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
83.04    86.96
X  Z
8.5
If all possible samples of the same size n=100 are taken, 95% of them will include the
true population mean annual income within the interval developed. Thus you are 95
percent confident that this sample is one that does correctly estimate the true mean annual
income.
8.7
If the population mean annual income is $71,000, the confidence interval estimate stated
in Problem 8.5 is correct because it contains $71,000.
8.11
X t
S
24
 75  2.0301
n
36
8.17
(a)
X t
(b)
(c)
No, a grade of 200 is in the interval.
It is not unusual. A tread-wear index of 210 for a particular tire is only 0.69
Standard deviation above the sample mean of 195.3.
n
8.35
8.47
(a)
n
= 85  1.96 
8
64
8.1
66.8796    83.1204
S
21.4
 195.3  2.1098 
n
18
Z 2  2 2.582 100 2
= 166.41

e2
202
p = 315/500 = 0.63
pZ
184.6581    205.9419
Use n = 167
p 1  p 
0.63 1  0.63
 0.63  1.96
n
500
0.5877    0.6723
(b)
You are 95% confident that the proportion of companies who informally
monitored social networking sites to stay on top of information related to their
company is somewhere between 0.5877 and 0.6723.
(c)
n
Z 2 1    1.962  0.631  0.63

= 8,954.44
e2
0.012
Use n = 8,955
8.51
Using PHStat,
Confidence Interval Estimate for the Total
Difference
Data
Population Size
Sample Size
Confidence Level
10000
200
95%
Intermediate Calculations
Sum of Differences
Average Difference in Sample
Total Difference
Standard Deviation of Differences
FPC Factor
Standard Error of the Total Diff.
Degrees of Freedom
t Value
Interval Half Width
200.63
1.00315
10031.5
4.998502
0.989999
3499.126
199
1.971957
6900.128
Confidence Interval
Interval Lower Limit
Interval Upper Limit
3131.37
16931.63
N  D  N t 
SD
n
N n
4.9985 10, 000  200
 10, 000 1.00315  10, 000 1.972 

N 1
10, 000  1
200
3131.37  Total Difference in the Population  16931.63