1. [10 pts]
A clinical trial was designed to test a drug believed to decrease blood clotting time. Subjects were randomized into two
groups of 20 subjects each. One group was given the drug; the other group was given a placebo and served as a control.
The researchers wish to determine whether the drug decreases clotting time.
a) Suppose clotting time is not normally distributed for each treatment group.
Hypotheses Ho,Ha
Ho: central tendency for drug group clotting time = central tendency for control group clotting time [mu1 = mu2]
Ha: central tendency for drug group clotting time < central tendency for control group clotting time [mu1 < mu2]
Test Rank-Sum [Mann Whitney]
b) Suppose clotting time is normally distributed; what test would you use?
Test 2 sample-t procedures
2. [10 pts]
A drug company claims that its headache remedy stops the average headache in 14 minutes. To test this claim, an
independent testing agency selected a random sample of 20 headache patients from an outpatient clinic and asked them
to take the remedy when they have a headache and to record the length of time (in minutes) until the headache
disappears. The testing agency wants to know whether the length of time until headaches disappear when taking the
remedy tends to be greater than 14 minutes.
a) Suppose the lengths of time until headaches disappear are normally distributed.
Hypotheses Ho,Ha
Ho: central tendency for the remedy time to relief = 14 minutes [mu = 14]
Ha: central tendency for the remedy time to relief > 14 minutes [mu > 14]
Test: 1 sample-t procedures
b) Suppose the lengths of time until headaches disappear are not normally distributed; what test would you use?
Test sign test
3. [8 pts]
According to the National Restaurant Association, hamburgers are the number one selling fast-food in the United States.
An economist studying the fast-food buying habits of Americans paid graduate students to stand outside McDonald’s
restaurants and ask departing customers whether they spent more than $3.25 on hamburger products for their lunch.
Twenty answered yes & fifty said no … Is there sufficient evidence to conclude that the median amount spent for
hamburgers at McDonald’s is less than $3.25?
Hypotheses Ho,Ha
Ho: median amount spent for hamburgers at McDonald’s is $3.25
Ha: median amount spent for hamburgers at McDonald’s is less than $3.25
Test sign test
Page 2
Part I. continued … For the setting described below give:
 the null and alternative hypotheses and
 the statistical test you would use. NAME THE TEST. Do NOT actually perform the test.
4. [10 pts]
The intraoccular pressure in glaucoma patients is often reduced by treatment with adrenaline. To compare a new
synthetic drug with adrenaline, seven patients were treated with both drugs, one eye with adrenaline and one eye with the
new drug. The reduction in pressure in each eye was recorded. The researchers want to know if the new drug (1) is more
effective in reducing pressure than adrenaline (2).
a) Suppose the differences in pressure reduction are not normally distributed.
Hypotheses Ho,Ha
Ho: central tendency difference in reduction in pressure for new – adrenaline = 0
Ha: central tendency difference in reduction in pressure for new – adrenaline > 0
[new drug is more effective in reducing pressure]
Test signed rank for differences or sign test for differences
b) Suppose the differences in pressure reduction are normally distributed; what test would you use?
Test paired-t procedures
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Page 3
Part II. Essay, Analysis, … Show All Work
5. [20 pts]
A researcher wants to conduct a study to test whether listening to soothing music for 20 minutes helps to reduce diastolic
blood pressure in patients with high blood pressure, compared to simply sitting quietly in a noise-free environment for 20
minutes. One hundred patients with high blood pressure at a large medical clinic are available to participate in this study.
a) Would you suggest 2-sample or paired data for this study to compare these two treatments. Why?
Each subject will receive both treatments, with a suitable length of time between treatments. The order of the treatments
will be randomly assigned to the subjects. For example, for each patient flip a coin to determine which treatment will be
administered first. Measure diastolic blood pressure, then have the subject sit quietly for 20 minutes in either a noise-free
environment or in a room where soothing music is played, depending on which treatment was selected at random (based
on the coin flip). At the end of the 20 minutes, measure diastolic blood pressure again and compute its change (after −
before). After a suitable period of time, repeat with the other treatment.
When the data have been collected, the difference (music − noise-free) in the change in diastolic blood pressure will be
computed for each subject, and then a paired t-test will be run to see if the mean difference is significantly greater than
zero.
b) The null hypothesis for this study is that there is no difference in the mean reduction of diastolic blood pressure for the
two treatments and the alternative hypothesis is that the mean reduction in diastolic blood pressure is greater for the music
treatment. If the null hypothesis is rejected, the clinic will offer this music therapy as a free service to their patients with
high blood pressure. Describe Type I and Type II errors and the consequences of each in the context of this study.
Type I error: Concluding that soothing music does reduce mean diastolic blood pressure compared to sitting quietly,
when in fact it does not. The consequence of this type of error is that the clinic will offer music therapy when it is not
effective.
Type II error: Soothing music does reduce diastolic blood pressure compared to sitting quietly, but we fail to detect this
and conclude that it does not. The consequence of this type of error is that the clinic will choose not to offer music
therapy when it would have been effective.
Which type of error is more serious? A case can be made for either type of error, and the student can take either side as
long as a reasonable justification is given. For example, the student can say a Type I error is more serious because it will
cost the clinic money with no benefit, or the student can say that a Type II error is more serious because the clinic will
miss an opportunity to improve the health and well-being of its patients.
Page 4
6. (12 pts)
The developers of a training program designed to improve manual dexterity claim that people who complete the 6-week
program will increase their manual dexterity. A random sample of 12 people enrolled in the training program was elected.
A measure of each person’s dexterity on a scale from 1 (lowest) to 9 (highest) was recorded just before the start of and
just after the completion of the 6-week program. The experiment resulted in the following data and analysis …
Person
Before
Program
After
Program
A
B
C
D
E
F
G
H
I
J
K
L
6.7
5.4
7.0
6.6
6.9
7.2
5.5
7.1
7.9
5.9
8.4
6.5
7.8
5.9
7.6
6.6
7.6
7.7
6.0
7.0
7.8
6.4
8.7
6.5
Paired T-Test for After - Before
After
Before
Difference
N
12
12
12
Mean
7.133
6.758
0.375
StDev
0.861
0.887
0.367
SE Mean
0.248
0.256
0.106
T-Test of mean difference = 0 (vs > 0):
T.S. T-Value =
_0.375/.106 = 3.54_ P-Value = 0.002
Does the data provide evidence to indicate the mean manual dexterity got people who have completed the 6-week training
program has significantly increased? Test using   0.1…
State the hypotheses, Ho & Ha
Ho: mean difference dexterity After – Before = 0
Ha: mean difference dexterity After – Before > 0
Report the value of the test statistic: __3.54__ Report the value of the p-value: __0.002__
State a conclusion in the context of the problem – Justify Your Answer.
p-value = 0.002 is small – below 0.1 --strong evidence the mean manual dexterity for people who have completed the 6-week training program has increased.
Page 5
7. (10 pts)
Sleep researchers know that some people are early birds, preferring to go to bed by 10 pm and arise by 7 am, while others
are night owls, preferring to go to bed after 11 pm and arise after 8am. A study was done to compare dream recall for early
birds and night owls. One hundred people of each of the two types were selected at random and asked to record their
dreams for one week. The researchers believe that night owls have better recall [more dreams] than do early birds
Below gives the results. Perform the appropriate test at the 5% significance level.
Two-Sample T-Test and CI
Sample
Night Owls
Early Birds
N
100
100
Mean
9.55
7.26
StDev
5.88
6.94
SE Mean
0.59
0.69
Difference = mu (Night Owls) - mu (Early Birds)
T-Test of difference = 0 (vs >): T.S. T-Value = 2.52 P-Value = 0.006
DF = 192
Report the value of the test statistic: __2.52__ Report the p-value: __0.006__
What do you conclude about the researchers’ beliefs – Justify Your Answer.
p-value = 0.006 is small – strong evidence to indicate that night owls have better recall [more dreams] than do early birds.
Page 6
8. (10 pts)
A state highway patrol was interested in knowing whether frequent patrolling of highways substantially reduces the number
of speeders. Two similar interstate highways were selected for the study – one heavily patrolled and the other only
occasionally patrolled. After 1 month, random samples of 100 cars were chosen on each highway, and the number of cars
exceeding the speed limit was recorded. This process was repeated on 5 randomly selected days. The data and a
signed-rank analysis are given below …
Highway 1
Heavily Patrolled
35
36
25
54
63
Day
1
2
3
4
5
Highway 2
Occasionally Patrolled
60
40
48
38
62
Difference
Highway1 – Highway2
-25
-4
-23
16
1
rank = 5 rank = 2 rank = 4 rank = 3 +
rank = 1 +
Wilcoxon Signed Rank Test: diff
Test of median = 0.000000 versus median < 0.000000
diff
N
N for
Test
5
5
Wilcoxon
Statistic
_T+ = 4
T- = 11_
P
Estimated
Median
0.209
-4.50
Fill in the blank is the above output and state a conclusion in context at the 10% significance level – Justify Your Answer.
p-value = 0.209 bigger than 0.10 --insufficient evidence to indicate that frequent patrolling of highways reduces the number of speeders.
9. (10 pts)
A study of ammonia levels near the exit ramp of a highway tunnel resulted in daily ammonia concentrations in ppm on
eight randomly selected days during afternoon drive-time. Use the following data and sign test analysis to determine if the
median daily ammonia concentration exceeds 1.5 ppm …
1.37
1.41
1.42
1.51
1.51
1.52
1.53
1.55
1.58
1.60
1.61
1.63
Sign Test for Median: PPM
Sign test of median =
PPM
1.500 versus > 1.500
N
Below
Equal
Above
P
12
__3__
__0__
__9__
0.0730
Median
1.525
Fill in the blanks in the above output and state a conclusion in context at the 5% significance level – Justify Your Answer.
p-value = 0.0730 bigger than 0.05 --insufficient evidence to indicate the median daily ammonia concentration exceeds 1.5 ppm.