252y0561s 11/14/05
ECO252 QBA2
SECOND EXAM
Nov 8-9, 2005
TAKE HOME SECTION
Name: ________Key_________________
Note that complete solutions for 1) and 2) were only worked out for two versions of these problems.
III. Neatness Counts! Show your work! Always state your hypotheses and conclusions clearly.
(19+ points)
1) A state is trying to figure out whether the background on highway signs makes a difference. In order to
do this two samples of 15 individuals are shown a number of slides rapidly. The slides have either a green
or a red background. You are trying to find out whether there is a difference between the number of slides
correctly read between those with a red or a green background. To do so you will compare the mean or
median as appropriate to the distribution. To personalize the data, look at the third digit from the end to
decide what red data you will use. Call the column that you pick rj and compute a column called dj with the
formula dj = green – rj. (Example: Seymour Butz’s student number is 976512, so he picks column 5 and
used d5 = green – r5.) Tell me what column you are using! If you compare means state your hypotheses
both in terms of 1 and  2 and in terms of D  1   2 .
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
green
8
10
6
5
9
7
3
7
7
3
6
6
8
3
10
r0
9
10
9
4
9
9
8
7
5
6
10
8
8
9
9
r1
5
9
5
7
6
7
9
8
7
5
7
5
8
7
7
r2
7
10
11
9
8
8
8
7
9
9
7
7
10
7
6
r3
6
12
7
10
12
10
9
11
9
8
7
6
11
13
8
r4
5
6
5
6
5
11
8
8
7
9
10
7
7
6
5
r5
8
9
6
10
7
7
5
6
11
9
10
7
7
8
8
r6
9
7
6
7
8
7
6
6
7
7
5
7
10
7
6
r7
7
9
10
11
11
6
11
6
9
7
7
7
9
8
8
r8
10
9
7
10
6
6
7
8
8
7
9
11
6
8
7
r9
5
6
6
7
10
10
10
13
9
6
8
6
7
8
10
Minitab computed some basic statistics from the data which will help you in some parts of this problem.
Variable
green
r0
r1
r2
r3
r4
r5
r6
r7
r8
r9
N
15
15
15
15
15
15
15
15
15
15
15
Mean
6.533
8.000
6.800
8.200
9.267
7.000
7.867
7.000
8.400
7.933
8.067
SE Mean
0.601
0.458
0.355
0.368
0.581
0.488
0.435
0.324
0.456
0.408
0.573
StDev
2.326
1.773
1.373
1.424
2.251
1.890
1.685
1.254
1.765
1.580
2.219
Minimum
3.000
4.000
5.000
6.000
6.000
5.000
5.000
5.000
6.000
6.000
5.000
Q1
5.000
7.000
5.000
7.000
7.000
5.000
7.000
6.000
7.000
7.000
6.000
Median
7.000
9.000
7.000
8.000
9.000
7.000
8.000
7.000
8.000
8.000
8.000
Q3 Maximum
8.000 10.000
9.000 10.000
8.000
9.000
9.000 11.000
11.000 13.000
8.000 11.000
9.000 11.000
7.000 10.000
10.000 11.000
9.000 11.000
10.000 13.000
a) Display the numbers that you are using in columns and compute a sample mean and sample standard
deviation for the d column. (1)
In this problem assume that the red and green data are two independent samples. Use a confidence level of
95%.
b) Assume that you believe that the normal distribution does not apply to the data and compare the means or
medians as appropriate. (4)
c) You suspect that the data has the Normal distribution. Test to see if the Normal distribution applies. Use
a test that I taught you. (3)
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252y0561s 11/14/05
d) You decide that the Normal distribution applies to the data, but do not know if the variances are equal.
Test them for equality. (1)
e) You conclude that the underlying distributions are Normal and that the population variances are equal.
Compare the means or medians as appropriate. Use a test ratio, critical value or a confidence interval (4) or
all three (6).
[15]
f) (Extra credit) You conclude that the underlying distributions are Normal and that the population
variances are not equal. Compare the means or medians as appropriate. Use a test ratio, critical value or a
confidence interval (5) or all three (7)
2) In fact the data on the previous page applies to a single sample of 15 individuals. That is the first line of
your worksheet tells you how the first person in the sample did when showed the same slides with red or
green backgrounds. This applies to a) and b) in this question. Use a confidence level of 95%.
a) Assume that you believe that the normal distribution does not apply to the data and compare the means or
medians as appropriate. (3)
b) You assume that the data has the Normal distribution. Compare the means or medians as appropriate. (3)
c) For any part of one of these problems (tell me which one!), compute a confidence interval that you would
use to compare means if your alternate hypothesis was H 1 :  2  1 . (2)
[23]
d) For the same part as you used in c), find a p-value for the null hypothesis. (2) [25]
These results are all supposed to look to me as you did them by hand. But what I don’t know won’t hurt me.
If you want to check your results by computer, you might try to use the following Minitab routine. If you
put green in C1 and label columns with headings like rj, dj and dsqj (Seymour called his green, r5, d5 and
dsq5.) The routine below with appropriate changes to rj, dj and dsqj, will compute much of the stuff above,
though not in the right order. Note that to do a Wilcoxon signed rank test by hand, you will have to drop all
zeroes from the d column.
Computations for comparing c1 and rj
MTB >
MTB >
MTB >
MTB >
MTB >
MTB >
MTB >
SUBC>
MTB >
MTB >
MTB >
SUBC>
MTB >
SUBC>
MTB >
SUBC>
let dj = c1 – rj
let dsqj = rj *rj
print c1 rj dj dsqj
describe c1 rj dj
sum dj
ssq dj
TwoSample c1 rj;
Pooled.
TwoSample c1 rj.
Paired c1 'rj'.
VarTest c1 'rj';
Unstacked.
WTest 0.0 'dj';
Alternative 0.
Mann-Whitney 95.0 c1 'rj';
Alternative 0.
If you want to fake the calculations for the Mann-Whitney test, try this.
Procedure for setting up Mann-Whitney Test
#c1 is green, c2 is rj, c3 is difference.
# Mann Whitney Test
MTB > Stack c1 c2 c5;
SUBC>
Subscripts c6.
MTB > Rank c5 c7.
MTB > Unstack (c7);
SUBC>
Subscripts c6;
SUBC>
After;
SUBC>
VarNames.
MTB > sum c8
MTB > sum c9
MTB > print c1 c8 c2 c9 #The rest is up to you.
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252y0561s 11/14/05
If you want to fake the calculations for the Wilcoxon signed rank test, try this. Unfortunately, I know no
good way to remove the zeros or change the signs except by hand.
Procedure for setting up Wilcoxon Signed Rank Test
#c1 is green, c2 is rj, c3 is difference.
MTB > Let c3 = c1-c2
#Maybe you already did this.
# Wilcoxon signed rank test
MTB > let c10 = c3
MTB > #remove zeroes from c10. (Just use delete on the cells with zeros.)
MTB > #Notice that n has gotten smaller.
MTB > let c11 = abs(c10)
MTB > rank c11 c12
MTB > let c13 = c12
MTB > #change signs in c13 to signs in c10.
MTB > let c14 = c13 *c10
#Check on signs. All should be positive.
# Aside from this consider c14 garbage.
MTB > print c10 c11 c12 c13 #You now have the four columns that I computed in
MTB > #the examples. The totals are up to you.
3) The results of a Gallup phone survey appear below. Consumers were asked if they objected to having
their medical records shared with different types of organizations. Results follow.
The proportion in a sample of 1000 who objected to sharing with insurance companies was p1  .820 .
The proportion in a sample of 1000 who objected to sharing with pharmacies was p 2  .590
The proportion in a sample of 1000 who objected to sharing with medical researchers was p3  .670
Personalize the data by using the second to last digit of your student number, call it d . Multiply it by .001.
Call the result .00d – If the second to last number is zero, use .00d = .010. Add .00d to .820 and subtract
.00d from .670. . (Example: Seymour Butz’s student number is 976532, so he adds .003 to .820 and gets
.823 and subtracts .003 from .670, getting .667. Hr leaves .590 alone.
a) Is the proportion of people who object different for different institutions?   .01 . (3)
b) If appropriate, use the Marascuilo procedure to determine which organizations are different. Discuss. (3)
[31]
3
252y0561s 11/14/05
————— 10/31/2005 10:56:18 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > WOpen "C:\Documents and Settings\rbove\My Documents\Minitab\252x05062.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0506-2.MTW'
Worksheet was saved on Mon Oct 31 2005
Results for: 252x0506-2.MTW
MTB > exec '252compgr'
Executing from file: 252compgr.MTB
MTB > print c1 r0 r1 r2 r3 r4 r5 r6 r7 r8 r9
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
green
8
10
6
5
9
7
3
7
7
3
6
6
8
3
10
r0
9
10
9
4
9
9
8
7
5
6
10
8
8
9
9
r1
5
9
5
7
6
7
9
8
7
5
7
5
8
7
7
r2
7
10
11
9
8
8
8
7
9
9
7
7
10
7
6
r3
6
12
7
10
12
10
9
11
9
8
7
6
11
13
8
r4
5
6
5
6
5
11
8
8
7
9
10
7
7
6
5
r5
8
9
6
10
7
7
5
6
11
9
10
7
7
8
8
r6
9
7
6
7
8
7
6
6
7
7
5
7
10
7
6
r7
7
9
10
11
11
6
11
6
9
7
7
7
9
8
8
r8
10
9
7
10
6
6
7
8
8
7
9
11
6
8
7
r9
5
6
6
7
10
10
10
13
9
6
8
6
7
8
10
Paired Samples
Method D4
Independent Samples
Methods D1- D3
Location - Distribution not
Normal. Compare medians.
Method D5b
Method D5a
Proportions
Method D6b
Method D6a
Location - Normal distribution.
Compare means.
Variability - Normal distribution.
Compare variances.
Method D7
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252y0561s 11/14/05
Interval for
Confidence
Interval
Hypotheses
Difference
between Two
Means (
known)
Method D1
D  d z 2  d
H 0 : D  D0 *
Difference
between Two
Means (
unknown,
variances
assumed equal)
Method D2
Difference
between Two
Means(
unknown,
variances
assumed
unequal)
Method D3
 12
d 
n1
 22

n2
z
d
H 1 : D  D0 ,
1
1

n1 n 2
sd  s p
D  1   2
DF  n1  n2  2
H 0 : D  D0 *
D  d  t 2 s d
s12 s 22

n1 n2
sd 
DF 
H 1 : D  D0 ,
 s12 s22 



n

 1 n2 
sˆ 2p 
t
D  1   2

1
FDF1 , DF2
2
Method D7
Difference
between Two
Means (paired
data.)
Method D4
d cv  D0  t  2 s d
d  D0
sd
n1  1s12  n2  1s22
n1  n2  2
d cv  D0  t  2 s d
d  D0
sd
2
   
s12
2
n1
n1  1
s p 
t
2
s 22
n2
n2  1
p1 q1 p 2 q 2

n1
n2
H 0 : p  p 0
H 1 : p  p 0
z
p  p 0
 p
p 0  p 01  p 02
If p  0
or p 0  0
 p 
 22 s22 DF1 , DF2

F

 12 s12 .5  .5  2 
DF1  n1  1
H0 : 12   22
H1 : 12   22
F DF2 , DF1 

 2

.5  .5   2    or
1  
2

d  x1  x 2
df  n  1 where
n1  n 2  n
 1
1 


n
n
1
2 

 p  p 0 q 0 
p0 
n1 p1  n 2 p 2
n1  n 2
s12
s 22
and
DF2  n 2  1
D  d t  2 s d
s p
F DF1 , DF2 
pcv  p0  z 2  p
If p  0
p 01q 01 p 02 q 02

n1
n2
Or use
1 , DF2
F1DF
 2
d cv  D0  z d
d  D0
D  1   2
H 0 : D  D0 *
D  d  t 2 s d
p  p1  p 2
Ratio of Variances
Critical Value
d  x1  x 2
p  p  z 2 s p
Difference
between
proportions
q  1 p
Method D6a
H 1 : D  D0 ,
Test Ratio
H 0 : D  D0 *
H 1 : D  D0 ,
D  1   2
t
d  D0
sd
sd 
s 22
s12
d cv  D0 t  2 s d
sd
n
5
252y0561s 11/14/05
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
green
8
10
6
5
9
7
3
7
7
3
6
6
8
3
10
r0
9
10
9
4
9
9
8
7
5
6
10
8
8
9
9
d0
-1
0
-3
1
0
-2
-5
0
2
-3
-4
-2
0
-6
1
dsq0
81
100
81
16
81
81
64
49
25
36
100
64
64
81
81
Descriptive Statistics: green, r0, d0
Variable
green
r0
d0
N
15
15
15
N*
0
0
0
Variable
green
r0
d0
Maximum
10.000
10.000
2.000
Mean
6.533
8.000
-1.467
SE Mean
0.601
0.458
0.608
StDev
2.326
1.773
2.356
Minimum
3.000
4.000
-6.000
Q1
5.000
7.000
-3.000
Median
7.000
9.000
-1.000
Q3
8.000
9.000
0.000
Sum of d0
Sum of d0 = -22
Sum of Squares of d0
Sum of squares (uncorrected) of d0 = 110
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252y0561s 11/14/05
Two-Sample T-Test and CI: green, r0
Two-sample T for green vs r0
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r0
15 8.00
1.77
0.46
Difference = mu (green) - mu (r0)
Estimate for difference: -1.46667
95% CI for difference: (-3.01340, 0.08006)
T-Test of difference = 0 (vs not =): T-Value = -1.94
Both use Pooled StDev = 2.0679
P-Value = 0.062
DF = 28
P-Value = 0.063
DF = 26
Two-Sample T-Test and CI: green, r0
Two-sample T for green vs r0
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r0
15 8.00
1.77
0.46
Difference = mu (green) - mu (r0)
Estimate for difference: -1.46667
95% CI for difference: (-3.01877, 0.08544)
T-Test of difference = 0 (vs not =): T-Value = -1.94
Paired T-Test and CI: green, r0
Paired T for green - r0
N
Mean
StDev SE Mean
green
15
6.53333 2.32584 0.60053
r0
15
8.00000 1.77281 0.45774
Difference 15 -1.46667 2.35635 0.60841
95% CI for mean difference: (-2.77157, -0.16176)
T-Test of mean difference = 0 (vs not = 0): T-Value = -2.41
P-Value = 0.030
Test for Equal Variances: green, r0
95% Bonferroni confidence intervals for standard deviations
N
Lower
StDev
Upper
green 15 1.63236 2.32584 3.94217
r0 15 1.24423 1.77281 3.00482
F-Test (normal distribution)
Test statistic = 1.72, p-value = 0.321
Levene's Test (any continuous distribution)
Test statistic = 0.91, p-value = 0.347
Wilcoxon Signed Rank Test: d0
Test of median = 0.000000 versus median not = 0.000000
N
for
Wilcoxon
Estimated
N Test Statistic
P
Median
d0 15
11
9.0 0.037
-1.500
Mann-Whitney Test and CI: green, r0
green
r0
N
15
15
Median
7.000
9.000
Point estimate for ETA1-ETA2 is -2.000
95.4 Percent CI for ETA1-ETA2 is (-2.999,-0.000)
W = 188.5
Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.0712
The test is significant at 0.0677 (adjusted for ties)
7
252y0561s 11/14/05
a) Display the numbers that you are using in columns and compute a sample mean and sample standard
deviation for the d column. (1)
Solution:
Row green r0 d0 dsq0
d 2  110 .
d  22 and
nd  15,
1
8
9 -1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
10
6
5
9
7
3
7
7
3
6
6
8
3
10
10
9
4
9
9
8
7
5
6
10
8
8
9
9
0
-3
1
0
-2
-5
0
2
-3
-4
-2
0
-6
1
-22

d
So d 
0
9
1
0
4
25
0
2
9
16
4
0
36
1
110
nd
s d2 
d
2


 22
 1.467
15
 nd 2
nd 1
110  15 1.467 2 77.7187

 5.5513
14
14
s d  5.5513  2.3561

sd 
5.5513
 0.37009  0.608
15
In this problem assume that the red and green data are two independent samples.
b) Assume that you believe that the normal distribution does not apply to the data and compare the means or
medians as appropriate. (4)
Solution: If the parent distribution is not Normal, we can use a Wilcoxon-Mann-Whitney test of the
equality of two medians.
If we use a Wilcoxon-Mann-Whitney rank test, we get the following.
Row
x1
r1
x2
r2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
8
10
6
5
9
7
3
7
7
3
6
6
8
3
10
17.0
28.5
8.5
5.5
23.0
12.5
2.0
12.5
12.5
2.0
8.5
8.5
17.0
2.0
28.5
188.5
9
10
9
4
9
9
8
7
5
6
10
8
8
9
9
23.0
28.5
23.0
4.0
23.0
23.0
17.0
12.5
5.5
8.5
28.5
17.0
17.0
23.0
23.0
276.5
n1  15, and n 2  15 . Note that n1  n 2  15  15  30 and the sum of the first 30 numbers is
30 31
 465 , so that, as a check on my ranking 188.5 + 276.5 = 465.
2
According to the outline, for values of n1 and n2 that are too large for the tables, W , the smaller of the two
n1  n2  1 and variance  W2
 16 n2 W . If
W  W
the significance level is 5% and the test is two-sided, we reject our null hypothesis if z 
does not
W
lie between z .025  1.960 . We have n1  15 and n 2  15 . W  188 .5 is the smaller of the rank sums.
rank sums, has the Normal distribution with mean W 
1
2 n1
W  1 2 n1 n1  n2  1  1 2 1515  15  1  7.5 31  232 .5 and  W2  16 n2 W
232 .5  581 .25. Thus
W  W
188 .5  232 .5
1936
 3.3308  1.83 .
581
.25
581 .25
Since this is between the critical values, do not reject the null hypothesis of equal medians.

1
6 15
z
W


8
252y0561s 11/14/05 (Page layout view!)
We could also say Pvalue  2Pz  1.83  2.5  .4664   2.0336   .0672  .05 .
Since the p-value is above the significance level, do not reject the null hypothesis of equal medians.
c) You suspect that the data has the Normal distribution. Test to see if the Normal distribution applies to
your rj. Use a test that I taught you. (3)
Solution: Assume   .10 H0 : Normal The only practical method is the Lilliefors method.
The numbers must be in order before we begin computing cumulative probabilities! From above, remember
xx
that x  8.0000 and s  1.773 . We compute z 
. (This is really a t .) Fe is the cumulative
s
distribution, gotten from the Normal table by adding or subtracting 0.5. For example if x  4 ,
4  8.0000
z
 2.26 . F 2.26   .5  P2.26  z  0  .5  .4881  .0119 . (In order to simulate student
1.773
work, my Lilliefors routine rounds all values of z .) Fo comes from the fact that there are 15 numbers, so
that each number is one fifteenth of the distribution.
For   .05 and n  15 the critical value from the Lilliefors table is 0.220. Since the largest
deviation here is 0.1789, we do not reject H 0 .
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
x
4
5
6
7
8
8
8
9
9
9
9
9
9
10
10
z
-2.26
-1.69
-1.13
-0.56
0.00
0.00
0.00
0.56
0.56
0.56
0.56
0.56
0.56
1.13
1.13
O
Ocum
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Fo
Fe
0.06667
0.13333
0.20000
0.26667
0.33333
0.40000
0.46667
0.53333
0.60000
0.66667
0.73333
0.80000
0.86667
0.93333
1.00000
0.011911
0.045514
0.129238
0.287740
0.500000
0.500000
0.500000
0.712260
0.712260
0.712260
0.712260
0.712260
0.712260
0.870762
0.870762
D
0.054756
0.087819
0.070762
0.021073
0.166667
0.100000
0.033333
0.178927
0.112260
0.045594
0.021073
0.087740
0.154406
0.062571
0.129238
d) You decide that the Normal distribution applies to the data, but do not know if the variances are equal.
Test them for equality. (1)
Solution: From the Minitab printout s12  2.3262  5.4103 and s 22  1.773 2  3.1435 , n1  15 and
n2  15 .
s 2 3.1435


14,14  2.98 . We
F 14,14  22 
 0.5810 for a two sided test compare this to F n1 1, n2 1  F.025
2
5
.
4103
s1
should also compare F 14,14 
s12
s 22

1
14,14  2.98 . Since both ratios are below
 1.721 against F.025
.5810
their critical values, we cannot reject the null hypothesis. We can also do this problem by a confidence
interval. In “Confidence Limits and Hypothesis testing for Variances,” in the syllabus supplement, the
s2
 2 s2
1
formula given is 12 n 1, n 1  12  12 F( n2 1, n1 1) , which becomes
s 2 F 1 2
 2 s2 2
2
 2 5.4103
5.4103 1
 12 
2.98 or
3.1435 2.98  2 3.1435
hypothesis.
0.58 
 12
 22
 5.13 . Since this includes 1 we cannot reject the null
9
252y0561s 11/14/05 (Page layout view!)
e) You conclude that the underlying distributions are Normal and that the population variances are equal.
Compare the means or medians as appropriate. Use a test ratio, critical value or a confidence interval (4) or
all three (6)
Solution:   .05
From the computer printout n1  15, x1  6.533 , and s12  2.3262  5.4102 .
From the computer printout n 2  15, x 2  8.000 , and s 22  1.773 2  3.1435 .
H 0 : 1   2 , H1 : 1   2 ,
d  x1  x 2  6.533  8.000  1.467 and   .05 . If we assume
n  1s12  n2  1s 22 155.4102   153.1435 

that the variances are equal s p2  1

n1  n 2  2
30
5.4102  3.1435
 4.2769 , sˆ p  2.068 so that
2
 1
1 
1 1
2
  4.2769     4.2769    0.5703 and
s d2  sˆ 2p  
 15 15 
 15 
 n1 n 2 

1
  1
s d  s p2  
 n1 n 2



  0.5703  0.7552 .

28
and df  n1  n 2  2  15  15  2  28 . t .025
 2.048 .
28
s d  1.467  2.048 0.7552   1.467  1.547 .
Confidence Interval: D  d  t  2 s d  -1.467  t 025
Since 1.547 is larger than 1.467, the interval includes zero, so that we cannot reject the null
hypothesis. Make a diagram: Show an almost Normal curve with a center at -1.467. The shaded
interval runs from -3.014 to 0.080. Make a mark to show that zero is in it.
d  D0  1.467

 1.942 Make a diagram: Show an almost Normal curve with
Test Ratio: t 
sd
0.7552
28
28
 2.048 and  t .025
 2.048 . Since the computed value
a center at zero and critical values at t .025
of t is between these, do not reject the null hypothesis. The ‘reject’ regions are the area below
-2.048 and the area above 2.048.
28
28
 1.701 and t .025
 2.048 . This means Pt  1.701  .05 and
To find a p-value, note that t .05

Pt  2.048   .025 . This means .025  Pt  1.942   .05 . But if we have a 2-sided test and a
negative value of t , p-value is defined as 2Pt  1.942  so .05  pvalue  .10 . Since this is
above   .05 do not reject the null hypothesis.
Critical Value: d cv  D0  t  2 s d  1.547 . Make a diagram: Show an almost Normal curve
with a center at zero and critical values at -1.547 and 1.547. Since the computed value of
d  x1  x 2  1.467 is between these critical values, do not reject the null hypothesis.
[15]
10
252y0561s 11/14/05 (Page layout view!)
f) (Extra credit) You conclude that the underlying distributions are Normal and that the population
variances are not equal. Compare the means or medians as appropriate. Use a test ratio, critical value or a
confidence interval (5) or all three (7)
s
Solution: From the computer printout n1  15, x1  6.533 , s12  2.3262  5.4102 and 1  0.601 .
n1
From the computer printout n 2  15, x 2  8.000 , s 22  1.773 2  3.1435 and
s2
 0.458 .
n2
H 0 : 1   2 , H1 : 1   2 , d  x1  x 2  6.533  8.000  1.467 and   .05 .
If we do not assume equal variances, use the following worksheet:
sx21 
s12
 0.6012  0. 3612
n1
s x22 
s 22
 0.458 2  0.2089
n2
s d2 
s12 s 22

n1 n 2
2
 s12 
 
 n1 
0.3612 2
 

 0.009319
n1  1
14
= 0.5701


  s2 s2 2
  1  2 
  n1 n 2 
df  
2
2
  s2 
 s 22 
1 




 n2 
  n1 
 


n

1
n2 1
 1
s12 s 22

 0.5701  0.7550
n1 n 2
sd 
2
 s22 
 
2
n 
 2   0.2089  0.003117
n2  1
14




2

s d2

0.5702 2



2
2
 0.009319  0.003117
 s x2

s x22
1


 n1 1 n 2  1


 
   
 0.325128

 26 .1440
 0.012436

26
 2.056 .
Round this down and use 26 degrees of freedom. t .025


26
s d  1.467  2.056 0.7550   1.467  1.552 .
Confidence Interval: D  d  t  2 s d  -1.467  t 025
Since 1.552 is larger than 1.467, the interval includes zero, so that we cannot reject the null
hypothesis. Make a diagram: Show an almost Normal curve with a center at -1.467. The shaded
interval runs from -3.019 to 0.085. Make a mark to show that zero is in it.
d  D0  1.467

 1.943 Make a diagram: Show an almost Normal curve with
Test Ratio: t 
sd
0.7550
26
26
 2.056 and  t .025
 2.056 . Since the computed value
a center at zero and critical values at t .025
of t is between these, do not reject the null hypothesis. The ‘reject’ regions are the area below
-2.056 and the area above 2.0056.
26
26
 2.056 . This means Pt  1.706   .05 and
 1.706 and t .025
To find a p-value, note that t .05
Pt  2.056   .025 . This means .025  Pt  1.943   .05 . But if we have a 2-sided test and a

negative value of t , p-value is defined as 2Pt  1.943  so .05  pvalue  .10 . Since this is
above   .05 do not reject the null hypothesis.
Critical Value: d cv  D0  t  2 s d  1.552 . Make a diagram: Show an almost Normal curve
with a center at zero and critical values at -1.552 and 1.552. Since the computed value of
d  x1  x 2  1.467 is between these critical values, do not reject the null hypothesis.
11
252y0561s 11/14/05 (Page layout view!)
2. Assume that the data in the previous problem is paired. Use a confidence level of 95%.
a) Assume that you believe that the normal distribution does not apply to the data and compare the means or
medians as appropriate. (3)
Solution: The Wilcoxon signed Rank test is a test for equality of medians when the data is paired. The Sign
Test for paired data is a simpler test to use in this situation, but it is less powerful. As in many tests for
measures of central tendency with paired data, the original numbers are discarded, and the differences
between the pairs are used. If there are n pairs, these are ranked according to absolute value from 1 to n ,
either top to bottom or bottom to top. After replacing tied absolute values with their average rank, each rank
is marked with a + or – sign and two rank sums are taken, T  and T  . The smaller of these is compared
with Table 7.
Row
d*
1
2
3
4
5
6
7
8
9
10
11
-1
-3
1
-2
-5
2
-3
-4
-2
-6
1
d * rank signed rank
1
3
1
2
5
2
3
4
2
6
1
2.0
7.5
2.0
5.0
10.0
5.0
7.5
9.0
5.0
11.0
2.0
-2.0
-7.5
2.0
-5.0
-10.0
5.0
-7.5
-9.0
-5.0
-11.0
2.0
1112 
 66 .
2
  .05 . For a two-sided test the table value in the .025 column is 11. We reject the null hypothesis if our
smaller T is below 11. It is, so we reject the null hypothesis.
n  11, T   9 and T   57 . These sum to 66. As a check, the sum of the first 11 numbers is
b) You assume that the data has the Normal distribution. Compare the means or medians as appropriate. (3)
In 1a) we found nd  15, d  1.467 ,
s d2
d

2
 nd 2
nd 1
 5.5513 , s d  2.3561 and
s d  0.608 .
H 0 : 1   2 , H1 : 1   2 , d  x1  x 2  6.533  8.000  1.467 and   .05 .
14
df  n d  1  15  1  14 . t .025
 2.145


Confidence Interval: D  d  t  2 s d  -1.467  t 14
05 s d  1.467  2.145 0.608   1.467  1.304 .
Since 1.304 is not larger than 1.467, the interval does not include zero, so that can reject the null
hypothesis. Make a diagram: Show an almost Normal curve with a center at -1.467. The shaded
interval runs from -2.771 to -1.163. Make a mark to show that zero is not in it.
d  D0  1.467

 2.413 Make a diagram: Show an almost Normal curve with
Test Ratio: t 
sd
0.608
14
14
 2.145 . Since the computed value
 2.145 and  t .025
a center at zero and critical values at t .025
of t is not between these, reject the null hypothesis. The ‘reject’ regions are the area below -2.145
and the area above 2.145.
14
14
 2.624 and t .025
 2.145 . This means Pt  2.624   .01 and
To find a p-value, note that t .01
Pt  2.145   .025

. This means .01  Pt  2.413   .025 . But if we have a 2-sided test and a
negative value of t , p-value is defined as 2Pt  2.413  so .02  pvalue  .05 . Since this is
below   .05 reject the null hypothesis.
Critical Value: d cv  D0  t  2 s d  1.304 . Make a diagram: Show an almost Normal curve
with a center at zero and critical values at -1.304 and 1.304. Since the computed value of
d  x1  x 2  1.467 is not between these critical values, reject the null hypothesis.
12
252y0561s 11/14/05 (Page layout view!)
c) For any part of one of these problems (tell me which one!), compute a confidence interval that you would
use to compare means if your alternate hypothesis was H 1 :  2  1 . (2)
[23]
Solution: H 0 : 1   2 , H1 : 1   2 or H 0 : 1   2  0, H1 : 1   2  0 or if D  1   2 ,
H 0 : D  0, H1 : D  0 . A two sided confidence interval is D  d  t  2 s d but a 1-sided interval must go
in the direction of the alternative hypothesis, so it would be D  d  t s d
In 1e) D  d  t  2 s d  -1.467
28
 t 025
sd
Remember   .05 .
 1.467  2.048 0.7552   1.467  1.547 becomes
28
D  d  t  2 s d  -1.467  t 05
s d  1.467  1.7010.7552   1.467  2.631 or D  1.164 .
26
s d  1.467  2.056 0.7550   1.467  1.552 becomes
In 1f) D  d  t  2 s d  -1.467  t 025
26
D  d  t  2 s d  -1.467  t 05
s d  1.467  1.706 0.7550   1.467  1.288 or D  1.179 .
In 2b) D  d  t  2 s d  -1.467  t 14
05 s d  1.467  2.145 0.608   1.467  1.304 becomes
D  d  t  2 s d  -1.467  t 14
05 s d  1.467  1.7610.608   1.467  1.071 or D  0.396
Note that, like the original tests, we can only reject the null hypothesis in 2b).
d) For the same part as you used in c), find a p-value for the null hypothesis. (2) [25]
This was done under ‘test ratio’ in 1e), 1f) and 2b).
13
252y0561s 11/14/05 (Page layout view!)
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
green
8
10
6
5
9
7
3
7
7
3
6
6
8
3
10
r1
5
9
5
7
6
7
9
8
7
5
7
5
8
7
7
d1
3
1
1
-2
3
0
-6
-1
0
-2
-1
1
0
-4
3
dsq1
9
1
1
4
9
0
36
1
0
4
1
1
0
16
9
Descriptive Statistics: green, r1, d1
Variable
green
r1
d1
Variable
green
r1
d1
N N*
Mean
15
0
6.533
15
0
6.800
15
0 -0.267
Maximum
10.000
9.000
3.000
SE Mean
0.601
0.355
0.658
StDev
2.326
1.373
2.549
Minimum
3.000
5.000
-6.000
Q1
5.000
5.000
-2.000
Median
7.000
7.000
0.000
Q3
8.000
8.000
1.000
Sum of d1
Sum of d1 = -4
Sum of Squares of d1
Sum of squares (uncorrected) of d1 = 92
14
252y0561s 11/14/05 (Page layout view!)
Two-Sample T-Test and CI: green, r1
Two-sample T for green vs r1
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r1
15 6.80
1.37
0.35
Difference = mu (green) - mu (r1)
Estimate for difference: -0.266667
95% CI for difference: (-1.695200, 1.161867)
T-Test of difference = 0 (vs not =): T-Value = -0.38
Both use Pooled StDev = 1.9099
P-Value = 0.705
DF = 28
P-Value = 0.706
DF = 22
Two-Sample T-Test and CI: green, r1
Two-sample T for green vs r1
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r1
15 6.80
1.37
0.35
Difference = mu (green) - mu (r1)
Estimate for difference: -0.266667
95% CI for difference: (-1.712960, 1.179626)
T-Test of difference = 0 (vs not =): T-Value = -0.38
Paired T-Test and CI: green, r1
Paired T for green - r1
N
Mean
StDev
SE Mean
green
15
6.53333
2.32584
0.60053
r1
15
6.80000
1.37321
0.35456
Difference 15 -0.266667 2.548576 0.658039
95% CI for mean difference: (-1.678021, 1.144688)
T-Test of mean difference = 0 (vs not = 0): T-Value = -0.41
P-Value = 0.691
Test for Equal Variances: green, r1
95% Bonferroni confidence intervals for standard deviations
N
Lower
StDev
Upper
green 15 1.63236 2.32584 3.94217
r1 15 0.96377 1.37321 2.32752
F-Test (normal distribution)
Test statistic = 2.87, p-value = 0.058
Levene's Test (any continuous distribution)
Test statistic = 3.17, p-value = 0.086
Wilcoxon Signed Rank Test: d1
Test of median = 0.000000 versus median not = 0.000000
N
for
Wilcoxon
N Test Statistic
P Estimated Median
d1 15
12
36.0 0.845
0.000000000
Mann-Whitney Test and CI: green, r1
N Median
green 15
7.000
r1
15
7.000
Point estimate for ETA1-ETA2 is 0.000
95.4 Percent CI for ETA1-ETA2 is (-2.000,1.001)
W = 227.5
Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.8519
The test is significant at 0.8491 (adjusted for ties)
15
252y0561s 11/14/05 (Page layout view!)
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
green
8
10
6
5
9
7
3
7
7
3
6
6
8
3
10
r2
7
10
11
9
8
8
8
7
9
9
7
7
10
7
6
d2
1
0
-5
-4
1
-1
-5
0
-2
-6
-1
-1
-2
-4
4
dsq2
1
0
25
16
1
1
25
0
4
36
1
1
4
16
16
Descriptive Statistics: green, r2, d2
Variable
green
r2
d2
Variable
green
r2
d2
N N*
Mean
15
0
6.533
15
0
8.200
15
0 -1.667
Maximum
10.000
11.000
4.000
SE Mean
0.601
0.368
0.708
StDev
2.326
1.424
2.743
Minimum
3.000
6.000
-6.000
Q1
5.000
7.000
-4.000
Median
7.000
8.000
-1.000
Q3
8.000
9.000
0.000
Sum of d2
Sum of d2 = -25
Sum of Squares of d2
Sum of squares (uncorrected) of d2 = 147
16
252y0561s 11/14/05 (Page layout view!)
Two-Sample T-Test and CI: green, r2
Two-sample T for green vs r2
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r2
15 8.20
1.42
0.37
Difference = mu (green) - mu (r2)
Estimate for difference: -1.66667
95% CI for difference: (-3.10912, -0.22421)
T-Test of difference = 0 (vs not =): T-Value = -2.37
Both use Pooled StDev = 1.9285
P-Value = 0.025
DF = 28
P-Value = 0.027
DF = 23
Two-Sample T-Test and CI: green, r2
Two-sample T for green vs r2
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r2
15 8.20
1.42
0.37
Difference = mu (green) - mu (r2)
Estimate for difference: -1.66667
95% CI for difference: (-3.12338, -0.20995)
T-Test of difference = 0 (vs not =): T-Value = -2.37
Paired T-Test and CI: green, r2
Paired T for green - r2
N
Mean
StDev SE Mean
green
15
6.53333 2.32584 0.60053
r2
15
8.20000 1.42428 0.36775
Difference 15 -1.66667 2.74296 0.70823
95% CI for mean difference: (-3.18567, -0.14767)
T-Test of mean difference = 0 (vs not = 0): T-Value = -2.35
P-Value = 0.034
Test for Equal Variances: green, r2
95% Bonferroni confidence intervals for standard deviations
N
Lower
StDev
Upper
green 15 1.63236 2.32584 3.94217
r2 15 0.99961 1.42428 2.41408
F-Test (normal distribution)
Test statistic = 2.67, p-value = 0.077
Levene's Test (any continuous distribution)
Test statistic = 2.33, p-value = 0.138
Wilcoxon Signed Rank Test: d2
Test of median = 0.000000 versus median not = 0.000000
N
for
Wilcoxon
Estimated
N Test Statistic
P
Median
d2 15
13
15.0 0.036
-1.500
Mann-Whitney Test and CI: green, r2
N Median
green 15
7.000
r2
15
8.000
Point estimate for ETA1-ETA2 is -1.000
95.4 Percent CI for ETA1-ETA2 is (-3.000,0.001)
W = 183.5
Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.0443
The test is significant at 0.0410 (adjusted for ties)
17
252y0561s 11/14/05 (Page layout view!)
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
green
8
10
6
5
9
7
3
7
7
3
6
6
8
3
10
r3
6
12
7
10
12
10
9
11
9
8
7
6
11
13
8
d3
2
-2
-1
-5
-3
-3
-6
-4
-2
-5
-1
0
-3
-10
2
dsq3
4
4
1
25
9
9
36
16
4
25
1
0
9
100
4
Descriptive Statistics: green, r3, d3
Variable
green
r3
d3
Variable
green
r3
d3
N N*
Mean
15
0
6.533
15
0
9.267
15
0 -2.733
Maximum
10.000
13.000
2.000
SE Mean
0.601
0.581
0.802
StDev
2.326
2.251
3.105
Minimum
3.000
6.000
-10.000
Q1
5.000
7.000
-5.000
Median
7.000
9.000
-3.000
Q3
8.000
11.000
-1.000
Sum of d3
Sum of d3 = -41
Sum of Squares of d3
Sum of squares (uncorrected) of d3 = 247
18
252y0561s 11/14/05 (Page layout view!)
Two-Sample T-Test and CI: green, r3
Two-sample T for green vs r3
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r3
15 9.27
2.25
0.58
Difference = mu (green) - mu (r3)
Estimate for difference: -2.73333
95% CI for difference: (-4.44521, -1.02146)
T-Test of difference = 0 (vs not =): T-Value = -3.27
Both use Pooled StDev = 2.2887
P-Value = 0.003
DF = 28
P-Value = 0.003
DF = 27
Two-Sample T-Test and CI: green, r3
Two-sample T for green vs r3
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r3
15 9.27
2.25
0.58
Difference = mu (green) - mu (r3)
Estimate for difference: -2.73333
95% CI for difference: (-4.44807, -1.01860)
T-Test of difference = 0 (vs not =): T-Value = -3.27
Paired T-Test and CI: green, r3
Paired T for green - r3
N
Mean
StDev SE Mean
green
15
6.53333 2.32584 0.60053
r3
15
9.26667 2.25093 0.58119
Difference 15 -2.73333 3.10453 0.80159
95% CI for mean difference: (-4.45256, -1.01410)
T-Test of mean difference = 0 (vs not = 0): T-Value = -3.41
P-Value = 0.004
Test for Equal Variances: green, r3
95% Bonferroni confidence intervals for standard deviations
N
Lower
StDev
Upper
green 15 1.63236 2.32584 3.94217
r3 15 1.57979 2.25093 3.81520
F-Test (normal distribution)
Test statistic = 1.07, p-value = 0.904
Levene's Test (any continuous distribution)
Test statistic = 0.02, p-value = 0.892
Wilcoxon Signed Rank Test: d3
Test of median = 0.000000 versus median not = 0.000000
N
for
Wilcoxon
Estimated
N Test Statistic
P
Median
d3 15
14
9.0 0.007
-2.500
Mann-Whitney Test and CI: green, r3
green
r3
N
15
15
Median
7.000
9.000
Point estimate for ETA1-ETA2 is -3.000
95.4 Percent CI for ETA1-ETA2 is (-5.001,-1.000)
W = 167.0
Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.0070
The test is significant at 0.0066 (adjusted for ties)
19
252y0561s 11/14/05 (Page layout view!)
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
green
8
10
6
5
9
7
3
7
7
3
6
6
8
3
10
r4
5
6
5
6
5
11
8
8
7
9
10
7
7
6
5
d4
3
4
1
-1
4
-4
-5
-1
0
-6
-4
-1
1
-3
5
dsq4
9
16
1
1
16
16
25
1
0
36
16
1
1
9
25
Descriptive Statistics: green, r4, d4
Variable
green
r4
d4
Variable
green
r4
d4
N N*
Mean
15
0
6.533
15
0
7.000
15
0 -0.467
Maximum
10.000
11.000
5.000
SE Mean
0.601
0.488
0.899
StDev
2.326
1.890
3.482
Minimum
3.000
5.000
-6.000
Q1
5.000
5.000
-4.000
Median
7.000
7.000
-1.000
Q3
8.000
8.000
3.000
Sum of d4
Sum of d4 = -7
Sum of Squares of d4
Sum of squares (uncorrected) of d4 = 173
20
252y0561s 11/14/05 (Page layout view!)
Two-Sample T-Test and CI: green, r4
Two-sample T for green vs r4
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r4
15 7.00
1.89
0.49
Difference = mu (green) - mu (r4)
Estimate for difference: -0.466667
95% CI for difference: (-2.051676, 1.118343)
T-Test of difference = 0 (vs not =): T-Value = -0.60
Both use Pooled StDev = 2.1191
P-Value = 0.551
DF = 28
P-Value = 0.552
DF = 26
Two-Sample T-Test and CI: green, r4
Two-sample T for green vs r4
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r4
15 7.00
1.89
0.49
Difference = mu (green) - mu (r4)
Estimate for difference: -0.466667
95% CI for difference: (-2.057187, 1.123854)
T-Test of difference = 0 (vs not =): T-Value = -0.60
Paired T-Test and CI: green, r4
Paired T for green - r4
N
Mean
StDev
SE Mean
green
15
6.53333
2.32584
0.60053
r4
15
7.00000
1.88982
0.48795
Difference 15 -0.466667 3.481926 0.899029
95% CI for mean difference: (-2.394893, 1.461560)
T-Test of mean difference = 0 (vs not = 0): T-Value = -0.52
P-Value = 0.612
Test for Equal Variances: green, r4
95% Bonferroni confidence intervals for standard deviations
N
Lower
StDev
Upper
green 15 1.63236 2.32584 3.94217
r4 15 1.32635 1.88982 3.20315
F-Test (normal distribution)
Test statistic = 1.51, p-value = 0.447
Levene's Test (any continuous distribution)
Test statistic = 0.48, p-value = 0.492
Wilcoxon Signed Rank Test: d4
Test of median = 0.000000 versus median not = 0.000000
N
for
Wilcoxon
Estimated
N Test Statistic
P
Median
d4 15
14
44.0 0.616
-0.5000
Mann-Whitney Test and CI: green, r4
N Median
green 15
7.000
r4
15
7.000
Point estimate for ETA1-ETA2 is -0.000
95.4 Percent CI for ETA1-ETA2 is (-2.001,1.000)
W = 225.5
Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.7875
The test is significant at 0.7849 (adjusted for ties)
21
252y0561s 11/14/05 (Page layout view!)
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
green
8
10
6
5
9
7
3
7
7
3
6
6
8
3
10
r5
8
9
6
10
7
7
5
6
11
9
10
7
7
8
8
d5
0
1
0
-5
2
0
-2
1
-4
-6
-4
-1
1
-5
2
dsq5
0
1
0
25
4
0
4
1
16
36
16
1
1
25
4
Descriptive Statistics: green, r5, d5
Variable
green
r5
d5
Variable
green
r5
d5
N N*
Mean
15
0
6.533
15
0
7.867
15
0 -1.333
Maximum
10.000
11.000
2.000
SE Mean
0.601
0.435
0.715
StDev
2.326
1.685
2.769
Minimum
3.000
5.000
-6.000
Q1
5.000
7.000
-4.000
Median
7.000
8.000
0.000
Q3
8.000
9.000
1.000
Sum of d5
Sum of d5 = -20
Sum of Squares of d5
Sum of squares (uncorrected) of d5 = 134
22
252y0561s 11/14/05 (Page layout view!)
Two-Sample T-Test and CI: green, r5
Two-sample T for green vs r5
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r5
15 7.87
1.68
0.43
Difference = mu (green) - mu (r5)
Estimate for difference: -1.33333
95% CI for difference: (-2.85225, 0.18559)
T-Test of difference = 0 (vs not =): T-Value = -1.80
Both use Pooled StDev = 2.0307
P-Value = 0.083
DF = 28
P-Value = 0.084
DF = 25
Two-Sample T-Test and CI: green, r5
Two-sample T for green vs r5
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r5
15 7.87
1.68
0.43
Difference = mu (green) - mu (r5)
Estimate for difference: -1.33333
95% CI for difference: (-2.86051, 0.19384)
T-Test of difference = 0 (vs not =): T-Value = -1.80
Paired T-Test and CI: green, r5
Paired T for green - r5
N
Mean
StDev SE Mean
green
15
6.53333 2.32584 0.60053
r5
15
7.86667 1.68466 0.43498
Difference 15 -1.33333 2.76887 0.71492
95% CI for mean difference: (-2.86668, 0.20002)
T-Test of mean difference = 0 (vs not = 0): T-Value = -1.87
P-Value = 0.083
Test for Equal Variances: green, r5
95% Bonferroni confidence intervals for standard deviations
N
Lower
StDev
Upper
green 15 1.63236 2.32584 3.94217
r5 15 1.18236 1.68466 2.85542
F-Test (normal distribution)
Test statistic = 1.91, p-value = 0.240
Levene's Test (any continuous distribution)
Test statistic = 1.05, p-value = 0.315
Wilcoxon Signed Rank Test: d5
Test of median = 0.000000 versus median not = 0.000000
N
for
Wilcoxon
Estimated
N Test Statistic
P
Median
d5 15
12
19.5 0.136
-1.500
Mann-Whitney Test and CI: green, r5
green
r5
N
15
15
Median
7.000
8.000
Point estimate for ETA1-ETA2 is -1.000
95.4 Percent CI for ETA1-ETA2 is (-2.999,-0.000)
W = 195.5
Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.1300
The test is significant at 0.1251 (adjusted for ties)
23
252y0561s 11/14/05 (Page layout view!)
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
green
8
10
6
5
9
7
3
7
7
3
6
6
8
3
10
r6
9
7
6
7
8
7
6
6
7
7
5
7
10
7
6
d6
-1
3
0
-2
1
0
-3
1
0
-4
1
-1
-2
-4
4
dsq6
1
9
0
4
1
0
9
1
0
16
1
1
4
16
16
Descriptive Statistics: green, r6, d6
Variable
green
r6
d6
Variable
green
r6
d6
N N*
Mean
15
0
6.533
15
0
7.000
15
0 -0.467
Maximum
10.000
10.000
4.000
SE Mean
0.601
0.324
0.601
StDev
2.326
1.254
2.326
Minimum
3.000
5.000
-4.000
Q1
5.000
6.000
-2.000
Median
7.000
7.000
0.000
Q3
8.000
7.000
1.000
Sum of d6
Sum of d6 = -7
Sum of Squares of d6
Sum of squares (uncorrected) of d6 = 79
24
252y0561s 11/14/05 (Page layout view!)
Two-Sample T-Test and CI: green, r6
Two-sample
N
green 15
r6
15
T for green vs r6
Mean StDev SE Mean
6.53
2.33
0.60
7.00
1.25
0.32
Difference = mu (green) - mu (r6)
Estimate for difference: -0.466667
95% CI for difference: (-1.864090, 0.930757)
T-Test of difference = 0 (vs not =): T-Value = -0.68
Both use Pooled StDev = 1.8683
P-Value = 0.500
DF = 28
P-Value = 0.501
DF = 21
Two-Sample T-Test and CI: green, r6
Two-sample T for green vs r6
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r6
15 7.00
1.25
0.32
Difference = mu (green) - mu (r6)
Estimate for difference: -0.466667
95% CI for difference: (-1.885379, 0.952046)
T-Test of difference = 0 (vs not =): T-Value = -0.68
Paired T-Test and CI: green, r6
Paired T for green - r6
N
Mean
StDev
SE Mean
green
15
6.53333
2.32584
0.60053
r6
15
7.00000
1.25357
0.32367
Difference 15 -0.466667 2.325838 0.600529
95% CI for mean difference: (-1.754673, 0.821340)
T-Test of mean difference = 0 (vs not = 0): T-Value = -0.78
P-Value = 0.450
Test for Equal Variances: green, r6
95% Bonferroni confidence intervals for standard deviations
N
Lower
StDev
Upper
green 15 1.63236 2.32584 3.94217
r6 15 0.87980 1.25357 2.12473
F-Test (normal distribution)
Test statistic = 3.44, p-value = 0.027
Levene's Test (any continuous distribution)
Test statistic = 4.91, p-value = 0.035
Wilcoxon Signed Rank Test: d6
Test of median = 0.000000 versus median not = 0.000000
N
for
Wilcoxon
Estimated
N Test Statistic
P
Median
d6 15
12
28.5 0.433
-0.5000
Mann-Whitney Test and CI: green, r6
N Median
green 15
7.000
r6
15
7.000
Point estimate for ETA1-ETA2 is 0.000
95.4 Percent CI for ETA1-ETA2 is (-2.001,1.000)
W = 222.5
Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.6936
The test is significant at 0.6857 (adjusted for ties)
25
252y0561s 11/14/05 (Page layout view!)
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
green
8
10
6
5
9
7
3
7
7
3
6
6
8
3
10
r7
7
9
10
11
11
6
11
6
9
7
7
7
9
8
8
d7
1
1
-4
-6
-2
1
-8
1
-2
-4
-1
-1
-1
-5
2
dsq7
1
1
16
36
4
1
64
1
4
16
1
1
1
25
4
Descriptive Statistics: green, r7, d7
Variable
green
r7
d7
Variable
green
r7
d7
N N*
Mean
15
0
6.533
15
0
8.400
15
0 -1.867
Maximum
10.000
11.000
2.000
SE Mean
0.601
0.456
0.768
StDev
2.326
1.765
2.973
Minimum
3.000
6.000
-8.000
Q1
5.000
7.000
-4.000
Median
7.000
8.000
-1.000
Q3
8.000
10.000
1.000
Sum of d7
Sum of d7 = -28
Sum of Squares of d7
Sum of squares (uncorrected) of d7 = 176
26
252y0561s 11/14/05 (Page layout view!)
Two-Sample T-Test and CI: green, r7
Two-sample T for green vs r7
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r7
15 8.40
1.76
0.46
Difference = mu (green) - mu (r7)
Estimate for difference: -1.86667
95% CI for difference: (-3.41081, -0.32252)
T-Test of difference = 0 (vs not =): T-Value = -2.48
Both use Pooled StDev = 2.0644
P-Value = 0.020
DF = 28
P-Value = 0.020
DF = 26
Two-Sample T-Test and CI: green, r7
Two-sample T for green vs r7
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r7
15 8.40
1.76
0.46
Difference = mu (green) - mu (r7)
Estimate for difference: -1.86667
95% CI for difference: (-3.41618, -0.31715)
T-Test of difference = 0 (vs not =): T-Value = -2.48
Paired T-Test and CI: green, r7
Paired T for green - r7
N
Mean
StDev SE Mean
green
15
6.53333 2.32584 0.60053
r7
15
8.40000 1.76473 0.45565
Difference 15 -1.86667 2.97289 0.76760
95% CI for mean difference: (-3.51300, -0.22033)
T-Test of mean difference = 0 (vs not = 0): T-Value = -2.43
P-Value = 0.029
Test for Equal Variances: green, r7
95% Bonferroni confidence intervals for standard deviations
N
Lower
StDev
Upper
green 15 1.63236 2.32584 3.94217
r7 15 1.23856 1.76473 2.99113
F-Test (normal distribution)
Test statistic = 1.74, p-value = 0.313
Levene's Test (any continuous distribution)
Test statistic = 0.53, p-value = 0.473
Wilcoxon Signed Rank Test: d7
Test of median = 0.000000 versus median not = 0.000000
N
for
Wilcoxon
Estimated
N Test Statistic
P
Median
d7 15
15
25.0 0.050
-1.500
Mann-Whitney Test and CI: green, r7
N Median
green 15
7.000
r7
15
8.000
Point estimate for ETA1-ETA2 is -2.000
95.4 Percent CI for ETA1-ETA2 is (-4.000,-0.000)
W = 181.5
Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.0362
The test is significant at 0.0340 (adjusted for ties)
27
252y0561s 11/14/05 (Page layout view!)
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
green
8
10
6
5
9
7
3
7
7
3
6
6
8
3
10
r8
10
9
7
10
6
6
7
8
8
7
9
11
6
8
7
d8
-2
1
-1
-5
3
1
-4
-1
-1
-4
-3
-5
2
-5
3
dsq8
4
1
1
25
9
1
16
1
1
16
9
25
4
25
9
Descriptive Statistics: green, r8, d8
Variable
green
r8
d8
Variable
green
r8
d8
N N*
Mean
15
0
6.533
15
0
7.933
15
0 -1.400
Maximum
10.000
11.000
3.000
SE Mean
0.601
0.408
0.748
StDev
2.326
1.580
2.898
Minimum
3.000
6.000
-5.000
Q1
5.000
7.000
-4.000
Median
7.000
8.000
-1.000
Q3
8.000
9.000
1.000
Sum of d8
Sum of d8 = -21
Sum of Squares of d8
Sum of squares (uncorrected) of d8 = 147
28
252y0561s 11/14/05 (Page layout view!)
Two-Sample T-Test and CI: green, r8
Two-sample T for green vs r8
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r8
15 7.93
1.58
0.41
Difference = mu (green) - mu (r8)
Estimate for difference: -1.40000
95% CI for difference: (-2.88701, 0.08701)
T-Test of difference = 0 (vs not =): T-Value = -1.93
Both use Pooled StDev = 1.9881
P-Value = 0.064
DF = 28
P-Value = 0.066
DF = 24
Two-Sample T-Test and CI: green, r8
Two-sample T for green vs r8
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r8
15 7.93
1.58
0.41
Difference = mu (green) - mu (r8)
Estimate for difference: -1.40000
95% CI for difference: (-2.89826, 0.09826)
T-Test of difference = 0 (vs not =): T-Value = -1.93
Paired T-Test and CI: green, r8
Paired T for green - r8
N
Mean
StDev SE Mean
green
15
6.53333 2.32584 0.60053
r8
15
7.93333 1.57963 0.40786
Difference 15 -1.40000 2.89828 0.74833
95% CI for mean difference: (-3.00501, 0.20501)
T-Test of mean difference = 0 (vs not = 0): T-Value = -1.87
P-Value = 0.082
Test for Equal Variances: green, r8
95% Bonferroni confidence intervals for standard deviations
N
Lower
StDev
Upper
green 15 1.63236 2.32584 3.94217
r8 15 1.10865 1.57963 2.67739
F-Test (normal distribution)
Test statistic = 2.17, p-value = 0.160
Levene's Test (any continuous distribution)
Test statistic = 1.45, p-value = 0.239
Wilcoxon Signed Rank Test: d8
Test of median = 0.000000 versus median not = 0.000000
N
for
Wilcoxon
Estimated
N Test Statistic
P
Median
d8 15
15
30.5 0.100
-1.500
Mann-Whitney Test and CI: green, r8
N Median
green 15
7.000
r8
15
8.000
Point estimate for ETA1-ETA2 is -1.000
95.4 Percent CI for ETA1-ETA2 is (-3.000,0.001)
W = 193.5
Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.1103
The test is significant at 0.1052 (adjusted for ties)
29
252y0561s 11/14/05 (Page layout view!)
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
green
8
10
6
5
9
7
3
7
7
3
6
6
8
3
10
r9
5
6
6
7
10
10
10
13
9
6
8
6
7
8
10
d9
3
4
0
-2
-1
-3
-7
-6
-2
-3
-2
0
1
-5
0
dsq9
9
16
0
4
1
9
49
36
4
9
4
0
1
25
0
Descriptive Statistics: green, r9, d9
Variable
green
r9
d9
Variable
green
r9
d9
N N*
Mean
15
0
6.533
15
0
8.067
15
0 -1.533
Maximum
10.000
13.000
4.000
SE Mean
0.601
0.573
0.792
StDev
2.326
2.219
3.067
Minimum
3.000
5.000
-7.000
Q1
5.000
6.000
-3.000
Median
7.000
8.000
-2.000
Q3
8.000
10.000
0.000
Sum of d9
Sum of d9 = -23
Sum of Squares of d9
Sum of squares (uncorrected) of d9 = 167
30
252y0561s 11/14/05 (Page layout view!)
Two-Sample T-Test and CI: green, r9
Two-sample T for green vs r9
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r9
15 8.07
2.22
0.57
Difference = mu (green) - mu (r9)
Estimate for difference: -1.53333
95% CI for difference: (-3.23350, 0.16683)
T-Test of difference = 0 (vs not =): T-Value = -1.85
Both use Pooled StDev = 2.2730
P-Value = 0.075
DF = 28
P-Value = 0.076
DF = 27
Two-Sample T-Test and CI: green, r9
Two-sample T for green vs r9
N Mean StDev SE Mean
green 15 6.53
2.33
0.60
r9
15 8.07
2.22
0.57
Difference = mu (green) - mu (r9)
Estimate for difference: -1.53333
95% CI for difference: (-3.23634, 0.16967)
T-Test of difference = 0 (vs not =): T-Value = -1.85
Paired T-Test and CI: green, r9
Paired T for green - r9
N
Mean
StDev SE Mean
green
15
6.53333 2.32584 0.60053
r9
15
8.06667 2.21897 0.57293
Difference 15 -1.53333 3.06749 0.79202
95% CI for mean difference: (-3.23206, 0.16539)
T-Test of mean difference = 0 (vs not = 0): T-Value = -1.94
P-Value = 0.073
Test for Equal Variances: green, r9
95% Bonferroni confidence intervals for standard deviations
N
Lower
StDev
Upper
green 15 1.63236 2.32584 3.94217
r9 15 1.55736 2.21897 3.76103
F-Test (normal distribution)
Test statistic = 1.10, p-value = 0.863
Levene's Test (any continuous distribution)
Test statistic = 0.00, p-value = 1.000
Wilcoxon Signed Rank Test: d9
Test of median = 0.000000 versus median not = 0.000000
N
for
Wilcoxon
Estimated
N Test Statistic
P
Median
d9 15
12
17.5 0.099
-1.500
Mann-Whitney Test and CI: green, r9
N Median
green 15
7.000
r9
15
8.000
Point estimate for ETA1-ETA2 is -1.000
95.4 Percent CI for ETA1-ETA2 is (-3.000,0.000)
W = 197.0
Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.1466
The test is significant at 0.1408 (adjusted for ties)
31
252y0561s 11/14/05 (Page layout view!)
Original Version of Problem 3 - Exercise 12.18 [12.23 in 9th]: The results of a Gallup phone survey
appear below. Consumers were asked if they objected to having their medical records shared with different
types of organizations. Results follow.
O
Ins Cos Pharm Research

Yes
820
590
670 


No  180
410
330 
a) Is the proportion of people who object different for different institutions?   .05 .
b) If appropriate, use the Marascuilo procedure to determine which organizations are different. Discuss.
Solution: a) We are testing H 0 : Homogeneity or H 0 : p1  p2  p3 , where p1 is the proportion saying
‘yes’ to an insurance company, p 2 is the proportion saying ‘yes’ to a pharmacy, etc.
O
Yes
Ins Cos



Pharm
Research
820
590
180
1000
410
1000
670 

330 
1000
Total
pr
2080
.6933
The row proportions are gotten by
.3067
1.0000
2080
dividing row totals into the overall total, for example .6933 
. We now get our expected table by
3000
using the row proportions to multiply the column totals, for example we replace 820 by .6933 1000 
No
Total
E
 693 .3 . The expected array is
Yes
No
Total
920
3000
Ins Cos



The formula for the chi-squared statistic is  2 
Pharm
693 .3
306.7
1000

Research
693.3
306.7
1000
O  E 2
E
or  2 
693.3 

306.7 
1000

Total
pr
2080
.6933
920
3000
.3067
1.0000
O2
 n . The first of these two
E
formulas is shown below.
820
590
 .820 p2 
 .590
1000
1000
670
pq
.820 .180 
pq
.590 .410 
p3 
 .670 and the variances 1 1 
 .0001476 , 2 2 
 .0002419 ,
1000
n1
1000
n2
1000
For the three column proportions saying ‘yes,’ we have p1 
p3q3 .670 .330 

 .0002211 . We get the following table …………………………….
n3
1000
3) The results of a Gallup phone survey appear below. Consumers were asked if they objected to having
their medical records shared with different types of organizations. Results follow.
The proportion in a sample of 1000 who objected to sharing with insurance companies was p1  .820 .
The proportion in a sample of 1000 who objected to sharing with pharmacies was p 2  .590
The proportion in a sample of 1000 who objected to sharing with medical researchers was p3  .670
Personalize the data by using the second to last digit of your student number, call it d . Multiply it by .001.
Call the result .00d – If the second to last number is zero, use .00d = .010. Add .00d to .820 and subtract
.00d from .670. . (Example: Seymour Butz’s student number is 976532, so he adds .003 to .820 and gets
.823 and subtracts .003 from .670, getting .667. He leaves .590 alone.
a) Is the proportion of people who object different for different institutions?   .01 . (3)
b) If appropriate, use the Marascuilo procedure to determine which organizations are different. Discuss. (3)
[31]
32
252y0561s 11/14/05 (Page layout view!)
a) We are testing H 0 : Homogeneity or H 0 : p1  p2  p3 , where p1 is the proportion
saying ‘yes’ to an insurance company, p 2 is the proportion saying ‘yes’ to a pharmacy, etc.
Solution:
Version 0 p1  .820 , p 2  .590 and p3  .670 .
O
Ins Cos
Pharm
Research
Total
pr

Yes
820
590
670
2080
.6933

No
180
410
330 
920
.3067
Total
1000
1000
1000
3000 1.0000
This E table is the same for all versions of this problem. The row proportions are gotten by dividing row
2080
totals into the overall total, for example .6933 
. We now get our expected table by using the row
3000
proportions to multiply the column totals, for example we replace 820 by .6933 1000 



E
Ins Cos



Yes
No
Total
Pharm
693 .3
306.7
1000
Research
693.3
306.7
1000
693.3 

306.7 
1000
The formula for the chi-squared statistic is  2 

Total
pr
2080
.6933
920
3000
.3067
1.0000
O  E 2
or  2 

O2
 n . In the display below,
E
E
O2
O2
D is O  E , Dsq/E is
, Osq/E is
, K6 is the sum
and the two items chisq and chisq1
E
E
E
O  E 2 and  2  O 2  n . n is computed from both the O and E columns as a check
are  2 
E
E
and sumD is the sum of the D column, which should be zero. All of the versions of this problem result in the
rejection of the null hypothesis.
O  E 2



————— 11/4/2005 11:17:32 PM ————————————————————
Welcome to Minitab, press F1 for help.
Results for: 252x0506-00.MTW
MTB > exec '252chisq'
Executing from file: 252chisq.MTB
Data Display
Row
1
2
3
4
5
6
O
820
180
590
410
670
330
E
693.3
306.7
693.3
306.7
693.3
306.7
D
-126.7
126.7
103.3
-103.3
23.3
-23.3
Dsq
16052.9
16052.9
10670.9
10670.9
542.9
542.9
Dsq/E
23.1543
52.3407
15.3914
34.7926
0.7831
1.7701
Osq/E
969.854
105.641
502.091
548.093
647.483
355.070
Data Display
n
ne
sumD
chisq1
chisq
K6
3000.00
3000.00
-0.000000000
128.232
128.232
3128.23
2
calc
 128.24 . The degrees of freedom for this application are r  1c  1  2  13  1  12  2 .
2
2
Since   .05, we compare the calculated chi-square with  2 .01  9.2103 . Since our calc
is larger than
the table value we reject H 0 .
MTB > exec '252marisc'
Executing from file: 252marisc.MTB
33
252y0561s 11/14/05 (Page layout view!)
Executing from file: 252marisc1.MTB
Data Display
Row
1
2
3
4
5
6
O
820
180
590
410
670
330
E
693.3
306.7
693.3
306.7
693.3
306.7
pnq
0.82
0.18
0.59
0.41
0.67
0.33
pq/n
0.0001476
0.0001476
0.0002419
0.0002419
0.0002211
0.0002211
b) The Marascuilo procedure says that, for 2 by c tests, if (i) equality is rejected and
 
(ii) p a  p b   2 s p , where a and b represent 2 groups, the chi - squared has c  1 degrees of
freedom and the standard deviation is s p 
p a q a pb qb

, you can say that you have a significant
na
nb
difference between p a and p b .
820
590
 .820 p 2 
 .590
1000
1000
pq
.820 .180 
670
p3 
 .670 (q i  1  p i ) and the variances 1 1 
 .0001476 ,
1000
n1
1000
For the three column proportions saying ‘yes,’ we have p1 
p q
p 2 q 2 .590 .410 
.670 .330 

 .0002419 , 3 3 
 .0002211 . The Minitab table appears in all of the
n2
1000
n3
1000
solutions. The interpretation is simple, For example, the first two rows give the contents of the O array
above. The first column is represented by the first two rows of this table. In the p, q column of this table
are the observed p' s and q' s as computed above. After the corresponding p' s and q' s , the variance
pi qi
is printed twice. This should make it extremely easy to check the Marascuilo for any version of this
ni
problem.
1
2
3
4
5
6
O
E
820
180
590
410
670
330
693.3
306.7
693.3
306.7
693.3
306.7
p, q
0.82
0.18
0.59
0.41
0.67
0.33
pq
n
0.0001476
0.0001476
0.0002419
0.0002419
0.0002211
0.0002211
2
We get the following table using  2 .01  9.2103 in the formula.
Pair
Critical Range
2 sp 
pa  pb
1 to 2
9.2103 .0001476  .0002419   .060
.82  .59  .23
2 to 3
9.2103 .0002419  .0002211   .065
.59  .67  .08
1 to 3
9.2103 .0001476  .0002211   .058
.82  .67  .15
Since, in every case, the difference between the proportions exceeds the critical range, we can say that there
is a significant difference between each pair of proportions.
34
252y0561s 11/14/05 (Page layout view!)
Version 1 p1  .821 , p 2  .590 and p3  .669 .
O
Yes
No
Total



Ins Cos
821
179
1000
Pharm Research
590
669 

410
331 
1000
1000
Total
pr
2080
.6933
920
.3067
3000 1.0000
Results for: 252x0506-01.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\252x050601.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0506-01.MTW'
MTB > exec '252chisq'
Executing from file: 252chisq.MTB
Data Display
Row
1
2
3
4
5
6
O
821
179
590
410
669
331
E
693.3
306.7
693.3
306.7
693.3
306.7
D
-127.7
127.7
103.3
-103.3
24.3
-24.3
Dsq
16307.3
16307.3
10670.9
10670.9
590.5
590.5
Dsq/E
23.5213
53.1702
15.3914
34.7926
0.8517
1.9253
C6
972.221
104.470
502.091
548.093
645.552
357.225
Data Display
n
ne
sumD
chisq1
chisq
K6
3000.00
3000.00
-0.000000000
129.652
129.652
3129.65
MTB > exec '252marisc'
Executing from file: 252marisc.MTB
Executing from file: 252marisc1.MTB
Data Display
Row
1
2
3
4
5
6
O
821
179
590
410
669
331
E
693.3
306.7
693.3
306.7
693.3
306.7
pnq
0.821
0.179
0.590
0.410
0.669
0.331
pq/n
0.0001470
0.0001470
0.0002419
0.0002419
0.0002214
0.0002214
Version 2 p1  .822 , p 2  .590 and p3  .668 .
O
Yes
No
Total



Ins Cos
822
178
1000
Pharm Research
590
668 

410
332 
1000
1000
Total
pr
2080
.6933
920
.3067
3000 1.0000
Results for: 252x0506-02.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\252x0506MTB > Save "C:\Documents and Settings\rbove\My Documents\Minitab\252x050602.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0506-02.MTW'
Existing file replaced.
35
252y0561s 11/14/05 (Page layout view!)
MTB > exec '252chisq'
Executing from file: 252chisq.MTB
Data Display
Row
1
2
3
4
5
6
O
822
178
590
410
668
332
E
693.3
306.7
693.3
306.7
693.3
306.7
D
-128.7
128.7
103.3
-103.3
25.3
-25.3
Dsq
16563.7
16563.7
10670.9
10670.9
640.1
640.1
Dsq/E
23.8911
54.0062
15.3914
34.7926
0.9233
2.0870
C6
974.591
103.306
502.091
548.093
643.623
359.387
Data Display
n
ne
sumD
chisq1
chisq
K6
3000.00
3000.00
-0.000000000
131.092
131.092
3131.09
MTB > exec '252marisc'
Executing from file: 252marisc.MTB
Executing from file: 252marisc1.MTB
Data Display
Row
1
2
3
4
5
6
O
822
178
590
410
668
332
E
693.3
306.7
693.3
306.7
693.3
306.7
pnq
0.822
0.178
0.590
0.410
0.668
0.332
pq/n
0.0001463
0.0001463
0.0002419
0.0002419
0.0002218
0.0002218
Version 3 p1  .823 , p 2  .590 and p3  .667 .
O
Yes
No
Total
Ins Cos
 823

 177
1000
Pharm Research
590
667 

410
333 
1000
1000
Total
pr
2080
.6933
920
.3067
3000 1.0000
Results for: 252x0506-03.MTW
MTB > Save "C:\Documents and Settings\rbove\My Documents\Minitab\252x050603.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0506-03.MTW'
Existing file replaced.
MTB > exec '252chisq'
Executing from file: 252chisq.MTB
* NOTE * Duplicate name(s) (Osq/E) made unique by adding _n.
Data Display
Row
1
2
3
4
5
6
O
823
177
590
410
667
333
E
693.3
306.7
693.3
306.7
693.3
306.7
D
-129.7
129.7
103.3
-103.3
26.3
-26.3
Dsq
16822.1
16822.1
10670.9
10670.9
691.7
691.7
Dsq/E
24.2638
54.8487
15.3914
34.7926
0.9977
2.2553
Osq/E_1
976.964
102.149
502.091
548.093
641.698
361.555
Data Display
36
252y0561s 11/14/05 (Page layout view!)
n
ne
sumD
chisq1
chisq
K6
3000.00
3000.00
-0.000000000
132.549
132.549
3132.55
MTB > exec '252marisc'
Executing from file: 252marisc.MTB
Executing from file: 252marisc1.MTB
Data Display
Row
1
2
3
4
5
6
O
823
177
590
410
667
333
E
693.3
306.7
693.3
306.7
693.3
306.7
pnq
0.823
0.177
0.590
0.410
0.667
0.333
pq/n
0.0001457
0.0001457
0.0002419
0.0002419
0.0002221
0.0002221
Version 4 p1  .824 , p 2  .590 and p3  .666 .
O
Yes
No
Total
Ins Cos
 824

 176
1000
Pharm Research
590
666 

410
334 
1000
1000
Total
pr
2080
.6933
920
.3067
3000 1.0000
Results for: 252x0506-04.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\252x050604.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0506-04.MTW'
MTB > exec '252chisq'
Executing from file: 252chisq.MTB
* NOTE * Duplicate name(s) (Osq/E) made unique by adding _n.
Data Display
Row
1
2
3
4
5
6
O
824
176
590
410
666
334
E
693.3
306.7
693.3
306.7
693.3
306.7
D
-130.7
130.7
103.3
-103.3
27.3
-27.3
Dsq
17082.5
17082.5
10670.9
10670.9
745.3
745.3
Dsq/E
24.6394
55.6977
15.3914
34.7926
1.0750
2.4300
Osq/E_1
979.339
100.998
502.091
548.093
639.775
363.730
Data Display
n
ne
sumD
chisq1
chisq
K6
3000.00
3000.00
-0.000000000
134.026
134.026
3134.03
MTB > exec '252marisc'
Executing from file: 252marisc.MTB
Executing from file: 252marisc1.MTB
37
252y0561s 11/14/05 (Page layout view!)
Data Display
Row
1
2
3
4
5
6
O
824
176
590
410
666
334
E
693.3
306.7
693.3
306.7
693.3
306.7
pnq
0.824
0.176
0.590
0.410
0.666
0.334
pq/n
0.0001450
0.0001450
0.0002419
0.0002419
0.0002224
0.0002224
Version 5 p1  .825 , p 2  .590 and p3  .665 .
O
Yes
No
Total



Ins Cos
825
175
1000
Pharm Research
590
665 

410
335 
1000
1000
Total
pr
2080
.6933
920
.3067
3000 1.0000
Results for: 252x0506-05.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\252x050605.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0506-05.MTW'
MTB > exec '252chisq'
Executing from file: 252chisq.MTB
* NOTE * Duplicate name(s) (Osq/E) made unique by adding _n.
Data Display
Row
1
2
3
4
5
6
O
825
175
590
410
665
335
E
693.3
306.7
693.3
306.7
693.3
306.7
D
-131.7
131.7
103.3
-103.3
28.3
-28.3
Dsq
17344.9
17344.9
10670.9
10670.9
800.9
800.9
Dsq/E
25.0179
56.5533
15.3914
34.7926
1.1552
2.6113
Osq/E_1
981.718
99.853
502.091
548.093
637.855
365.911
Data Display
n
ne
sumD
chisq1
chisq
K6
3000.00
3000.00
-0.000000000
135.522
135.522
3135.52
MTB > exec '252marisc'
Executing from file: 252marisc.MTB
Executing from file: 252marisc1.MTB
Data Display
Row
1
2
3
4
5
6
O
825
175
590
410
665
335
E
693.3
306.7
693.3
306.7
693.3
306.7
pnq
0.825
0.175
0.590
0.410
0.665
0.335
pq/n
0.0001444
0.0001444
0.0002419
0.0002419
0.0002228
0.0002228
38
252y0561s 11/14/05 (Page layout view!)
Version 6 p1  .826 , p 2  .590 and p3  .664 .
O
Yes
No
Total



Ins Cos
826
174
1000
Pharm Research
590
664 

410
336 
1000
1000
Total
pr
2080
.6933
920
.3067
3000 1.0000
Results for: 252x0506-06.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\252x050606.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0506-06.MTW'
MTB > Save "C:\Documents and Settings\rbove\My Documents\Minitab\252x050606.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0506-06.MTW'
Existing file replaced.
MTB > exec '252chisq'
Executing from file: 252chisq.MTB
Data Display
Row
1
2
3
4
5
6
O
826
174
590
410
664
336
E
693.3
306.7
693.3
306.7
693.3
306.7
D
-132.7
132.7
103.3
-103.3
29.3
-29.3
Dsq
17609.3
17609.3
10670.9
10670.9
858.5
858.5
Dsq/E
25.3992
57.4154
15.3914
34.7926
1.2383
2.7991
Osq/E
984.099
98.715
502.091
548.093
635.938
368.099
Data Display
n
ne
sumD
chisq1
chisq
K6
3000.00
3000.00
-0.000000000
137.036
137.036
3137.04
Executing from file: 252marisc.MTB
Executing from file: 252marisc1.MTB
Data Display
Row
1
2
3
4
5
6
O
826
174
590
410
664
336
E
693.3
306.7
693.3
306.7
693.3
306.7
pnq
0.826
0.174
0.590
0.410
0.664
0.336
pq/n
0.0001437
0.0001437
0.0002419
0.0002419
0.0002231
0.0002231
Version 7: p1  .827 , p 2  .590 and p3  .663 .
O
Ins Cos Pharm Research
 827
Yes
590
663 


No
410
337 
 173
Total
1000
1000
1000
Total
pr
2080
.6933
920
.3067
3000 1.0000
Results for: 252x0506-07.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\252x050607.MTW";
SUBC>
Replace.
39
252y0561s 11/14/05 (Page layout view!)
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0506-07.MTW'
MTB > exec '252chisq'
Executing from file: 252chisq.MTB
Data Display
Row
1
2
3
4
5
6
O
827
173
590
410
663
337
E
693.3
306.7
693.3
306.7
693.3
306.7
D
-133.7
133.7
103.3
-103.3
30.3
-30.3
Dsq
17875.7
17875.7
10670.9
10670.9
918.1
918.1
Dsq/E
25.7835
58.2840
15.3914
34.7926
1.3242
2.9934
Osq/E
986.483
97.584
502.091
548.093
634.024
370.293
Data Display
n
ne
sumD
chisq1
chisq
K6
3000.00
3000.00
-0.000000000
138.569
138.569
3138.57
MTB > exec '252marisc'
Executing from file: 252marisc.MTB
Executing from file: 252marisc1.MTB
Data Display
Row
1
2
3
4
5
6
O
827
173
590
410
663
337
Version 8
O
Yes
No
Total
E
693.3
306.7
693.3
306.7
693.3
306.7
pnq
0.827
0.173
0.590
0.410
0.663
0.337
pq/n
0.0001431
0.0001431
0.0002419
0.0002419
0.0002234
0.0002234
p1  .828 , p 2  .590 and p3  .662 .
Ins Cos Pharm Research
Total
pr
 828

590
662
2080
.6933


172
410
338
920
.3067


1000
1000
1000
3000 1.0000
Results for: 252x0506-08.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\252x050608.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0506-08.MTW'
MTB > exec '252chisq'
Executing from file: 252chisq.MTB
Data Display
Row
1
2
3
4
5
6
O
828
172
590
410
662
338
E
693.3
306.7
693.3
306.7
693.3
306.7
D
-134.7
134.7
103.3
-103.3
31.3
-31.3
Dsq
18144.1
18144.1
10670.9
10670.9
979.7
979.7
Dsq/E
26.1706
59.1591
15.3914
34.7926
1.4131
3.1943
Osq/E
988.871
96.459
502.091
548.093
632.113
372.494
40
252y0561s 11/14/05 (Page layout view!)
Data Display
n
ne
sumD
chisq1
chisq
K6
3000.00
3000.00
-0.000000000
140.121
140.121
3140.12
MTB > exec '252marisc'
Executing from file: 252marisc.MTB
Executing from file: 252marisc1.MTB
Data Display
Row
1
2
3
4
5
6
O
828
172
590
410
662
338
E
693.3
306.7
693.3
306.7
693.3
306.7
pnq
0.828
0.172
0.590
0.410
0.662
0.338
pq/n
0.0001424
0.0001424
0.0002419
0.0002419
0.0002238
0.0002238
MTB > Save "C:\Documents and Settings\rbove\My Documents\Minitab\252x050608.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0506-08.MTW'
Existing file replaced.
Version 9 p1  .829 , p 2  .590 and p3  .661 .
O
Yes
No
Total



Ins Cos
829
171
1000
Pharm Research
590
661 

410
339 
1000
1000
Total
pr
2080
.6933
920
.3067
3000 1.0000
Results for: 252x0506-09.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\252x050609.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0506-09.MTW'
MTB > Save "C:\Documents and Settings\rbove\My Documents\Minitab\252x050609.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0506-09.MTW'
Existing file replaced.
MTB > exec '252chisq'
Executing from file: 252chisq.MTB
Data Display
Row
1
2
3
4
5
6
O
829
171
590
410
661
339
E
693.3
306.7
693.3
306.7
693.3
306.7
D
-135.7
135.7
103.3
-103.3
32.3
-32.3
Dsq
18414.5
18414.5
10670.9
10670.9
1043.3
1043.3
Dsq/E
26.5606
60.0407
15.3914
34.7926
1.5048
3.4017
Osq/E
991.261
95.341
502.091
548.093
630.205
374.702
41
252y0561s 11/14/05 (Page layout view!)
Data Display
n
ne
sumD
chisq1
chisq
K6
3000.00
3000.00
-0.000000000
141.692
141.692
3141.69
MTB > exec '252marisc'
Executing from file: 252marisc.MTB
Executing from file: 252marisc1.MTB
Data Display
Row
1
2
3
4
5
6
O
829
171
590
410
661
339
E
693.3
306.7
693.3
306.7
693.3
306.7
pnq
0.829
0.171
0.590
0.410
0.661
0.339
pq/n
0.0001418
0.0001418
0.0002419
0.0002419
0.0002241
0.0002241
Version 10 p1  .830 , p 2  .590 , p3  .660
O
Yes
No
Total
E
Yes
No
Total
Ins Cos Pharm Research
830
590
660 

170
410
340 
1000
1000
1000
Ins Cos Pharm Research

693 .3
693.3
693.3 


306.7
306.7 
 306.7
1000
1000
1000



Total
pr
2080
.6933
920
.3067
3000 1.0000
Total
pr
2080
.6933
920
.3067
3000 1.0000
Results for: 252x0506-10.MTW
MTB > Save "C:\Documents and Settings\rbove\My Documents\Minitab\252x050610.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0506-10.MTW'
Existing file replaced.
MTB > exec '252chisq'
Executing from file: 252chisq.MTB
Data Display
Row
1
2
3
4
5
6
O
830
170
590
410
660
340
E
693.3
306.7
693.3
306.7
693.3
306.7
D
-136.7
136.7
103.3
-103.3
33.3
-33.3
Dsq
18686.9
18686.9
10670.9
10670.9
1108.9
1108.9
Dsq/E
26.9535
60.9289
15.3914
34.7926
1.5994
3.6156
Osq/E
993.654
94.229
502.091
548.093
628.299
376.916
Data Display
n
ne
sumD
chisq1
chisq
K6
3000.00
3000.00
-0.000000000
143.281
143.281
3143.28
MTB > exec '252marisc'
Executing from file: 252marisc.MTB
42
252y0561s 11/14/05 (Page layout view!)
Executing from file: 252marisc1.MTB
Data Display
Row
1
2
3
4
5
6
O
830
170
590
410
660
340
E
693.3
306.7
693.3
306.7
693.3
306.7
pnq
0.83
0.17
0.59
0.41
0.66
0.34
pq/n
0.0001411
0.0001411
0.0002419
0.0002419
0.0002244
0.0002244
2
We get the following table using  2 .01  9.2103 in the formula.
Pair
Critical Range
2 sp 
pa  pb
1 to 2
9.2103 .0001411  .0002419   .059
.83  .59  .24
2 to 3
9.2103 .0002419  .0002244   .066
.59  .66  .07
1 to 3
9.2103 .0001411  .0002244   .058
.83  .66  .17
Since, in every case, the difference between the proportions exceeds the critical range, we can say that there
is a significant difference between each pair of proportions.
43