ECO252 QBA2
SECOND HOUR EXAM
In-class materials
Nov 8, 2005
Exhibit 1
The director of the MBA program of a state university wanted to know if a one week orientation would
change the proportion among potential incoming students who would perceive the program as being
good. Given below is the result from 215 students’ view of the program before and after the
orientation.
Before the Orientation
Good
Not Good
Total
1.
After the Orientation
Good
Not Good
93
37
71
14
164
51
Total
130
85
215
Referring to Exhibit 1, which test should she use?
2
a)  -test for difference in proportions
b) Z-test for difference in proportions
c) McNemar test for difference in proportions
d) Wilcoxon rank sum test
ANSWER:
c
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: McNemar test, assumption
In Method D6b, the McNemar Test, we compare two proportions taken from the same sample. Assume that
question 2
question 1
yes no
two different questions are asked of the same group with the following responses.
yes
 x11 x12 
x

no
 21 x 22 
question1
In this case
good
not
question 2
good not
93 37 
 71 14 


H 0 : p1  p 2
If we wish to test 
,where p1 is the proportion saying ‘yes’ before and p 2 is the proportion
H 1 : p1  p 2
x  x 21
37  71
 34
 34



 3.2717 (The test is valid only if
saying ‘yes’ after, let z  12
x12  x 21
37  71
108 10 .3923
x12  x 21  10 .)
2.
Referring to Exhibit 1, what is the null hypothesis?
.
3.
Referring to Exhibit 1, what is the value of the test statistic using a 1% level of significance?
ANSWER:
3.37 or 3.37
TYPE: PR DIFFICULTY: Moderate
1
KEYWORD: McNemar test, test statistic
4.
Referring to Table Exhibit 1, what is the p-value of the test statistic using a 5% level of
significance?
ANSWER:
0.0008
TYPE: PR DIFFICULTY: Moderate
KEYWORD: McNemar test, p-value
5.
Referring to Table Exhibit 1, what should be the director’s conclusion?   .04 
a) There is sufficient evidence that the proportion of potential incoming students who
perceive the program as being good is the same before and after the orientation.
b) There is insufficient evidence that the proportion of potential incoming students who
perceive the program as being good is the same before and after the orientation.
c) There is sufficient evidence that the proportion of potential incoming students who
perceive the program as being good is not the same before and after the orientation.
d) There is insufficient evidence that the proportion of potential incoming students who
perceive the program as being good is not the same before and after the orientation.
ANSWER:
c
TYPE: TF DIFFICULTY: Moderate
KEYWORD: McNemar test, conclusion
Exhibit 2 (Problem D7)
In a study of sleep gotten with a sleeping pill and with a placebo the results were (Keller, Warren, Bartel,
2nd ed. p. 354)
d
x1
x2
Pill
Placebo
difference
7.3
6.8
.5
8.5
7.9
.6
6.4
6.0
.4
9.0
8.4
.6
6.9
6.5
.4
x1  7.620 x 2  7.120 d  0.500
s12  1.197 s 22  0.997 s d2  0.010
We want to see if the means or medians, as appropriate, are different.
Assume that these are independent samples from population with a Normal distribution and that  12   22
Assume these are paired samples from a Normal distribution.
6.
Referring to Exhibit 1, what should be the degrees of freedom for this test?
a) DF  4
b) DF  8
c) DF  9
2
 s12 s 22 



 n1 n 2 
0.4388 2



 7.9341 (Rounded to 7.)
d) DF 
2
2
0.2394 2  0.1994 2
 s12 
 s 22 
 
 
4
4
 n1 
 n2 
 
 

n1  1
n2 1
2
e)
Degrees of freedom are irrelevant because we are using a (Mann-Whitney-) Wilcoxon
rank sum test.
f) Degrees of freedom are irrelevant because we are using a Wilcoxon signed rank test.
g) We do not have enough information to answer this question. (You must explain what
information is missing)
7.
Referring to Exhibit 1, in the formula D  d  t sd , we should use the following.
2
a)
b)
c)
sd 
sd
0.010
 .002  0.447
n2
5
s
s2
s d  1  2  0.4388  0.6624
n1 n 2
sˆ2p 

n1  1s12  n2  1s22
s d  sˆ p
41.197   40.947 

 1.097 , which is used to compute
n1  n2  2
8
1
1
1 1

 1.097     0.439  0.662
n1 n 2
5 5
d) The standard error is irrelevant because we are using a (Mann-Whitney-) Wilcoxon
signed rank test.
e) The standard error is irrelevant because we are using a (Mann-Whitney-) Wilcoxon
signed rank test.
f) We do not have enough information to answer this question. (You must explain what
information is missing)
8.
Referring to Exhibit 1, assume that the correct alternate hypothesis is 1   2 , that use of the
formula t 
d  D0 x1  x 2   1   2 

is correct, that t  2.263 and that we there are 27
sd
sd
degrees of freedom (Assume that all of these are correct, even though it is very unlikely!), we
should do the following.
27 
27 
 2.052 and t .01
 2.473 so that the p-value is between .05 and .02.
Note t .025
a) Reject the null hypothesis only if the significance level is a value below .01
b) Reject the null hypothesis if the significance level is any value below .02
c) Reject the null hypothesis if the significance level is any value above .025
d) Reject the null hypothesis if the significance is any value above .05
e) Reject the null hypothesis only if the significance level is above .10
f) None of the above.
9.
Referring to Exhibit 1, assume that the correct alternate hypothesis is 1   2 , that use of the
formula t 
d  D0 x1  x 2   1   2 

is correct, that t  1.877 and that there are 27
sd
sd
degrees of freedom (Assume that all of these are correct, even though it is very unlikely!), we
should do the following.
27 
27 
 2.052 , so that the pvalue is between .10 and .05
 1.703 and t .025
Note t .05
a) Reject the null hypothesis only if the significance level is a value below .025
b) Reject the null hypothesis if the significance level is any value below .025
3
c)
d)
e)
f)
Reject the null hypothesis if the significance level is any value above .025
Reject the null hypothesis if the significance is any value above .05
Reject the null hypothesis if the significance level is any value above .10
None of the above.
10. You are having a part produced in two different machines. x1 is 201 randomly selected data
points that represent the length of parts from machine one, x 2 is 501 randomly selected data
points that represent the length of parts from machine two. You want to test your suspicion that
parts from machine two are more variable in length than parts from machine one (This is the same
as saying that machine 1 is more reliable than machine 2). Test this suspicion after stating your
hypotheses. Your sample means are 25.593 inches for machine 1 and 25.592 for machine 2.
Sample standard deviations are 8.379 for machine 1 and 9.293 for machine 2.
Solution:
H 0 :   
 H 0 :  12   22
 12
 22
1
2
or
.
In
terms
of
the
variance
ratio
or
, the alternate


 22
 12
 H 1 :  12   22
H 1 :  1   2
hypothesis rules, so H 0 :
 22
 12
 1 and H 1 :
 22
 12
 1.
Since you are comparing variances, use Method D7. Compare the ratio
s 22
s12
against F .
2
s22  9.293 

  1.230 . This has the F distribution with 500 and 200 degrees of freedom. From
s12  8.379 
the table F 500, 200  1.22 . Since the computed F is larger than the table F, reject the null
.05
hypothesis.
11. You are having a part produced in two different machines. x1 is 101 randomly selected data
points that represent the length of parts from machine one, x 2 is 126 randomly selected data
points that represent the length of parts from machine two. You want to test your suspicion that
parts from machine one are more variable in length than parts from machine two (This is the same
as saying that machine 2 is more reliable than machine 1). Test this suspicion after stating your
hypotheses. Your sample means are 25.593 inches for machine 1 and 25.592 for machine 2.
Sample standard deviations are 8.379 for machine 1 and 6.964 for machine 2.
Solution:
H 0 :   
 H 0 :  22   12
2
2
2
1
or 
. In terms of the variance ratio 12 or 22 , the alternate

2
1
 H 1 :  22   12
H 1 :  2   1
hypothesis rules, so H 0 :
 12
 22
 1 and H 1 :
 12
 22
 1.
Since you are comparing variances, use Method D7. Compare the ratio
s 22
s12
against F .
2
s12  8.379 

  1.453 . This has the F distribution with 100 and 125 degrees of freedom. From
s22  6.964 
the table F 100,125  1.36 . Since the computed F is larger than the table F, reject the null
.05
hypothesis.
4
12. (Extra credit) compute a confidence interval for the ratio of the two variances in the previous
problem.
Solution 1: From the outline
 22 s 22 ( n1 1, n2 1) s22  9.293 


F
. 2 
  1.230 has
s1  8.379 
s12 Fn2 1,n1 1  12 s12 2
s 22
2
1
2
the F distribution with 500 and 200 degrees of freedom. So,
2
( 200, 500)
500, 200  1.27 , F 200,500 must be between
. F.025
.025
500, 200   2  1.230 F.025
F.025
1
F 200,400  1.27 and F 200,1000  1.23 , probably about 1.26. The interval thus becomes
1.230 
2
1
.025
.025
1.230 
 22
2
1
 2  1.230 1.26  or 0.968  22  1.55 .
1.27  1
1
Solution 2: From the outline
s12
s 22
 12 s12 ( n2 1, n1 1) s12  8.379 


F
. 2 
  1.453


s2  6.964 
F n1 1, n2 1  22 s 22 2
2
1
2
has the F distribution with 100 and 125 degrees of freedom.

(125, 100)
100,125  1.45 . F 125,100 must be between
 12  1.453 F.025
F.025
.
.025

100,125
2
F.025
F 100,100  1.48 and F 200,100  1.42 , probably about 1.46. The interval thus becomes
1.453 
2
1
.025
.025
1.453 
1

1.45
 12
 22
 1.453 1.46  or 1.00 
 12
 22
 2.12 .
13. A researcher randomly samples female (Sample 1) and male graduates of an MBA program. The
figures represent starting salaries. The Minitab command used here is a pull-down command that
is equivalent to the two-sample command that you used.
MTB > TwoT 18 48266.7 13577.63 12 55000 11741.29;
SUBC>
Alternative -1.
Two-Sample T-Test and CI
Sample
1
2
N
18
12
Mean
48267
55000
StDev
13578
11741
SE Mean
3200
3389
Difference = mu (1) - mu (2)
Estimate for difference: -6733.30
95% upper bound for difference: 1229.26
T-Test of difference = 0 (vs <): T-Value = -1.44
DF = 25
P-Value = 0.081
a) Turn in your first computer assignment only. (2)
b) For this assignment make 3 curves with centers at zero and with the t ratio marked by a vertical
line. Indicate, using the three diagrams and information from the printout, what the p-value would
be if the null hypothesis was
pvalue  1  .081  .919
(i) H 0 : 1   2
(ii) H 0 : 1   2
pvalue  .081
pvalue  2.081   .161
(iii) H 0 : 1   2
Give me a number for the p-value.
Diagram: It says T-Value = -1.44 so make three diagrams with an almost Normal curve and
a mean at zero. (i) .919 is the area above -1.44, (ii) .081 is the area below -.144, (iii) .161 is the
area above +1.44 and the area below -1.44.
(1.5)
5
c) Using the style that I use for null hypotheses, which of the three hypotheses in b) is the null
hypothesis used here and would it be rejected if the confidence level was .05?
(1)
Solution: It says T-Test of difference = 0 (vs <). This says H 1 : 1   2 , so the null
hypothesis is H 0 : 1   2 . P-value is above .05, so do not reject the null hypothesis.
d) Nothing in the command told Minitab which method to use. Of the four methods you learned to
compare two means, which one did Minitab use.
Solution: D3.
A real estate company is comparing the amount of time people live in Smallville (Sample 1) with
the amount of time they live in Metropolis (Sample 2). The figures here indicate duration in
months. The Minitab command used here is a pull-down command that is equivalent to the twosample command that you used.
MTB > TwoT 100 35 30 150 50 32.4037;
SUBC>
Alternative -1.
Two-Sample T-Test and CI
Sample
1
2
N
100
150
Mean
35.0
50.0
StDev
30.0
32.4
SE Mean
3.0
2.6
Difference = mu (1) - mu (2)
Estimate for difference: -15.0000
95% upper bound for difference: -8.3931
T-Test of difference = 0 (vs <): T-Value = -3.75
DF = 223
P-Value = 0.000
a) Turn in your first computer assignment only. (2)
b) For this assignment make 3 curves with centers at zero and with the t ratio marked by a vertical
line. Indicate, using the three diagrams and information from the printout, what the p-value would
be if the null hypothesis was
(i) H 0 : 1   2
pvalue  1  0  1
(ii) H 0 : 1   2
pvalue  0
pvalue  20  0
(iii) H 0 : 1   2
Give me a number for the p-value.
Diagram: It says T-Value = -3.75 so make three diagrams with an almost Normal curve and
a mean at zero. (i) 1 is the area above -3.75, (ii) 0 is the area below -3.75, (iii) 0 is the area above
+3.75 and the area below -3.75.
(1.5)
c) Using the style that I use for null hypotheses, which of the three hypotheses in b) is the null
hypothesis used here and would it be rejected if the confidence level was .05? (1)
Solution: It says T-Test of difference = 0 (vs <). This says H 1 : 1   2 , so the null
hypothesis is H 0 : 1   2 . P-value is below .05, so reject the null hypothesis.
d) Nothing in the command told Minitab which method to use. Of the four methods you learned to
compare two means, which one did Minitab use?
Solution: D3.
14. Problem 10.13 b in text. The manager suspects that the amount of stuff going into cans packed by
machines 1 and 2 is different. Assume that variances are equal.
x1  8.005 x 2  7.997 s1  0.012 s 2  0.005 n1  11 n2  16
H0:
1  2
1  2
where Populations: 1 = Line A, 2 = Line B
H1:
Decision rule: DF. = 12. If |t| > 2.1788, reject H0.
6
Test statistic:
t
X
1
 X 2    1  2 
S12 S22

n1 n2

8.005  7.997   0
0.0122 0.0052

11
16
 2.0899
0.012 2
0.005 2
 .00000156
 .0000131
16
11
Since t = 2.0899 < 2.1788 or p-value = 0.0586 > 0.05, do not reject H0. There is not
sufficient evidence of a difference in the mean weight of cans filled on the two lines.
The following problem is an easier version of a problem in the text.
Revised version 1: A pet food canning factory produces 8 oz cans of cat food. The manager suspects that
the amount of cat food put into the cans by machine 1 is significantly larger than that put
in by machine 2. A sample of output is taken with the results below.
x1  8.005 x 2  7.997 s1  0.048 s 2  0.015 n1  176 n 2  144
a) What are the managers’s null and alternate hypotheses?
d  D0
b) You will use a test ratio of the form
to test the hypothesis. Find s d .
sd
Warning: Be accurate! s d2 is roughly the size of .0000175. If you start rounding
excessively, your answers will be completely wrong. If you absolutely cannot do this
section, say so and use .000175, which is very wrong.
c) Compute the test ratio and find a p-value for your result.
d) If the manager had, instead, suspected that a larger amount was going into cans filled
by machine 2, what would the p-value be?
Solution: a) H 1 : 1   2 , so H 0 : 1   2 . Note: d  8.005  7.997  0.008
0.015 2
 .00000156
144
s12 s 22

n1 n1
b)
0.048 2
 .0000131
176

0.048 2 0.015 2

 .0000131  .00000156  .0000147  .003828
176
144
sd 
8.005  7.997
 2.0899 pvalue  Pz  2.09   .5  .4817  .0183
.003828
d) 1-.0183 = .9817.
Revised version 2: A pet food canning factory produces 8 oz cans of cat food. The manager suspects that
the amount of cat food put into the cans by machine 2 is significantly larger than that put
in by machine 1. A sample of output is taken with the results below.
x1  8.012 x 2  8.020 s1  0.048 s 2  0.015 n1  176 n 2  144
a) What are the managers’s null and alternate hypotheses?
d  D0
b) You will use a test ratio of the form
to test the hypothesis. Find s d .
sd
c) z 
Warning: Be accurate! s d2 is roughly the size of .0000175. If you start rounding
excessively, your answers will be completely wrong. If you absolutely cannot do this
section, say so and use .000175, which is very wrong.
c) Compute the test ratio and find a p-value for your result.
d) If the manager had, instead, suspected that a larger amount was going into cans filled
by machine 1, what would the p-value be?
Solution: a) H 1 : 1   2 , so H 1 : 1   2 Note: d  8.012  8.020  0.008
7
b)

0.048 2
 .0000131
176
0.015 2
 .00000156 s d 
144
s12 s 22

n1 n1
0.048 2 0.015 2

 .0000131  .00000156  .0000147  .003828
176
144
8.012  8.020
 2.0899 pvalue  Pz  2.09   .5  .4817  .0183
.003828
d) 1-.0183 = .9817.
c) z 
8
15. Problem 12.53 in text. Over 104 weeks, the following number of mortgages were approved by a
bank. Do the results below follow a Poisson distribution?
H0 : The number of commercial mortgages approved per week follows a Poison distribution.
H1 : The number of commercial mortgages approved per week does not follow a Poison distribution.
7
X 
Use
m
j 1
j
fj

n
  2.1058
219
 2.1058
104
Number Approved/Week
0
1
2
3
4
5
6
7
8 or more
Total
13
25
32
17
9
6
1
1
0
104
Combine the last three classes:
Number Approved/Week
0
1
2
3
4
5
6 or more
Total

2
k  p 1

k
 f0  fe 
fe
f0
f0
13
25
32
17
9
6
2
104
P X 
mj f j
fe
0.121752
0.256382
0.269940
0.189477
0.099749
0.042010
0.014744
0.004435
0.001511
1.000000
0
25
64
51
36
30
6
7
0
219
fe
12.66221
26.66368
28.07378
19.70564
10.37388
4.369
2.15181
104
 f0  fe 
12.66221
26.66368
28.07378
19.70564
10.37388
4.369
1.533351
0.461269
0.15719
104
2
/ fe
0.009011434
0.103805808
0.549095235
0.371491012
0.181951945
0.608872099
0.010710218
1.834937752
2
 1.8349
2
2
crit
 5,0.01
=15.0863
Since 1.8349 < 15.0863, do not reject H0. At the 1% level of significance, there is not enough
evidence to reject the null hypothesis that the distribution of commercial mortgages approved per
week follows a Poisson distribution.
9
Revised version 1:
Over 104 weeks, the following numbers of mortgages were approved by a bank. Do the results below
follow a Poisson distribution?
a) Find the average number of mortgages approved per week? Hint: The original version of the
problem came up with a mean of 2.1058 approvals per week. You will come out with something closer
to a mean that you actually can find in your tables (1)
b) If the data followed an appropriate Poisson distribution exactly, what would the numbers of weeks
be with 0, 1, 2, 3, 4, 5, 6, 7 and 8 or more approvals be? (3)
c) Use a statistical test to compare the number of actual approvals with the distribution you found in
the last section. (3)
d) If all this sounds like too much work, guess the mean and compare the observed data with a Poisson
distribution with the mean that you guessed. Do not use a Chi-squared method in d. (5)
Number Approved x 
0
1
2
3
4
5
6
7
8 or more
Total
Number Approved/Week
0
1
2
3
4
5
6
7
8 or more
Total
mean 
O
13
25
31
17
8
5
1
1
0
104
f0
16
25
31
17
8
5
1
1
0
104
xO
0
25
62
51
32
25
6
7
0
208
mj f j
0
25
62
51
32
25
6
7
0
208
P X 
fe
0.121752
0.256382
0.269940
0.189477
0.099749
0.042010
0.014744
0.004435
0.001511
1.000000
12.66221
26.66368
28.07378
19.70564
10.37388
4.369
1.533351
0.461269
0.15719
104
208
2
104
Combine the last three classes:
Number Approved/Week
0
1
2
3
4
5
6 or more
Total
f0
16
25
31
17
8
5
1
104
fe
12.66221
26.66368
28.07378
19.70564
10.37388
4.369
2.15181
104
 f0  fe 
2
/ fe
0.009011434
0.103805808
0.549095235
0.371491012
0.181951945
0.608872099
0.010710218
1.834937752
10

2
k  p 1

k
 f0  fe 
fe
2
 1.8349
2
2
crit
 5,0.01
=15.0863
Since 1.8349 < 15.0863, do not reject H0. At the 1% level of significance, there is not enough
evidence to reject the null hypothesis that the distribution of commercial mortgages approved per
week follows a Poisson distribution.
11
Revised version 2:
Over 105 weeks, the following numbers of mortgages were approved by a bank. Do the results below
follow a Poisson distribution?
a) Find the average number of mortgages approved per week? Hint: The original version of the
problem came up with a mean of 2.1058 approvals per week. You will come out with something closer
to a mean that you actually can find in your tables (1)
b) If the data followed an appropriate Poisson distribution exactly, what would the numbers of weeks
be with 0, 1, 2, 3, 4, 5, 6, 7 and 8 or more approvals be? (3)
c) Use a statistical test to compare the number of actual approvals with the distribution you found in
the last section.
d) If all this sounds like too much work, guess the mean and compare the observed data with a Poisson
distribution with the mean that you guessed. Do not use a Chi-squared method in d.
Number Approved x 
0
1
2
3
4
5
6
7
8 or more
Total
O
23
24
30
16
6
4
1
1
0
105
Number Approved/Week
0
1
2
3
4
5
6
7
8 or more
Total
mean 
xO
0
24
60
48
24
25
6
7
0
189
f0
23
24
30
16
6
4
1
1
0
105
mj f j
0
24
60
48
24
25
6
7
0
189
P X 
fe
0.121752
0.256382
0.269940
0.189477
0.099749
0.042010
0.014744
0.004435
0.001511
1.000000
12.66221
26.66368
28.07378
19.70564
10.37388
4.369
1.533351
0.461269
0.15719
104
189
 1.80
105
Combine the last three classes:
Number Approved/Week
0
1
2
3
4
5
6 or more
Total
f0
16
25
31
17
8
5
1
104
fe
12.66221
26.66368
28.07378
19.70564
10.37388
4.369
2.15181
104
 f0  fe 
2
/ fe
0.009011434
0.103805808
0.549095235
0.371491012
0.181951945
0.608872099
0.010710218
1.834937752
12

2
k  p 1

k
 f0  fe 
fe
2
 1.8349
2
2
crit
 5,0.01
=15.0863
Since 1.8349 < 15.0863, do not reject H0. At the 1% level of significance, there is not enough
evidence to reject the null hypothesis that the distribution of commercial mortgages approved per
week follows a Poisson distribution.
13
Chi-squared Computations
————— 11/8/2005 11:59:53 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > #Version 1
MTB > WOpen "C:\Documents and Settings\rbove\My Documents\Minitab\notmuch.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My
Documents\Minitab\notmuch.MTW'
Worksheet was saved on Thu Apr 14 2005
Results for: notmuch.MTW
MTB > Execute "C:\Documents and Settings\rbove\My
Documents\Minitab\252chisq.mtb" 1.
Executing from file: C:\Documents and Settings\rbove\My
Documents\Minitab\252chisq.mtb
Data Display
Row
1
2
3
4
5
6
O
16
25
31
17
8
7
E
14.0748
28.1498
28.1498
18.7665
9.3833
5.4758
D
-1.92516
3.14978
-2.85022
1.76649
1.38330
-1.52419
Dsq
3.70624
9.92114
8.12373
3.12048
1.91351
2.32316
Dsq/E
0.263324
0.352441
0.288589
0.166279
0.203927
0.424259
Osq/E
18.1885
22.2027
34.1388
15.3998
6.8206
8.9485
Data Display
n
ne
sumD
chisq1
chisq
K6
104.000
104.000
0.000000000
1.69882
1.69882
105.699
MTB > print c2 c7
Data Display
Row
1
2
3
4
5
6
E
14.0748
28.1498
28.1498
18.7665
9.3833
5.4758
f
0.135335
0.270671
0.270671
0.180447
0.090224
0.052652
Results for: 252x0506-11.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\252x050611.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0506-11.MTW'
14
MTB > #version 2
MTB > exec '252chisq'
Executing from file: 252chisq.MTB
Data Display
Row
1
2
3
4
5
6
O
23
24
30
16
6
6
E
17.3564
31.2415
28.1173
16.8705
7.5917
3.8226
D
-5.64361
7.24149
-1.88268
0.87045
1.59171
-2.17737
Dsq
31.8503
52.4392
3.5445
0.7577
2.5335
4.7409
Dsq/E
1.83507
1.67851
0.12606
0.04491
0.33372
1.24023
Osq/E
30.4787
18.4370
32.0087
15.1745
4.7420
9.4176
Data Display
n
ne
sumD
chisq1
chisq
K6
105.000
105.000
0.000000000
5.25851
5.25851
110.259
MTB > print c2 c7
Data Display
Row
1
2
3
4
5
6
E
17.3564
31.2415
28.1173
16.8705
7.5917
3.8226
f
0.165299
0.297538
0.267784
0.160671
0.072302
0.036406
Results for: 252x0506-12.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\252x050612.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0506-12.MTW'
15