Chapter 11 solution.docx - Chapter 11 11.7 Rejection region...

Chapter 11
11.7 Rejection region: z <
575 . 2
005 .
   z
or z >
005 .
z
= 2.575
00 . 1
100 / 200
1000 980
n /
x
z  



 

p-value = 2P(Z < –1.00) = 2(.1587) = .3174
There is not enough evidence to infer that
 
1000.
11.11 Rejection region: z >
01 .
z
= 2.33
00 . 5
100 / 20
70 80
n /
x
z 



 

p-value = p(z > 5.00) = 0
There is enough evidence to infer that

> 70.

11.15 a.
00 . 1
25 / 5
20 21
n /
x
z 



 

p-value = 2P(Z > 1.00) = 2(1 – .8413) = .3174
b.
00 . 2
25 / 5
20 22
n /
x
z 



 

p-value = 2P(Z > 2.00) = 2(1 – .9772) = .0456
c.
00 . 3
25 / 5
20 23
n /
x
z 



 

p-value = 2P(Z > 3.00) = 2(1 – .9987) = .0026
d. The value of the test statistic increases and the p-value decreases.
11.17 a.
00 . 4
100 / 25
1000 990
n /
x
z  



 

p-value = P(Z < –4.00) = 0
b.
00 . 2
100 / 50
1000 990
n /
x
z  



 

p-value = P(Z < –2.00) = .0228

c.
00 . 1
100 / 100
1000 990
n /
x
z  



 

p-value = P(Z < –1.00) = .1587
d. d. The value of the test statistic increases and the p-value increases.
11.29
 : H
0
= 50
 : H
1
> 50
89 . 3
18 / 10
50 17 . 59
n /
x
z 



 

p-value = P(Z > 3.89) = 0