Problem Solving
11/08/2007
http://www.rossmanchance.com/applets/Confsim/Confsim.html
1. T/F The sample mean of an SRS equals its corresponding population mean.
False. A sample mean is usually a random variable, and its corresponding population mean is a
fixed constant.
2. T/F The length of a 95% CI is longer than a 90%CI based on the same SRS for the same
parameter.
False. The way to determine CI is not unique. But, if the two CIs are both obtained by using
(mean±zs*SD), the statement is true.
3. T/F A 95% CI, obtained from a valid SRS, for the corresponding population mean can capture
the sample mean approximately 95% of the time.
False. 95%CI for a population mean can capture the population mean approximately 95% of the
time. Usually, the sample mean should be always included in a CI for a population mean.
4. T/F In terms of a given valid SRS, there exists only one valid 95% CI for the corresponding
population mean.
False. We may have two-sided CI or one-sided CI.
9.55 The article “Selected Characteristics of High-Risk Students and Their Enrollment
Persistence” (Journal of College Student Development [1994]: 54-60) examined factors that affect
whether students stay in college. The following summary statistics are based on data from a
sample of 44 students who did not return to college after the first quarter (the non-persisters) and a
sample of 257 students who did return (the persisters):
Number of Hours Worked per week during the First Quarter
Mean
Standard Deviation
Non-persisters
25.62
14.41
Persisters
18.10
15.31
a.
Suppose the sample of 44 non-persisters is a SRS from the population of all
non-persisters. With z critical value, construct a 98% CI and a 90%CI for the mean
numbers of hours worked per week for non-persisters.
98% CI: (meannon ± z .99 SDnon) =
14.41 

 25.62  2.326 *
 =(20.57, 30.67)
44 

90% CI: (meannon ± z .95 SDnon) =
14.41 

 25.62  1.645 *
 =(22.05, 29.19)
44 

b. Based on the 98%CI in (a), do you think that the mean numbers of hours worked per
week for non-persisters is greater than 21?
No, we do not. Since the 98% CI contains 21, we do not have enough evidence to claim
that the mean numbers of hours worked per week for non-persisters is greater than 21 in
terms of the 98% CI.
c.
Find the 1% percentile for the mean numbers of hours worked per week for
non-persisters.
1% percentile: (meannon + z .01 SDnon) =
14.41 

 25.62  2.326 *
 = 20.57, which is
44 

the lower bound of the 98% CI obtained in (a).
d. Based on the 90%CI in (a), do you think that the mean numbers of hours worked per
week for non-persisters is greater than 21?
Yes, we do. Since the 90% CI is greater than 21, we can claim that the mean numbers of
hours worked per week for non-persisters is greater than 21with 90% Confidence.
e. Does the answer for (b) agree with the one for (d)? Why?
No, the two answers are apparently different. The reason for this difference is that the
confidence level in (c) is smaller than that in (b).
f. Suppose the sample of 257 persisters is a SRS from the population of all persisters. With
z critical value, construct a 98% CI for the mean numbers of hours worked per week for
persisters.

98% CI: (meanper ± z .99 SDper) = 18.10  2.326 *

15.31 
 =(15.88, 20.32)
257 
g. Why the 98%CI for persisters is narrower than the one for non-persisters?
The standard deviation of the sample mean for persisters is smaller than that for
non-persisters, because the sample size of persisters is much larger than that for the
non-persisters but their sample standard deviations are similar.
h. Suppose the two samples are both SRS with respect to the corresponding populations.
With z critical value, construct a 98% CI for the difference of the mean numbers of hours
worked per week for the two populations.
2
2
 14.41   15.31 
 =2.373.
The SD for the difference is about: SDdiff = 
  
 44   257 
The mean difference of the two sample means is 25.62-18.10 = 7.52.
98% CI for the difference: (meandiff ± z .99 SDdiff)
= 7.52  2.326 * 2.373 =(2, 13.04)
i.
j.
Based on the 98% CI obtained in (h), can you claim that you are 100% sure that the
difference of the mean numbers of hours worked per week for the two populations is
significant or not zero.
No, we can not. Although zero is not covered by the CI for the difference, we can not say
“100% sure” since we have only 98% confidence about the statement.
Redo (a) by using t critical value (with degree of freedom = 43), and compare your new
98%CI with the one you obtained in (a). Which one is better? Why?
t.99,df  43  2.416 .
98%CI on t critical value: (meannon ± t.99,df  43 SDnon) =

=  25.62  2.416 *

14.41 
 =(20.37, 30.87).
44 
The CI with t critical value is better, since we used sample standard deviation instead of
the population standard deviation. Based on the sample standard deviation, the CI with t critical
value is more accurate.