Statistics 104
Solutions to Sample problems for Chapter 10
Problem:
1. A major medical center in the Northeastern U.S. conducted a study looking at blood
cholesterol levels and incidence of heart attack. Below are data from 16 people who had
a heart attack and 20 people who did not have a heart attack.
242
318
224
310
(1) Heart Attack
186
266
294
282
276
262
280
248
206
234
360
258
182
198
178
162
(2) No Heart Attack
222
198
192
188
166
204
202
164
230
182
218
170
238
182
186
200
a) Is this an experiment or an observational study? Explain briefly.
This is an observational study. No factor, or variable, is being manipulated.
The cholesterol levels are being observed for the two naturally occurring
groups.
b) Compute 5-number summaries for each group.
Group
Minimum
(1) Heart Attack
(2) No Heart Attack
186
162
Lower
Quartile
238
180
Median
264
190
Upper
Quartile
288
203
Maximum
360
238
c) Construct side-by-side box plots. Compare the two groups in terms of center and
spread.
Side-by-side box plots indicate that the heart attack group members have a
higher central level of cholesterol and more spread in their cholesterol values
than the no heart attack group.
400
Cholesterol
350
300
250
200
150
100
(2) No Heart Attack
(1) Heart Attack
Group
1
d) Describe how the individuals in the study need to be selected in order for the
randomization condition to be met.
The 16 individuals should be selected at random from a population of
individuals who have experienced heart attacks. The 20 individuals should be
selected at random from a population of individuals who have not experienced a
heart attack.
Below are summary statistics for the two groups.
(1) Heart Attack
(2) No Heart Attack
y1  265.375
s1  43.645
n1  16
y2  193.1
s2  21.623
n2  20
e) Is there sufficient evidence to indicate that the mean cholesterol for people who have
had a heart attack is greater than that for people who have not had a heart attack?
Perform the appropriate test of hypothesis. Use df = 20 and α=0.05. Note: the
normal distribution condition is satisfied.
H 0 : 1  2
H A : 1  2
t
 y1  y2   0 
s12 s22

n1 n2
265.375  193.1
43.6452  21.6232
16

72.275
 6.056
11.9345
20
Using Table T and 20 degrees of freedom, the one tail probability (P-value) is
less than 0.001. Because the P-value is so small, we reject the null hypothesis.
People who have had a heart attack have a population mean cholesterol level
that is higher than that of people who have not had a heart attack.
f) Construct a 95% confidence interval for the difference in population mean
cholesterol levels. Give and interpretation of this interval.
 y1  y 2   t *
s12 s22

n1 n2
t *  2.086
72.275  2.08611.9345
72.275  24.895
47.38 to 97.17
We are 95% confident that the population mean cholesterol for people who have
had a heart attack is from 47.38 to 97.17 points higher than that for people who
have not had a heart attack.
2
g) Do the results of the hypothesis test in e) and those of the confidence interval in f)
agree? Explain briefly.
Yes. The test of hypothesis indicated that the population mean cholesterol level
of people who have had a heart attack is higher than that of people who have not
had a heart attack. The confidence interval confirms this and says that the
population mean cholesterol level of people who have had a heart attack can be
from 47.4 to 97.2 points higher than that of people who have not had a heart
attack.
2. An article in the Journal of the American Medical Association examined whether the true
body temperature is 98.6 degrees Fahrenheit and if there are differences between men and
women in terms of body temperature. JMP was used to analyze the data on separate
random samples of 65 men and 65 women. The output is on the next page.
a) Give the sample means and sample standard deviations for the two groups, men (M)
and women (F).
Sample Mean
Sample Std Dev
Women (F)
98.394 oF
0.7435 oF
Men (M)
98.105 oF
0.6988 oF
b) Compare the two samples in terms of center and spread.
Women, on average have a slightly higher body temperature than men. Women
also have slightly more variability in their body temperatures compared to men.
c) Report the values of the 95% confidence interval for the difference between the
population mean body temperatures for men and women. According to this interval
could the difference in population mean body temperatures be zero? Explain briefly.
Upper CL Dif = 0.53965
Lower CL Dif = 0.03881
We are 95% confident that the population mean body temperature for women is
from 0.039 oF to 0.540 oF higher than the population mean body temperature for
men.
d) Test the hypothesis that the difference in population mean body temperatures is zero
against and alternative that the difference is not zero. Be sure to include all the steps
for a test of hypothesis.
Step 1: Conditions
There is a random sample of 65 women and a separate random sample of 65
men.
The shapes of the data distributions for both women and men are symmetric
and mounded in the middle. This supports the normal distribution condition.
Step 2: Hypotheses
H0 : F   M
H A : F   M
3
Step 3: Test statistic and P-value
t Ratio = 2.28543
P-value = Prob > |t| = 0.0239
Step 4: Decision
Reject the null hypothesis because the P-value is small (< 0.05)
Step 5: Conclusion
The population mean body temperature for women is different from the
population mean body temperature for men.
e) Are the results of the test of hypothesis consistent with those for the confidence
interval in c)? Explain briefly.
Yes. The test indicates that there is a difference between the population mean
body temperatures for men and women and the confidence interval says that
that difference could be from from 0.039 oF to 0.540 oF.
4
Women (F)
Men (M)
Body Temperature oF
100.0%
75.0%
50.0%
25.0%
0.0%
Women (F)
maximum
100.8
quartile
98.8
median
98.4
quartile
98
minimum
96.4
Mean
98.393846
Std Dev
0.7434878
N
65
Body Temperature oF
100.0%
75.0%
50.0%
25.0%
0.0%
Men (M)
maximum
quartile
median
quartile
minimum
Mean
Std Dev
N
Difference
Std Err Dif
Upper CL Dif
Lower CL Dif
Confidence
99.5
98.6
98.1
97.6
96.3
98.104615
0.6987558
65
0.28923
0.12655
0.53965
0.03881
0.95
t Ratio
DF
Prob > |t|
Prob > t
Prob < t
2.28543
127.5103
0.0239*
0.0120*
0.9880
5