Chapter 7 Section 2
Homework Set A
7.61 Comparison of blood lipid levels in males and females. A recent study at Baylor University investigated
the lipid levels in a cohort of sedentary university students.23 A total of 108 students volunteered for the study
and met the eligibility criteria. The following table summarizes the blood lipid levels, in milligrams per deciliter
(mg/dl), of the participants broken down by gender:
Females (n = 71)
Total Cholesterol
LDL
Males (n = 37)
x
s
x
s
173.70
34.79
171.81
33.24
96.38
29.78
109.44
31.05
61.62
13.75
46.47
7.94
HDL
(a) Is it appropriate to use the two-sample t procedures that we studied in this section to analyze these data for
gender differences? Give reasons for your answer.
(b) Describe appropriate null and alternative hypotheses for comparing male and female total cholesterol levels.
Alright, this problem is very short on details. Thus, I will use the fall back procedure, which assumes there
is no difference in the cholesterol levels for men and women.
H0: women = men
Ha: women ≠ men
(c) Carry out the significance test. Report the test statistic with the degrees of freedom and the P-value. Write a
short summary of your conclusion.
Now the sample standard deviations are nearly the same 34.79 versus 33.24, thus assuming equal standard
deviations is acceptable, men = women.
However, I will run both with pooling and without pooling so you can see the difference.
No Pooling.
t=
173.70 - 171.81
34.79 2 33.24 2
+
71
37
= 0.2760
P(x-barfemale –xbarmen> 1.89) = P(t > 0.2760)
= 0.3921
using the conservative
37 – 1 degrees of
freedom.
Thus the p-value is 2(0.3921) = 0.7842. Clearly there is no evidence of a difference in cholesterol level
between men and women, assuming a model that says there is no difference.
What occurs if we pool?
sp2
t=
(71-1)34.792 + (37 -1)33.24 2
=
= 1174.532
71 + 37 - 2
173.70 -171.81
1 
 1
1174.53  +

 71 37 
= 0.2720
You can see the t-stat is nearly the same.
But the real difference is going to be that the degrees of freedom will be higher. Instead of going
conservative I can now use the degrees of freedom associated with pooling; n1 + n2 – 2. Why?
Look by saying that you know men = women you have essentially said that the information you have is
better than if you can not assume men = women. Thus, the degrees of freedom allow us to make a judgment
on how good is the data.
P(x-barfemale –xbarmen> 1.89) = P(t > 0.2720) = 0.3931
Again the result is the same. 2(0.3931) = 0.7862 is the p-value.
(d) Find a 95% confidence interval for the difference between the two means. Compare the information given by
the interval with the information given by the significance test.
I will use the non-pooled significance test, thus the degrees of freedom will be the conservative of 36.
(173.70 – 171.81) ±2.0281
34.792 33.242
+
71
37
1.89 ± 2.0281(6.849)
(-12.00, 15.78) Notice this interval contains 0, which means
no difference in the means. Thus, this interval cannot rule
out that there is no difference between the means, as is
shown by having positive and negative differences.
(e) The participants in this study were all taking an introductory health class. To what extent do you think the
results can be generalized to other populations?
7.65 Dust exposure at work. Exposure to dust at work can lead to lung disease later in life. One study measured
the workplace exposure of tunnel construction workers.25 Part of the study compared 115 drill and blast workers
with 220 outdoor concrete workers. Total dust exposure was measured in milligram years per cubic meter
(mg.y/m3). The mean exposure for the drill and blast workers was 18.0 mg.y/m3 with a standard deviation of 7.8
mg.y/m3. For the outdoor concrete workers, the corresponding values were 6.5 mg.y/m3 and 3.4 mg.y/m3.
(a) The sample included all workers for a tunnel construction company who received medical examinations as
part of routine health checkups. Discuss the extent to which you think these results apply to other similar types of
workers.
(b) Use a 95% confidence interval to describe the difference in the exposures. Write a sentence that gives the
interval and provides the meaning of 95% confidence.
(c) Test the null hypothesis that the exposures for these two types of workers are the same. Justify your choice of
a one-sided or two-sided alternative, Report the test statistic, the degrees of freedom, and the P-value. Give a
short summary of your conclusion.
(d) The authors of the article describing these results note that the distributions are somewhat' skewed. Do you
think that this fact makes vour analysis invalid? Give reasons for your answer.
7.71 More basic concepts. For each of the following, answer the question and give a short explanation of your
reasoning.
(a) A significance test for comparing two means gave t = -3.69 with 9 degrees of freedom. Can you reject the null
hypothesis that the ’s are equal versus the two-sided alternative at the 5% significance level?
P(t < -3.69) = 0.0025. Thus, the p-value = 2(0.0025) = 0.005 Yes the result is significant at 5%.
(b) Answer part (a) for the one-sided alternative that the difference in means is negative.
P(t < -3.69) = 0.0025.
7.72 Effect of the confidence level. Assume x1 = 100, x 2 = 120, s1= 10, s2= 12, n1 = 50, and n2 = 50. Find a
95% confidence interval for the difference in the corresponding values of . Does this interval include more or
fewer values than a 99% confidence interval? Explain your answer.
The 95% confidence interval has a t* value equal to
2.0096.
102 122
+
(120 – 100) ±2.0096
50 50
20 ± 2.0096(2.2091)
(15.56, 24.44)
Immediately after baking:
47.62
49.79
If I did a 99% confidence interval, that
Three days after baking:
21.25
22.34
would mean that I am 99% certain that
the mean difference is in the interval
created. The increase in certainty is done by increasing the margin of error. Thus, the interval includes
more differences in sample averages.
7.76 Effect of storage time on vitamin C content. Does bread lose its vitamins when stored? Small loaves of
bread -were prepared with flour that was fortified with a fixed amount of vitamins. After baking, the vitamin C
content of two loaves was measured. Another two loaves were baked at the same time, stored for three days, and
then the vitamin C content was measured. The units are milligrams per hundred grams of flour (mg/100 g).28 Here
are the data:
Immediately after baking: 47.62, 49.79.
Three days after baking: 21.25, 22.34.
(a) When bread is stored, does it lose vitamin C? To answer this question, perform a two-sample t test for these
data. Be sure to state your hypotheses, the test statistic with degrees of freedom, and the P-value (show all your
work).
Assume that there is no Vitamin C lost.
Then the null hypothesis is
H0: immediate = 3-day
Ha: immediate > 3-day I used one sided since we are interesting in the case of losing the vitamin content.
Summary statistics:
Column
n Mean Variance Std. Dev. Std. Err.
Immediate 2 48.705
2.35445 1.5344217
1.085
3-days
0.59405 0.7707464
0.545
2 21.795
Here is a situation in which we should not use a pooled statistic. The standard deviations are too far apart.
Test Statistic: t =
48.705  21.795
1.5342 0.7712

2
2
= 22.17. P(xbarimmediate – xbar3day > 26.91) = P(t > 22.16)
= 0.0039. Our p-value is
0.0039.
Hypothesis test results:
μ1 : mean of Immediate
μ2 : mean of 3-days
μ1 - μ2 : mean difference
H0 : μ1 - μ2 = 0
HA : μ1 - μ2 > 0
(without pooled variances)
Difference Sample Mean Std. Err.
μ1 - μ2
DF
T-Stat
P-value
26.91 1.214187 1.4744174 22.16298 0.0039
(b) Give a 90% confidence interval for the amount of vitamin C lost.
I will use the conservative degrees of freedom of 1, or since the computer already did the work, I could use
1.4744174 both give t* = 6.192
(48.705 – 21.795)± 6.192
26.91 ± 6.192(1.214187)
(19.42, 34.40)
1.5342 0.7712

2
2
7.79 Are the samples too small? Refer to Exercises 7.76 and 7.78. Some people claim that significance test with
very small samples never lead to rejection of the null hypothesis. Discuss this claim using the results of these two
exercises.