1
Questions 1 to 5 are based on the following information:
A pharmaceutical company is running a test comparing two
forms of advertising for the same product, one of which involves
a television campaign (Method 1) and the other a print
campaign (Method 2). Market sectors are randomly assigned to
receive a particular form of advertising, and product sales are
recorded during the month following the ad campaigns. The
following summary statistics are obtained:
Type
of
Advertising
Television:
Print:
Number of
markets
80
120
Mean sales
125
150
Standard
deviation
24.5
46.1
Question 1
A point estimate for the difference in population mean sales
( 1   2 ) between the two forms of advertising is:
(A) 275
(B) 40
(C) 25
(D) -25
(E) -40
(1)
Question 2
The test statistic to conduct a preliminary test of the equality of
population variances is:
s12 24.52
F 2 
s2 46.12
(A) 0.273
(B) 0.531
(C) 0.729
(D) 1.372
(E) 0.282
(1)
2
Question 3
2
2
Given the critical values to test H 0 : 1   2 versus
H1 :  12   22
are F0.975,(79,119)  1.45 and F0.0255,(79,119)  0.658 , your
conclusion is
We reject H and conclude that there is no significant
(A) evidence to show that 12   22 .
0
We do not reject H and conclude that there is no
significant evidence to show that 12   22 .
0
(B)
H 0 and conclude that there is significant
(C) We reject
2
2
evidence to show that 1   2 .
We do not reject H and conclude that there is significant
(D) evidence to show that 12   22 .
0
(E)
Cannot yet make a decision.
(2)
Question 4
The test statistic, to test if there is a significant difference in
product sales by method of advertisement is:
Z
X
1
 X2  0
s12 s22

n1 n2
(A) -4.979

125  150
24.52 46.12

80
120
(B) -1.960
(C) 36.232
(D) 30.087
(E) -1.001
(2)
3
Question 5
A 95% confidence interval estimate for the difference in the
mean product sales between method 1 and method 2 is:
(A)  25  (1.645)  24.5  46.1
(24.5)2 (46.1)2

(B)  25  (1.645) 
80
120
80
(C)
(E)
25  (1.645) 
(24.5)2 (46.1)2

80
120
2
 24.5   46.1
 25  (1.96)  

 
 80   120 
(D)
 25  (1.96) 
120
(24.5)2 (46.1)2

80
120
2
(1)
Questions 6 to 9 are based on the following information:
An entomologist conducted an experiment to see if wounding a
tomato plant would induce changes that improve its defence
against insect attack. She grew larvae of the tobacco hornworm
on wounded plants and control plants. The accompanying table
shows the weight (mg) of the larvae after 7 days of growth.
Assume that the data are normally distributed and the two
population variances are equal.
Control (Group 1)
n:
X:
Standard deviation (S):
18
37.96
11.14
Wounded (Group 2
16
28.66
9.02
Note: Better defence implies less larvae
Question 6
The pooled estimate of the common standard deviation is:
Sp 
(n1  1) s12  (n2  1) s22
(17)11.14 2  (15)9.02 2

n1  n2  2
32
(A) 102.73 (B) 11.979 (C) 10.201 (D)
104.066
(E)
103.987
(2)
4
Question 7
The null and alternative hypothesis for this experiment is:
(A)H 0 : X1  X 2 (B)H 0 : 1   2 (C)H 0 : 1   2 (D)H 0 : 1   2 (E)H 0 : X1  X 2
H 1 : X1  X 2
H 1 : 1   2
H 1 : 1   2
H 1 : 1   2
H 1 : X1  X 2
(1)
Question 8
The value of the test statistic is:
t
X
1
Sp
(A) 0.912
 X 2  0
1 1

n1 n2
(B) 2.653

37.96  28.66  0
10.201
(C) 2.037
1 1

18 16
(D) 0.757 (E) None of A - (2)
D
Question 9
Assume the calculated test statistic was 2.32. The p-value that
you are going to report in your research article is:
(A) 0.9706 (B)
(C)
(D) 0.02<p(E) 0.01<p(2)
0.0294 0.0588 value<0.05
value<0.025
5
Questions 10 to 11 are based on the following information:
A pharmaceutical company claims that a new formula it has
prepared could improve the growth of malnourished babies,
more than a generic brand. The weight gains of 14 babies in the
month before they were given the new formula (Before) and the
month after they were given the new formula (After) were
recorded and given in the following table
Babies
1
2
3
4
5
6
7
8
9
10 11 12 13 14
Before
175 132 218 151 200 219 234 199 236 248 206 179 214 249
After
142 211 337 262 302 195 253 149 187 211 176 214 206 179
Differences -33 79 119 111 102 -24 19 -50 -49 -37 -30 35
-8 -70
( di )
The summary statistics on the difference scores ( d i ) are
14
 d i  164
i 1
sd  65.978
Assume that the difference scores ( d i ) are normally distributed.
Question 10
The null and alternative hypotheses to test if there is any
improvement in the babies’ growth because of the use of the
new formula are:
(A) H0 : d  0 vs. H1 : d  0
(B) H 0 :  d  0 vs. H1 :  d  0
(C)
(E)
H 0 :  d  0 vs. H1 :  d  0
(D)
H 0 :  d  0 vs. H1 :  d  0
H 0 :  d  0 vs. H1 :  d  0
(1)
Question 11
The value of the calculated test statistic is:
X  d 0 164 / 14  0
t d

sd
65.978
14
n
(A) 0.177 (B) 0.664 (C) 1.506 (D) 2.160
(E) 1.96
(2)
6
Questions 12
We wish to design a study to compare two antihypertensive
medications. Two outcome variables will be considered: systolic
and diastolic blood pressure. How many subjects would be
required to detect with 80% power and 5% level of significance
a difference of 10 units in mean systolic blood pressures
between two groups? Assume that the standard deviation in
systolic blood pressure is 25.
1   2 10
ES 

 0.4

25
 Z1  Z1 
2
ni  2

ES

 1.96  0.84 
 2

0
.
4


(A) 25
subjects
per group
(B) 69
subjects
per group
(C) 98
subjects
per group
(D) 50
subjects
per group




2
2
(E) 138
subjects
per group
(2)
Questions 13 to 17 are based on the following information:
A study is conducted to determine if the percent of women who
receive financial aid in undergraduate study is different from the
percent of men who receive financial aid in undergraduate
study. A random sample of undergraduates revealed the
following results:
Sample size:
Number receiving aid:
Women
Men
(Group 1) (Group 2)
250
300
200
180
7
Question 13
A point estimate for the difference in proportions of women and
men who received financial aid is:
200 180
pˆ 1  pˆ 2 

 0.8  0.6  0.2
250 300
(A) 0.20
(B) 0.6
(C) 0.69
(D) 0.80
(E) None
of these
(1)
Question 14
The standard error of the point estimate of the difference in
proportions is:
pˆ 1 (1  pˆ 1 ) pˆ 2 (1  pˆ 2 )
0.8 * 0.2 0.6 * 0.4



n1
n2
250
300
(A) 0.0023 (B) 0.0014 (C) 0.0365 (D) 0.0379 (E) None
of these
(2)
Question 15
The test statistic value to conduct a test on H 0 : p1  p2
versus H1 : p1  p2 is
z
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
 n1 n2 
(A)
1.96

(B)
5.051
0.8  0.6
1 
 1
0.691(1  0.691)


 250 300 
pˆ 
(C) 6.542 (D)
X1  X 2

n1  n2
200180
250 300
 0.691
(E)
 1.96 or  1.96
 1.645 or  1.645
Question 16
A 95% confidence interval for the difference in the proportions
of women and men who received financial aid is:
(A)
1 
 1
0.69  (1.645)  (0.69)(0.31)


 250 300 
(B)
1 
 1
0.20  (1.96)  (0.69)(0.31)


 250 300 
(2)
8
(C)
1 
 1
0.69  (1.96)  (0.69)(0.31)


 250 300 
(E)
0.20  (1.645) 
(D)
0.20  (1.96) 
0.80(0.20) 0.60(0.40)

250
300
0.80(0.20) 0.60(0.40)

250
300
(1)
Question 17
The critical or rejection region of the test in Question 15 at 5%
level of significance is:
(A)
(B)
(C)
(D)
(E)
(1
 1.645 or  1.645
)
 1.96  Z  1.96
 1.96
 1.96
 1.96
or  1.96
Questions 18 to 20 are based on the following information:
A public health researcher believes that smoking affects the
gender of offspring. He records the gender of newborns that are
delivered in local hospitals over a one-year period. He also
interviews the parents of the newborns to determine their
degree of cigarette smoking. The following data are collected.
Offspring
Cigarette Smoking
Boys Girls Total
Neither parent smokes at
60
40 100
least a pack a day
One parent smokes at least
57
43 100
a pack a day
Both parents smoke at least
18
32
50
a pack a day
Total
135
115 250
9
The null and alternative hypotheses for the research are
H0 : There is no association between Cigarette smoking and
gender of offspring versus
H1 : There is association between Cigarette smoking and
gender of offspring.
Question 18
The expected frequency of “One parent smokes at least a pack
a day” and the offspring are girls ( E22 ) under the null hypothesis
is:
(A) 17.20 (B) 19.78 (C) 43
(D) 46
(E) 54
(1)
Question 19
The degrees of freedom for the
(A) 6
(B) 5
(C) 4
2-
test statistic is:
(D) 3
(E) 2
(1)
Question 20
The critical or rejection region of the test at 1% level of
significance is:
(A)  5.99 (B)  9.21 (C)  10.60
(D)  11.34
(E)  16.81
(2)
Questions 21 to 28 are based on the following information:
In order to test the yielding abilities of five different wheat
varieties, 20 equal-sized plots were allocated to a uniform site.
Each variety was planted in four plots, the distribution of the
varieties being random. The yields of grain in kg/plot were
measured, summary of the results were as follows:
Variety :
1
2
3
4
5
Mean (kg/plot) : 26.50 21.50 25.75 18.75 22.50
10
Variance ( si2 ) :
4.33
1.67
4.92
2.92
7.00
Assume that wheat yield is normally distributed and the
population variances of the five wheat varieties are equal. The
following partially completed analysis of variance table is
computed using the above summary statistics:
Source of Sum of
Degrees of Mean
variation
squares freedom
squares
F
Between
161.5
?
?
?
variety
?
?
?
Within
variety
Total
224.0
Question 21
Using the summary statistic from the above table, the overall
mean yield (i.e. X.. ) is:
(A) 28.75
(B) 23
(C) 5.75
(D) 115
(E) 92
(2)
Question 22
The within variety sum squares (i.e. SS w ) is:
(A) 4.167 (B) 15.000 (C) 40.375 (D) 62.500 (E) 161.5
(1)
Question 23
The between variety mean squares (i.e. S b2 ) is:
(A) 32.5
(B) 40.375 (C) 10.767 (D) 4.167
(2)
(E) 161.5
11
Question 24
The F – test statistic value to test the hypotheses
H 0 : 1   2   3   4   5
H 1 : Means not all equal.
(A) 0.103
(B) 2.584
, is:
(C) 3.06
(D) 9.689
(E) 3.80
(2)
Question 25
The numerator and denominator degrees of freedom for the Fstatistic value in Question 24 are:
(A) 4 and 19 (B) 4 and 15 (C) 5 and 15 (D) 5 and 19 (E) 4 and 16 (1)
Question 26
An appropriate conclusion to the test the hypotheses in
Question 24 given that the p-value < 0.001 is:
A
B
C
D
E
There is significant evidence, at   0.01 , to show that
the mean wheat yields under the five wheat varieties
are all equal.
There is insufficient evidence to indicate that there is
difference in the mean wheat yields among the five
wheat varieties at 5% level of significance.
There is significant evidence, at   0.01 , to show that
the mean wheat yields under the five wheat varieties
are not all equal.
There is no sufficient evidence to make comparison
on the mean yields among the five wheat varieties.
None of the above.
(2)
12
Question 27
The test statistic value to conduct the Scheffe pairwise
comparisons between the mean yields of varieties 1 and 4 is:
F
(X i  X j )2
1
1 

s

n n 
j 
 i
2
W

( X .i  X . j ) 2
1
1 

MS error

n n 
j 
 i
(A) 28.828 (B) 14.414 (C) 5.371
(D) 3.720
(E) 2.975
(2)
Question 28
The decision rule to conduct the test
H 0 : 1   3
H 1 : 1   3
using the Scheffe pairwise comparisons at 5% level of
significance is:
(A) Reject (B) Reject
H0
H0
F  3.06
F  15.3
(C) Reject
H0
F  12.24
(D) Reject
H0
F  11.6
(E) Reject
H0
F  2.90
(2)
13
Questions 29 to 33 are based on the following information:
A researcher interested in the relationship between the rate of
germination of warm-season forage grasses and temperature
has postulated a linear regression model in which the number
of seeds germinated per day is dependent (Y) on the average
daily temperature (X). The following data were collected by the
researcher.
Temperature Germinated Temperature Germinated
(0C)
Seed
(0C)
Seed
(X)
(number /
(X)
(number /
day)
day)
10
5
22
28
11
7
23
31
13
9
24
35
15
10
26
38
16
14
27
49
18
20
29
55
20
24
30
61
21
25
32
73
The following summary statistics obtained from the above data:
n  16 ,
 X
16
i 1
 X   696.94,
2
i
16
16
i 1
i 1
 X i  337 ,
 Y
16
i 1
 X
16
i 1
i
i
 Y   6301.00,
2
 X Yi  Y   2029.75
 Yi  484 ,
14
80
y = 2.9124x - 31.092
R² = 0.9382
70
60
50
40
30
20
10
0
0
5
10
15
20
25
30
35
-10
Question 29
An estimate of the slope of the linear-regression line relating the
number of seeds germinated per day to the average daily
temperature is:
(A) (B) 0.322 (C) 0.969 (D) 2.912 (E) None of the (2)
31.092
previous
Question 30
An estimate of the the Y-intercept of the linear-regression line
relating the number of seeds germinated per day to the average
daily temperature is:
(A) -31.092 (B)
(C) 0.969 (D) 2.912 (E) None of the (2)
0.322
previous
Question 31
The sample correlation coefficient (r) between the number of
seeds germinated per day and the average daily temperature
is:
(A) (B) 0.938 (C) 0.969 (D) 0.322 (E) None of the
0.969
previous
(2)
15
Question 32
The test statistic to test the hypotheses
, is:
(A) 3.744 (B) 14.561 (C)
(D)
15.566 20.592
versus
(E) None of the
previous
(1)
Question 33
The percent of variation in the number of seeds germinated per
day explained by the average daily temperature, is:
(A)
(B)
(C)
(D)
(E) None of the (1)
93.896% 96.900% 2.912%
87.980% previous
Questions 34 and 35 are based on the following
information:
In an investigation of possible brain damage due to alcoholism,
an X-ray procedure known as a computerized tomography (CT)
scan was used to measure brain densities in eleven chronic
alcoholics. For each alcoholic, a non-alcoholic control was
selected who matched the alcoholic on age, sex education and
other factors. The brain density measurements on the
alcoholics and the matched controls are reported in the
following table.
Pair
Alcoholic
Control
Differences
( di )
1
2
3
4
5
6
7
8
9
10
11
40.1 38.5 36.9 41.4 40.6 42.3 37.2 38.6 38.5 38.4 38.1
41.3 40.2 37.4 46.1 43.9 41.9 39.9 40.4 38.6 38.1 39.5
-1.2 -1.7 -0.5 -4.7 -3.3 0.4 -2.7 -1.8 -0.1 0.3 -1.4
Assume that the distribution of the differences are skew
(not normally distributed).
16
Question 34
The null and alternative hypothesis for this investigation is:
(A) H 0 : 1   2 versus H 1 : 1   2
(B) H 0 : 1   2 versus H 1 : 1   2
(C) H 0 : Medians are equal, versus
H 1 : Medians are not equal.
(D) H 0 : The medians are equal, versus
H 1 : The median brain density of the alcoholic group is less than that of Control group.
(E)
H 0 : The medians are equal, versus
H 1 : The median brain density of the alcoholic group is greater than that of Control group.
(1)
Question 35
The Wilcoxon signed-rank test, test statistic value using the
normal approximation is:
Z
n(n  1)
4
n(n  1)(2n  1)
24
(A) -5.824
T
(B) -1.022
(C) -0.482
(D) 1.022
(E) -2.490
(2)
END OF PAPER
17