COMPLETE
BUSINESS
STATISTICS
by
AMIR D. ACZEL
&
JAYAVEL SOUNDERPANDIAN
7th edition.
Prepared by Lloyd Jaisingh, Morehead State
University
Chapter 8
The Comparison of Two Populations
McGraw-Hill/Irwin
Copyright © 2009 by The McGraw-Hill Companies, Inc. All rights reserved.
8-2
8 The Comparison of Two Populations
• Using Statistics
• Paired-Observation Comparisons
• A Test for the Difference between Two Population Means Using
•
•
Independent Random Samples
A Large-Sample Test for the Difference between Two Population
Proportions
The F Distribution and a Test for the Equality of Two Population
Variances
8-3
8 LEARNING OBJECTIVES
After studying this chapter you should be able to:
•
•
•
•
•
•
•
Explain the need to compare two population parameters
Conduct a paired-difference test for the difference in population means
Conduct an independent-samples test for the difference in population means
Describe why a paired-difference test is better than independent-samples test
Conduct a test for difference in population proportions
Test whether two population variances are equal
Use templates to carry out all tests
8-4
8-1 Using Statistics
• Inferences about differences between parameters of two
populations


Paired-Observations
Observe the same group of persons or things



At two different times: “before” and “after”
Under two different sets of circumstances or “treatments”
Independent Samples
 Observe different groups of persons or things

At different times or under different sets of circumstances
8-5
8-2 Paired-Observation Comparisons
• Population parameters may differ at two different times or
under two different sets of circumstances or treatments
because:


The circumstances differ between times or treatments
The people or things in the different groups are themselves different
• By looking at paired-observations, we are able to minimize
the “between group” , extraneous variation.
8-6
Paired-Observation Comparisons
of Means
Test statistic for the paired - observatio ns test
D 
t
D
0 , df  n 1
S
D
n
D  sample average for the difference s
S  sample standard deviation for the difference s
D
n  sample size
 D  mean of the population of difference s under the null hypothesis
0
8-7
Example 8-1
A random sample of 16 viewers of Home Shopping Network was selected for an experiment. All viewers in the
sample had recorded the amount of money they spent shopping during the holiday season of the previous year.
The next year, these people were given access to the cable network and were asked to keep a record of their total
purchases during the holiday season. Home Shopping Network managers want to test the null hypothesis that
their service does not increase shopping volume, versus the alternative hypothesis that it does.
Shopper Previous
1
334
2
150
3
520
4
95
5
212
6
30
7
1055
8
300
9
85
10
129
11
40
12
440
13
610
14
208
15
880
16
25
Current
405
125
540
100
200
30
1200
265
90
206
18
489
590
310
995
75
Diff
71
-25
20
5
-12
0
145
-35
5
77
-22
49
-20
102
115
50
H0: D  0
H1: D > 0
df = (n-1) = (16-1) = 15
Test Statistic:
t 
D  D
0
sD
n
Critical Value: t0.05 = 1.753
Do not reject H0 if : t 1.753
Reject H0 if: t > 1.753
8-8
Example 8-1: Solution
D  D
32.81  0
0
t

 2.354
sD
55.75
t = 2.354 > 1.753, so H0 is rejected and we conclude that
there is evidence that shopping volume by network
viewers has increased, with a p-value between 0.01 an
0.025. The Template output gives a more exact p-value
of 0.0163. See the next slide for the output.
16
n
t Distribution: df=15
0.4
f(t)
0.3
0.2
Nonrejection
Region
0.1
Rejection
Region
0.0
-5
0
1.753
= t0.05
5
2.131
= t0.025
2.602
= t0.01
2.354=
test
statistic
t
8-9
Example 8-1: Using the Template for
Testing Paired Differences
Decision: Reject the null hypothesis
8-10
Example 8-1: Using Minitab for Testing
Paired Differences
Decision: Reject the null hypothesis, P-value < 0.05
8-11
Example 8-2
It has recently been asserted that returns on stocks may change once a story about a company appears in The Wall
Street Journal column “Heard on the Street.” An investments analyst collects a random sample of 50 stocks that
were recommended as winners by the editor of “Heard on the Street,” and proceeds to conduct a two-tailed test of
whether or not the annualized return on stocks recommended in the column differs between the month before and
the month after the recommendation. For each stock the analysts computes the return before and the return after
the event, and computes the difference in the two return figures. He then computes the average and standard
deviation of the differences.
H0: D  0
H1: D > 0
D  D
0.1  0
0
z 

 14 .14
sD
0.05
n = 50
D = 0.1%
sD = 0.05%
Test Statistic:
n
50
p - value: p ( z  14.14 )  0
D 
z
D
0
s
D
n
This test result is highly significant,
and H 0 may be rejected at any reasonable
level of significance.
8-12
Confidence Intervals for Paired
Observations
A (1- ) 100% confidence interval for the mean difference D :
D  t 2 sD
n
where t 2 is the value of the t distributi on with (n -1) degrees of freedom
that cuts off an area of  2 to its right, When the sample size is large,
we may approximat e t 2 with z 2.
8-13
Confidence Intervals for Paired
Observations – Example 8-2
95% confidence interval for the data in Example 8  2 :
D  z sD  0.1  1.96 0.05  0.1  (1.96)(.0071)
n
50
 0.1  0.014  [0.086, 0.114]
2
Note that this confidence interval
does not include the value 0.
8-14
Hypothesis Test & Confidence Interval for
Example 8-2 - Using the Template
Decision: Reject the null hypothesis.
Confidence Interval
8-15
8-3 A Test for the Difference between Two Population
Means Using Independent Random Samples
• When paired data cannot be obtained, use independent
random samples drawn at different times or under different
circumstances.

Large sample test if:



Both n1 30 and n2 30 (Central Limit Theorem), or
Both populations are normal and 1 and 2 are both known
Small sample test if:

Both populations are normal and 1 and 2 are unknown
8-16
Comparisons of Two Population
Means: Testing Situations
•
I: Difference between two population means is 0

1= 2


•
II: Difference between two population means is less than 0

1 2


•
H0: 1 -2 = 0
H1: 1 -2  0
H0: 1 -2  0
H1: 1 -2  0
III: Difference between two population means is less than D

1  2+D


H0: 1 -2  D
H1: 1 -2  D
8-17
Comparisons of Two Population
Means: Testing Situations
•
IV: Difference between two population means is greater

1 2
H0: 1 -2  0
 H1: 1 -2 <0
V: Difference between two population means is greater than D

•
than 0

1  2+ D


H0: 1 -2  D
H1: 1 -2 < D
8-18
Comparisons of Two Population
Means: Test Statistic
Large-sample test statistic for the difference between two
population means:
z
( x  x )  (   )
1
2

1
2
1
n
1

2

0
2
2
n
2
The term (1- 2)0 is the difference between 1 an 2 under the
null hypothesis. Is is equal to zero in situations I , II and IV, and it
is equal to the prespecified value D in situations III and V. The
term in the denominator is the standard deviation of the difference
between the two sample means (it relies on the assumption that
the two samples are independent).
8-19
Two-Tailed Test for Equality of Two
Population Means: Example 8-3
Is there evidence to conclude that the average monthly charge in the entire population of American Express Gold
Card members is different from the average monthly charge in the entire population of Preferred Visa
cardholders?
Population1 : Preferred Visa
H
n = 1200
0
:  0
1
2
H :  0
1
1
2
1
x = 452
1
 = 212
z 
1
Population 2 : Gold Card
( x  x )  (   )
1
2
1
2 0  ( 452  523)  0
2
2
2
2


212
185
1  2

1200
800
n
n
1
2
 71

80.2346

 71
 7.926
8.96
n = 800
2
x = 523
p - value : p(z < -7.926)  0
2
 = 185
2
H
0
is rejected at any common level of significan ce
8-20
Example 8-3: Carrying Out the Test
Standard Normal Distribution
0.4
f(z)
0.3
0.2
0.1
0.0
-z0.01=-2.576
Rejection
Region
Test Statistic=-7.926
0
Nonrejection
Region
z
z0.01=2.576
Rejection
Region
Since the value of the test
statistic is far below the lower
critical point, the null
hypothesis may be rejected,
and we may conclude that
there is a statistically
significant difference between
the average monthly charges
of Gold Card and Preferred
Visa cardholders.
8-21
Example 8-3: Using the Template
Decision: reject the null hypothesis.
8-22
Example 8-4
Is there evidence to substantiate Duracell’s claim that their batteries last, on average, at least 45 minutes longer
than Energizer batteries of the same size?
Population1 : Duracell
H :     45
0 1
2
H :     45
1 1
2
n = 100
1
x = 308
1
 = 84
1
Population 2 : Energizer
( x  x )  (   )
2
1
2 0  (308  254)  45
z 1
2
2
2
2


84
67
1  2

100
100
n
n
1
2

9
115.45

9
 0.838
10.75
n = 100
2
x = 254
2
 = 67
2
p - value : p(z > 0.838) = 0.201
H may not be rejected at any common
0
level of significan ce
8-23
Example 8-4 – Using the Template
Is there evidence to substantiate Duracell’s claim that their batteries last, on average, at least 45 minutes longer
than Energizer batteries of the same size?
P-value
8-24
Confidence Intervals for the Difference
between Two Population Means
A large-sample (1-)100% confidence interval for the difference
between two population means, 1- 2 , using independent random
samples:
(x  x )  z
1
2

2
2
2

1  2
n
n
1
2

A 95% confidence interval using the data in example 8-3:
(x  x )  z
1
2

2
2
2
2 1852

212
1  2  (523  452)  1.96

 [53.44,88.56]
1200
800
n
n
1
2

8-25
A Test for the Difference between Two Population
Means: Assuming Equal Population Variances

If we might assume that the population variances 12 and 22 are equal (even though
unknown), then the two sample variances, s12 and s22, provide two separate estimators of
the common population variance. Combining the two separate estimates into a pooled
estimate should give us a better estimate than either sample variance by itself.
** * * * * **
**
x1
Deviation from the
mean. One for each
sample data point.
}
}
Deviation from the
mean. One for each
sample data point.
* *
* *
Sample 1
From sample 1 we get the estimate s12 with
(n1-1) degrees of freedom.
* ** *
* ** * *
x2
** *
*
Sample 2
From sample 2 we get the estimate s22 with
(n2-1) degrees of freedom.
From both samples together we get a pooled estimate, sp2 , with (n1-1) + (n2-1) = (n1+ n2 -2)
total degrees of freedom.
8-26
Pooled Estimate of the Population
Variance
A pooled estimate of the common population variance, based on a sample
variance s12 from a sample of size n1 and a sample variance s22 from a sample
of size n2 is given by:
2
2
(
n

1
)
s

(
n

1
)
s
1
2
2
s2p  1
n1  n2  2
The degrees of freedom associated with this estimator is:
df = (n1+ n2-2)
The pooled estimate of the variance is a weighted average of the two
individual sample variances, with weights proportional to the sizes of the two
samples. That is, larger weight is given to the variance from the larger
sample.
8-27
Using the Pooled Estimate of the
Population Variance
The estimate of the standard deviation of (x1  x 2 ) is given by:
1 
2 1
sp 


 n1 n 2 
Test statistic for the difference between two population means, assuming equal
population variances:
(x1  x 2 )  (  1   2 ) 0
t=
1
2 1
sp 


n
n
 1 2
where (  1   2 ) 0 is the difference between the two population means under the null
hypothesis (zero or some other number D).
The number of degrees of freedom of the test statistic is df = ( n1  n2  2 ) (the
2
number of degrees of freedom associated with s p , the pooled estimate of the
population variance.
8-28
Example 8-5
Do the data provide sufficient evidence to conclude that average percentage increase in the CPI differs when oil
sells at these two different prices?
Population 1: Oil price = $66.00
n = 14
1
x = 0.317%
1
s = 0.12%
1
Population 2 : Oil price = $58.00
n2 = 9
x 2 = 0.21%
s 2 = 0.11%
df = (n  n  2)  (14  9  2)  21
1 2
H 0 : 1   2  0
H1:  1   2  0
( x1  x 2 )  (  1   2 ) 0
t 
 ( n1  1) s12  ( n2  1) s22   1 1 

  
n

n

2

  n1 n2 
1
2
0.107
0.107


 2.154
0
.
0497
0.00247
Critical point: t
= 2.080
0.025
H 0 may be rejected at the 5% level of significance
8-29
Example 8-5: Using the Template
Do the data provide sufficient evidence to conclude that average percentage increase in the CPI differs when oil
sells at these two different prices?
Decision: reject the null hypothesis.
8-30
Example 8-5: Using Minitab
Do the data provide sufficient evidence to conclude that average percentage increase in the CPI differs when oil
sells at these two different prices?
Decision: reject the null hypothesis; p-value = 0.043.
8-31
Example 8-6
The manufacturers of compact disk players want
to test whether a small price reduction is enough
to increase sales of their product. Is there
evidence that the small price reduction is enough
to increase sales of compact disk players?
H :  0
0
2
1
H :  0
1
2
1
t
Population 1: Before Reduction
n 1 = 15
x 1 = $6598

s1 = $844
Population 2: After Reduction
n 2 = 12
x 2 = $6870
s 2 = $669

( x  x )  (   )
2
1
2
1 0
2
2
 ( n  1) s  ( n  1) s  1 1
 1
1
2
2 


 n n
n n 2
1
2

 1 2




( 6870  6598)  0
 (14)8442  (11)6692  1 1 

  

 15 12 
15  12  2


272

89375.25
272
 0.91
298.96
Critical point : t
= 1.316
0.10
df = (n  n  2 )  (15  12  2 )  25
1
2
H may not be rejected even at the 10% level of significan ce
0
8-32
Example 8-6: Continued
t Distribution: df =25
0.4
f(t)
0.3
0.2
0.1
0.0
-5
-4
-3
-2
-1
Nonrejection
Region
0
1
2
3
4
t0.10=1.316
Rejection
Region
Test Statistic=0.91
5
t
Since the test statistic is less
than t0.10, the null hypothesis
cannot be rejected at any
reasonable level of
significance. We conclude
that the price reduction does
not significantly affect sales.
8-33
Example 8-6: Using the Template
Decision: Do not reject the null hypothesis; p-value = 0.1858.
8-34
Example 8-6: Using Minitab
Decision: Do not reject the null hypothesis; p-value = 0.186.
8-35
Confidence Intervals Using the Pooled
Variance
A (1-) 100% confidence interval for the difference between two
population means, 1- 2 , using independent random samples and
assuming equal population variances:
( x1  x2 )  t

2 1

sp 


 n1 n2 
1
2
A 95% confidence interval using the data in Example 8-6:
( x1  x 2 )  t

2
2
sp
 1  1


 n1 n2 
 ( 6870  6598 )  2 .06 ( 595835)( 0.15)  [ 343.85,887 .85]
8-36
Confidence Intervals Using the Pooled Variance
and the Template-Example 8-6
NOTE: The MINITAB outputs have the confidence
Intervals included in the output as well.
Confidence Interval
8-37
8-4 A Large-Sample Test for the Difference
between Two Population Proportions
•
Hypothesized difference is zero
 I: Difference between two population proportions is 0
•
•
p1= p2
» H0: p1 -p2 = 0
» H1: p1 -p20
 II: Difference between two population proportions is less than 0
• p1 p2
» H0: p1 -p2  0
» H1: p1 -p2 > 0
Hypothesized difference is other than zero:
 III: Difference between two population proportions is less than D
•
p1 p2+D
» H0:p-p2  D
» H1: p1 -p2 > D
8-38
8-4 A Large-Sample Test for the Difference
between Two Population Proportions
•
•
Hypothesized difference is zero
 IV: Difference between two population proportions is greater than 0
• p1 p2
» H0: p1 -p2  0
» H1: p1 -p2 < 0
Hypothesized difference is other than zero:
 V: Difference between two population proportions is greater than D
•
p1  p2+D
» H0:p-p2  D
» H1: p1 -p2 < D
8-39
Comparisons of Two Population Proportions When the
Hypothesized Difference Is Zero: Test Statistic
When the population proportions are hypothesized to be equal, then a pooled estimator of
the proportion ( p ) may be used in calculating the test statistic.
A large-sample test statistic for the difference between two population
proportions, when the hypothesized difference is zero:
z
( p1  p 2 )  0
1 1

p(1  p)  
 n1 n2 
x1
x
where p1 
is the sample proportion in sample 1 and p 1  1 is the sample
n1
n1
proportion in sample 2. The symbol p stands for the combined sample
proportion in both samples, considered as a single sample. That is:
pˆ 
x x
n n
1
1
1
2
8-40
Comparisons of Two Population Proportions When the
Hypothesized Difference Is Zero: Example 8-7
Carry out a two-tailed test of the equality of banks’ share of the car loan market in 1980 and 1995.
H 0 : p1  p 2  0
H 1: p1  p 2  0
Population 1: 1980
n1 = 100
x1 = 53
z 
p 1 = 0.53
p (1 
Population 2: 1995

n 2 = 100
p 2 = 0.43
p 
n1  n 2
1
p ) 
 n1
0.10

0.004992
x 2 = 43
x1 + x 2
( p1  p 2 )  0
Critical point: z

53  43
100  100
 0.48



n2 
1
0.10

0.53  0.43
 1  1 
 100 100
(.48)(.52) 
 1.415
0.07065
= 1.645
0.05
H 0 may not be rejected even at a 10%
level of significance.
8-41
Example 8-7: Carrying Out the Test
Standard Normal Distribution
0.4
f(z)
0.3
0.2
0.1
0.0
-z0.05=-1.645
Rejection
Region
0
z
z0.05=1.645
Nonrejection
Region
Test Statistic=1.415
Rejection
Region
Since the value of the test
statistic is within the
nonrejection region, even at a
10% level of significance, we
may conclude that there is no
statistically significant
difference between banks’
shares of car loans in 1980
and 1995.
8-42
Example 8-7: Using the Template
Decision: Do not reject the null hypothesis; p-value = 0.157.
8-43
Example 8-7: Using Minitab
Decision: Do not reject the null hypothesis; p-value = 0.157.
8-44
Comparisons of Two Population Proportions When the
Hypothesized Difference Is Not Zero: Example 8-8
Carry out a one-tailed test to determine whether the population proportion of traveler’s check buyers who buy at
least $2500 in checks when sweepstakes prizes are offered as at least 10% higher than the proportion of such
buyers when no sweepstakes are on.
n1 = 300
H 0 : p1  p 2  0.10
H 1 : p1  p 2  0.10
x1 = 120
z
Population 1: With Sweepstakes
p 1 = 0.40
Population 2: No Sweepstakes
n 2 = 700
x 2 = 140
p 2 = 0.20

( p1  p 2 )  D
 p (1  p )
1
 1
 n1 

p (1  p ) 
2
2 
n2
( 0.40  0.20)  0.10
 ( 0.40)( 0.60) ( 0.20)(.80) 



700
 300

Critical point: z



0.10
 3.118
0.03207
= 3.09
0.001
H 0 may be rejected at any common level of significance.
8-45
Example 8-8: Carrying Out the Test
Standard Normal Distribution
0.4
f(z)
0.3
0.2
0.1
0.0
0
Nonrejection
Region
z
z0.001=3.09
Rejection
Region
Test Statistic=3.118
Since the value of the test
statistic is above the critical
point, even for a level of
significance as small as 0.001,
the null hypothesis may be
rejected, and we may conclude
that the proportion of
customers buying at least
$2500 of travelers checks is at
least 10% higher when
sweepstakes are on.
8-46
Example 8-8: Using the Template
Decision: Reject the null hypothesis; p-value = 0.0009.
8-47
Example 8-8: Using Minitab
Decision: Reject the null hypothesis; p-value = 0.001.
8-48
Confidence Intervals for the Difference
between Two Population Proportions
A (1-) 100% large-sample confidence interval for the difference
between two population proportions:
( p 1  p 2 )  z

 p (1  p )
1
 1
 n1 

p (1  p ) 
2
2 
n2


2
A 95% confidence interval using the data in example 8-8:
 p1 (1  p1 ) p 2 (1  p 2 ) 

  ( 0.4  0.2)  1.96 ( 0.4 )( 0.6)  ( 0.2)( 0.8)
( p1  p 2 )  z

n2
300
700

 n1

2
 0.2  (1.96)( 0.0321)  0.2  0.063  [ 0.137 ,0.263]
8-49
Example 8-8 – Using the Template
Confidence Interval
8-50
Example 8-8 – Using Minitab
NOTE: In order to use Minitab to construct the confidence interval, you will have to
Make sure that the “Not Equal” Alternative option is selected.
8-51
8-5 The F Distribution and a Test for
Equality of Two Population Variances
The F distribution is the distribution of the ratio of two chi-square random variables
that are independent of each other, each of which is divided by its own degrees of
freedom.
An F random variable with k1 and k2 degrees of freedom:
Fk1, k 2 
12 k1
 2
 2 k2
8-52
The F Distribution
F Distributions with different Degrees of Freedom
f(F)
• The F random variable cannot
be negative, so it is bound by
zero on the left.
• The F distribution is skewed to
the right.
• The F distribution is identified
the number of degrees of
freedom in the numerator, k1,
and the number of degrees of
freedom in the denominator,
k2 .
F(25,30)
1.0
F(10,15)
0.5
F(5,6)
0.0
0
1
2
3
4
5
F
8-53
Using the Table of the F Distribution
Critical Points of the F Distribution Cutting Off a
Right-Tail Area of 0.05
1
2
3
4
5
6
7
8
9
k2
1 161.4 199.5 215.7 224.6 230.2 234.0 236.8 238.9 240.5
2 18.51 19.00 19.16 19.25 19.30 19.33 19.35 19.37 19.38
3 10.13 9.55 9.28 9.12 9.01 8.94 8.89 8.85 8.81
4
7.71 6.94 6.59 6.39 6.26 6.16 6.09 6.04 6.00
5
6.61 5.79 5.41 5.19 5.05 4.95 4.88 4.82 4.77
6
5.99 5.14 4.76 4.53 4.39 4.28 4.21 4.15 4.10
7
5.59 4.74 4.35 4.12 3.97 3.87 3.79 3.73 3.68
8
5.32 4.46 4.07 3.84 3.69 3.58 3.50 3.44 3.39
9
5.12 4.26 3.86 3.63 3.48 3.37 3.29 3.23 3.18
10
4.96 4.10 3.71 3.48 3.33 3.22 3.14 3.07 3.02
11
4.84 3.98 3.59 3.36 3.20 3.09 3.01
3.01 2.95 2.90
12
4.75 3.89 3.49 3.26 3.11 3.00 2.91 2.85 2.80
13
4.67 3.81 3.41 3.18 3.03 2.92 2.83 2.77 2.71
14
4.60 3.74 3.34 3.11 2.96 2.85 2.76 2.70 2.65
15
4.54 3.68 3.29 3.06 2.90 2.79 2.71 2.64 2.59
0.7
0.6
0.5
f(F)
k1
F Distribution with 7 and 11 Degrees of Freedom
0.4
0.3
0.2
0.1
F
0.0
0
The left-hand critical point to go along with F(k1,k2) is given by:
1
2
3
4
5
F0.05=3.01
1
F k 2 ,k 1
Where F(k1,k2) is the right-hand critical point for an F random variable with the reverse
number of degrees of freedom.
8-54
Critical Points of the F Distribution:
F(6, 9),  = 0.10
F Distribution with 6 and 9 Degrees of Freedom
0.7
0.05
0.90
0.6
The right-hand critical point read
directly from the table of the F
distribution is:
f(F)
0.5
F(6,9) =3.37
0.4
0.3
0.05
0.2
0.1
0.0
0
1
F0.95=(1/4.10)=0.2439
2
3
4
F0.05=3.37
5
F
The corresponding left-hand critical
point is given by:
1
1

 0.2439
F 9 , 6 410
.
8-55
Test Statistic for the Equality of Two
Population Variances
Test statistic for the equality of the variances of two normally
distributed populations:
F n 1,n 1
1
2
s12
 2
s2
 I: Two-Tailed Test
1 = 2
• H0: 1 = 2
• H1: 2
 II: One-Tailed Test
• 12
• H0: 1  2
• H1: 1  2
•
8-56
Example 8-9
The economist wants to test whether or not the event (interceptions and prosecution of insider
traders) has decreased the variance of prices of stocks.
Population1 : Before
n = 25
1
s 2  9 .3
1
Population 2 : After
n
= 24
2
s 2  3 .0
2
24,23

 2.01
  0.01
F
24,23
H1: 
1
2
1


2
2
21
2
2
s2
9.3
1
F
 F


 3.1
3.0
n1  1, n 2  1
24,23
s2
2
  0.05
F
H 0: 
2
 2.70



H 0 may be rejected at a 1% level of significance.
8-57
Example 8-9: Solution
Distribution with 24 and 23 Degrees of Freedom
0.7
0.6
f(F)
0.5
0.4
0.3
0.2
0.1
F
0.0
0
1
2
F0.01=2.7
3
4
5
Test Statistic=3.1
Since the value of the test
statistic is above the critical
point, even for a level of
significance as small as 0.01,
the null hypothesis may be
rejected, and we may conclude
that the variance of stock
prices is reduced after the
interception and prosecution
of inside traders.
8-58
Example 8-9: Solution Using the
Template
Decision: Reject the null hypothesis; p-value = 0.0042.
8-59
Example 8-10: Testing the Equality of
Variances for Example 8-5
Population 1 Population 2
n = 14
1
n =9
2
2
2
s  0.12
1
2
2
s  0.11
2
  0.05
F
13,8
 3.28
  0.10
F
13,8
 2.50
2
2
H :  
0 1
2
2
2
H :  
1 1
2
s2
0.122
1
F
F


 119
.
2
2
n1  1, n2  1 13,8 s 0.11
2
H may not be rejected at the 10% level of significance.
0
8-60
Example 8-10: Solution
F Distribution with 13 and 8 Degrees of Freedom
0.7
0.10
0.80
0.6
f(F)
0.5
0.4
0.3
0.10
0.2
0.1
0.0
0
1
F0.90=(1/2.20)=0.4545
Test Statistic=1.19
2
3
4
F0.10=3.28
5
F
Since the value of the test
statistic is between the critical
points, even for a 20% level of
significance, we can not reject
the null hypothesis. We
conclude the two population
variances are equal.
8-61
Template to test for the Difference between Two
Population Variances: Example 8-10
Decision: Do not reject the null hypothesis; p-value = 0.8304;
Assume equal variances..
8-62
Example 8-10: Using Minitab to Test for
Equal Variances
Confidence intervals overlap with
sample estimates in both – assume
Equal variances.
Test for Equal Variances
F-Test
Test Statistic
P-Value
1
Decision: Do not reject
the null hypothesis;
p-value = 0.830;
Assume equal variances.
2
0.05
0.10
0.15
0.20
95% Bonferroni Confidence Intervals for StDevs
0.25
1.19
0.830
8-63
The F Distribution Template
8-64
The Template for Testing Equality of
Variances with data
Do not reject the
Null hypothesis for
Equality of variances
Since P-value = 0.6882
8-65
Using Minitab to test for the Equality of
Variances with data
Do not reject the
null hypothesis for
equality of variances
since P-values are
large for both the
F-test and Levine’s test.
8-66
Using Minitab to test for the Equality of
Variances with data
Test for Equal Variances for Data
F-Test
Test Statistic
P-Value
1
0.81
0.688
Sample
Lev ene's Test
Test Statistic
P-Value
2
200
300
400
500
95% Bonferroni Confidence Intervals for StDevs
600
Sample
1
2
0
200
400
600
Data
800
1000
1200
0.07
0.799
Do not reject the
null hypothesis for
equality of variances
since the confidence
intervals for the
standard deviations
overlap.