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CHAPTER - 4 Sampling Distributions and Tests of Hypothesis
4.1 Concepts of Population and Samples:
4.1.1 Population
A group of individuals under investigation is called a
Population.
In Statistical analysis, the word population refers to general
magnitude and the study of
variation with respect to one or more characteristics relating
to the individuals in a
group is known as population. Thus, the population is nothing
but an aggregate study of
group of objects, living or non-living, at a particular time
period of the given
geographical location.
The population is further classified into two types: namely,
Finite population, which
constitutes a finite or countable finite number of objects, for
example: the no. of
engineering colleges under ANU, the no. of sugar cane industries
in AP state, etc., On the
other hand an infinite population which consists of an infinite
no. of objects, for example:
the no of independent trials of an unbiased coin, etc.,
4.1.2 Sample and Sample Size
A finite subset of observations chosen from a given population
is known as a sample and
the number of observations in the sample is known as sample
size.
Classification of Samples: Samples are classified into two
ways.
Large Sample: If the size of the sample 30 the sample is said to
be large sample.
Small Sample: If the size of the sample < 30 the sample is
said to be small
sample.
4.1.3 Methods of drawing samples
There are two different methods of selecting sample from the
given population.
a) Lottery Method: Under this method, all the members
(observations) of the
given population are numbered serially from 1, 2,.., N on an
uniform slips and
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put into a box. A required number of members (observations) are
selected from
these thoroughly shuffled slips, is known as a sample.
This method is so simple and easy to apply and understand, if
and only if the
population size is small. When the population size is large or
sufficiently large
this method is not applicable. Moreover this method is not
reliable as since
different persons may get different samples from same set of
population and
sample, which we are selected, may not be a true representation
for the
population characteristics.
b) Random Numbers Method: This is most important and widely
applied method
to select a sample from the given population (finite or
infinite). The method
consists of the following steps:
1 Let us consider a population of size N observations, say 1, 2,
3, ...., N.
2 The population size N may be a single digit (0,1, .., 9) or a
two digit
(00,01,, 99) or a three digit number etc.,
3 A required sample of size n is selected from the well defined
random
numbers tables given by Fisher and Yates tables, by neglecting
the starting
and the ending number in the population, by moving in any
direction, i.e.,
horizontally or vertically down words or diagonally.
4 The sample by this method is a true representative of the
population
characteristics.
4.1.4 Random Sample (Finite Population)
Let us consider (x1, x2, .. , xn) be a sample of n observations
from a finite population
and is said to be a random sample, if it satisfies the following
two conditions:
1 All the sample observations are independent
2 All the sample observations are known chance of being selected
in the
sample (i.e., N
n)
4.1.5 Random Sample (Infinite Population)
Let us consider (x1, x2, .. , xn) be a sample of n observations
from an infinite
population and is said to be a random sample, if it satisfies
the following two conditions:
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80
1 All the sample observations are independent
2 All the sample observations are satisfies the probability
function associated
with the given population.
4.2 Notation and Terminology
Population mean
Population variance 2
Population s.d.
Sample mean x
Sample variance s2
Sample s.d. s
4.3 Sampling
This is one of the methods of studying the population and this
is most commonly applied
to the data. Under this method, the population is studied on the
basis of the
information available in the sample, which is selected from the
given population.
4.4 Parameter and Statistic
Any statistical constant of the Population observations is
called the Parameter. In other
words, Parameter is a single constant value, which represents
the entire population
characteristics. Whereas any statistical constant of the Sample
observations is called a
Statistic. In other words, Statistic is a mathematical constant
value, which represents
the sample of observations. Statistic is known as a parameter
estimate.
4.5 Sampling Distribution
A series of statistic values are arranged in the form of a
frequency distribution is known
as a sampling distribution.
More precisely, a sample of size n is selected from a given
population of size N. Then
the number of possible samples will be nNC = k, say. Now
calculate a statistic such
as mean, median, mode, standard deviation, etc., on each of the
k samples, then we will
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get a series of k statistic values. These statistic values are
arrange in the form of a
frequency table is known as sampling distribution.
Suppose if we calculate the arithmetic mean on k samples then we
will get sampling
distribution of sample mean, if we calculate the s. d., on k
samples then we will get
sampling distribution of sample s. d., etc.,
4.6 Standard Error
The standard deviation of sampling distribution is known as
standard error.
4.7 The Sampling Distribution of the Mean ( Known)
If we want to draw a sample of size n (without replacement) from
a finite population of
size N, the total no. of possible samples are C(N,n) =
= k(say)
Example 4.1
A population consists of 4 observations 10, 20, 30, 40 write all
the possible samples of
size 2 and then construct the sampling distribution (i) about
mean (ii) about variance (iii)
also show that the mean of sample means is equal to the
population mean
Solution
Given population observations are 10, 20, 30, 40
A sample of size 2 can be drawn in C(4,2) = 6 ways
Sample distribution about mean and Varience:
S.No. Sample observations Sample mean Sample Variance
1) 10, 20 1 = 15 S1
2 = 25
2) 10, 30 2 = 20 S2
2 = 100
3) 10,40 3 = 25 S3
2 = 225
4) 20, 30 4 = 25 S42 = 25
5) 20, 40 5 = 30 S52 = 100
6) 30, 40 6 = 35 S62 = 35
x
x
x
x
x
x
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82
(iii) Mean of Sample means = (15+20+25+25+30+35)/6 = 25= x
Population mean = 10+20+30+40 = 25
x
4.8 The Central Limit Theorem
If repeated samples of size n are drawn from any infinite
population with mean and
variance 2, and n is large (n 30), the distribution of , the
sample mean, is
approximately normal, with mean (i.e. ( )E x ) and variance 2/n
(i.e. 2
( )V xn
),
and this approximation becomes better as n becomes larger.
Note
As in the previous illustrating example, we can see the
following modifications:
(i) If the population is finite, 2
( ) (1 )n
V xN n
; where (1-n/N) is known as
the finite population correction factor. When N is very big, the
factor is
equal to 1.
(ii) If n is small, say less than 30, the sampling distribution
is not so normal. A
t-distribution will be used (discussed later).
In the above example, N=4, n=2, 2
( ) (1 )n
V xN n
=(1-2/4)(1.6667/2) = 0.4167. If the
population is big (or the sample is drawn with replacement),
then 2
( )V xn
=1.6667/2=0.8333.
In this course we assume a big population or sampling with
replacement.
Example 4.2
An electrical firm manufactures light bulbs that have a length
of life that is
approximately normal distributed with mean equal to 800 hours
and a standard
deviation of 40 hours. Find the probability that a random sample
of 16 bulbs will have
an average life of less than 775 hours.
x
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83
Solution
Let X be the average life of the 16 bulbs. X N( x 800, n
x
22
16
402)
16
40
800775775)775( ZP
XPXP
x
x
x
x
00621.0)5.2( ZP
Example 4.3
The mean IQ scores of all students attending a college is 110
with a standard deviation
of 10. (a) If the IQ scores are normally distribution, what is
the probability that the
score of any one student is greater than 112? (b) What is the
probability that the
mean score in a random sample of 36 students is greater than
112? (c) What is the
probability that the mean score in a random sample of 100
students is greater than 112?
Solution
(a) Let X be the student's IQ score. X N(110, 210 )
4207.0)2.0(10
110112)112(
ZPZPXP
(b) Let 1X be the mean score of a sample of 36 students.
1X N( ,110 x 36
10222 n
x
)
1151.0)2.1(
36
10
110112)112( 1
ZPZPXP
(c) Let 2X be the mean score of a sample of 100 students.
2X N( ,110 100
1022
n
)
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84
228.0.0)2(
100
10
110112)112( 2
ZPZPXP
4.9 Theory of Estimation:
4.9.1 Estimation
The process of estimating the unknown values of the population
parameters on the
basis of information contained in the sample is known as
estimation.
4.9.2 Estimator and Estimate
Let us consider (x1, x2, .. , xn) be a random sample of size n
is chosen to from a
population, for which the parameter is an unknown value.
Any statistic or mathematical function that is defined on this
sample is known as an
estimator and the particular numerical value, which is assigned
to the estimator, is
known as an estimate.
The followings are some examples
Estimator Population parameter
1. x
2. s2 2
There are two important properties for an estimator, namely,
unbiasedness and
efficiency.
An estimator, for example, x , is unbiased if and only if ( )E x
.
Efficiency: The efficiency of an estimator, for example, x , is
given by ( )V x . The smaller
the ( )V x , the more accurate will be the x as an
estimator.
4.9.3 Point Estimation and Interval Estimation
Let us consider be an estimator to the parameter of the given
population.
A point estimate is a single-value estimate of a population
parameter, for example
^ ^
;x P p . In other words, if the estimator is a single value in
the entire real number
scale then such estimation is known as point estimation.
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85
An interval estimate of a population parameter gives an interval
that may contain the
true value of the parameter with a certain probability (i.e.
confidence); for example,
Pr( ) 0.99.a b In other words, if the estimator provides an
interval of values
between which the true values of the population parameter will
exists is known as
Interval Estimation. The interval of values is known as
Confidence Interval and the limit
is a confidence limits.
For a point estimate, both the accuracy and reliability of the
estimation are unknown.
For an interval estimate, the width of the interval gives the
accuracy and the probability
gives the reliability of the estimation.
Example 4.4
(a) The mean and standard deviation for the quality point
averages of a random
sample of 36 college seniors are calculated to be 2.6 and 0.3,
respectively. Find a
95% confidence interval for the mean of the entire senior
class.
(b) How large a sample is required in (a) if we want to be 95%
confident of is off by
less than 0.05?
Solution
Let be the mean of the entire senior class.
Given: n = 36, 6.2x , s = 0.3, (1 - ) = 0.95 05.0
(a) A 95% confidence interval estimate for the is
36
3.096.16.2
36
3.096.16.2
025.0025.0
nzx
nzx
698.2502.2
(b) Let 1n be the required sample size.
To be 95% confident that if off by less than 0.05 would
imply
05.03.0
96.105.0
11
025.0
nnz
13930.13805.0
)3.0)(96.1(2
1
n
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Note
The following table gives the summary for constructing (1 )%
confidence interval for
mean (single mean and difference of means) for large and small
samples
Estimating Conditions Formula
Mean Large samples (n 30)
OR is known nZX
nZX
22
Mean Small samples and
unknown n
stX
n
stX
2,
2,
v=n-1
Difference of means Large sample OR 1 and
2 are known 2
2
1
2
1
221
)(
nnZXX
Difference of means
Small sample & 1 and
2 are unknown, assume
21
212
,21
11)()(
nnstXX p
221 nn ,
pooled estimate of sample standard deviation:
2
)1()1(
21
2
22
2
11
nn
snsns p
Difference of means
Small sample & 1 and
2 are unknown,
assume 1 2
2 2
1 21 2 ,
21 2
( )s s
X X tn n
1
)/(
1
)/(
)//(
2
2
2
2
2
1
2
1
2
1
2
2
2
21
2
1
n
ns
n
ns
nsns
Difference of means Paired observations , / 2d
v
sd t
n ; 1 2d x x
and v=n-1
Example 4.5
The contents of seven similar containers of sulfuric acid are
9.8, 10.2, 10.4, 9.8, 10.0,
10.2 and 9.6 liters. Find a 95% confidence interval for the mean
of all such containers,
assuming an approximate normal distribution.
Solution Let be the mean of all such containers.
Given: n = 7 70x 48.7002x
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87
107
70
n
xx 08.0
6
77048.700
1
)( 2222
n
n
xx
s
s = 2828.008.0 ; (1 - ) = 0.95 05.0 , 447.2025.0,6 t
A 95% confidence interval estimate for the is
7
2828.0447.210
7
2828.0447.210025.0,6025.0,6
n
stx
n
stx
262.10738.9
Example 4.6
A standardized chemistry test was given to 50 girls and 75 boys.
The girls made an
average grade of 76 with a standard deviation of 6, while the
boys made an average
grade of 82 with a standard deviation of 8. Find a 96%
confidence interval for the
difference 1 and 2, where 1 is the mean score of all boys and 2
is the mean score of
all girls who might take this test.
Solution
Given: 751 n , 502 n , 821 x , 81 s , 762 x , 62 s ,
04.096.)1(
A 96% confidence interval for 21 is:
2
2
2
1
2
102.02121
2
2
2
1
2
102.021
)(
)(
nnzxx
nnzxx
50
6
75
805.2)7682(
50
6
75
805.2)7682(
22
21
22
,
301 n & 302 n , so 11 s & 22 s
57.843.3 21
Example 4.7
In a batch chemical process, two catalysts are being compared
for their effect on the
output of the process reaction. A sample of 12 batches is
prepared using catalyst 1 and
a sample of 10 batches was obtained using catalyst 2. The 12
batches for which catalyst
1 was used gave an average yield of 85 with a sample standard
deviation of 4, while the
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average for the second sample gave an average of 81 and a sample
standard deviation of
5. Find a 90% confidence interval for the difference between the
population means,
assuming the populations are approximately normally distributed
with equal variances.
Solution
Let 1 and 2 be the mean population yield using catalyst 1 and
catalyst 2, respectively.
Given: 121 n , 102 n , 851 x , 41 s , 812 x , 52 s ,
10.090.)1( , 2021012221 nn , 725.105.0,20 t
pooled estimate of sample standard deviation 2
)1()1(
21
2
22
2
11
nn
snsns p
478.421012
5)110(4)112( 22
A 90% confidence interval for 21 is:
21
05.0,202121
21
05.0,2021
11)()(
11)()(
nnstxx
nnstxx pp
10
1
12
1)478.4)(725.1()8185(
10
1
12
1)478.4)(725.1()8185( 21
31.769.0 21
Example 4.8
The weight of 10 adults selected randomly before and after a
certain new diet was
introduced was recorded as follows:
Adult Before ( 1x ) After ( 2x ) Difference
1 76 81 -5
2 60 52 8
3 85 87 -2
4 58 70 -12
5 91 86 5
6 75 77 -2
7 82 90 -8
8 64 63 1
9 79 85 -6
10 88 83 5
Find a 98% confidence interval for the mean difference in
weight.
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89
Solution
idd
n
= -1.6 2
2( ( 1.6))
40.71
i
d
ds
n
For v = n-1 = 9; 0.01 2.821t .
A 98% confidence interval is 6.38
1.6 (2,821)10
That is 7.29 4.09d
Exercise 4.1
4.1.1 What is the value of the finite population correction
factor for when (i) n = 10
and N = 1000; (ii) n = 5 and N = 200?
4.1.2 How many different samples of size n=2 can be chosen from
a finite population of
size (i) N=25; (ii) N=6?
4.1.3 When we sample from an infinite population, what happens
to the standard error
of the mean if the sample size is (a) increased from 50 to 200
(b) decreased from
225 to 25?
4.1.4 The mean of a random sample size n=25 is used to estimate
the mean of an infinite
population with the s.d. = 2.4. What can we assert about the
probability that the
error will be less than 1.2, if we use the central limit
theorem?
4.1.5 A random sample of size 100 is taken from an infinite
population having the mean
= 76 and the variance of 2 = 256. What is the probability that x
will be between
75 and 78?
4.1.6 A random sample of size n = 36 is taken from an infinite
population with mean =
63 and the variance of 2 = 81. What can we assert about the
probability of getting
a sample mean greater than 66.75, if we use the central limit
theorem?
4.1.7 A research worker wants to determine the average time it
takes a mechanic to
rotate the tires of a car and she wants to be able to assert
with 95% confidence
that the mean of her sample is off by at most 0.50 minute. If
she can presume
from past experience that = 1.6 minutes, how large a sample will
she have to
take?
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90
4.1.8 If we want to determine the average mechanical aptitude of
a large group of
workers, how large a random sample will we need to be able to
assert with
probability 0.95 that the sample mean will not differ from the
true mean by more
than 3.0 points? Assume that it is known from past experience
that = 20.0.
4.1.9 What is the maximum error one can expect to make with
probability 0.90 when
using the mean of a random sample of size n = 64 to estimate the
mean of a
population with 2 = 2.56?
4.1.10 A random sample of 40 drums of a chemical, drawn from
among 200 such drums
whose weights can be expected to have the = 12.2, has a mean
weight of 240.8
pounds. If we estimate the mean weight of all 200 drums as 240.8
pounds, what
can we assert with 99% confidence about the maximum error?
4.1.11 To estimate the average time it takes to assemble a
certain computer
component, the industrial engineer at an electronics firm timed
40 technicians in
the performance of this task, getting a mean of 12.73 minutes
with a standard
deviation of 2.06 minutes. Construct a 98% confidence interval
for the true
average time it takes to assemble the computer component.
4.1.12 Ten bearings made by certain process have a mean diameter
of 0.5060 cm with a
standard deviation of 0.0040 cm. Assuming that the data may be
looked upon as a
random sample from a normal population, construct 95% and 99%
confidence
interval for the actual average diameter of bearings made by
this process.
4.10 Statistical Hypothesis
Any statement regarding the characteristics of the population or
the parameter of the
population or the nature of the population probability
distribution is known as a
Hypothesis.
For example: i) the ratio of women for every 1000 men is 937 in
India, ii) the literacy rate
in India is 56.7%, iii) there is substantial growth in prices of
the infrastructure goods in
the past five years in India., and so on.
Further the hypothesis is classified into two types namely:
Simple hypothesis any
hypothesis which specifies the population completely or
absolutely is known as simple
hypothesis, otherwise the hypothesis is said to be Composite or
Comparative
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91
hypothesis. In the above example (i) & (ii) are the simple
hypothesis and (iii) is the
composite hypothesis.
4.11 Test of Hypothesis
The Test of Hypothesis is nothing but a two-action decision
problem, either to accept
the hypothesis or to reject the hypothesis.
4.12 Test of Significance
The test of significance is to test the deviation between
i) the parameter and statistic, or
ii) the difference between two independent statistics, is just
because of
chance variation or assignable variation.
4.13 Null Hypothesis
Any statement or hypothesis, which gives no difference between
the parameter and a
statistic or no differences in two independent statistics, is
known as Null Hypothesis.
According to Prof. Fisher, Null hypothesis is a hypothesis,
which is tested for possible
rejection under the assumption that it is true. The null
hypothesis is always denoted by H0:
4.14 Alternative Hypothesis
Any statement or hypothesis, which gives a difference between
the parameter and a
statistic or difference in between two independent statistics,
is known as an alternative
hypothesis. The alternative hypothesis is always denoted by
H1
4.15 Errors in Sampling
There are four possible situations in any testing of hypothesis
procedure and are given in the following table:
0H is correct 0H is wrong
Accept 0H Correct decision Type II Error
Reject 0H Type II error Correct decision
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92
In the above table, there are two types of errors occurred in
sampling, while testing an
hypothesis problem, which are given by
Type I Error: Rejecting the Null Hypothesis H0, when it is
true.
Type II Error: Accepting the Null Hypothesis H0, when it is
false.
4.16 Sizes of Errors
The probabilities of occurring the errors are known as size of
the errors and are
respectively given by:
Size of Type-I Error = Prob. {Type I Error} = Prob. {Reject H0 |
H0} =
Size of Type-II Error = Prob. {Type II Error} = Prob. {Accept H0
| H1} =
Here , is known as Level of significance and , is known as power
of a test.
4.17 Critical Region and Critical Value
The region of rejecting the null hypothesis is known as critical
region and is denoted by
. Generally, the critical region is also known as the level of
significance, and is fixed in
advance and is either 5% or 1%.
A value, which separates out the acceptance region and rejection
region, is known as
critical value.
4.17.1 Significant or Critical Values
The following table represents the significant values at
different levels of significance.
Z Test Level of Significance (%)
1% 5% 10%
Two-tailed Test 2.58 1.96 1.645
Right-tailed Test 2.33 1.645 1.25
Left-tailed Test -2.33 -1.645 -1.25
Note:
The significant values for small sampling techniques ( i.e.
using t-Test, F-Test and Chi-
Square (2) Test) depends upon the size of the sample.
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93
4.18 Hypothesis Testing Terminology
1. Null hypothesis, H0: A hypothesis that is held to be true
until very strong
evidence to the contrary is obtained. H0 : 0
2. Alternative hypothesis, H1: It is a hypothesis that is
complement to the null
hypothesis. Hence it will be accepted if the null hypothesis is
rejected.
1 0:H (two-tail test)
1 0:H (One-tail test)
1 0:H (One-tail test)
In the one-tail test we have some expectation about the
direction of the error
when the null hypothesis is wrong, while in the two-tail test we
dont have such
expectation.
3. Test statistics is the value, based on the sample, used to
determine whether the
null hypothesis should be rejected or accepted.
4. Critical region is a region in which if the test statistic
falls the null hypothesis will
be rejected.
5. Types of error
(a) Type I error: Reject H0 when H0 is true
(b) Type II error: Accept H0 when Ha is true
6. The significance level, is the probability of committing a
Type I error, i.e.,
P(Type I error) = .
The probability of committing a Type II error is ; i.e., P(Type
II error) = .
4.19 Basic Steps in Testing Hypothesis
1. Formulate the null hypothesis.
2. Formulate the alternative hypothesis.
3. Specify the level of significance to be used.
4. Select the appropriate test statistic and establish the
critical region.
5. Compute the value of the test statistic.
6. Conclusion: Reject H0 if the statistic has a value in the
critical region, otherwise
accept H0.
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94
4.20 Tests Concerning Means
The tests concerning means (for large and small samples) are
summarized in the
following table.
H0 Conditions Test statistic
0 Large samples (n 30) OR is known n
xz
0
Small samples and unknown ns
xt 0
with 1 n
021 d Large samples OR 1 and
2 are known
2
2
2
1
2
1
021 )(
nn
dxxz
Small sample & 1 and 2
are unknown, assume
21
21
021
11
)(
nns
dxxt
p
with 221 nn and
2
)1()1(
21
2
22
2
112
nn
snsnsp
if 1 = 2 but unknown
Small sample & 1 and 2
are unknown,
assume 1 2
2
2
2
1
2
1
021 )(
n
s
n
s
dxxt
with
1
)/(
1
)/(
)//(
2
2
2
2
2
1
2
1
2
1
2
2
2
21
2
1
n
ns
n
ns
nsns
021 d Paired observations
ns
ddt
d
0 with 1 n
Example 4.9
A manufacturer of sports equipment has developed a new synthetic
fishing line that he
claims has a mean breaking strength of 8 kilograms with a
standard deviation of 0.5
kilogram. Test the hypothesis that = 8 kilograms against the
alternative that 8
kilograms if a random sample of 50 lines is tested and found to
have a mean breaking
strength of 7.8 kilograms. Use a 0.01 level of significance.
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95
Solution
Null hypothesis: 8 kilograms
Alternative hypothesis: 8 kilograms (Two tailed test
alternative)
Level of significance: 0.01
Critical region: Z > 58.2005.0 z or Z < 58.2005.0 z
Computation: n = 50 8.7x 5.0
828.2
505.0
88.7
n
xz
Conclusion: As the sample z (= -2.828) falls inside the critical
region, so reject the null
hypothesis at 0.01 level of significance and conclude that is
significantly smaller than
8 kilograms.
Example 4.10
The average length of time for students to register for fall
classes at a certain college has
been 50 minutes with a standard deviation of 10 minutes. A new
registration procedure
using modern computing machines is being tried. If a random
sample of 12 students had
an average registration time of 42 minutes with a standard
deviation of 11.9 minutes
under the new system, test the hypothesis that the population
mean is now less than
50, using a level of significance of (1) 0.05, and (2) 0.01.
Assume the population of times
to be normal.
Solution
Let be the population mean time for students to register in the
new registration
procedure.
Null hypothesis: 50 minutes
Alternative hypothesis: 50 minutes (left tailed test
alternative)
Level of significance: 0.05
Critical region: (n = 12 < 30; and the new is unknown, so
t-test should be used)
degree of freedom ( ) = n -1 = 12 -1 =11
-
96
t < 796.105.0,11 t
Computation:
n = 12 42x s = 11.9
329.2
129.11
5042
ns
xt
Conclusion:
(1) As sample t (= -2.329) falls inside the critical region, so
reject the null hypothesis
at 0.05 level of significance and conclude that is significantly
smaller than 50
minutes.
(2) Identical with those of (1) except the critical region would
be replaced by:
718.201.0,11 tt
and the corresponding conclusion would be changed as
follows:
As sample t (= -2.329) falls outside the critical region, so
reject the alternative
hypothesis at 0.01 level of significance and conclude that is
not highly
significantly smaller than 50 minutes.
Example 4.11
An experiment was performed to compare the abrasive wear of two
different laminated
materials. Twelve pieces of material 1 were tested, by exposing
each piece to a machine
measuring wear. Ten pieces of material 2 were similarly tested.
In each case, the depth
of wear was observed. The samples of material 1 gave an average
(coded) wear of 85
units with a standard deviation of 4, while the samples of
material 2 gave an average of
81 and a standard deviation of 5. Test the hypothesis that the
two types of material
exhibit the same mean abrasive wear at the 0.10 level of
significance. Assume the
populations to be approximately normal with equal variances.
Solution:
Let 1 and 2 be the mean abrasive wear of material 1 and 2
respectively.
Null hypothesis: 21 , i.e. 021
Alternative hypothesis: 21 , i.e. 021
Level of significance: 0.10
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97
Critical region: (As both 1n and 2n are smaller than 30 and
their standard deviations are
unknown, so t-test has to be used.)
2021012221 nn , 725.105.0,20 tt or 725.105.0,20 tt
Computation:
121 n 851 x 41 s
102 n 812 x 52 s
05.2021012
5)9(4)11(
2
)1()1( 22
21
2
22
2
112
nn
snsns p
086.2
10
1
12
105.20
0)8185(
11
)()(
21
2121
nns
xxt
p
Conclusion: As the sample t (=2.086) falls inside the critical
region, so reject the null
hypothesis at 0.10 level of significance and conclude that the
mean abrasive wear of
material 1 is significantly higher than that of the material
2.
Example 4.12
Five samples of a ferrous-type substance are to be used to
determine if there is a
difference between a laboratory chemical analysis and an X-ray
fluorescence analysis of
the iron content. Each sample was split into two sub-samples and
the two types of
analysis were applied. Following are the coded data showing the
iron content analysis:
Sample
Analysis 1 2 3 4 5
x-ray 2.0 2.0 2.3 2.1 2.4
Chemical 2.2 1.9 2.5 2.3 2.4
Solution:
Assuming the populations normal, test at the 0.05 level of
significance whether the two
methods of analysis give, on the average, the same result.
Let 1 and 2 be the mean iron content determined by the
laboratory chemical
analysis and X-ray fluorescence analysis respectively; and
D be the mean of the population of differences of paired
measurements.
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98
Null hypothesis: 21 or 0D
Alternative hypothesis: 21 or 0D
Level of significance: 0.05
Critical region: (As n = 5 < 30, so t-test should be
used.)
776.2025.0,4 tt or 776.2025.0,4 tt
Computation:
Sample
Analysis 1 2 3 4 5
x-ray 2.0 2.0 2.3 2.1 2.4
Chemical 2.2 1.9 2.5 2.3 2.4
id -0.2 0.1 -0.2 -0.2 0
2
id 0.04 0.01 0.04 0.04 0
5.05
1
i
id 13.05
1
2 i
id
1.05
5.0
5
dd
02.0)4)(5(
)5.0()13.0)(5(
)1(
222
2
nn
ddnsd
5811.1
502.0
0)1.0(
n
s
dt
d
D
Conclusion: As the sample t (=-1.5811) falls outside the
critical region, so reject the
alternative hypothesis at 0.05 level of significance and
conclude that there is no
significant difference in the mean iron content determined by
the above two analyses.
4.21 Chi-square Tests
The various tests of significance studied earlier such that as
Z-test, t-test, F-test was
based on the assumption that the samples were drawn from normal
population. Under
this assumption the various statistics were normally
distributed. Since the procedure of
testing the significance requires the knowledge about the type
of population or
parameters of population from which random samples have been
drawn, these tests are
-
99
known as parametric tests.
But there are many practical situations in which the assumption
of any kind about the
distribution of population or its parameter is not possible to
make. The alternative
technique where no assumption about the distribution or about
parameters of
population is made is known as non-parametric tests. Chi-square
test is an example of
the non parametric test. Chi-square distribution is a
distribution free test.
Chi-square distribution was first discovered by Helmert in 1876
and later independently
by Karl Pearson in 1900. The range of chi-square distribution is
0 to .
The square of a standard normal variate is known as chi-square
variate with one degree
of freedom. Symbolically we can write it as
test is used for enumeration data (the data obtained by
enumeration or counting is
called enumeration data. For example, number of blue flowers,
number of intelligent
boys, number of curled leaves, etc.,) which generally relate to
discrete variable where as
t-test and standard normal deviate tests are used for measure
mental data which
generally relate to continuous variable.
test can be used to know whether the given objects are
segregating in a theoretical
ratio or whether the two attributes are independent in a
contingency table.
a) Conditions for the validity of test:
The validity of -test of goodness of fit between theoretical and
observed, the
following conditions must be satisfied.
i) The sample observations should be independent
ii) Constraints on the cell frequencies, if any, should be
linear oi =ei
iii) N, the total frequency should be reasonably large, say
greater than 50
-
100
iv) If any theoretical (expected) cell frequency is < 5, then
for the application of chi-
square test it is pooled with the preceding or succeeding
frequency so that the pooled
frequency is more than 5 and finally adjust for the d.f. lost in
pooling.
b) Applications of Chi-square Test:
1. testing the independence of attributes
2. to test the goodness of fit
3. testing of linkage in genetic problems
4. comparison of sample variance with population variance
5. testing the homogeneity of variances
6. testing the homogeneity of correlation coefficient
4.21.1 Goodness-of-fit Test
A test to determine if a population has a specified theoretical
distribution. The test is
based on how good a fit we have between the frequency of
occurrence of observations
in an observed sample and the expected frequencies obtained from
the hypothesized
distribution.
A goodness-of-fit test between observed and expected frequencies
is based on the
quantity test2
2
( )O E
E
i i
i
where test2 is a value of the random variable whose sampling
distribution is
approximated very closed by the Chi-square distribution,
Oi is the observed frequency of cell i,
Ei is the expected frequency of cell i.
The number of degrees of freedom in a Chi-square goodness-of-fit
test is equal to the
number of cells minus the number of quantities obtained from the
observed data that
are used in the calculations of the expected frequencies.
For a level of significance equal to , test2 2 constitutes the
critical region. The
decision criterion described here should not be used unless each
of the expected
frequencies is at least equal to 5.
-
101
Example 4.13 Consider the tossing of a die 120 times.
Faces
1 2 3 4 5 6
Observed 20 22 17 18 19 24 Expected
By comparing the observed frequencies with the expected
frequencies, one has to
decide whether the die is fair die or not.
Null hypothesis: the die is a fair die, i.e. 6
1)( iXP for i = 1, 2, 3, 4, 5, and 6
Alternative hypothesis: the die is not a fair die
Level of significance: 0.05
Critical region: 5161 n ; 07.112 05.0,52
Computation:
Expected value = 20)6
1(120)( iXnP
I 1 2 3 4 5 6
Observed )( iO 20 22 17 18 19 24
Expected )( iE 20 20 20 20 20 20
ii EO 0 2 -3 -2 -1 4
6
1
2
2 )(
i i
ii
E
EO
7.120
4
20
)1(
20
)2(
20
)3(
20
2
20
0 222222
Conclusion: As the sample 2 (= 1.7) falls outside the critical
region, so reject the
alternative hypothesis and conclude that the die is a fair
die.
4.21.2 Test for Independence
The Chi-square test procedure can also be used to test the
hypothesis of independence
of two variables/attributes. The observed frequencies of two
variables are entered in a
two-way classification table, or contingency table.
The expected frequency of the cell in the ith row and jth column
in the contingency table.
-
102
The degrees of freedom for the contingency table is equal to (r
1) (c 1) where r is the
number of rows and c is the number of columns in the table.
Example 4.14
Suppose that we wish to study the relationship between grade
point average and
appearance of 150 students of a college, whose analysis are
given below. Use the level
of significance is 0.05
Grade Point Average
Appearance 1 2 3 4 Totals
Attractive 14 ( ) 11 ( ) 10 ( ) 5 ( ) 40 Ordinary 10 ( ) 16 ( )
16 ( ) 14 ( ) 56 Unattractive 3 ( ) 4 ( ) 7 ( ) 10 ( ) 24
Totals 27 31 33 29 120
Null hypothesis: There is no relationship between grade point
average and appearance.
That is, the two characteristics are independent.
Alternative hypothesis: There is a relationship between grade
point average and
appearance. That is, the two characteristics are not
independent.
Level of significance: 0.05
Critical region: 2 05.0,2
, where = (r -1)(c - 1)
Computation:
Grade Point Average
Appearance 1 2 3 4 Totals
Attractive 14
(9)
11
(10.33)
10
(11)
5
(9.67)
40
Ordinary 10 ( 12.6)
16 ( 14.47)
16 (15.4)
14 (13.53)
56
Unattractive 3
(5.4)
4
(6.2)
7
(6.6)
10
(5.8)
24
Totals 27 31 33 29 120
67.9
)67.95(
11
)1110(
33.10
)33.1011(
9
)914()( 222222
E
EO
53.13
)53.1314(
4.15
)4.1516(
47.14
)47.1416(
6.12
)6.1210( 2222
-
103
818.108.5
)8.510(
6.6
)6.67(
2.6
)2.64(
4.5
)4.53( 2222
Conclusion: Since 596.122 05.0,6 , so sample 2 (=10.818) falls
outside the critical
region. So reject the alternative hypothesis and conclude that
there is no evidence to
support there is relationship between grade point average and
appearance.
4.21.3 Test for Homogeneity
To test the hypothesis that several population proportions are
equal.
The approach for the test of homogeneity is the same as for the
test of independence of
variables/attributes.
Example 4.15
A study of the purchase decisions for 3 stock portfolio
managers, A, B, and C was
conducted to compare the rates of stock purchases that resulted
in profits over a time
period that was less than or equal to one year. One hundred
randomly selected
purchases obtained for each of the managers showed the results
given in the table. Do
these data provide evidence of differences among the rates of
successful purchases for
the three portfolio managers? Test with .05.0
Result
Manager
A B C
Purchases show profit 63 71 55
Purchase do no show profit 37 29 45
Total 100 100 100
Null hypothesis: the rates of stock purchases that resulted in
profit were the same for
the three stock portfolio managers
Alternative: their rates were not all the same
Level of significance: 0.05
Critical region: 2)13)(12( ; 991.52 05.0,22
Computation:
-
104
Result
Manager
A B C Total
Purchases show profit 63
(63) 71
(63) 55
(63) 189
Purchase do no show profit 37
(37) 29
(37) 45
(37) 111
Total 100 100 100 300
49.537
)3745(
37
)3729(
37
)3737(
63
)6355(
63
)6371(
63
)6363( 2222222
Conclusion: As the sample 2 (= 5.49) falls outside the critical
region so reject the
alternative hypothesis and conclude that there is no sufficient
evidence to support the
rates of purchases resulted in profit of the three portfolio
managers were different.
Exercise 4.2
4.2.1 According to the norms established for a mechanical
aptitude test, persons who
are 18 years old should average 73.2 with standard deviation of
8.6. If 45
randomly selected persons of that age averaged 76.7, test the
null hypothesis =
73.2 against the alternative hypothesis > 73.2 at the 0.01
level of significance.
4.2.2 In 64 randomly selected hours of production, the mean and
standard deviation of
the number of acceptable pieces produced by an automatic
stamping machine is
1,038 and 146 respectively. At the 0.05 level of significance
does this enable us to
reject the null hypothesis = 1000 against the alternative
hypothesis >1000?
4.2.3 A random sample of 6 steel beams has a mean compressive
strength of 58,392psi,
with a standard deviation of 648psi. Use this information and
the level of
significance of = 0.05 to test whether the true average
compressive strength of
the steel beam from which this sample came is 58,000psi. Assume
normality.
4.2.4 Five measurements of tar content of a certain kind of
cigarette yielded 14.5, 14.2,
14.4, 14.3 and 14.6 mg per cigarette. Show that the difference
between the mean
of this sample, = 14.4 and the average tar claimed by the
manufacturer = 14.0,
is significant at = 0.05. Assume normality.
4.2.5 A manufacturer of gunpowder has developed a new powder
which is designed to
produce a muzzle velocity equal to 3000 ft./sec. Eight shells
are loaded with the
charge and the muzzle velocities measured. The resulting
velocities are as: 3005,
-
105
2935, 2965, 2995, 3905, 2935 and 2905. Do these data present
sufficient evidence
to indicate that the average velocity differs from
3000ft/sec?
4.2.6 An investigation of two kinds of photocopying equipment
showed that 71 failures
of the first kind of equipment took on the average 83.2 minutes
to repair with a
standard deviation of 19.3 minutes, while 75 failures of the
second kind of
equipment took on the average 90.8 minutes to repair with a
standard deviation of
21.4 minutes. Test the null hypothesis (the hypothesis that on
the
average it takes an equal amount of time o repair either kind of
equipment)
against the alternative hypothesis at the level of significance
of =
0.05
4.2.7 Suppose that we want to investigate whether on the average
men earn more than
$20 per week more than women in a certain industry. If sample
data show that 60
men earn on the average per week with a standard deviation of s1
=
$15.60, while 60 women earn on the average per week with a
standard deviation of s2 = $18.20, what can we conclude at the
0.01 level of
significance?
4.2.8 Measuring specimens of nylon yarn taken from two spinning
machines, it was
found that 8 specimens from the first machine had a mean denier
of 9.67 with a
standard deviation of 1.81, while 10 specimens from the second
machine has a
mean denier of 7.43 with a standard deviation of 1.48. Assuming
that the
populations sampled are normal and have the same variance, test
the null
hypothesis against the alternative hypothesis at the
level of significance of = 0.05.
4.2.9 The following are the Brinell hardness values obtained for
samples of two
magnesium alloys:
Alloy 1: 66.3 63.5 64.9 61.8 64.3 64.7 65.1 64.5 68.4 63.2
Alloy 2: 71.3 60.4 62.6 63.9 68.8 70.1 64.8 68.9 65.8 66.2
Use the 0.05 level of significance to test null hypothesis
against the
alternative hypothesis .
4.2.10 A genetical law says that children having one parent of
blood group M and other
parent of blood group N will always be one of the three groups
M, MN and N;
and that the average number of children in these three groups
will be in the ratio
-
106
1:2:1. The report on an experiment states as follows: Of 162
children having
one M Parent, and one N parent, 28% were found to be of group M,
42% of
group MN and the rest of them are group N. Do the data in the
report conform
to the expected genetic ratio 1:2:1? (Given value of Chi-square
at 2, 0.05 is 5.991)
4.2.11 A chemical extraction plant processes sea water to
collect sodium chloride and
magnesium. It is known that sea water contains sodium chloride,
magnesium
and other elements in the ratio of 62:4:34. A sample of 200
tones of sea water
has resulted in 130 tones of sodium chloride, 6 tones of
magnesium and 64 tones
of other elements. Are these data consistent with the known
composition of sea
water at 5% level?
4.2.12 The following table shows the number of aircraft
accidents that occurred during
the six days of a week. Find whether the accidents are uniformly
distributed over
the week? Given value of 2 = 11.070 with 2, 0.05
Day Mon Tue Wed Thu Fri Sat Total
No. of Accidents
14 18 12 11 15 14 84
4.2.13 200 digits were chosen at random from a set of tables.
The frequencies of the
digits were as follows:
Digit 0 1 2 3 4 5 6 7 8 9 Total
F 18 19 23 21 16 25 22 20 21 15 200
Use Chi-square test; to assess the correctness of the hypothesis
that the digits
were distributed in equal number in the table, given the value
of chi-square at 9,
0.05 is 16.9
4.2.14 In an experiment on immunization of cattle from
tuberculosis, the following
results were obtained:
Affected Unaffected
Inoculated 12 28
Not Inoculated 13 7
Examine the effect of vaccine in controlling the incidence of
the disease.
4.2.15 Can vaccination be regarded as a preventive measure of
small-pox as evidenced
by the following data: Of 1,482 persons exposed to small pox in
a locality, 368 in
-
107
all were attacked. Of these 1,482 persons, 343 were vaccinated
and of these,
only 35 were attacked.
4.2.16 Out of 8000 graduates in a city 800 are females; out of
1600 graduate employees
120 are females. Use 2 test to determine if any distinction is
made in
appointment on the basis of gender. (Value of 2 is 3.84 for
1d.f., at 0.05 level)
