252soln2 (3/17/00)
PROBLEM: Downing and Clark, Chapter 17, Computational Problem 15. Twenty people rank two political
candidates(A, B) on a scale of 1-10. Test the null hypothesis that people have no preference between the
candidates. Data is shown in columns A and B below.   .10 
Solution: Because this is preference data, we cannot assume that it has the normal distribution. Because it
H :   2
is paired, use the Wilcoxon Signed Rank Test.  0 1
or the null hypothesis is simply 'similar
H 1 : 1   2
distributions.'
Person
A x1 
B
x 2  d  x1  x 2 d  x1  x2 Rank r  Corrected Rank r *
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
2
5
3
7
4
8
9
8
7
5
6
7
8
8
9
9
8
4
10
8
8
6
5
8
5
4
8
9
8
6
5
4
5
6
6
7
9
6
7
5
-6
-1
-2
-1
-1
4
1
-1
-1
-1
1
3
3
2
3
2
-1
-2
3
3
6
1
2
1
1
4
1
1
1
1
1
3
3
2
3
2
1
2
3
3
20
1
10
2
3
19
4
5
6
7
8
14
15
11
16
12
9
13
17
18
20511.55519+
5+
5555+
16+
16+
11.5+
16+
11.5+
511.516+
16+
To explain the calculation of corrected ranks we need the table below. Because of the presence of numbers
of equal magnitude, the number in ' Corrected Rank r * ' is the average of the numbers in ' Rank r  .'
Rank r 
Corrected Rank r *
d  x1  x 2
1
1 2 3 4 5 6 7 8 9
5
2
10 11 12 13
11.5
3
14 15 16 17 18
16
4
19
19
6
20
20
If we sum the corrected ranks we get T   132 for those with a + sign and T   78 for those with a sign. The smaller of these is designated TL  78. (Check: T    T   132  78  210 . This should be
equal to the sum of the first 20 numbers, which is 2021 2  210 .) If we use Table 7 " Critical Values of
TL in the Wilcoxon Signed Rank Sum Test …..," we use the .05 column for a 2-sided test with   .10 . For
n  20 the critical value is 60. Since 78 is above 60, do not reject H 0 .
For values of n above 15, TL , the smaller of T  and T  , has the normal distribution and may
be used here with  T  1 4 nn  1  1 4 20 21  105 and variance
 T2  16 2n  1T  16 41105  717.5. If the significance level is 10% and the test is two-sided, we
1
252soln2 (3/17/00)
reject
z
our
TL   T
T
null

hypothesis
78  105
717 .5
if
z
TL   T
T
does
not
lie
between
 z 2   z.05  1.645.
 1.007 . Since z this is between 1.645 , we do not reject H 0 .
PROBLEM: Downing and Clark, Chapter 17, Computational Problem 9. Defense budgets are given for two
countries over two decades. Check to see that they have the same distribution. Data is given in columns x1
and x 2 below.   .10  Note - It is not totally clear to me that the Wilcoxon-Mann-Whitney test is
appropriate in this case, because the data in each row may come from a single year a paired data may be
more appropriate, and it is not clear in general why a nonparametric procedure is appropriate. For this
reason, Computational problem 11 may be a better example of when to use this method.
Solution: Because the data are assumed to be independent random samples and thus not paired use the
H 0 : 1   2
Wilcoxon-Mann-Whitney Rank Sum Test. 
or the null hypothesis is simply 'similar
H 1 : 1   2
distributions.' n1  10 and n2  10 . In the table below, two pairs of ranking columns are given, first r1
and r2 , then r1 * and r2 * . r1 and r2 represent bottom to top ranking, while r1 * and r2 * represent
bottom to top ranking. Both are equally valid because both samples are of equal size, so the 'rank from the
extreme of the smaller sample' rule doesn't hold .
r2 *
x1
r1
r1 *
x2
r2
10
3
18
9
2
19
12
4
17
8
1
20
15
7
14
17
9
12
16
8
13
14
6
15
18
10
11
13
5
16
22
12
9
19
11
10
26
15
6
24
13
8
28
17
4
25
14
7
30
19
2
27
16
5
29
18
3
31
20
1
113
97
97
113
So the sums of the ranks are SR1  113, SR2  97 or SR1 *  97, SR2 *  113 . Check - Note that these two
rank sums must add to the sum of the first n1  n2  10  10  20  n numbers, and that this is
nn  1 20 21

 210 , and that SR1  SR2  97  113  210 .
2
2
The smaller of SR1 and SR2 is called W . This can be compared against the critical values for TL and TU
in Table 6b. For n1  10 , n2  10 and a 2-tailed test with   .10 , TL  83 and TU  127 . Since
W  97 (no matter whether we ranked from the top or from the bottom), it is between these values and we
cannot reject the null hypothesis.
For values of n1 and n 2 that are too large for the tables, W has the normal distribution with mean
W  1 2 n1 n1  n2  1  1 2 1010  10  1  105 and variance  W2  16 n2 W  16 10105  175.
W  W 97  105
z

 0.605 . If we wish to take a p-value approach
W
175
p  value  2Pz  0.605   .5  .2274  .2726 . Since our p-value is larger than the significance level,
we cannot reject the null hypothesis. Though the text uses this method in this problem, strictly speaking, it
should be limited to cases where n2  20, so that we are better off using the table.
2
252soln2 (3/17/00)
9,14,16,16,18,19,22,23,25,26 ,
PROBLEM D.4a a)If our sample consists of the numbers
test the
 H :   15
hypotheses  0
by computing x   for each value of x and using the magnitude and sign of the
 H 1 :   15
results to rank them and perform a Wilcoxon signed rank test.   .05 
H :   2
b) For the following data, test the hypotheses  0 1
on the following paired samples
H 1 : 1   2
 x1

x2
09 14 16 16 18 19 22 23 25 78
using a Wilcoxon signed rank test.   .05  .
14 10 08 14 13 16 12 40 13 24
Solution: a)
d  x  0
x
9
14
16
16
18
19
22
23
25
26
d  x 
6
1
1
1
3
4
7
8
10
11
-6
-1
1
1
3
4
7
8
10
11
Rank r 
6
1
2
3
4
5
7
8
9
10
Corrected Rank r *
622+
2+
4+
5+
7+
8+
9+
10+
 H 0 :   15
so d  x  0  x  15 . If we sum the corrected ranks we get T   47 for those with a +

 H 1 :   15
sign and T   8 for those with a
sign. The smaller of these is designated TL  8. (Check:
T   T   47  8  55 . This should be equal to the sum of the first 10 numbers, which is 1011  55 .)
-
2
If we use Table 7 " Critical Values of TL in the Wilcoxon Signed Rank Sum Test …..," we use the .025
column for a 2-sided test with   .05 . For n  10 the critical value is 8. Since this is equal to our value of
TL , reject H 0 .
b)
Observation
1
2
3
4
5
6
7
8
9
10
x1
9
14
16
16
18
19
22
23
25
78
x2
14
10
8
14
13
16
12
40
13
24
d  x1  x 2
-5
4
8
2
5
3
10
-17
12
54
d  x1  x 2
5
4
8
2
5
3
10
17
12
54
Rank r 
4
3
6
1
5
2
7
9
8
10
Corrected Rank r *
4.53+
6+
1+
4.5+
2+
7+
98+
10+
3
252soln2 (3/17/00)
H 0 : 1   2
. If we sum the corrected ranks we get T   41 .5 for those with a + sign and T   13.5 for

H 1 : 1   2
those with a - sign. The smaller of these is designated TL  13 .5. (Check: T    T   41.5  13.5  55 .
This should be equal to the sum of the first 10 numbers, which is 1011  55 .) If we use Table 7 "
2
Critical Values of TL in the Wilcoxon Signed Rank Sum Test …..," we use the .025 column for a 2-sided
test with   .05 . For n  10 the critical value is 8. Since our value of TL is above the critical value, do not
reject H 0 .
PROBLEM D.5 We have the following data for returns on two stocks:
Stock A 7, 8, -5, 9, 11 nA = 5
Stock B 6, 7, 0, 4, 9, 15 nB = 6
a. Find a 95% confidence interval for
 A2
 B2
b. Test the following at a 95% level:
H 0 :  A2   B2
H 1 :  A2   B2
Solution: The formulas for this problem are given in “Confidence limits and Hypothesis Testing for
Variances” in the Syllabus Supplement and summarized on the formula pages.
a) From the data above we can compute s A2  40.00 and s B2  25.36 . The formula given is
s 22
s12
 22 s 22 ( n1 1, n2 1)


F
. If we let Stock A be x 2 , and Stock B be x1 , then we can state


F n2 1,n1 1  12 s12 2
1
2
that DF1  n B  1  5 , DF1  n A  1  4 and
confidence
interval
formula
( 5, 4 )
( 4,5)
F.025
 9.36 and F.025
 7.39 ,
0.213 
 A2
 B2
get 0.462 
b)
get
s12
s A2
s B2
1.577 
so

s A2
s B2

40 .00
 1.577 . Substitute these values in the
25 .36
 A2 s A2 (5, 4)


F.025 .
F4,5   B2 s B2
1
From
the
F
table
2
2
1
 A2  1.577 9.36 
7.39  B
,
which
becomes
 14 .76 . If we wish an interval for the standard deviations, we can take the square roots to
A
B
 3.842 .
We are testing
F DF1 , DF2 
and
s 22
H 0 :  A2   B2
H1 :  A2   B2 . According to the syllabus supplement, test
s12
where DF1  n1  1 ,
s22
DF2  n 2  1 and s12 is the larger of the two variances
according to the alternate hypothesis. Since this is a 1-sided test with   .05 ,
s A2
s B2

40 .00
 1.577 is
25 .36
4,5  5.19 . Since the ratio is smaller than the F, we accept H .
thus compared to F.05
0
4
252soln2 (3/17/00)
PROBLEM D.6 In a study of sleep gotten with a sleeping pill and with a placebo the results were (Keller,
Warren, Bartel, 2nd ed. p. 354)
Pill
x1
Placebo
x2
7.3
8.5
6.4
9.0
6.9
6.8
7.9
6.0
8.4
6.5
x1  7.620
difference
d
.5
.6
.4
.6
.4
x 2  7.120 d  0.500 s12  1.197
s 22  0.997 s d2  0.010
a. Assume that these are independent samples from a normal distribution and that  12   22
(Test if  12   22 ).
b. Assume that these are independent samples and that  12   22 . Optional
c. Assume these are paired samples.
In each case do (i) a 99% confidence interval for 1-2 , (ii) test if 1=2 . (iii) In case a) test
if  12   22 .
d. Redo part a assuming that the parent population is not normal.
e. Redo part c(ii) assuming that the parent distribution is not normal.
Solution: Assume   .01 .
a) Assume that these are independent samples from a normal distribution and that  12   22 (Test if
 12   22 ).
From the Syllabus supplement:
Interval for
Confidence
Interval
Difference
  d  t  2 sd
Between Two
1 1
Means (
sd  s p

Unknown,
n1 n2
Variances
Assumed equal)
DF  n1  n2  2
(i)
Hypotheses
Test Ratio
H 0 :   0 *
H 1:   0
  1   2
t
sˆ 2p 
Critical Value
d 0
sd
d cv   0  t  2 s d
n1  1s12  n2  1s22
n1  n2  2
Confidence interval: In the case of equal variances we used a pooled variance,
n  1s12  n2  1s 22 41.197   40.947 
sˆ 2p  1

 1.097 . This is used to compute
n1  n 2  1
8
s d  sˆ p
1
1


n1 n 2
d  x1  x 2


and
1.097  1  1  
5
  1   2
8
t  tn1  n2 1  t .005
 3.355 ,
2
0.439  0.662 .
5
,
the
becomes
equation
Since
we
can
  d  t  sd
2
1   2  x1  x 2   t 2 s d
say
,
that
where
or
1   2  0.500  3.355 0.662   0.500  2.221
5
252soln2 (3/17/00)
(ii)
H 0 :   0 H1 :   0 or H 0 : 1   2
H1 : 1   2 . If we use a test ratio,
d   0  x1  x 2   1   2  0.500  0


 0.755 .
sd
sd
0.662
8
 t .005
 3.355 , we accept H 0 . If we use
t
Since
a
this
critical
is
between
value
instead,
d cv   0  t 2 sd  0  3.355 0.062   2.221 . Since d  0.500 is between these critical
values, we accept H 0 .
(iii)
H 0 :  12   22
We are testing
test F
DF1 , DF2
H1 :  12   22 . According to the syllabus supplement,
s12
s22
DF2 , DF1
 2 and F
 2 , where DF1  n1  1 and DF2  n 2  1 .
s2
s1
s 22 0.997
1.197
or

 0.833 , so we

1
.
201
s12 1.197
s 22 0.997
4,4 , But it's not
accept H 0 . (Actually we should be checking against F4,4  F.005
4,4   9.60 and is larger than
F.01
s12

2
4, 4  must be larger than F 4, 4  .
available on the table. A check of the table shows that F.005
.01
4, 4  , it also must be less than F 4, 4  .
So if 1.201 is less than F.005
.01
b) Assume that these are independent samples and that  12   22 .
From the Syllabus supplement:
Difference
H 0 :   0 *
  d  t  2 sd
Between Two
H 1:   0
Means(
s12 s22
  1   2
sd 

Unknown,
n1 n2
Variances
2
 s12 s22 
Assumed
  
n
n2 
1
Unequal)
DF  
t
d cv   0  t  2 sd
d  0
sd
   
s12
2
s 22
n1
n1  1
(i)
2
n2
n2  1
(Optional) Confidence interval: In the case of unequal variances we use the Satterthwaite
method.
s12 1.197
s 22 0.997
s2 s2

 0.2394 ,

 0.1994 , so 1  2  0.2394  0.1994  0.4388 .
n1
5
n2
5
n1 n 2
If we use this in the degrees of freedom formula, we find
DF 
 s12 s 22 



 n1 n 2 


2
2
2
 s12 
 s 22 
 
 
 n1 
 n2 
 
 

n1  1
n2 1

0.4388 2
0.2394 2  0.1994 2
4
 7.9341 . We round this down to get
4
7 degrees of freedom. This is used with s d 
s12 s 22

 0.4388  0.6624 . Since we
n1 n 2
can say that d  x1  x 2 and   1   2 , the equation
  d  t  sd
2
, where
6
252soln2 (3/17/00)
7
t  t .005
 3.499 , becomes 1   2  x1  x 2   t s d or 1   2  0.500  3.499 0.662 
2
 0.500  2.318
(ii)
H 0 :   0 H1 :   0 or H 0 : 1   2
H1 : 1   2 . If we use a test ratio,
d   0  x1  x 2   1   2  0.500  0


 0.755 .
sd
sd
0.6624
7
 t .005
 3.499 , we accept H 0 . If we use
t
Since
a
this
critical
is
between
value
instead,
d cv   0  t 2 sd  0  3.499 0.0624   2.318 . Since d  0.500 is between these critical
values, we accept H 0 .
c)
Assume these are paired samples.
(i)
Confidence interval: In the case of paired data, we act as if we have only n  n1  n 2
4
 4.604 and s d 
pairs. DF  n  1  4 . t  t .005
d  x1  x 2 and   1   2 , the equation
sd
0.010

n
 .002  0.447 .
5
  d  t  sd
2
, becomes
1   2  x1  x 2   t 2 s d or 1   2  0.500  4.604 0.0447   0.500  0.206
(ii)
H 0 :   0 H1 :   0 or H 0 : 1   2
H1 : 1   2 . If we use a test ratio,
d   0  x1  x 2   1   2  0.500  0


 11 .18 . Since this is not between
sd
sd
0.0447
 t 4   4.604 , we reject H . If we use a critical value instead,
t
.005
0
d cv   0  t 2 sd  0  4.604 0.0447   0.206 . Since d  0.500 is not between these
critical values, we reject H 0 .
d) Redo part a(ii) assuming that the parent population is not normal.
Since the parent population is not
x1 r1
normal and the data represents two independent
7.3
5
samples we do a Wilcoxon rank sum test. To do
8.5
2
this we rank the ten numbers from 1 to ten
6.4
9
starting at the extreme end of the smallest
9.0
1
sample. Since the samples are of the same size
6.9
6
we arbitrarily pick x1 as the smaller sample and
23
note that 9 is the largest number in both samples
so that is where we start our ranking. Since we
are working with non normal items, our
hypotheses
are
stated
as
H0 : 1   2 H1 : 1   2
x2
r1
6.8
7
7.9
4
6.0 10
8.4
3
6.5
8
32
d
.5
.6
.4
.6
.4
From the above n1  n2  5 , and the sums of the ranks are SR1  23 and SR2  32 . W is the smaller of
the two rank sums and is 23. To check our rank sums note that n1  n2  n  10 and that if the rank sums
nn  1
10 11
 55 , so the ranking seems correct. If we go
. In this case 23  32 
2
2
to Table 5 in the syllabus supplement, we find that the p-value for W  23 is .210. Since this is a 2-sided
test it should be doubled to .410. In any case, it is above   .01 , so accept H 0 . For a 5% test Table 6
could be used.
are correct, SR1  SR2 
7
252soln2 (3/17/00)
e)
Redo part c(ii) assuming that the parent distribution is not normal.
Since the parent population is not
normal and the data represents paired samples
we would prefer to do a Wilcoxon signed rank
test of the hypotheses H0 : 1   2 H1 : 1   2 .
To do this we take the values of d  x1  x 2 and
replace them with their absolute values d . We
rank the n values from 1 to n . To compute
corrected ranks we add + or - according to the
sign in d and replace all ties with average ranks.
x1
x2
d
7.3
8.5
6.4
9.
6.9
6.8
7.9
6.0
8.4
6.5
.5
.6
.4
.6
.4
d
.5
.6
.4
.6
.4
rank corrected rank
3
4
2
5
1
+3.0
+4.5
+1.5
+4.5
+1.5
For example, ranks 4 and 5 are both replaced
with 4.5, their average, because they correspond
to identical values (.6) of d .
We next compute T  and T  , the sums of the positive and negative ranks. In this case
T   3.0  4.5  1.5  4.5  1.5  15 , while T   0. Our check on the ranking is that the sum of the numbers
nn  1 56 
nn  1

 15 , which, as it should be, is the sum of
is 1 to n is
, In this case, since n  5 ,
2
2
2
T  and T  . We call the smaller of T  and T  , in this case 0, TL , and look it up on Table 7 in the
syllabus supplement. Unfortunately for n  5 , there are no appropriate value, so we cannot reject H 0 .
A second choice test here would be a sign test. We use a binomial table to find out the probability
of getting 5 (or more) positive differences in 5 tries, assuming that the probability is .5. From the binomial
table this probability is .0313, but to make this into a p-value for a 2-sided test, we must double it to .0626.
Since   .01 is less than the p-value, we must accept H 0 , though, if we were working with a higher
significance level we could reject it.
8