A Review of Chapters 8 and 9
Answers to Section Three Exercises
A Review of Chapters 8 and 9
1.
5.
9.
10.
D
A
2.
6.
D
A
3.
7.
D
B
4.
8.
C
C



8.8  8.6   = Pr {z  0.59} = 0.5000  0.2224 = 0.2776
Pr  z 

2.0



35 



1100

1210

  = Pr {z   1.69} = 0.5000  0.4545 = 0.0455
Pr  z 

325


25


 20 
.
 40 
11.
The 98% confidence interval is from 152.33 up to 167.67, found by 160  2.46 
12.
The 90% confidence interval is from 2.4826 up to 2.8174, found by 2.65  1.664 
13.
 115.5 
985.5  2.571
 for an interval from 864.27 up to 1106.73.
 6 
14.
Since they are not randomly selected, it is not reasonable to use these data.
15.
 35 
240  2.131
 for an interval from 221.4 up to 258.6. Since 250 is included in the
 16 
 0.45 
.
 20 
confidence interval, there is no evidence of increased production.
2
2
16.
 (1.96)(0.9) 
n
  311.2 use 312.
0.1


18.
(0.6)(0.4)
; 0.532 up to 0.668
0.6  1.96
200
20.
0.25  2.33
21.
 2.33 
n  (0.4)(0.6) 
  1448
 0.03 
17.
 (1.96)(25) 
n
  150
4


19.
 2.33 
n  (0.08)(0.92) 
  999
 0.02 
2
(0.25)(0.75)
; from 0.205 up to 0.295. Yes, the proportion is less than 0.30.
500
2
Review 3-1
A Review of Chapters 8 and 9
SOLUTION TO CASE I
CENTURY NATIONAL BANK
Twenty-six of the 60, or 0.4333 of the customers use a debit card. A 95 percent confidence interval is
computed as follows:
0.4333(1  0.4333)
0.4333  1.96
 0.4333  0.1254
60
The confidence limits are from 0.3079 up to 0.5587. The value of 0.50 is in the interval. Hence we
cannot conclude that more than 0.50 of the customers use the card.
To find out if the mean account balance is now less than $1600, we conduct the following test of
hypothesis:
Ho:   1600 H1:  < 1600 Ho is rejected if z < 1.65
1499.90  1600
z
 1.299 Ho is not rejected. We cannot conclude that the mean is less than $1600.
596.7 / 60
To look at the mean number of ATM transactions, we conduct the following tests:
Ho:   10
H1:  > 10
Using the 0.05 significance level, Ho is rejected if z > 1.65
10.30  10
z
 0.54 Ho is not rejected. We cannot conclude that the mean number of transactions is
4.295 / 60
more than 10. The p-value is 0.2946
Can we conclude that it is more than 9?
Ho:   9
H1:  > 9
Reject Ho if z > 1.65
10.30  9
z
 2.345 Ho is rejected. We can conclude that the mean number of ATM transactions is
4.295 / 60
more than nine per month.
SOLUTION TO CASE II
KLEIN MARKETING RESEARCH
The mean cost for the 35 observations at the Marsh Store is $131.17 with a standard deviation of $4.89.
The mean cost at Kazmaier’s is $132.69, with a standard deviation of $4.99, for a sample of 38
observations. We conduct the following test of hypothesis:
Ho: m = k
H1: m  k
131.17  132.69
z
 1.31
(4.89)2 (4.99)2

35
38
For a two-tailed test the p-value is 0.1902, found by 2(0.5000 – 0.4049). We conclude that there is not a
difference in the prices.
Review 3-2
A Review of Chapters 8 and 9
SOLUTION TO CASE III
LEVENSON BROTHERS DEPARTMENT STORE
First conduct a test of hypothesis regarding the mean unpaid balance. The mean for the 50 customers is
$1074 with a standard deviation of $968.
Ho:   3900
H1:  < 3900
Reject Ho if z < 1.65, using  = 0.05
1074  3900
z
 20.6 The mean balance for the Levenson customer is less than $3900.
968 / 50
Regarding the fraction who pay their balance at the end of the month.
Ho:   0.33
H1:  > 0.33
Reject Ho if z > 1.65, using  = 0.05
0.36  0.33
z
 0.45
0.33(1  0.33)
50
Cannot reject Ho. Cannot conclude that the proportion who pays their balance is different from 0.33.
Review 3-3
A Review of Chapters 8 and 9
PRACTICE TEST
Part I
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
random sample
sampling error
standard error
become smaller
sample statistic
confidence interval
population size
proportion
positively skewed
0.5
Part II
11  12.2
 1.81
2.3
12
1.
0.0351, found by 0.5000-0.4649. The corresponding z 
2.
a.
b.
c.
The population mean is unknown.
9.3 years, which is the sample mean
0.3922, found by 2/√26
d.
The confidence interval is from 8.63 up to 9.97, found by 9.3  1.708
 2 

 26 
2
3.
 2.33 
2675, found by .271  .27 

 .02 
4.
The confidence interval is from 0.5459 up to 0.7341, found by .64  1.96
Review 3-4
.641  .64
100