Chapter 11
2 
3.
(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
Actual p-value = .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
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
11.
H0:  2 = .009216
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
Actual p-value = .2191
p-value > .05, do not reject H0. Cannot conclude the variance in interest rates has changed.
11 - 1
13. a. F.05 = 3.33
b. F.025 = 2.76
c. F.01 = 4.50
d.
19.
F.10  1.94
H 0 : 12   22
H a : 12   22
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
Actual 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.
11 - 2
Statistical Inference about Means and Proportions with Two Populations
21. a.
( xi  x )2
n 1
s2 
2
sNov
= 9664
2
sDec
= 19,238 (Population 1 since s 2 is larger)
b.
H 0 : 12   22
H a : 12   22
F
s12 19, 238

 1.99
9664
s22
Degrees of freedom 9 and 9
Using F table, area in tail is greater than .10
Two-tail p-value is greater than .20
Actual p-value = .3197
p-value > .05, do not reject H0. There is no evidence that the population variances differ.
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
Actual p-value = .5086
p-value > .10, do not reject H0
11 - 3
30. a. Try n = 15
2
 .025
= 26.119
2
 .975
= 5.629 (14 degrees of freedom)
(14)(64)
(14)(64)
2 
26.119
5.629
2
34.3    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
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
Actual p-value = .1091
p-value > .10, do not reject H0. Cannot conclude that there is a difference between the population variances.
11 - 4
Statistical Inference about Means and Proportions with Two Populations
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
Actual p-value = .0689
p-value  .10, reject H0. Conclude that the population variances are not equal.
11 - 5