Math 370 HW 6 - Homework 6 Solution 4-43 (a, c), 4-46, 4-54...

1
Homework 6 Solution
4-43 (a, c), 4-46, 4-54 (a, c), 4-55 (a, c, d), 4-56, 4-57
4-43. a) 1) The parameter of interest is the true mean compressive strength,  .
2) H
0
:  = 3500
3) H
1
:   3500
4)
z
x
n
0

 
 /
5) Reject H
0
if z
0
<  z
 /2
where  z
0.005
=  2.58 or z
0
> z
 /2
where z
0.005
= 2.58
6)
42 . 3255 x 
,  = 31.62
79 . 26
12 / 62 . 31
3500 42 . 3255
z
0
 


7) Because  26.79 <  2.58, reject the null hypothesis and conclude the mean
compressive strength is significantly different from 3500 at  = 0.01.
b) Smallest level of significance = P-value = 2[1 

(26.84) ]= 2[1  1] = 0
The smallest level of significance at which we are willing to reject the null hypothesis
is approximately 0.
c) z
 /2
= z
0.025
= 1.96
x z
n
x z
n






 
 






0 025 0 025 . .











  









12
62 . 31
96 . 1 42 . 3255
12
62 . 31
96 . 1 42 . 3255 
3237.53    3273.31
With 95% confidence, the true mean compressive strength is between 3237.53 psi and
3273.31psi.

2
d) z
 /2
= z
0.005
= 2.58
x z
n
x z
n






   






0 005 0 005 . .











  









12
62 . 31
58 . 2 42 . 3255
12
62 . 31
58 . 2 42 . 3255 
3231.96    3278.88