Chapter 11
Inferences About Population Variances
Learning Objectives
1.
Understand the importance of variance in a decision-making situation.
2
Understand the role of statistical inference in developing conclusions about the variance of a single
population.
3.
Know that the sampling distribution of (n - 1) s2/ 2 has a chi-square distribution and be able to use
this result to develop a confidence interval estimate of  2.
4.
Be able to compute p-values using the chi-square distribution.
5.
Know how to test hypotheses involving  2.
6.
Understand the role of statistical inference in developing conclusions about two population variances.
7.
Know that the sampling distribution of s12 / s22 has an F distribution and be able to use this result to
test hypotheses involving two population variances.
8.
Be able to compute p-values using the F distribution.
11 - 1
Chapter 11
Solutions:
1.
a. 11.070
b. 27.488
c. 9.591
d. 23.209
e. 9.390
2.
s2 = 25
2
2
a. With 19 degrees of freedom  .05
= 30.144 and  .95
= 10.117
19(25)
19(25)
2 
30.144
10.117
15.76   2  46.95
2
2
b. With 19 degrees of freedom  .025
= 32.852 and  .975
= 8.907
19(25)
19(25)
2 
32.852
8.907
14.46   2  53.33
c. 3.8    7.3
3.
2 
(n  1) s 2
 02

(16  1)(9.5) 2
 27.08
50
Degrees of Freedom = (16 - 1) = 15
Using  2 table, p-value is between .025 and .05
Exact p-value using  2 =27.08 is .0281
p-value  .05, reject H0
Critical value approach
2
 .05
= 24.996
Reject H0 if  2  24.996
27.08 > 24.996, reject H0
11 - 2
Inferences About Population Variances
4.
a. n = 18
s2 = .36
2
 .05
= 27.587
2
 .95
= 8.672 (17 degrees of freedom)
17(.36)
17(.36)
2 
27.587
8.672
.22   2  .71
b. .47    .84
5.
a.
s2 
( xi  x )2
 31.07
n 1
s  31.07  5.57
b.
2
 .025
= 16.013
2
 .975
= 1.690
(8  1)(31.07)
(8  1)(31.07)
2 
16.013
1.690
13.6   2  128.7
c. 3.7    11.3
6.
a.
x
 xi 4.41

 .2205
n
12
s2 
( xi  x )2
 47.9500
n 1
s  47.9500  6.92
b.
2
 .025
= 32.852
2
 .975
= 8.907
(20  1)(47.95)
(20  1)(47.95)
2 
32.852
8.907
27.73   2  102.28
5.27   10.11
The 95% confidence interval for the population standard deviation ranges from a quarterly standard
deviation of 5.27% to 10.11%.
11 - 3
Chapter 11
7.
a.
s2 
( xi  x )2
 57.7230
n 1
s  57.7230  7.60
b.
2
With 11 df,  .025
= 21.920
2
 .975
= 3.816
(12  1)(57.7230)
(12  1)(57.7230)
2 
21.920
3.816
28.9668   2  166.3923
c. 5.38    12.90
8.
( xi  x )2
.0929

 .00845
n 1
12  1
a.
s2 
b.
s  .00845  .092
c. 11 degrees of freedom
2
 .025
= 21.920
2
 .975
= 3.816
(12  1).00845
(12  1).00845
2 
21.920
3.816
.0042   2  .0244
.065   .156
9.
H0:  2 .0004
Ha:  2 .0004
2 
(n  1) s 2

2
0

(30  1)(.0005)
 36.25
.0004
Degrees of freedom = n - 1 = 29
Using  2 table, p-value is greater than .10
Exact p-value using  2 = 36.25 is .1664
p-value > .05, do not reject H0. The product specification does not appear to be violated.
10.
 2 18.2)2 = 331.24
H0:  2 331.24
11 - 4
Inferences About Population Variances
Ha:  2 331.24
2 
(n  1) s 2
 02

(36  1)(22.2) 2
 52.075
(18.2)2
Degrees of freedom = n - 1 = 35
Using  2 table, p-value is between .025 and .05
Exact p-value corresponding to  2 = 52.075 is .0317
p-value  .05, reject H0. The standard deviation for Vanguard PRIMECAP fund is greater than the
average standard deviation for large cap mutual funds.
H0:  2 = .009216
11.
Ha:  2  .009216
2 
(n  1) s 2

2

(20  1)(.114) 2
 26.79
.009216
Degrees of freedom = n - 1 = 19
Using  2 table, area in upper tail is greater than .10
Two-tail p-value is greater than .20
Exact p-value corresponding to  2 = 26.79 is .2193
p-value > .05, do not reject H0. Cannot conclude the variance in interest rates has changed.
12. a.
s2 
( xi  x )2
 .8106
n 1
b. H0:  2 .94
Ha:  2 .94
2 
(n  1) s 2

2
0

(12  1)(.8106)
 9.49
.94
Degrees of freedom = n - 1 = 11
Using  2 table, area in tail is greater than .10
Two-tail p-value is greater than .20
Exact p-value corresponding to  2 = 9.49 is .8465
p-value > .05, cannot reject H0.
11 - 5
Chapter 11
13. a. F.05 = 3.33
b. F.025 = 2.76
c. F.01 = 4.50
d.
14. a.
F.10  1.94
F
s12 5.8

 2.4
s22 2.4
Degrees of freedom 15 and 20
Using F table, p-value is between .025 and .05
Exact p-value corresponding to F = 2.4 is .0345
p-value  .05, reject H0. Conclude  12   22
b. F.05 = 2.20
Reject H0 if F  2.20
2.4  2.20, reject H0. Conclude  12   22
15. a. Larger sample variance is s12
F
s12 8.2

 2.05
4
s22
Degrees of freedom 20 and 25
Using F table, area in tail is between .025 and .05
Two-tail p-value is between .05 and .10
Exact p-value corresponding to F = 2.05 is .0904
p-value > .05, do not reject H0.
b. Since we have a two-tailed test
F / 2  F.025  2.30
Reject H0 if F  2.30
2.05 < 2.30, do not reject H0
16.
H 0 : 12   22
H a : 12   22
11 - 6
Inferences About Population Variances
s12 942

 2.63
s22 582
Degrees of freedom 25 and 29
F
Using F table, p-value is less than .01
Exact p-value corresponding to F = 2.63 is .0067
p-value  .01, reject H0. Conclude adults have a greater variance in online times.
17. a. Population 1 is 4 year old automobiles
H 0 : 12   22
H a : 12   22
b.
F
s12 1702

 2.89
s22 1002
Degrees of freedom 25 and 24
Using F table, p-value is less than .01
Exact p-value corresponding to F = 2.89 is .0057
p-value  .01, reject H0. Conclude that 4 year old automobiles have a larger variance in annual repair
costs compared to 2 year old automobiles. This is expected due to the fact that older automobiles are
more likely to have some more expensive repairs which lead to greater variance in the annual repair
costs.
18.
H 0 : 12   22
H a : 12   22
F
s12 4.27 2

 3.54
s22 2.27 2
Degrees of freedom 9 and 6
Using F table, area in tail is between .05 and .10
Two-tail p-value is between .10 and .20
Exact p-value corresponding to F = 3.54 is .1378
p-value > .05, do not reject H0. Cannot conclude any difference between variances of the two
industries.
19.
H 0 : 12   22
H a : 12   22
11 - 7
Chapter 11
s12  .0489
s22  .0059
F
s12 .0489

 8.28
s22 .0059
Degrees of freedom 24 and 21
Using F table, area in tail is less than .01
Two-tail p-value is less than .02
Exact p-value  0
p-value  .05, reject H0. The process variances are significantly different. Machine 1 offers the best
opportunity for process quality improvements.
Note that the sample means are similar with the mean bag weights of approximately 3.3 grams.
However, the process variances are significantly different.
20.
H 0 : 12   22
H a : 12   22
F
s12 11.1

 5.29
2.1
s22
Degrees of freedom 25 and 24
Using F table, area in tail is less than .01
Two-tail p-value is less than .02
Exact p-value  0
p-value  .05, reject H0. The population variances are not equal for seniors and managers.
21.
Consider the Small cap fund as population 1 and the large cap fund as population 2.
H 0 : 12   22
H a : 12   22
F
s12 (13.03) 2

 2.15
s22
(8.89) 2
Degrees of freedom 25 and 25
Upper-tail p-value is the area to the right of the test statistic.
11 - 8
Inferences About Population Variances
From the F table, the p-value is between .025 and .05
Exact p-value corresponding to F = 2.15 is .0306
p-value ≤ .05, reject H0. The population variance and standard deviation for the small cap growth
fund are larger. Financial analysts would say the small cap growth fund is riskier.
22. a. Population 1 - Wet pavement.
H 0 : 12   22
H a : 12   22
F
s12 322

 4.00
s22 162
Degrees of freedom 15 and 15
Using F table, p-value is less than .01
Exact p-value corresponding to F = 4.00 is .0054
p-value  .05, reject H0. Conclude that there is greater variability in stopping distances on wet
pavement.
b. Drive carefully on wet pavement because of the uncertainty in stopping distances.
23. a. s2 = (30) 2 = 900
b.
2
2
 .05
= 30.144 and  .95
= 10.117 (19 degrees of freedom)
(19)(900)
(19)(900)
2 
30.144
10.117
567   2  1690
c. 23.8    41.1
24.
With 12 degrees of freedom,
2
 .025
= 23.337
2
 .975
= 4.404
(12)(14.95)2
(12)(14.95)2
2 
23.337
4.404
114.9   2  609
10.72    24.68
11 - 9
Chapter 11
25. a.
b.
x
xi
 260.16
n
s2 
( xi  x )2
 4996.8
n 1
s  4996.79  70.69
c.
2
 .025
= 32.852
2
 .975
= 8.907 (19 degrees of freedom)
(20  1)(4996.8)
(20  1)(4996.8)
2 
32.852
8.907
2890   2  10,659
53.76    103.24
26. a. H0:  2 .0001
Ha:  2 .0001
2 
(n  1) s 2
2

(15  1)(.014) 2
 27.44
.0001
Degrees of freedom = n - 1 = 14
Using  2 table, p- value is between .01 and .025
Exact p-value corresponding to  2 = 27.44 is .0169
p-value  .10, reject H0. Variance exceeds maximum variance requirement.
b.
2
 .05
= 23.685
2
 .95
= 6.571 (14 degrees of freedom)
(14)(.014)2
(14)(.014)2
2 
23.685
6.571
.00012   2  .00042
27.
H0:  2 .02
Ha:  2 .02
2 
(n  1) s 2

2
0

(41  1)(.16) 2
 51.20
.02
Degrees of freedom = n - 1 = 40
11 - 10
Inferences About Population Variances
Using  2 table, p- value is greater than .10
Exact p-value corresponding to  2 = 51.20 is .1104
p-value > .05, do not reject H0. The population variance does not appear to be exceeding the
standard.
28.
H0:  2 
Ha:  2 
2 
(n  1) s 2

2
0

(22  1)(1.5)
 31.50
1
Degrees of freedom = n - 1 = 21
Using  2 table, p-value is between .05 and .10
Exact p-value corresponding to  2 = 31.50 is .0657
p-value  .10, reject H0. Conclude that  2 > 1.
29.
s2 
( xi  x )2 101.56

 12.69
n 1
9 1
H0:  2 = 10
Ha:  2  10
2 
(n  1) s 2

2

(9  1)(12.69)
 10.16
10
Degrees of freedom = n - 1 = 8
Using  2 table, area in tail is greater than .10
Two-tail p- value is greater than .20
Exact p-value corresponding to  2 = 10.16 is .5080
p-value > .10, do not reject H0
30. a. Try n = 15
2
 .025
= 26.119
2
 .975
= 5.629 (14 degrees of freedom)
11 - 11
Chapter 11
(14)(64)
(14)(64)
2 
26.119
5.629
34.3   2  159.2
5.86    12.62
A sample size of 15 was used.
b. n = 25; expect the width of the interval to be smaller.
2
 .05
= 39.364
2
 .975
= 12.401 (24 degrees of freedom)
(24)(8)2
(24)(8)2
2 
39.364
12.401
39.02   2  126.86
6.25    11.13
31.
H 0 : 12   22
H a : 12   22
Population 1 is NASDAQ
F
s12 15.82

 4.00
s22
7.92
Degrees of freedom 9 and 9
Using F table, area in tail is between .025 and .05
Two-tail p-value is between .05 and .10
Exact p-value corresponding to F = 4.00 is .0510
p-value > .05, do not reject H0. Cannot conclude population variances differ. But with such a low pvalue, a larger sample is recommended.
32.
H 0 : 12   22
H a : 12   22
Use critical value approach since F tables do not have 351 and 72 degrees of freedom.
F.025 = 1.47
Reject H0 if F  1.47
11 - 12
Inferences About Population Variances
F
s12 .9402

 1.39
s22 .797 2
F < 1.47, do not reject H0. We are not able to conclude students who complete the course and
students who drop out have different variances of grade point averages.
33.
H 0 : 12   22
H a : 12   22
Population 1 has the larger sample variance.
F
s12 5.4

 2.35
s22 2.3
Degrees of freedom 15 and 15
Using F table, area in tail is between .05 and .10
Two-tail p-value is between .10 and .20
Exact p-value corresponding to F = 2.35 is .1087
p-value > .10, do not reject H0. Cannot conclude that there is a difference between the population
variances.
34.
H 0 : 12   22
H a : 12   22
F
s12 25

 2.08
s22 12
Degrees of freedom 30 and 24
Using F table, area in tail is between .025 and .05
Two-tail p-value is between .05 and .10
Exact p-value corresponding to F = 2.08 is .0695
p-value  .10, reject H0. Conclude that the population variances are not equal.
11 - 13