝑏∑𝑥ଷ+𝑐∑𝑥ସ −−−−(∗)
subs values in eq (*), we get
145=5𝑎+10𝑐−−(1)
46=10𝑏−−(2) ⇒𝑏=4.6
292=10𝑎+34𝑐−−(3)
multiply eq (1) by 2, we get 290=10 𝑎+20𝑐−−(4)
subtracting eq (4) from eq (3), we get 2=14𝑐⇒𝑐=ଶ
ଵସ=0.1428
substituting c = 0.1428 in eq (1), we get 𝑎=28.7144
∴𝑦=28.7144+4.6𝑥+ 0.1428𝑥ଶ−−−(5)
is the required equation of the parabola.
To get the production of milk for the year 1995, we put x = 8 in (5)
∴𝑦=28.7144+4.6(8)+0.1428(8)ଶ =74.6536
To get the production of milk for the year 1982, we put x = -5 in (5)
∴𝑦= 28.7144+4.6(−5)+0.1428(−5)ଶ =9.2844
12.7.3 Non-Linear Relationship :
Fitting of a power Curve:
Suppose we want to fit a curve whose equation is of the form 𝑦=𝑎⋅𝑥
to variables x and y on which n paired observations are available. Then we
first rewrite the equation of the curve as
log𝑦=log𝑎+𝑏log𝑥
which can be written in the linear form 𝑌=𝐴+𝑏𝑋 where Y = log y, A =
log a and X = log x. Therefore this curve can be fitted by the method of
least squares solving the following normal equations.
∑𝑌=𝑛𝐴+𝑏∑𝑋 ,∑𝑋𝑌 =𝐴∑𝑋+𝑏∑𝑋ଶ
Once we get the value of A, we can find a = antilog (A). The method of
fitting is explained in the following example.
Eg: 1) Fit a curve of the form 𝑦=𝑎⋅𝑥 for the following data
x 1 2 4 5 7 10 y 2.1 4.9 20.8 32.7 60 131 munotes.in
Page 198
198 Solution: n = 6
We can write the given curve in the form of 𝑌=𝐴+𝑏𝑋 where Y = log y,
A = log a, X = log x
x y X = log x Y = log y XY X2 1 2.1 0 0.3222 0 0 2 4.9 0.3010 0.6902 0.2078 0.0906 4 20.8 0.6021 1.3181 0.7936 0.3625 5 32.7 0.6990 1.5142 1.0586 0.4886 7 60 0.8451 1.7782 1.5028 0.7142 10 131 1 2.1173 2.1173 1 Total 3.4472 7.7405 5.6801 2.6559
The given curve is of the form 𝑦=𝑎𝑥, therefore the normal equations
are given by
∑𝑌=𝑛𝐴+𝑏∑𝑋 ,∑𝑋𝑌 =𝐴∑𝑋+𝑏∑𝑋ଶ −−−(1)
substituting the values in eq (1), we get
7.7405=6𝐴+3.4472𝑏 −−−(2)
5.6801= 3.4472𝐴+2.6559𝑏 −−−−(3)
multiplying eq (1) by 3.4472 and eq (2) by 6, then subtracting the two eq.,
we get
4.0522𝑏= 7.3975⇒𝑏=7.3975
4.0522=1.8256
substituting b = 1.8256 in eq (1), we get a = 0 02412
Now a = antilog(A) = antilog(0.2412) = 1.7426
∴𝑦=1.7426𝑥ଵ.଼ଶହ
is the required equation.
Fitting of the curve 𝒚=𝒂𝒆𝒃𝒙
We first try to convert the equation 𝑦=𝑎𝑒௫ to a linear equation by
applying logarithms on both sides with respect to base 10. By applying
logarithms on both sides, we get
log𝑦=log𝑎+𝑏𝑥𝑙𝑜𝑔ଵ𝑒
Let Y = log y, A = log a, logଵ𝑒=0.4343𝑏. Then the equation of the
curve reduces to a linear equation of the form
𝑌=𝐴+𝐵𝑥
By using the least square method the constants A and b can be found by
solving the normal equations
∑𝑌=𝑛𝐴+𝐵∑𝑥,∑𝑥𝑌=𝐴∑𝑥+𝐵∑𝑥ଶ
munotes.in
Page 199
199 Once A and B are found, we can find a = antilog (A) and b =
.ସଷସଷ
Eg: 1) Fit a cur ve of the form 𝑦=𝑎𝑒௫ for the following data
x 0 2 4 6 8 y 3 55 1095 22000 442000
Solution : n =5
The given equation is of the form 𝑦=𝑎𝑒௫, therefore the normal
equations are given by
∑𝑌=𝑛𝐴+𝐵∑𝑥,∑𝑥𝑌=𝐴∑𝑥+𝐵∑𝑥ଶ −−−(1)
substituting the values in eq (1), we get
15.2447=5𝐴+20𝐵 −−−−(2)
86.8560=20𝐴+120𝐵 −−−−(3)
multiplying eq (1) by 4 and subtracting it from eq (2), we get
25.8772=40𝐵⇒𝐵=25.8772
40=0.6469
substituting B = 0.6469 in eq (1), we get
15.2447=5𝐴+20(0.6469)⇒𝐴=2.3067
5=0.4613
Now a = antilog (A) = antilog(0.4613) = 2.892
b =
.ସଷସଷ=.ସଽ
.ସଷସଷ=1.4895
∴𝑦=2.892𝑒ଵ.ସ଼ଽହ௫
is the required equation.
12.8 REGRESSION
Often, on the basis of sample data, we wish to estimate the value of a
variable Y corresponding to a given value of a variable x. This can be
accomplished by estimating the value of y from a least -squares curve that
fits the sample data. The resulting curve is called a regression curve of y
on x, since y is estimated from x.
x y Y = log y xY x2 0 3 0.4771 0 0 2 55 1.7404 3.4808 4 4 1095 3.0394 12.1576 16 6 22000 4.3424 26.0544 36 8 442000 5.6454 45.1632 64 Total 20 15.2447 86.8560 120 munotes.in
Page 200
200 If we want to estimate the value of x from a given value of y, we would
use a regression curve of x on y, which amounts to interchanging the
variables in the scatter diagram so that x is the dependent variable and y is
the independent variable. This is equivalent to replacing the vertical
deviations in the definition of the least -squares curve with horizontal
deviations.
In general, the regression line or curve of y on x is not the same as the
regression line or curve of x on y.
12.9 APPLICATIONS TO TIME SERIES
If the independent variable x is time, the data show the values of y at
various times. Data arranged according to time are called time series. The
regression line or curve of y on x in this case is often called atrend line or
trend curve and is often used for purposes of estimation, p rediction, or
forecasting.
12.10 PROBLEMS INVOLVING MORE THAN TWO VARIABLES
Problems involving more than two variables can be treated in a manner
analogous to that for two variables. For example, there may be a
relationship between the three variables X, Y, and Z that can be described
by the equation
𝑧=𝑎+𝑏𝑥+𝑐𝑦
which is called a linear equation in variables x, y and z.
In a three dimensional rectangular coordinate system this equation
represents a plane, and the actual sample points
(𝑥ଵ,𝑦ଵ,𝑧ଵ),(𝑥ଶ,𝑦ଶ,𝑧ଶ),⋯⋯,(𝑥ே,𝑦ே,𝑧ே) may scatt er not too far from this
plane which we call an approximating plane.
By extension of the least square method, we can talk about a least square
plane approximating the data. If we are estimating z from the given value s
of x and y, this would be called a regression plane of z on x and y. The
normal equations corresponding to the least square plane are given by
∑𝑧=𝑎𝑁+𝑏∑𝑥+𝑐∑𝑦
∑𝑥𝑧 =𝑎∑𝑥+𝑏∑𝑥ଶ+𝑐∑𝑥𝑦
∑𝑦𝑧= 𝑎∑𝑦+𝑏∑𝑥𝑦+𝑐∑𝑦ଶ
Problems involving the estimation of a variable from two or more
variables are called problems of multiple regression which is not there in
the syllabus.
12.11 SUMMARY
In this chapter we learnt about curve fitting, various equations of
approximating curves, the straight line method. We also learnt about the munotes.in
Page 201
201 fitting of linear and non -linear curves using the technique of least square
which minimizes the sum of the square of the error.
12.12 EXERCISES
1) Fit a straight line of the form y = ax+ b using least square method x 0 1 2 3 y 2 5 8 11
2) Fit a straight line of the form y = a + bx x 10 12 13 16 17 20 25 y 19 22 24 27 29 33 37
3) Fit a straight line of the form y = a + bx x 1 3 5 7 9 11 13 15 17 y 10 15 20 27 31 35 30 35 40
Also estimate y when x = 21.
4) Fit a parabola of the form 𝑦=𝑎+𝑏𝑥+𝑐𝑥ଶ using least square
method x 0 1 2 3 4 y 1 0 3 10 21
5) Fit a parabola of the form 𝑦=𝑎+𝑏𝑥+𝑐𝑥ଶ x 0 1 2 3 4 y 1 1.8 1.3 2.5 6.3
6) Fit a parabola using least square method 𝑦=𝑎𝑥+𝑏𝑥ଶ x 1 2 3 4 5 y 1.8 5.1 8.9 14.1 19.8
7) Fit a curve of the form 𝑦=𝑎𝑥 for the following data x 1 2 3 4 y 0.7 0.86 0.97 1.06
8) Fit a curve of the form 𝑦=𝑎𝑒௫ for the following x 0 2 4 y 5.012 10 31.62
9) Fit a curve of the form 𝑦=𝑎𝑒௫ for the following x 1 2 3 4 5 y 1.230 2.042 3.162 3.981 5.624 munotes.in
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202
10) The number y of bacteria per unit volume present in a culture after x
hours is given in the following table. Fit a of the form 𝑦=𝑎𝑏௫
using least square method for the following data. x 0 1 2 3 4 5 6 y 30 45 60 90 130 190 275
11) Fit a curve of the form 𝑦=𝑎𝑏௫ for the following x 1 2 3 y 8.3 15.4 33.1
12) The population of a state at ten yearly intervals is given below. Fit a
curve of the form 𝑦=𝑎𝑏௫ using least square method and also
estimate the population for the year 1961. year 1881 1891 1901 1911 1921 1931 1941 1951 population in millions 3.9 5.3 7.3 9.6 12.9 17.1 23.2 30.5
13) Fit a straight line of the form y = a + bx for the following data x 5 4 3 2 1 y 1 2 3 4 5
14) Fit a straight line of the form y = a + bx for the following data x 3 5 7 9 11 y 2.3 2.6 2.8 3.2 3.5
15) Fit a straight line to the following data on production Year 1996 1997 1998 1999 2000 Production 40 50 62 58 60
16) Fit a straight line to the following data on profit Year 1992 1993 1994 1995 1996 1997 1998 1999 Profit 38 40 65 72 69 60 87 95
17) Fit a second degree parabolic equation of the form 𝑦=𝑎+𝑏𝑥+
𝑐𝑥ଶ x 12 10 8 6 4 2 y 6 5 4 3 2 1
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203 12.14 SOLUTION TO EXERCISES Q. No. Solution Q. No. Solution 1 2, 3 2 a = 7.7044, b = 1.213 3 a=11.43, b=1.73 4 1, -3, 2 5 1.42, -1.07, 0.55 6 a= 1.44 , b = 0.51 7 𝑦=4.642𝑒.ସ௫ 8 𝑦=4.642𝑒.ସ௫ 9 𝑦=0.7762𝑒.ସଶ௫ 10 𝑦=30(1.306)௫ 11 𝑦=8.099(1.997)௫ 12 𝑦=11.074(1.0298)௫ିଵଽଵ 13 6, -1 14 1.83, 0.15 15 54, 4.8 16 62.0833, 7.3333 17 0, 0.5, 0
12.15 LOGARITHM TABLES
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205
Note: The logarithmic tables are directly taken from internet and inserted
in picture format.
12.13 REFERENCE FOR FURTHER READING
Following books are recommended for further reading:
Statistics by Murray R, Spiegel, Larry J. Stephens, Mcgraw Hill
International Publisher, 4th edition
Fundamental of Mathematical Statistics by S. C. Gupta and V. K.
Kapoor, Sultan Chand and Sons publisher, 11th edition
Descriptive Statistics by R. J. Shah
*****
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206 13
CORRELATION THEORY
Unit Structure
13.0 Objectives
13.1 Introduction
13.2 Correlation
13.2.1 Scatter Diagram and Linear Correlation
13.2.2 Coefficient of Correlation
13.2.3 Coefficient of Rank Correlation
13.2.4 The Least Square Regression Lines
13.3 The Least Square Regression Lines
13.3.1 Regression
13.3.2Least Square Method
13.3.3 Regression Lines and Regression Coefficients
13.4 Standard Error of Estimate
13.5 Explained and Unexplained Variation
13.6 Coefficient of determination
13.7 Summary
13.8 Exercises
13.9 Solution to exercises
13.10 Reference for further reading
13.0 OBJECTIVES
This Chapter would make you understand about the following concepts
about correlation and regression:
Scatter Diagram
Linear Correlation
Least square regression
Explained and unexplained variation
Regression Lines
Correlation of Time series and Attributes
13.1 INTRODUCTION
By now we know how to find averages and dispersion of a distribution.
Involving one variable. These measures give a complete idea about the
structure of a distribution.
Sometimes it is necessary to know the relationship between two
variables. For instance , a businessman would like to know the effect of
production on price. If the two are related, he would like to know the munotes.in
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207 nature of the relationship and to use that knowledge to his benefit. If we
get heights and weights of a group of students we can see tha t in general
taller students will be heavier, though there will be some exceptions. As a
man’s income increases he spends more. That is, there will be some
relation between income and expenditure of any person. For such data we
would like to find answer fo r the following questions:
i) Are the two variables related?
ii) If they are related, how?
iii) To what extent they are related?
Correlation helps to find out answers to these questions.
If two variables vary together in the same direction or in opposite
directions, they are said to be correlated. That is if as X increases, Y
increases consistently, we say that X and Y are positively related i.e the
variables are directly related with each other. In this case the values of X
and Y for a particular individual have rou ghly the same relative position
among their respective distributions, i.e if X is far above mean of X, then
corresponding Y will be above mean of Y. If X is near to mean of X then
Y will be near to mean of Y. We know that if the weight of the parents is
above average then the son/daughter also will be having more weight.
There are some variables which are negatively correlated where, as X
increases, Y decreases and as X decreases, Y increases i.e the variables are
inversely proportional to each other , e.g Price increases as the supply
decreases, that is, as the commodity becomes scare, the price increases.
If the change in one variable is proportional to the change in the other, the
two variables are said to perfectly correlated.
If the two variables are not related to each other, we say the two variables
have zero correlation , e.g length of the hair of an individual and the I.Q
level of the same individual.
13.2 CORRELATION
The various method of finding whether two given variables are realted or
not are:
Scatter diagram
Coefficient of Correlation
13.2.1 Scatter Diagram :
Scatter Diagram can be obtained by simply plotting the points on a graph
where the two variables, say x a nd y are taken along x – axis and y – axis
respectively.
Eg: A manager of a firm may want to appoint salesman for promoting his
sales. When he gets a number of applications, he conducts an aptitude
test and selects on the basis of their results. But afte r employing them, munotes.in
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208 he wants to know whether there is any relation between the actual
sales and the marks in the aptitude test.
The data for seven salesman are as follows: Salesman 1 2 3 4 5 6 7 Aptitude test score 47 49 60 55 59 70 83 Actual sales in ‘000 in Rs. 70 69 80 75 75 87 90
Solution:
Here the test scores are plotted on the X –axis and actual sales on the
Y –axis.
Here we can see that as X increases, Y also increases therefore the two
variables are positively related. The businessman knows by this that it is
useful to select salesman on the basis of the aptitude test conducted in this
case.
Various possible patterns are shown here below
6065707580859095
4555657585Actual Sales(in rs. '000)
Test Scores SCATTER DIAGRAM
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209 Types of Correlation:
Perfect Positive Correlation: When all plotted values lie exactly on a
straight line and the line runs from lower left to upper right corner it is
called perfect positive correlation. Here r = 1
Perfect Negative Correlation: When all plotted values lie exactly on a
straigh t line and the line runs from upper left to lower right corner it is
called perfect negative correlation . Here r = -1
Positive Correlation: If the value of two variables deviate in the same
direction. It is known as Positive correlation or direct correlation.
Negative Correlation: If the value of two variables deviate in the opposite
direction. It is also known as the Negative correlation or inverse
correlation.
No Correlation: The variables are independent i.e there is no relation
between the two variables i.e r = 0.
13.2.2 Coefficient of Correlation :
The things discussed in the previous section gives us the direction of
existence of correlation. But we also require to find exact numerical
measurement for the degree or extent of correlation. It is useful to have a
numerical measure, which is independent of units of the original data, so
that the two variables can be compared. For this we calculate the
coefficient of correlation. It is denoted by 𝑟.
Definition: The coefficient of Correlation denoted by 𝑟 and name after
Karl Pearson is defined as
𝑟=𝑟௫,௬= ∑൫(𝑥− 𝑥̅)(𝑦− 𝑦ത)൯
𝑁𝜎௫𝜎௬
𝜎௫= ඨቆ∑(𝑥−𝑥̅)ଶ
𝑁ቇ ,𝜎௬= ඨ൬∑𝑦−𝑦ത
𝑁൰
where there are N pairs.
This is also called Product Moment Coefficient of Correlation.
Covariance of x and y is defined as 𝑐𝑜𝑣(𝑥,𝑦)=∑(௫ି ௫̅)(௬ି௬ത)
ே
∴𝑟=𝑐𝑜𝑣(𝑥,𝑦)
𝜎௫𝜎௬
The formula of r can be simplified as 𝑟=∑௫௬ ି ே௫̅௬ത
ට∑௫మ – ே௫̅మට∑௬మ – ே௬തమ
=∑𝑥𝑦−∑௫∑௬
ே
ට∑𝑥ଶ−(∑௫)మ
ேට∑𝑦ଶ−(∑௬)మ
ே
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210 Properties of Correlation:
1) −1≤𝑟≤1
2) 𝑟=1,𝑝𝑒𝑟𝑓𝑒𝑐𝑡+𝑣𝑒𝑐𝑜𝑟𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛
3) 𝑟= −1,𝑝𝑒𝑟𝑓𝑒𝑐𝑡−𝑣𝑒𝑐𝑜𝑟𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛
4) If 0 < r < 1, the correlation is positive
5) If -1 < r < 0, the correlation is negative
6) r = 0, no correlation
7) r is a pure number and is not affected by change of origin and scale
in magnitude.
𝑖.𝑒𝑖𝑓𝑢=𝑥−𝑎
𝑏 ,𝑣=𝑦−𝑐
𝑑𝑡ℎ𝑒𝑛𝑟௫௬ =𝑏𝑑
|𝑏||𝑑|𝑟௨௩
a) If b and d are of same sign then 𝑟௫௬=𝑟௨௩
b) If b and d are of opposite signs then 𝑟௫௬ = −𝑟௨௩
8) If y = ax + b, 𝑟௫௬ =1 𝑖𝑓𝑎>0 𝑎𝑛𝑑𝑟௫௬= −1,𝑖𝑓𝑎<0
Eg 1 : 𝑟௫௬=0.6, find 𝑟௨௩if 2u – 3x + 4 = 0, 4v – 16y + 11 = 0
solution:2𝑢−3𝑥+4=0⇒𝑢=ଷ௫ିସ
ଶ=௫ – ర
య
మ
య⇒𝑏=ଶ
ଷ
4𝑣−16𝑦 +11=0⇒𝑣=16𝑦−11
4=𝑦−ଵଵ
ଵ
ସ
ଵ⇒𝑑=4
16
𝑏 𝑎𝑛𝑑 𝑑 𝑎𝑟𝑒 𝑜𝑓 𝑠𝑎𝑚𝑒 𝑠𝑖𝑔𝑛𝑠∴𝑟௨௩=𝑟௫௬ =0.6
Eg 2: 𝑟௫௬=0.6, find 𝑟௨௩if 2u + 3x + 4 = 0, 4v – 16y + 11 = 0
solution:2𝑢+3𝑥+4=0⇒𝑢=−௫ ା ర
య
మ
య⇒𝑏= −ଶ
ଷ,
4𝑣−16𝑦+11=0⇒𝑣=𝑦−ଵଵ
ଵ
ସ
ଵ⇒𝑑=4
16
𝑏 𝑎𝑛𝑑 𝑑 𝑎𝑟𝑒 𝑜𝑓 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑔𝑛𝑠∴𝑟௨௩= −𝑟௫௬ = −0.6
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211 Eg 3 : If 2x + y = 3 what is the value of r xy.
solution: 2𝑥+𝑦=3⇒𝑦= −2𝑥+3 ⇒𝑎= −2<0⇒𝑟௫௬= −1
Eg4: Calculate the karlpearson’s coefficient of correlation given
𝑐𝑜𝑣(𝑥,𝑦)= −15,𝜎௫=5,𝜎௬ = 4.
solution: 𝑟=௩(௫,௬)
ఙೣఙ = −ଵହ
ହ×ସ= −ଵହ
ଶ= −ଷ
ସ=−0.75
Eg 5: Calculate the coefficient of correlation for the following: x -2 -1 0 1 2 y 4 1 0 1 4 solution: 𝑁=5,∑𝑥=0,∑𝑦=10,∑(𝑥−𝑥̅)(𝑦−𝑦ത)=0,𝜎௫=ටଵ
ହ=
√2,𝜎௬=ටଵସ
ହ= √2.8
𝑟=∑(𝑥− 𝑥̅)(𝑦− 𝑦ത)
𝑁𝜎௫𝜎௬=0
5⋅√2⋅√2.8=0
Remark: If there is no correlation between the two variables, r = 0 but the
converse is not true. In this example the values are related by the equation
𝑦=𝑥ଶbut the observed value of coefficient of correlation is zero.
Eg 6: Calculate the coefficient of correlation for the following: x 1 2 3 4 5 6 7 8 9 10 y 2 4 9 7 10 5 14 16 2 20
solution: N = 10 Total x 1 2 3 4 5 6 7 8 9 10 55 y 2 4 9 7 10 5 14 16 2 20 89 xy 2 8 27 28 50 30 98 128 18 200 589 𝑥ଶ 1 4 9 16 25 36 49 64 81 100 385 𝑦ଶ 4 16 81 49 100 25 196 256 4 400 1131
𝑟 =∑𝑥𝑦 − ∑௫∑௬
ට∑𝑥ଶ − (∑௫)మ
ට∑𝑦ଶ − (∑௬)మ
= ቀ589−ହହ×଼ଽ
ଵቁ
ට385−(ହହ)మ
ଵට1131−(଼ଽ)మ
ଵ
𝑟=(589−489.5)
√82.5×√338.9= 99.5
(9.08)(18.41)=0.6
Eg 7: Find the coefficient of correlation for the following:
x 60 50 45 47 53 70 75 57 73 48 y 30 29 29 28 29 35 40 32 35 28 munotes.in
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212 solution: N = 10 , we take a = 60, c = 35(assumed mean) Total x 60 50 45 47 53 70 75 57 73 48 y 30 29 29 28 29 35 40 32 35 28 u = x – 60 0 -10 -15 -13 -7 10 15 -3 13 -12 -22 v = y – 35 -5 -6 -6 -7 -6 0 5 -3 0 -7 -35 uv 0 60 90 91 42 0 75 9 0 84 451 u2 0 100 125 169 49 100 225 9 169 144 1190 v2 25 36 36 49 36 0 25 9 0 49 265
here b and d are same i.e 1 ∴𝑟௫௬=𝑟௨௩=∑௨௩ ି ∑ೠ∑ೡ
ಿ
ට∑௨మ ି (∑ೠ)మ
ಿට∑௩మ ି ∑ೡమ
ಿ
=451 − (ିଶଶ)(ିଷହ)
ଵ
ට1190 − (ିଶଶ)మ
ଵට265 − (ିଷହ)మ
ଵ = 374
√1141.6√142.5 =0.927
Eg 8: Calculate the coefficient of correlation for the following: x 5 10 10 15 15 20 25 30 y 15 17 17 19 21 21 19 17
solution: N = 8 , we take a = 20, c = 19, b = 5, d = 2 Total x 5 10 10 15 15 20 25 30 y 15 17 17 19 21 21 19 17 𝑢=𝑥−205 -3 -2 -2 -1 -1 0 1 2 -6 𝑣=𝑦−192 -2 -1 -1 0 1 1 0 -1 -3 uv 6 2 2 0 -1 0 0 -2 7 u2 9 4 4 1 1 0 1 4 24 v2 4 1 1 0 1 1 0 1 9
Since b and d are of same signs, 𝑟௫௬=𝑟௨௩= ∑௨௩ ି ∑ೠ∑ೡ
ಿ
ට∑௨మ ି (∑ೠ)మ
ಿට∑௩మ ି ∑ೡమ
ಿ
𝑟௫௬=𝑟௨௩=7 − (ି)(ିଷ)
଼
ට24 − (ି)మ
଼ට9 − (ିଷ)మ
଼ = 7 − 2.25
√19.5√7.87 =0.38
Eg 9: From the data given below find the number of items n, r =
0.5,∑(𝑥−𝑥̅)(𝑦−𝑦ത) =120,𝜎௬=8,∑(𝑥−𝑥̅)ଶ =90.
solution: 𝑟=∑(௫ି௫̅)(௬ି௬ത)
ఙೣఙ ,𝜎௫ =ටቀ∑(௫ି௫̅)మ
ቁ ,𝜎௬=ට∑(௬ି௬ത)మ
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213 0.5 =120
𝑛×ටቀଽ
ቁ×8⇒0.5 =120
ට𝑛ଶ×ଽ
×8
0.5 =120
√90𝑛×8⇒√90𝑛=120
0.5×8=30
𝑠𝑞𝑢𝑎𝑟𝑖𝑛𝑔 𝑤𝑒 𝑔𝑒𝑡,90𝑛=900⇒𝑛=10
Eg 10: Calculate the coefficient of correlation between x and y. x y no.of observations 15 15 arithmetic mean 25 18 standard deviation 3.01 3.03 sum of squares of the deviation from arithmetic mean 136 138 ∑(𝑥−𝑥̅)(𝑦−𝑦ത)=122
solution: 𝑟=∑(௫ି௫̅)(௬ି௬ത)
ఙೣఙ ,𝜎௫ =ටቀ∑(௫ି௫̅)మ
ቁ ,𝜎௬=ට∑(௬ି௬ത)మ
𝑟=122
15×ටଵଷ
ଵହ×ටଵଷ଼
ଵହ=122
15×3.01×3.03=0.89
Eg 11: n = 20, r = 0.3, 𝑥̅ =15,𝑦ത=20,𝜎௫=4,𝜎௬ =5. One pair (27, 30)
was wrongly taken as (17, 35). Find corrected value of r.
solution: 𝑥 ഥ=15,∑௫
=15⇒∑𝑥=15×20 =300
corrected value 𝑥̅=ଷିଵାଶ
ଶ =ଷଵ
ଶ=15.5
𝑦ത=20,∑𝑦
𝑛=20⇒∑𝑦=20×20=400
corrected value 𝑦ത=ସିଷହାଷ
ଶ=ଷଽହ
ଶ=19.75
𝜎௫=4⇒(𝜎௫)ଶ=16⇒∑𝑥ଶ
𝑛−(𝑥̅)ଶ=16
∑𝑥ଶ=(16+ 225)×20=4820
corrected value of ∑𝑥ଶ =4820+27×27 −17×17=5260
corrected value of 𝜎௫ =ටହଶ
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214 𝜎௬=5⇒൫𝜎௬൯ଶ=25⇒∑𝑦ଶ
𝑛−(𝑦ത)ଶ=25
∑𝑦ଶ=(25+400)×20=8500
corrected value of ∑𝑦ଶ=8500+30×30−35×35=8175
corrected value of 𝜎௬=ට଼ହ
ଶ−(19.75)ଶ=5.91
𝑟=∑(𝑥−𝑥̅)(𝑦−𝑦ത)
𝑛𝜎௫𝜎௬⇒0.3=∑(𝑥−𝑥̅)(𝑦−𝑦ത)
20×4×5
⇒∑(𝑥−𝑥̅)(𝑦−𝑦ത)=120
corrected value of
∑(𝑥−𝑥̅)(𝑦−𝑦ത)
=120+(27− 17.5)(30−19.75)
−(17−15)(35−20)
=120+9.5×10.25−(2)×(15)=187.375=187.38
𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑟 =187.38
20 × 4.77×5.91=0.33
13.2.3 Coefficient of Rank Correlation :
In certain types of characteristics it is not possible to get numerical
measurements, but we can rank the indivi duals in order according to our
own judgment, e.g. beauty, smartness. If two persons rank a given group
of individuals and we have to find how far the two judges agree with each
other, the technique of rank correlation can be used. In some cases though
actual measurements are available we may be interested in only the ranks,
that is the relative position of an individual in the group. Here also the
rank correlation is used.
The f ormula for Spearman’s rank correlation coefficient is
𝑅=1−∑ௗమ
ே(ேమ ି ଵ) where d = difference between the ranks of the same
individual, N = number of individuals.
Remark: R follows the same property as r
Eg 1: The ranks according to judges in a beauty contest are R1 1 2 3 4 5 6 R2 4 1 2 3 6 5 Find the coefficient of rank correlation.
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215 Solution: N = 6 Total R1(d1) 1 2 3 4 5 6 R2(d2) 4 1 2 3 6 5 d = d1 - d2 -3 1 1 1 -1 1 d2 9 1 1 1 1 1 14
𝑅=1−6∑𝑑ଶ
𝑁(𝑁ଶ−1) = 1− 6×14
6(36−1 ) = 1−14
35 =21
35=3
5
= 0.6.
Eg 2: Find spearman’s rank correlation coefficient between cost and sales
for the following cost 39 65 62 90 82 75 25 98 36 78 sales 47 53 58 86 62 68 60 91 51 84
Solution: N = 10 Total cost (X) 39 65 62 90 82 75 25 98 36 78 sales (Y) 47 53 58 86 62 68 60 91 51 84 d1 8 6 7 2 3 5 10 1 9 4 d2 10 8 7 2 5 4 6 1 9 3 d = d1 – d2 -2 -2 0 0 -2 1 4 0 0 1 d2 4 4 0 0 4 1 16 0 0 1 30
𝑅=1−6∑𝑑ଶ
𝑁(𝑁ଶ−1)= 1− 6×30
10(100−1) = 1−18
99=81
99=9
11
= 0.8182
Coefficient of Rank correlation when ranks are repeated
In the above example, the ranks were different for all the but in some
cases, two or more items may have the same numerical values and ranks
should be the same for these values. Suppose we give ranks 1,2,3 and the
next two values have to given same rank. In this case next two ranks are 4
and 5.T hese are to be distributed equally. Therefore both the individuals
will get the rank ସ ା ହ
ଶand the next one will get the rank as 6.
When there are groups getting the same rank, there is some adjustment in
the formula also. If 𝑚ଵ,𝑚ଶ,𝑚ଷ,… denotes the number of times the same
rank appear, the coefficient of Rank Correlation will be
𝑅=1−൛∑ௗమାிൟ
ே(ேమିଵ), where =ଵ
ଵଶ[(𝑚ଵଷ−𝑚ଵ)+(𝑚ଶଷ−𝑚ଶ)+
(𝑚ଷଶ−𝑚ଷ)+ ⋯⋯] , CF – Correction Factor
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216 Eg 1: Marks of 10 students in two I.Q tests carried out in two successive
months. Find the coefficient of correlation. marks test 1 75 60 60 73 55 57 53 72 65 69 marks test 2 65 65 64 70 58 60 58 68 63 65
Solution: N = 10 Total marks test 1 75 60 60 73 55 57 53 72 65 69 marks test 2 65 65 64 70 58 60 58 68 63 65 d1 1 6.5
{6} 6.5
{7} 2 9 8 10 3 5 4 d2 4 {3} 4
{4} 6 1 9.5
{9} 8 9.5 {10} 2 7 4 {5} d = d1 – d2 -3 2.5 0.5 1 -
0.5 0 0.5 1 2 0 d2 9 6.2
5 0.2
5 1 0.2
5 0 0.25 1 4 0 22
m1 = 2, {no. of times marks 60 is repeated in test 1}
m2 = 3, {no. of times marks 65 is repeated in test 2}
m3 = 2, {no. of times marks 58 is repeated in test 2}
𝐶𝐹=1
12{(𝑚ଵଷ−𝑚ଵ)+(𝑚ଶଷ−𝑚ଶ)+(𝑚ଷଷ−𝑚ଷ)} =1
12{6+24+6}
=1
12{36}=3
𝑅=1−6{∑𝑑ଶ+𝐶𝐹}
𝑛(𝑛ଶ−1)= 1−6{22+3}
10(99)= 1− 6×25
990= 1−15
99
=84
99= 0.8485
Eg 2: The coefficient of Rank correlation for certain data is found to 0.6.
If the sum of the squares of the differences is given to be 66, find the
number of observations.
Solution: Given R = 0.6, ∑𝑑ଶ=66,To find number of observations i.e to
find N
𝑅=1−6∑𝑑ଶ
𝑁(𝑁ଶ−1) ⇒ 0.6= 1− 6×66
𝑁(𝑁ଶ−1)
⇒6× 66
𝑁(𝑁ଶ−1) = 1− 0.6 munotes.in
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217 ⇒6×66
𝑁(𝑁ଶ−1) =0.4 ⇒𝑁(𝑁ଶ−1) =6×66
0.4
⇒(𝑁−1)𝑁(𝑁+1)=990
⇒(𝑁−1)𝑁(𝑁+1)=9×10×11 ⇒ 𝑁=10
13.3 THE LEAST SQUARE REGRESSION LINES
When we know that two given variables are correlated we try to establish
some relation between the two so that we can estimate the value of one of
the variables given the value of other e.g If we know that there is positive
correlation between the heights and weights of a group of individuals, we
can find an equation between the height and the weigh t. We can estimate
the weight of an individual belonging to the same population given his
height.
Correlation coefficient only determines whether the variables are related
and if so, how strong is the relationship. But it is not useful for prediction.
The equations used for prediction or estimation are known as regression
equations. With the help of regression analysis we establish a model which
expresses the functional relationship between the two variables. These are
also known as estimation equations.
13.3.1 Regression Lines :
We can find two straight lines which will be useful for estimating Y when
X is given. It is known as regression of Yon X. Here X is considered as
independent variable. For estimating X when Y is given. It is known as
regression of X on Y. Here Y is considered as independent variable. We fit
a straight line for the set of points given in the bivariate data.
13.3.2 Least Square Method :
We know that a degree one equation represents a straight line. Therefore
we take the equation of t he regression line of Y on X as 𝑌=𝑎+𝑏𝑋where
a and b are constants. The constant ‘a’ determines the point where the line
cuts the Y – axis and constant ‘b’ determines the slope of the line. The
method of least squares is used to determine the constants and we get a and
b by solving the following two equations simultaneously which are known
as normal equations.
∑𝑌=𝑁𝑎+𝑏 ∑𝑋 , ∑𝑋𝑌=𝑎∑𝑋+𝑏 ∑𝑋ଶ
Similarly, we take the regression of X on Y as 𝑋=𝑐+𝑑𝑌where c and d
are constants which can be determined by solving the following normal
equations.
∑𝑋=𝑁𝑐+𝑑∑𝑌 , ∑𝑋𝑌=𝑐∑𝑌 +𝑑 ∑𝑌ଶ
munotes.in
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218 Eg 1: Find the regression of Y on X and X on Y for the following data.
Also estimate Y when X = 7 and estimate X when Y = 16 X 1 2 3 4 5 Y 10 12 15 14 15
Solution: N = 5 Total X 1 2 3 4 5 15 Y 10 12 15 14 15 66 XY 10 24 45 56 75 210 X2 1 4 9 16 25 55 Y2 100 144 225 196 225 890 Normal Equation for regression of Y on X are given by
∑𝑦=𝑁𝑎+𝑏∑𝑋 𝑖.𝑒 66=5𝑎+15𝑏
∑𝑋𝑌=𝑎∑𝑋+𝑏∑𝑋ଶ 𝑖.𝑒 210=15𝑎+55𝑏
Solving the two equations simultaneously we get a = 9.6 and b = 1.2
∴ The regression line of Y on X is given by 𝑌=9.6+1.2𝑋 --- (1)
For estimating value of Y when X = 7, we put X = 7 in eq (1)
∴ The estimate will be Y = 9.6 + 1.2(7) = 18
Normal equations fo r regression of X on Y are given by
∑𝑋=𝑁𝑐+𝑑∑𝑌 𝑖 .𝑒 15=5𝑐+66𝑑
∑𝑋𝑌=𝑐∑𝑌 +𝑑∑𝑌ଶ 𝑖.𝑒 210=66𝑐+ 890𝑑
Solving the two equations simultaneously we get c = -5.4255 , d = 0.6383
∴ The regression line of X on Y is given by 𝑋= −5.4255 +0.6383𝑌 ---
(2)
For estimating value of X when Y = 16, we put Y = 16 in eq (2)
∴ The estimate will be X = -5.4255 + 0.6383(16) = 4.7873
13.3.3 Regression Lines and Regression Coefficients :
The regression of Y on X is given by
𝑌−𝑌ത =𝑏(𝑋−𝑋ത),𝑏=𝑟𝜎
𝜎
The regression of X on Y is given by
𝑋−𝑋ത =𝑏(𝑌−𝑌ത),𝑏 =𝑟𝜎
𝜎
where 𝑏 𝑎𝑛𝑑 𝑏 are known as regression coefficients.
Properties:
1) 𝑏=𝑟ఙೊ
ఙ ,𝑏=𝑟ఙ
ఙೊ
𝑟ଶ =𝑏×𝑏 ⇒𝑟 = ±ඥ𝑏×𝑏
𝑟>0 𝑖𝑓 𝑏&𝑏>0
𝑟<0 𝑖𝑓 𝑏&𝑏<0 munotes.in
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219 2) 𝑖𝑓 𝑢=𝑋−𝑎,𝑣=𝑌−𝑏 𝑡ℎ𝑒𝑛 𝑏=𝑏௨௩&𝑏 =𝑏௩௨
3) 𝑏 =𝑟ఙೊ
ఙ =௩(,)
ఙమ =∑ ି ∑∑ೊ
ಿ
∑మ ି (∑)మ
ಿ
4) 𝑏௫௬ =𝑟ఙ
ఙೊ =௩(,)
ఙೊమ =ቀ∑ ି ∑∑ೊ
ಿቁ
∑మ ି (∑ೊ)మ
ಿ
Eg: 1) The following data are given about the expenditure on clothes and
expenditure on entertainment. Average expenditure on clothes Rs. 300,
average expenditure on entertainment Rs. 100, S.D of expenditure on
clothes Rs. 20, S.D of expenditure on entertainment Rs. 15, coefficient of
correlation 0.78. Find the two regression equations.
Solution: Let the expenditure on clothes be denoted by x and the
expenditure on entertainment be denoted by y.
Given 𝑥̅=300,𝑦ത=100,𝜎௫=20,𝜎௬=15,𝑟 =0.78
∴𝑏௫௬ =𝑟𝜎௫
𝜎௬ =0.78×20
15=0.78×4
3 =1.04
& 𝑏௬௫=𝑟𝜎௬
𝜎௫ =0.78×15
20=0.78×3
4 =0.585
𝑅𝑒𝑔𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑦 𝑜𝑛 𝑥 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦
(𝑦−𝑦ത)=𝑏௬௫(𝑥−𝑥̅)⇒𝑦−100=0.585(𝑥−300)
−0.585𝑥+𝑦=100−175.5⇒𝑦=0.585𝑥−75.5
𝑅𝑒𝑔𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑥 𝑜𝑛 𝑦 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦
(𝑥−𝑥̅)=𝑏௫௬(𝑦−𝑦ത)⇒𝑥−300=1.04(𝑦−100)
𝑥=1.04𝑦+300−104⇒𝑥=1.04𝑦+196
Eg 2: If ∑𝑥=37,∑𝑦=71,∑𝑥𝑦=563,∑𝑥ଶ= 297,∑𝑦ଶ =1079,𝑛=
5. Find the two regression equations .
Solution: 𝑥̅=ଷ
ହ =7.4 ,𝑦ത=ଵ
ହ=14.2,𝑏௬௫= ∑௫௬ ି ∑ೣ∑
∑௫మ ି (∑ೣ)మ
=ቀହଷିయళ×ళభ
ఱቁ
ଶଽି(యళ)మ
ఱ
𝑏௬௫=563×5−37×71
297×5−37×37 =2815−2627
1485−1369=188
116=1.62
𝑏௫௬=∑𝑥𝑦−∑௫∑௬
∑𝑦ଶ−(∑௬)మ
= ቀ563−ଷ×ଵ
ହቁ
1079−(ଵ)మ
ହ
𝑏௫௬=563×5−37× 71
1079×5−71×71 =188
354=0.53 munotes.in
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220 𝑟𝑒𝑔𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑦 𝑜𝑛 𝑥
𝑦−𝑦ത=𝑏௬௫(𝑥 −𝑥̅)⇒𝑦−14.2=1.62×(𝑥−7.4)
𝑦=1.62𝑥+2.21
𝑟𝑒𝑔𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑥 𝑜𝑛 𝑦
𝑥−𝑥̅=𝑏௫௬(𝑦−𝑦ത)⇒𝑥−7.4=0.53×(𝑦−14.2)
𝑥=0.53𝑦−0.126
Eg 3: Find the two regression lines for the following data Price in Rs. 100 120 110 110 160 150 Demand(in units) 40 38 43 45 37 23 Also estimate the demand when price is 130 and the price when
demand is 30 units.
Solution: N=6 Total Price in Rs. (X) 100 120 110 110 160 150 750 Demand(in units) (Y) 40 38 43 45 37 23 226 XY 4000 4560 4730 4950 5920 3450 27610 X2 10000 14400 12100 12100 25600 22500 96700 Y2 1600 1444 1849 2025 1369 529 8816
𝑥̅=∑𝑥
𝑛=750
6=125,𝑦ത=∑𝑦
𝑛=226
6=37.67
𝑏௬௫ =∑𝑥𝑦−∑௫∑௬
∑𝑥ଶ−(∑௫)మ
=ቀ27610−ହ×ଶଶ
ቁ
96700−750×ହ
= −3840
17700= −0.22
𝑏௫௬=ቀ∑𝑥𝑦−∑௫∑௬
ቁ
∑𝑦ଶ−(∑௬)మ
= −3840
1820= −2.11
𝑟𝑒𝑔𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑦 𝑜𝑛 𝑥
𝑦−𝑦ത=𝑏௬௫(𝑥−𝑥̅)⇒𝑦−37.67= −0.22(𝑥−125)
𝑦= −0.22𝑥+65.17 −−− −(1)
𝑟𝑒𝑔𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑥 𝑜𝑛 𝑦
𝑥−𝑥̅=𝑏௫௬(𝑦−𝑦ത)⇒𝑥−125= −2.11(𝑦−37.67)
𝑥= −2.11𝑦+204.48 −−−−−(2) munotes.in
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221 𝑑𝑒𝑚𝑎𝑛𝑑 =?𝑤ℎ𝑒𝑛 𝑝𝑟𝑖𝑐𝑒=130,𝑤𝑒 𝑝𝑢𝑡 𝑥=130 𝑖𝑛 (1)
𝑦= −0.22×130+65.17 =36.57
𝑝𝑟𝑖𝑐𝑒= ?𝑤ℎ𝑒𝑛 𝑑𝑒𝑚𝑎𝑛𝑑 =30,𝑤𝑒 𝑝𝑢𝑡 𝑦=30 𝑖𝑛 (2)
𝑥= −2.11× 30 +204.48 =141.18
Eg 4: The two regression lines are given by 𝑥+2𝑦=5,2𝑥+3𝑦−8=
0,(𝜎௫)ଶ=12.Find the values of 𝑥ഥ,𝑦ത ,൫𝜎௬൯ଶ.
Solution:We solve the two regression equations simultaneously to find
𝑥̅&𝑦ത
𝑥+2𝑦=5 −−(1)
2𝑥+3𝑦−8=0 − −(2)
𝑀𝑢𝑙 𝑒𝑞 (1) 𝑏𝑦 2 𝑎𝑛𝑑 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑓𝑟𝑜𝑚 𝑒𝑞 (2) 𝑔𝑖𝑣𝑒𝑠 𝑦ത=2
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑦=2 𝑖𝑛 𝑒𝑞 (1) 𝑔𝑖𝑣𝑒𝑠 𝑥̅=1 ,
𝑥+ 2𝑦=5⇒𝑥= −2𝑦+5,𝑏௫௬ = −2
2𝑥+3𝑦−8=0⇒𝑦=൬−2
3൰𝑥+4,𝑏௬௫ = −2
3
𝑏௫௬×𝑏௬௫ =4
3>1 𝑛𝑜𝑡 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒
𝑟𝑒𝑔𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑜𝑓 𝑦 𝑜𝑛 𝑥 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 𝑥+2𝑦=5 𝑎𝑛𝑑 𝑥 𝑜𝑛 𝑦 𝑖𝑠
𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 2𝑥+3𝑦−8=0
𝑥+2𝑦=5 ⇒𝑦= −𝑥
2+5
2 ,𝑏௬௫ = −1
2
2𝑥+3𝑦=8⇒𝑥=൬−3
2൰𝑦+4 ,𝑏௫௬ = −3
2
𝑟ଶ=𝑏௬௫×𝑏௫௬⇒𝑟ଶ=3
4=0.75⇒ 𝑟= −ඨ3
4 = −0.87
𝑟ଶ=൫𝑏௬௫൯ଶ×൫𝜎௬൯ଶ
(𝜎௫)ଶ ⇒ 𝑟ଶ =1
4×൫𝜎௬൯ଶ
12⇒3
4=1
4×൫σ୷൯ଶ
12
⇒൫𝜎௬൯ଶ =0.75×48=36
13.4 STANDARD ERROR OF ESTIMATE
Ley y 1 be the value of y for given value of x, a measure of the scatter about
the regression line y on x is given by
𝑆௬,௫ =ඨ∑(𝑦−𝑦ଵ)ଶ
𝑛 munotes.in
Page 222
222 which is called the standard error of estimate of y on x.
Standard error of estimate of x on y
𝑆௫,௬ =ඨ∑(𝑥−𝑥ଵ)ଶ
𝑛
In general, 𝑆௬,௫≠𝑆௫,௬
Eg: 1) If the regression line of y on x is 𝑦=𝑎+𝑎ଵ𝑥. Show that
𝑆௬,௫ଶ =∑௬మିబ∑௬ିభ∑௫௬
.
solution:𝑦ଵ=𝑎+𝑎ଵ𝑥
𝑆௬,௫ଶ=∑(𝑦−𝑦ଵ)ଶ
𝑛 =∑(𝑦−𝑦ଵ)ଶ
𝑛=∑(𝑦ଶ−2𝑦𝑦ଵ+𝑦ଵଶ)
𝑛
=∑(𝑦ଶ−2𝑦(𝑎+𝑎ଵ𝑥)+(𝑎+𝑎ଵ𝑥)ଶ)
𝑛
=∑(𝑦ଶ−2𝑎𝑦−2𝑎ଵ𝑥𝑦+𝑎ଶ+2𝑎𝑎ଵ𝑥+𝑎ଵଶ𝑥ଶ)
𝑛
= ∑(𝑦ଶ−𝑎𝑦−𝑎ଵ𝑥𝑦−𝑎𝑦−𝑎ଵ𝑥𝑦+𝑎ଶ+𝑎𝑎ଵ𝑥+𝑎ଵଶ𝑥ଶ+𝑎𝑎ଵ𝑥)
𝑛
={∑𝑦(𝑦−𝑎−𝑎ଵ𝑥)−𝑎∑(𝑦−𝑎ଵ𝑥−𝑎)−𝑎ଵ∑𝑥(𝑦−𝑎ଵ𝑥−𝑎)}
𝑛
𝑦=𝑎+𝑎ଵ𝑥 𝑏𝑦 𝑙𝑒𝑎𝑠𝑡 𝑠𝑞𝑢𝑎𝑟𝑒 𝑚𝑒𝑡ℎ𝑜𝑑 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑎 𝑎𝑛𝑑 𝑎ଵ,
𝑤𝑒 𝑠𝑜𝑙𝑣𝑒 ∑𝑦=𝑎𝑛+𝑎ଵ∑𝑥 𝑎𝑛𝑑 ∑𝑥𝑦 =𝑎∑𝑥+𝑎ଵ∑𝑥ଶ
⇒∑(𝑦−𝑎−𝑎ଵ𝑥)=0 ,∑(𝑥𝑦−𝑎𝑥−𝑎ଵ𝑥ଶ)=0
𝑆௬,௫ଶ=∑𝑦(𝑦−𝑎−𝑎ଵ𝑥)
𝑛=∑𝑦ଶ−𝑎∑𝑦−𝑎ଵ∑𝑥𝑦
𝑛
Eg: 2) If the regression line of y on x is y = 35.82 + 0.476x and values of
x and y are x 65 63 67 64 68 62 70 y 68 66 68 65 69 66 68
Find 𝑆௬,௫ଶ.
solution: y = 35.82 + 0.476x x y y1= 35.82+0.476x (y –y1) (𝑦−𝑦ଵ)ଶ 65 68 66.76 1.24 1.538 63 66 65.808 0.192 0.037 67 68 67.712 0.288 0.083 64 65 66.284 -1.284 1.649 68 69 68.188 0.812 0.659 62 66 65.332 0.668 0.446 munotes.in
Page 223
223 70 68 69.14 -1.14 1.30 total 5.712
𝑆௬,௫ଶ =∑(𝑦−𝑦ଵ)ଶ
𝑛 =5.712
7=0.816
Eg: 3) Show that ∑(𝑦−𝑦ത)ଶ=∑(𝑦−𝑦ଵ)ଶ+∑(𝑦ଵ−𝑦ത)ଶ
solution: 𝑙ℎ𝑠=∑(𝑦−𝑦ത)ଶ =∑(𝑦−𝑦ଵ+𝑦ଵ−𝑦ത)ଶ
=∑{(𝑦−𝑦ଵ)ଶ+2(𝑦−𝑦ଵ)(𝑦ଵ−𝑦ത)+(𝑦ଵ−𝑦ത)ଶ}
=∑(𝑦−𝑦ଵ)ଶ+∑(𝑦ଵ−𝑦ത)ଶ+2∑(𝑦−𝑦ଵ)(𝑦ଵ−𝑦ത)
𝑛𝑜𝑤𝑡𝑜𝑝𝑟𝑜𝑣𝑒 ,∑(𝑦−𝑦ଵ)(𝑦ଵ−𝑦ത)=0
i.e ∑(𝑦−𝑦ଵ)(𝑦ଵ−𝑦ത)=∑(𝑦−𝑎−𝑎ଵ𝑥)(𝑎+𝑎ଵ𝑥−𝑦ത)
=𝑎∑(𝑦−𝑎−𝑎ଵ𝑥)+𝑎ଵ∑𝑥(𝑦−𝑎−𝑎ଵ𝑥)−𝑦ത∑(𝑦−𝑎−𝑎ଵ𝑥)=0
𝑦=𝑎+𝑎ଵ𝑥,𝑢𝑠𝑖𝑛𝑔𝑙𝑒𝑎𝑠𝑡𝑠𝑞𝑢𝑎𝑟𝑒𝑚𝑒𝑡 ℎ𝑜𝑑 we 𝑓𝑖𝑛𝑑𝑎𝑎𝑛𝑑𝑎ଵ,
𝑤𝑒𝑠𝑜𝑙𝑣𝑒 ∑𝑦=𝑎𝑛+𝑎ଵ∑𝑥𝑎𝑛𝑑 ∑𝑥𝑦 =𝑎∑𝑥+𝑎ଵ∑𝑥ଶ
⇒∑(𝑦−𝑎−𝑎ଵ𝑥)=0 ,∑(𝑥𝑦−𝑎𝑥−𝑎ଵ𝑥ଶ)=0
⇒ ∑(𝑦−𝑦ത)ଶ=∑(𝑦−𝑦ଵ)ଶ+∑(𝑦ଵ−𝑦ത)ଶ
13.5 EXPLAINED AND UNEXPLAINED VARIATION In the previous section we had proved,
∑(𝑦−𝑦ത)ଶ=∑(𝑦−𝑦ଵ)ଶ+∑(𝑦ଵ−𝑦ത)ଶ
∑(𝑦−𝑦ଵ)ଶ is called unexplained variation as the deviation (𝑦−𝑦ଵ)
behave randomly and have an unpredictable manner.
∑(𝑦ଵ−𝑦ത)ଶis called explained variation as the deviation (𝑦ଵ−𝑦ത)have a
definite pattern.
∑(𝑦−𝑦ത)ଶis called as total variation.
Eg: 1) For the following calculate total variation, explained variation and
unexplained variation, given the regression of on x as y = 35.82 + 0.476x
x 65 63 67 64 68 62 70 y 68 66 68 65 69 66 68 Solu tion: y = 35.82 + 0.476x, 𝑦ത=∑௬
= 67.143 x y y1= 35.82+0.476x (y –y1) (𝑦−𝑦ଵ)ଶ (𝑦ଵ−𝑦ത)ଶ 65 68 66.76 1.24 1.538 0.147 63 66 65.808 0.192 0.037 1.306 67 68 67.712 0.288 0.083 0.734 64 65 66.284 -1.284 1.649 4.592 68 69 68.188 0.812 0.659 3.448 munotes.in
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224
Unexplained variation= ∑(𝑦−𝑦ଵ)ଶ =5.712
Explained variation = ∑(𝑦ଵ−𝑦ത)ଶ= 12.267
Total variation = ∑(𝑦−𝑦ത)ଶ=∑(𝑦−𝑦ଵ)ଶ+∑(𝑦ଵ−𝑦ത)ଶ=17.979
13.6 COEFFICIENT OF DETERMINATION
The ratio of the explained variation to the total variation is called the
coefficient of determination.
It is given by ∑(௬భ ି ௬ത)మ
∑(௬ ି ௬ത)మ.
If there is zero explained variation (i.e the total variation is same as
unexplained variation) then coefficient of determination will be 0. If there
is zero unexplained variation (i.e the total variation is same as explained
variation) then coefficient of determination will be 1. Since the ratio is
always non negative, we denote it by 𝑟ଶ. 𝑟called as coefficient of
correlation is given by
𝑟= ± ඨ𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑 𝑣𝑎𝑟𝑖𝑎𝑡𝑖𝑜𝑛
𝑡𝑜𝑡𝑎𝑙 𝑣𝑎𝑟𝑖𝑎𝑡𝑖𝑜𝑛 = ± ඨ∑(𝑦ଵ−𝑦ത)ଶ
∑(𝑦− 𝑦ത)ଶ
varies from between -1 and +1. The + and – signs indicates po sitive and
negative linear correlation respectively. Note 𝑟is a dimensionless quantity,
i.e it does not depend on the units.
For the case of linear correlation, the quantity r is the same regardless of
whether X or Y is considered the independent variables. Thus r is a good
measure of the linear correlation between two variables.
Eg : 1) Show that 𝑠௬,௫ଶ =𝑠௬ଶ(1−𝑟ଶ).
Solu tion:𝑟= ±ට௫ௗ ௩௧
௧௧ ௩௧ = ±ට∑(௬భି௬ത)మ
∑(௬ି ௬ത)మ
𝑟ଶ=ቌඨ∑(𝑦ଵ− 𝑦ഥ)ଶ
∑(𝑦 − 𝑦ത)ଶቍଶ
=ቌඨ∑(𝑦 − 𝑦ത)ଶ − ∑(𝑦 − 𝑦ଵ)ଶ
∑(𝑦 − 𝑦ത)ଶቍଶ
⇒𝑟ଶ =ቌඨ1−ቆ∑(𝑦 − 𝑦ଵ)ଶ
∑(𝑦 − 𝑦ത)ଶቇቍଶ
=ቌඨ1−𝑠௬,௫ଶ
𝑆௬ଶቍଶ
⇒ 𝑠௬,௫ଶ =𝑠௬ଶ(1−𝑟ଶ)
62 66 65.332 0.668 0.446 1.306 70 68 69.14 -1.14 1.30 0.734 total 5.712 12.267 munotes.in
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225 Eg: 2) Given the regression of y on x as y = 35.82 + 0.476x. Find the
coefficient of determination and coefficient of correlation for the
following data x 65 63 67 64 68 62 70 y 68 66 68 65 69 66 68
Solu tion:y = 35.82 + 0.476x, 𝑦ത=∑௬
= 67.143
x y y1= 35.82+0.476x (y –y1) (𝑦−𝑦ଵ)ଶ (𝑦ଵ−𝑦ത)ଶ 65 68 66.76 1.24 1.538 0.147 63 66 65.808 0.192 0.037 1.306 67 68 67.712 0.288 0.083 0.734 64 65 66.284 -1.284 1.649 4.592 68 69 68.188 0.812 0.659 3.448 62 66 65.332 0.668 0.446 1.306 70 68 69.14 -1.14 1.30 0.734 total 5.712 12.267
Unexplained variation = ∑(𝑦−𝑦ଵ)ଶ =5.712
Explained variation = ∑(𝑦ଵ−𝑦ത)ଶ= 12.267
Total variation = ∑(𝑦−𝑦ത)ଶ=∑(𝑦−𝑦ଵ)ଶ+∑(𝑦ଵ−𝑦ത)ଶ=17.979
Coefficient of determination = ∑(௬భ ି ௬ത)మ
∑(௬ ି ௬ത)మ=ଵଶ.ଶ
ଵ.ଽଽ =0.682
Coefficient of correlation 𝑟= ±ට௫ௗ ௩௧
௧௧ ௩௧= ±√0.682
13.7 SUMMARY
In this chapter we have learnt about the scatter diagram which helps us to
find the nature and extent of relationship between the variables ,
Coefficient of correlation which is a numerical measure of nature an d
extent of relationship between two given variables whose values lies
between +1 and -1. We h ad also learnt about the coefficient of Rank
correlation which is used in cases where it is not possible to get numerical
measurements, but we can rank the individuals in order according to our
judgment. This chapter also deals with the least square regres sion lines of
y on x as well as x on y. This chapter also explains about the standard
error of estimate, Explained and unexplained variation and correlation of
time series.
13.8 EXERCISES 1) Following table gives the us the marks obtained by 6 students in mid
term exam and final semester exam. Find the coefficient of
correlation. munotes.in
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226 Mid term Exam 12 14 23 18 10 19 Final semester Exam 68 78 85 75 70 74
2) Given covariance = 27, r = 0.6, variance of y = 25. Find variance of
x.
3) Find the coefficient of correlation between the heights of male
students and female students height of male students 65 66 67 68 69 70 71 height of female students 67 68 66 69 72 72 69
4) If𝑟 =0.38,𝑐𝑜𝑣(𝑥,𝑦)=10.2,𝜎௫=16.𝐹𝑖𝑛𝑑 𝜎௬
5) Find r, if 𝑐𝑜𝑣(𝑥,𝑦) =6,𝜎௫ =2.45,𝜎௬=3.41.
6) Find r, given ∑(𝑥−𝑥̅)(𝑦−𝑦ത)=29,∑(𝑥−𝑥̅)ଶ =28,∑(𝑦−
𝑦ത)ଶ =42.
7) Ten competitors in Miss Universe are ranked by three judges in the
following order: J1 1 6 5 10 3 2 4 9 7 8 J2 3 5 8 4 7 10 2 1 6 9 J3 6 4 9 8 1 2 3 10 5 7 Using rank correlation coefficient determi ne which pair of judges
has the nearest approach to common tastes in beauty.
8) The coefficient of rank correlation for certain data is found to be 0.6.
If the sum of the squares of the differences is given to be 66. Find
the n umber of items in the group.
9) The ranks of 16 students in the subject of DBMS and CN are given
as follows. Calculate the rank coefficient of correlation. Rank in DBMS 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Rank in CN 1 10 3 4 5 7 2 6 8 11 15 9 14 12 16 13
10) The coefficient of rank correlation of marks obtained by 9 students
was calculated to be 0.4. It was later discovered that the value of the
difference between ranks for one student was written wrongly as 6
instead of 8. Find the correct value of coeff icient of rank correlation.
11) Find the two regression equations given 𝑥̅=8,𝑦ത=2000,𝜎௫=
2,𝜎௬=80,𝑟=0.7.
Also estimate y when x = 10 and estimate x when y = 2500.
10
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227 12) Find the two regression equations for the following data:
∑(𝑥−𝑥̅)(𝑦−𝑦ത)=135,∑(𝑥−𝑥̅)ଶ=96,∑(𝑦−𝑦ത)ଶ =206,∑𝑥
= 120,
∑𝑦=180,𝑛=5
13) n = 50, regression equation of marks in mathematics (Y) on the
marks in English(X) was 4Y - 5X = 8.Mean marks in English are 40.
The ratio of the two standard deviation 𝜎௬:𝜎௫=5:2. Find the
average marks in mathematics and coefficient of correlation between
the marks in the two subjects.
14) The two regression lines between x and y are given below 2𝑥+
3𝑦= 61,
𝑥+𝑦=25. Find𝑥̅,𝑦ത 𝑎𝑛𝑑 𝑟.
15) Find the regression line of profits on output from the following data
using least square method. Ouput(100 tons) 5 7 9 11 13 15 Profit per unit(Rs.) 1.7 2.4 2.8 3.4 3.7 4.4
16) For the following find total variation given the regression of y on x
as y = 35.82 + 0.476x x 65 63 67 64 68 62 70 66 68 67 69 71 y 68 66 68 65 69 66 68 65 71 67 68 70
13.9 SOLUTION TO EXERCISE Q. No. Solution Q. No. Solution 1 r = 0.81 2 variance of x = 81 3 r = 0.67 4 1.68 5 r = 0.72 6 r = 0.85 7 R12 = -0.212, R13 = 0.636, R23 = -0.297 8 n= 10 9 R = 0.8 10 R = 0.17 11 y = 28x +1776, x = 0.02y – 32, y = 2056, x = 18 12 y = 1.41x + 2.16, x = 0.66y +0.24 13 52, r = 0.5 14 14, 11, -0.82 15 y = 0.26x + 0.5 16 38.917
13.10 REFERENCES
Following books are recommended for further reading:
Statistics by Murray R, Spiegel, Larry J. Stephens, Mcgraw Hill
International Publisher, 4th edition
Fundamental of Mathematical Statistics by S. C. Gupta and V. K.
Kapoor, Sultan Chand and Sons publisher, 11th edition
Mathematical Statistics by J. N. Kapur and H. C. Saxena, S. Chand
publisher, 12th edition
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