Math 370 HW 7 - Homework 7 Solution 5-3 (a, c), 5-8, 5-19,...

1
Homework 7 Solution
5-3 (a, c), 5-8, 5-19, 5-20, 5-37, 5-38
5-3 a) 1) The parameter of interest is the difference in fill volume,
 
1 2

2) H
0
:  
1 2
0   or  
1 2

3) H
1
:
 
1 2
0  
or
 
1 2

4) The test statistic is
z
x x
n n
0
1 2 0
1
2
1
2
2
2

 

( ) 
 
5) Reject H
0
if z
0
<  z

/2
=  1.96 or z
0
> z

/2
= 1.96
6) x
1
 16.015 x
2
 16.005  = 0

1

0.03

2

0.025
n
1
= 10 n
2
= 10
81 . 0
10
) 025 . 0 (
10
) 03 . 0 (
0 ) 005 . 16 015 . 16 (
2 2
0


 
 z
7) P-value =
418 . 0 ) 7910 . 0 1 ( 2 )) 81 . 0 ( 1 ( 2     
Because the P-value is > 0.05, fail to reject the null hypothesis. There is not sufficient evidence to conclude that the
means of the two fill volumes are different at  = 0.05.
b) Power = 1   , where

   
 
 
















  

















 

 
z
n n
z
n n
/ / 2
0
1
2
1
2
2
2
2
0
1
2
1
2
2
2
=















   















 
10
) 025 . 0 (
10
) 02 . 0 (
04 . 0
96 . 1
10
) 025 . 0 (
10
) 02 . 0 (
04 . 0
96 . 1
2 2 2 2
=
       
91 . 5 99 . 1 95 . 3 96 . 1 95 . 3 96 . 1            
= 0.0233  0
= 0.0233

2
Power = 1  0.0233 = 0.977
c)
   
x x z
n n
x x z
n n
1 2 2
1
2
1
2
2
2
1 2 1 2 2
1
2
1
2
2
2
        
 
 
 
 
/ /
   
16 015 16 005 196
0 02
10
0 025
10
16 015 16 005 196
0 02
10
0 025
10
2 2
1 2
2 2
. . .
( . ) ( . )
. . .
( . ) ( . )
          
    00098 00298
1 2
. .  
With 95% confidence, the difference in the mean fill volumes is between  0.0098 and 0.0298. Because 0 is contained
in this interval, we can conclude there is no significant difference between the means.
d) Assume the sample sizes are to be equal, use  = 0.05,  = 0.01, and  = 0.04
 
 
 
 
, 08 . 2
) 04 . 0 (
) 025 . 0 ( ) 02 . 0 ( 33 . 2 96 . 1
z z
n
2
2 2 2
2
2
2
2
1
2
2 /

 

 


 
 
n = 11.79, use n
1
= n
2
= 12
5-8. x
1
 88.85 x
2
 92.54

1
2

1.5
