AMS572.01
Practice Midterm Exam
Fall, 2010 ♠♣♥♦
Name: ________________________________ ID: _____________________ Signature: _________________________
Instruction: This is a close book exam. Anyone who cheats in the exam shall receive a grade of F. Please provide
complete solutions for full credit. The exam goes from 12:502:10pm. Good luck! (*The real midterm will be given on
Monday, 10/25/2010, in class.)
1 (for all students in class). In a study of hypnotic suggestion, 10 male volunteers were randomly allocated to an
experimental group and a control group. Each subject participated in a twophase experimental session. In the
first phase, respiration was measured while the subject was awake and at rest. In the second phase, the subject
was told to imagine that he was performing muscular work, and respiration was measured again. For subjects in
the experimental group, hypnosis was induced between the first and second phases; thus, the suggestion to
imagine muscular work was “hypnotic suggestion” for experimental subjects and “waking suggestion” for
control subjects. The accompanying table shows the measurements of total ventilation (liters of air per minute
per square meter of body area) for all 10 subjects.
Experimental Group
Control Group
Subject
Rest
Work
Subject
Rest
Work
1
6
6
6
6
5
2
7
9
7
5
5
3
5
8
8
5
5
4
7
12
9
6
6
5
6
7
10
5
4
(a) Use suitable tests to investigate (Use α =.05 for each test. Please report the pvalue for each test and state the
assumption(s) of the test.)
(i)
the response of the experimental group to suggestion;
(ii)
the response of the control group to suggestion;
(iii)
the differences between the responses of the experimental and control groups.
(b) Please write up the entire SAS program necessary to answer questions raised in (a).
Please include the data
step as well as tests for testing for various assumptions.
Answer:
(a) Response = Work  Rest
(i) Inference on one population mean. Small sample.
,
2
.
2
1
=
x
,
9
.
1
1
=
s
5
1
=
n
0
:
1
0
=
μ
H
vs
0
:
1
a
H
56
.
2
5
/
9
.
1
0
2
.
2
/
0
1
1
1
0
≈

=

=
n
s
x
t
Since
132
.
2
56
.
2
05
.
0
,
4
0
=
≈
t
t
we reject H
0
at the significance level
0.05.
α
=
Since
132
.
2
56
.
2
776
.
2
05
.
0
,
4
0
025
.
0
,
4
=
≈
=
t
t
t
we can infer that
05
.
0
025
.
0
<

<
value
p
. The assumption
is that the response from the experimental group is normally distributed.
Note: if the normality assumption is not true, we will perform the nonparametric test – either the sign test or the
signedrank test.
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 Fall '10
 WeiZhu
 Statistics, Normal Distribution, Variance, significance level, Nonparametric statistics, Mann–Whitney U

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