Statistics 503, Divisions 1 and 2
Midterm Examination 2
Thursday, November 9, 2000
Time: 2 hours, 7:00 pm – 9:00 pm
Name (please print):
ANSWERS
Section
.
(1 = 10:30, 2 = 11:30)
Show all your work and calculations. Partial credit will be given for work that is partially correct.
Points will be deducted for false statements shown in the solution to a problem, even if the final answer
is correct. Numbers in the right-hand margin indicate the number of points for each question. If extra
space is needed, use the back of a page and indicate this clearly on the question page.
This exam is closed-book. You may not consult any notes or books during this exam. One formula
page, and tables for several distributions and binomial coefficients are provided separately. There is no
need to hand them in. Calculators are permitted.
Please sign below to indicate your agreement with the following honour code.
Honour code: I promise not to cheat on this exam. I will neither give nor receive any unauthorized
assistance. I promise not to share information about this exam with anyone who may be taking it at a
different time. I have not been told anything about the exam by someone who has already taken it.
Signature:
.
Question
1
Possible
16
2
23
3
15
4
14
5
15
6
17
Total
100
Actual
1
Date:
.
(16)
1. Short-answer questions (2 pts each). Fill in the blanks to make the following sentences correct, or
answer the question in the space provided. It is possible for more than one word to go in a single
blank space.
a) When the alternative hypothesis is ___false_________ and the null hypothesis is
(2)
___rejected__________ then a type I error has been made.
b) When the alternative hypothesis is ___true__________ and the null hypothesis is
(2)
_____not rejected_______ then a type II error has been made.
c) Either __decreasing_________ the sample size or __increasing______ the variance will
(2)
decrease the power of a hypothesis test.
d) In a directional two-independent-samples t-test, I would need a sample of size __24__ to achieve
(1)
80% power to detect an effect of size 0.95 with  = 0.01.
e) A t-test is performed with HA : 1 > 2 and a test statistic of ts = –1.24 is observed. Draw a sketch (2)
of the area corresponding to the P-value, and bracket the p-value when df = 12.
(Area under the t-distribution curve to the right of –1.24. ) Since the P-value for ts = +1.24 would be
bracketed between 0.1 and 0.2, the P-value for ts = –1.24 must satisfy 0.8 < P < 0.9.
f) If  = 0.02 and the P-value is 0.015 I should ___reject______ the null hypothesis.
(1)
g) The type of ANOVA studied in this class and performed on problem 2 is an appropriate analysis
(2)
for a _____________completely randomized _______________ design.
h) A(n) _experimental
_observational
OR
OR
double-blind__ study is usually the best kind to do, but sometimes a(n)
(2)
blind/open_________ must be done instead, for ethical or practical reasons.
i) Blocking on __extraneous_______________ variables is a good way to take them into account.
(2)
In this type of study design, subjects are first grouped into blocks and then __randomized_______
within each block.
j)
(Bonus question – if correct, score can replace any of the above. No penalty for getting it
wrong.) Give a clear definition of P-value.
The P-value is the probabillity, assuming the null hypothesis to be true, of observing a test statistic at
least as extreme as what was actually observed.
2
(2)
2. In a study of weight loss methods, efforts were made to change the body weights of 20 moderately (23)
overweight student volunteers by interventions involving diet, increased physical activity, or both.
Specifically, the 20 students were randomly divided into 4 groups of 5 students each, and the groups
each received a different treatment as shown. Their weight losses (before – after) are given in the
accompanying table.
Control Diet Activity Diet &
Only Only Activity
ANOVA Table
5
2
8
12
SS
df
MS
-2
8
0
6
3
4
2
15
Between
205.35
3
68.45
2
12
6
8
Within
185.20
16
11.58
0
4
2
10
n
5
5
5
5
Total
390.55
19
mean
1.6 6.0
3.6
10.2
SD
2.70 4.00
3.29
3.49
a) Complete the ANOVA table for these data. Do not
(10)
do any more work than is necessary. Use only 2 decimal places when writing your answers in the
table.
y
5  1.6  6.0  3.6  10.2 
20
 5.35; n *  20;
SSB  5 1.6  5.35   5  6.0  5.35   5  3.6  5.35   5 10.2  5.35   205.35; dfb  k  1  3
2
2
2
2
2
2
2
2
SSW  4  2.7    4.00    3.29    3.49    185.20; dfw  n *  k  20  4  16


205.35
185.20
MSB 
 68.45; MSW 
 11.58; SST  205.35  185.20  390.55; dft  3  16  19
3
16
b) Perform an appropriate hypothesis with significance level 0.01.
Do any of the treatments influence body weight?
Let 1, 2, 3, 4 be the mean body weight of moderately overweight students with no treatment, diet
only, activity only, and diet and activity, respectively.
H0: 1 = 2 = 3 = 4; mean body weight of students is the same under all four treatments
HA: students’ mean body weights are not all the same under the four treatments
Use a global F-test
Fs = MSB/MSW has an F-distribution with df 3,16 under H0
Test at level  = 0.01. Critical value is F.01 = 5.29. If Fs > 5.29 then will reject H0.
Fs 
68.45
 5.91 . 5.91 > 5.29 so reject H0.
11.58
This study provides evidence at the 0.01 significance level that mean body weights of moderately
overweight students are not all the same under the four treatments.
3
(13)
(15)
3. A certain breed of rat is known to show a mean weight gain of 65g during the first 3 months of life.
Sixteen rats of this breed are fed a new diet from birth until age 3 months. These 16 have a mean
weight gain of 60.75g with standard deviation 3.84g. Assuming that weight gain follows a normal
distribution, test whether the new diet has any effect on 3-month weight gain in this breed of rat.
Does the new diet have an effect on 3-month weight gain in this breed of rat?
Let  be the mean 3-month weight of this breed of rat on the new diet.
[Let 0 = 65g be the mean 3-month weight of this breed of rat on a standard diet.]
H0:  = 65; the mean 3-month weight of this breed of rat is unchanged on the new diet.
HA:   65; the mean 3-month weight of this breed of rat is different on the new diet.
Use a one-sample, non-directional t-test.
ts 
y  65
has a t-distribution with 16 – 1 = 15 df under H0.
SE
Test at level  = 0.05. Critical value is t.025 = 2.131. If ts > 2.131 or ts < –2.131 then will reject H0.
SE 
3.84
16
 0.96; ts 
 60.75  65 
0.96
 4.43
Since –4.43 < –2.131 we reject H0.
This study provides evidence at the 0.05 significance level that the mean 3-month weight of this breed
of rat is different on the new diet.
4
(14)
4. An experiment was performed to consider the effect of body mass on toxin metabolism.
Twenty-three yearling sheep were fed pingue (a toxin-producing weed). Of these, 13 died and the
remaining 10 survived. The weights of the sheep (in kg) just before they were fed pingue were as
shown.
n
mean SD
K
Died
21 25 27 27 28 29 29 29 30 30 32 34 34 13 28.8 3.53 62
Survived 20 26 26 28 29 30 30 33 34 34
10 29.0 4.32 68
a) Briefly justify your choice of which type of hypothesis test to perform.
(3)
We are not told to assume a normal distribution, and the sample size is small. Furthermore, the data
do not look normal, since there are many observations in the right tail (near 34) , while the left tail is
very spread out; thus the distribution seems skewed. Therefore I will use a Wilcoxon-Mann-Whitney
(WMW) test.
b) Perform an appropriate hypothesis test at level 0.05.
Does body mass have an effect on toxin metabolism?
H0: the body mass distribution of sheep who survive is the same as that of those who die after pingue
exposure.
H1: the body mass distribution of sheep who survive is different from that of those who die after pingue
exposure.
Use a WMW test (non-directional).
Us = max(K1, K2) follows the Mann-Whitney null distribution with n = 13, n’ = 10under H0.
Test at level 0.05. Critical value is U.05 = 97. If Us  97 then will reject H0.
Check: K1 + K2 =62 + 68 = 130 = 13x10 = n1 n2
Us = max(62, 68) = 68
68 < 97 so do not reject H0.
This study does not provide evidence at the 0.05 significance level that the body mass distribution of
sheep who survive is different from that of those who die after pingue exposure.
5
(11)
(15)
5. Twelve participants were chosen at random to
participate in a blood pressure study. Of interest is the
idea of postural hypotension, namely, that standing up
tends to lower blood pressure. The systolic blood
pressure of each participant was measured both while
standing and while supine (lying down). The results
are shown in the accompanying table. Assuming that
systolic blood pressure follows a normal distribution,
use these data to test whether blood pressure is
decreased while standing.
Subject Standing Supine Difference
1
132
136
-4
2
146
145
1
3
135
140
-5
4
141
147
-6
5
139
142
-3
6
162
160
2
7
128
137
-9
8
137
136
1
9
145
149
-4
10
151
158
-7
11
131
130
1
12
143
150
-7
mean
140.83 144.17
-3.33
s
9.49
9.12
3.75
Is blood pressure decreased while standing?
Let 1 be the mean systolic blood pressure of people while standing.
Let 2 be the mean systolic blood pressure of people while supine.
H0: 1 = 2 ; mean systolic blood pressure is the same whether standing or supine.
HA: 1 < 2 ; mean systolic blood pressure is lower while standing than while supine
Use a paired-sample, directional t-test.
ts 
 y1  y 2 
SEdiff
has a t-distribution with 12 – 1 = 11 df under H0.
Test at level  = 0.05; critical value is t.05 = 1.796. If ts < –1.796 then will reject H0.
SEdiff 
3.75
12
 1.08; ts 
3.33
 3.08
1.08
Since –3.08 < –1.796 we reject H0.
This study provides evidence at the 0.05 significance level that mean systolic blood pressure is lower
while standing than while supine.
6
(17)
6. Ten students were given a mathematics aptitude test in a quiet room. The
same students were given a similar test in a room with background music.
Their performances on the tests were as shown. Do not make any
assumptions about the distribution of the test scores.
a) Test an appropriate hypothesis at the 0.10 significance level.
Student Without With
Music Music
1
2
3
4
5
6
7
8
9
(14)
mean
114
121
141
102
99
114
127
150
129
112
122
136
107
96
109
121
146
127
122.7 120.4
Does background music affect math test scores?
Let  be the probability that a student will score higher without music.
H0:  = 0.5; a student is equally likely to score higher with music or without music.
HA:   0.5; a student is more likely to score higher in one of the two situations.
Use a non-directional sign test.
Bs = max(N+, N–) has a folded binomial (n = 9,  = 0.5) distribution under H0.
Test at level  = 0.01; critical value is 8. If Bs  8 then will reject H0.
N+ = 7, N– = 2, so Bs = 7
Since 7 < 8 we do not reject H0.
This study does not provide evidence at the 0.10 significance level that a student is more likely to score
higher in one of the two situations.
(3)
b) Calculate the P-value for this test.
With Y binomial(n = 9,  = 0.5),
P-value  2  Pr Y  7   2  Pr Y  7   Pr Y  8   Pr Y  9  
 9 
9
9
7
2
8
1
9
0
 2     0.5   0.5      0.5   0.5      0.5   0.5  
8
9
 7 

 2  36  9  1 0.5   0.1797
9
7