252y0622 11/8/06
ECO252 QBA2
SECOND HOUR EXAM
November 8, 2006
Name KEY
Circle Hour of Class Registered
MWF2, MWF3, TR12:30, TR2
Show your work! Make Diagrams! Exam is normed on 50 points. Answers without reasons are not
usually acceptable.
I. (8 points) Do all the following. Make diagrams!
x ~ N 19, 9 - If you are not using the supplement table, make sure that I know it.
35  19 
10  19
z
 P 1.00  z  1.78   P1.00  z  0  P0  z  1.78 
1. P10  x  35   P 
9 
 9
 .3413  .4625  .8038
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the area between -1.00 and 1.78. Because this is on both sides of zero we must add together the area
between -1.00 and zero and the area between zero and 1.78. If you wish, make a completely separate
diagram for x . Draw a Normal curve with a mean at 19. Indicate the mean by a vertical line! Shade the
area between 10 and 35. This area includes the mean (19), and areas to either side of it so we add together
these two areas.
2  19 

 Pz  1.89   P1.89  z  0  Pz  0  .4706  .5  .9706
2. Px  2  P  z 
9 

For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the entire area above -1.89. Because this is on both sides of zero we must add together the area
between -1.89 and zero and the entire area above zero. If you wish, make a completely separate diagram for
x . Draw a Normal curve with a mean at 19. Indicate the mean by a vertical line! Shade the entire area
above 2. This area includes the mean (19), and areas to either side of it so we add together these two areas.
19  19 
 0  19
z
 P 2.11  z  0   .4826
3. P0  x  19 .00   P 
9
9 

For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the area between -2.11 and zero. Because this is completely on the left of zero and touches zero, we
can simply look up our answer on the standardized Normal table. If you wish, make a completely separate
diagram for x . Draw a Normal curve with a mean at 19. Indicate the mean by a vertical line! Shade the
area between 0 and 19. This area includes the mean (19), but does not include any points to the right of the
mean, so that we neither add nor subtract.
x.065 (Do not try to use the t table to get this.) For z make a diagram. Draw a Normal curve with a
mean at 0. z .065 is the value of z with 6.5% of the distribution above it. Since 100 – 6.5 = 93.5, it is also
the 93.5th percentile. Since 50% of the standardized Normal distribution is below zero, your diagram should
show that the probability between z .065 and zero is 93.5% - 50% = 43.5% or P0  z  z.065   .4350 . The
closest we can come to this is P0  z  1.51  .4345 . (1.52 is also acceptable here.) So z .065  1.51 . To
4.
get from z .065 to x.065 , use the formula x    z , which is the opposite of z 
x
.

x  19  1.519  32.59 . If you wish, make a completely separate diagram for x . Draw a Normal curve
with a mean at 7. Show that 50% of the distribution is below the mean (7). If 6.5% of the distribution is
above z .065 , it must be above the mean and have 43.5% of the distribution between it and the mean.
32 .59  19 

Check: Px  32.59   P  z 
  Pz  1.51  Pz  0  P0  z  1.51
9


 .5  .4355  .0645  6.5%
1
252y0622 11/8/06
II. (22+ points) Do all the following? (2points each unless noted otherwise). Look them over first – the
Note the following:
1. This test is normed on 50 points, but there are more points possible including the take-home.
You are unlikely to finish the exam and might want to skip some questions.
2. A table identifying methods for comparing 2 samples is at the end of the exam.
3. If you answer ‘None of the above’ in any question, you should provide an alternative
answer and explain why. You may receive credit for this even if you are wrong.
4. Use a 5% significance level unless the question says otherwise.
5. Read problems carefully. A problem that looks like a problem on another exam may be
quite different.
computer problem is at the end.
1.
A company gives an exam to graduates of quality control programs in two plants samples of scores
are as follows:
Boston 78 82 66 69 85
Atlanta 50 66 69 85
a. Compute the sample variance for Boston – Show your work! The sample mean and standard deviation
for Atlanta are 67.50 and 14.34. (2)
b. Is there a significant difference between the scores in the two plants? You may assume that variances are
equal. State your hypotheses! (2)
Solution:
x12
x1
Row
1
2
3
4
5
78
82
66
69
85
380
x
1
6084
6724
4356
4761
7225
29150
 380 ,
x
2
1
x1  x1
x1  x 1 2
2
6
-10
-7
9
0
4
36
100
49
81
270
 29150 , n1  5
x1 
s12 

x
1
n
x
2
1
 x
1

380
 76 .00
5
 nx12
n 1
 x1 2
n 1


29150  576 .00 2
4
270 .00
 67 .5
4
s1  67 .50  8.216
Solution: The formula table gives us the formulas below. (Method D2 – Comparison of two means with
samples coming from populations with similar variances.)
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Difference
H 0 : D  D0 *
D  d  t 2 s d
d cv  D0  t  2 s d
d  D0
t
between Two
H 1 : D  D0 ,
sd
1 1
Means (
sd  s p

D




n1  1s12  n2  1s22
1
2
n n
2
unknown,
1
variances
assumed equal)
2
sˆ p 
n1  n2  2
DF  n1  n2  2
n1  5 , x1  76.00 s12  67.50 , n 2  4 , x 2  67 .50 and s12  14 .34 2  205 .64
 H 0 : 1   4
DF  n1  1  n2  1  4  3  7  n1  n2  2. Our hypotheses are 
or
 H 1 : 1   4
 H 0 : 1   2  0
H 0 : D  0
or if D  1   2 , 
. d  x1  x 2  76.00  67.50  8.50 .

H
:




0
2
 1 1
H 1 : D  0
2
252y0622 11/8/06
sˆ 2p 
n1  1s12  n2  1s 22
n1  n 4  2

467 .50  3205 .64  126 .703 . This is the pooled variance.
7
sˆ p  11.256 .
  .05 so t.7025  2.365 .
s d  sˆ p
 1
1
1
1 
1 1
 9 
  126 .703     126 .703    57.016  7.551 . Recall that

 sˆ 2p  
n1 n 2
5 4
 20 
 n1 n 2 
our alternate hypothesis is H 1 : D  0 so this is a two-sided test. You may have said
1 1
126 .703   
5 4
 126 .703 0.2000  0.2500   57 .016  7.551
Test Ratio: t 
x  x 2   10   20 
d  D0
.
or t  1
sd
sd
If this test ratio lies between  t.7025  2.365 , do not reject H 0 .
t
d  D0
8 .5  0

 1.126 . Make a
sd
7.551
diagram with zero in the middle showing shaded ‘reject’ regions below -2.365 and above 2.365. Since –
1.126 does not fall in the 'reject' region, do not reject H 0 . Or you can say that, since 1.126 falls between
7
t .05
 1.895 and t.7025  2.365 , for a one-sided test, .025  p  value  .05 . But this is a two-sided test so
that .05  p  value  .10 Since the p-value is above   .05 , do not reject H 0 .
Critical Value: d CV  D0  t  2 s d or x1  x 2 CV  10   20   t  2 s d . For a two-sided test we want two
critical values above and below D0  0. If d  x1  x 2 is between the critical values, do not reject H 0 .
d CV  D0  t  2 s d  0  2.365 7.551   17 .858 . Make a diagram with 0 in the middle showing shaded
‘reject’ regions below -17.858 and above 17.858. Since d  8.50 does not fall in the 'reject' region, do not
reject H 0 .
Confidence Interval: D  d  t  2 s d . We already know that t  s d  2.3657.551  17.858 so we can say
2
D  8.5  17.858 or 9.35  D  26.36 . Make a diagram with 8.5 in the middle. Represent the confidence
interval by shading the area between -9.35 and 26.36. Since zero is in this area, do not reject H 0 . Or
simply note that the error part of the confidence interval is larger that the sample mean difference so the
interval must include zero.
3
252y0622 11/8/06
Exhibit 1
The director of the MBA program of a state university wanted to know if a year of MBA work would
change the proportion among potential students who would perceive the program as being good. A
random sample of 215 incoming students is compared with a random sample of 215 second-year
students. Of the first group 164 students viewed the program as ‘good.’ Of second year students 130
viewed the program as ‘good.’ She wishes to see if the fraction that considered the program ‘good’ has
fallen.
2.
Referring to Exhibit 1, which test should she use? (2)
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
3. Referring to Exhibit 1, what is the null hypothesis? (2)
Solution: Solution: She wants to know if p 2  p1 which is the same as H 1 : p1  p 2 . So we have
H 0 : p1  p 2 or H 0 : p 2  p1
4. Referring to Exhibit 1, what is the value of the computed test statistic? (4)
[6]
H
:
p

p
H
:
p

p

0
H
:

p

0
 0 1


2
2
or  0 1
or if p  p1  p 2 ,  0
are implied. If you use

H
:
p

p
H
:
p

p

0
2
2
 1 1
 1 1
H 1 : p  0
H 0 : p  0
p  p 2  p1 we have 
H 1 : p  0
164
130
164  130
164  130 294
p1 
 .76279 p 2 
 .60465 p 
 0.15814 p 0 

 .68372
215
215
215
215  215 430
 2 
  .00201154  .04485
 215 
 p  .68372 .31627 
For the first definition z 
5.
p  p 0
 p

0.15814
 3.526
.04485
Referring to Table Exhibit 1, what is the p-value of the test statistic? (2)
[8]
Solution: The approximate p-value will be, since this is a right sided test
Pz  3.53  Pz  0  P0  z  3.53  .5  .4998  .0002
4
252y0622 11/8/06
Exhibit 2
(Dummeldinger) A researcher took a random sample of n graduates of MBA programs, which included
n1 women and n 2 men. Their starting salaries were recorded. Use 1 and/or  1 for population parameters
for women and  2 and/or  2 for men.   .10  The sample yields following data.
x 2  50200 .00
s 2  13557 .29
x1  48266 .70
s1  15678 .01
and
. n1  n 2  200 . The researcher wants to show that men have a higher mean starting salary
than women. Assume that these are independent samples. Hint: To preserve both our sanity, move the
decimal point three places to the left in both the means and standard deviations. You will be working
in thousands but will get the same value for the test ratio.
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 .
6.
Referring to Exhibit 2, if D  1   2 what is the null hypothesis? (1)
a) D  0
b) D  0
c) D  0
d) * D  0
e) D  0
f) None of the above. (Give the correct one!)
Explanation: It says ‘men have a higher mean starting salary than women.’ This is 1   2 or
D  0 . Because this does not contain an equality, it must be an alternate hypothesis!
7.
Referring to Exhibit 2, if we do not have a computer available, which of the following methods
would be most practical (and correct) for you to use? (2)
a) *z- test comparing two means
b) t- test comparing two means assuming equal variances
c) t- test comparing two means not assuming equal variances
d) t- test comparing two means for paired data
e) z- test comparing two proportions
f) None of the above . (Give the correct one!)
[17]
8.
Referring to Exhibit 2, assume that the correct alternate formula for a critical value is
d cv  D0  t  2 s d , where t can be replaced by z for one method. What is the value of sd ? (3)
[20]
Solution: The formula table says.
Difference
between Two
Means (
known)
D  d z 2  d
d 
 12
n1

 22
n2
H 0 : D  D0 *
H 1 : D  D0 ,
D  1   2
z
d  D0
d
d cv  D0  z d
d  x1  x 2
For large samples replace  by s .
d 
s12 s 22
15 .67801 2 13 .55729 2



 1.2290  0.9190  2.148 =1.4656
200
200
n1 n 2
5
252y0622 11/8/06
9.
Referring to Exhibit 2, and the previous problems, what is the value of the test ratio that we would
use to test your hypothesis? (2)
H 0 : 1   2
H0 : D  0
x1  48266 .70
Solution: Our hypotheses are
or
where D  1   2
and
s1  15678 .01
H 1 : 1   2
H1 : D  0
x 2  50200 .00
s 2  13557 .29
. n1  n 2  200 and d  48.26670  50.20000  1.9333 . This is a left sided test.
d  D0  1.9333
s12 s 22

 1.320
=1.4656 so z 

d
1.4656
n1 n 2
d 
10. Referring to Exhibit 2, and the previous problems, assuming that your null hypothesis is correct, do
you reject the null hypothesis? Why? (2)
[24]
Solution: z .05  1.645 , z .10  1.282 This is a left-sided test with   .10 .
If we use a test ratio the rejection region is below -1.282. Since our value of z is in the rejection region,
reject the null hypothesis! If you use a critical value, you want a critical value below zero so
d cv  D0  z d becomes d cv  0  1.2821.4656  1.8789 and our rejection region is below this point.
Since d  48.26670  50.20000  1.9333 is in the rejection zone reject the null hypothesis. The Minitab
run follows.
MTB > TwoT 200 48266.70 15678.01 200 50200 13557.29;
SUBC>
Alternative -1.
Two-Sample T-Test and CI
Sample
N
Mean StDev SE Mean
1
200 48267 15678
1109
2
200 50200 13557
959
Difference = mu (1) - mu (2)
Estimate for difference: -1933.30
95% upper bound for difference: 483.16
T-Test of difference = 0 (vs <): T-Value = -1.32
P-Value = 0.094
DF = 389
11. (Extra credit) compute a two-sided confidence interval for the ratio of the two variances in the
previous problem. (3)
Solution: The formulas given in the first article in the syllabus supplement is
s12
 12 s12 ( n2 1, n1 1)
s 22
 22 s 22 ( n1 1, n2 1)
1
1


F


F
or

s12 Fn2 1,n1 1  12 s12 2
s 22 Fn1 1, n2 1  22 s 22 2
2
x1  48266 .70
s1  15678 .01
s 22
s12

2
and
x 2  50200 .00
s 2  13557 .29
. n1  n 2  200
s1
s2
 1.1564 so 12  1.1564 2  1.3372
s2
s2
1
199,199 . This should be very close to F 200, 200  1.26
 0.7478 Because   .10 we need F.05
.05
1.3372
so the intervals are 1.3372
2
2
1
1
 12  1.3372 1.26  or 0.7478
 22  0.7478 1.26 
1.26  2
1.26  1
6
252y0622 11/8/06
Exhibit 3
You drive a New York avenue with 25 traffic lights on it. You suspect that this is equivalent to playing a
game with a constant probability of success (There are 25 tries.) You record the number of red lights you hit
on 50 trips up the avenue and from the data you come to the conclusion that the (sample) probability of
success is .5. You use a binomial table  p  .5, n  25  to figure out the probabilities of getting various
numbers of red lights on a single run and then you record the number of times you get these numbers of red
lights during the 50 trips up the avenue.
The left column is copied from my cumulative binomial table. The next column comes from
differencing the first column. The third column is the second column multiplied by 50. The fourth column
gives the numbers actually observed. The last two columns show the process of making the observed data
into a cumulative distribution.
E
O
Cum O Cum O
n
5.7380
6
6
.120
Px  9  0.11476
Px  9  0.11476
Px  10   0.21218
Px  11  0.34502
Px  12   0.50000
Px  13   0.65498
Px  14   0.78782
Px  15   0.88524
Px  25   1.00000
Px  10   0.09742
Px  11  0.13284
Px  12   0.15498
Px  13  0.15498
Px  14   0.13284
Px  15   0.09742
Px  16   0.11476
4.8710
8
14
.280
6.6420
6
20
.400
7.7490
4
24
.480
7.7490
6
30
.600
6.6420
6
36
.720
4.8710
8
44
.880
5.7380
6
50
1.00
12. Referring to Exhibit 3. The correct way to test for the distribution cited is: (2)
a) Kolmogorov-Smirnoff Test
b) Lilliefors Test
c) Chi-squared test with 8 degrees of freedom
d) Chi-squared test with 7 degrees of freedom
e) *Chi-squared test with 6 degrees of freedom
f) z test for a proportion
g) None of the above. (Give the correct one!)
[26]
Explanation: KS cannot be used because we have estimated a parameter from the data. 8 lines
would normally give us 7 degrees of freedom, but we lose a degree of freedom when we estimate a
parameter from the data.
.
13. Referring to Exhibit 3. Assume that your choice of method in the previous problem is correct.
Carry out the test. (4)
Solution: H 0 : Binomial
O  E 2 [30]O 2
IndivP
E
O CumO Fo
CumP
E O
E
E
Row
1
2
3
4
5
6
7
8
0.11476
0.21218
0.34502
0.50000
0.65498
0.78782
0.88524
1.00000
0.11470
0.09742
0.13284
0.15498
0.15498
0.13284
0.09742
0.11476
5.738 6
4.871 8
6.642 6
7.749 4
7.749 6
6.642 6
4.871 8
5.738 6
50.000 50
6
14
20
24
30
36
44
50
0.12
0.28
0.40
0.48
0.60
0.72
0.88
1.00
-0.262
-3.129
0.642
3.749
1.749
0.642
-3.129
-0.262
0.000
0.01196
2.00999
0.06205
1.81378
0.39476
0.06205
2.00999
0.01196
6.37655
6.2740
13.1390
5.4201
2.0648
4.6458
5.4201
13.1390
6.2740
56.3765
50.0000
6.3765
Notice that the 5% chi-squared on the table for 6 df is 12.5916.
I have computed chi-squared both ways and it is pretty evident that since our computed value is
less than the table value, we cannot reject the null hypothesis.
7
252y0622 11/8/06
14. An exit poll of 613 voters in Hotzeplotz yields the following results.
Conservative Moderate Liberal Sum
Voted Democrat 100
156
143
399
Voted Republican 127
72
15
214
Sum
227
228
158
a) A political expert has stated that the electorate is equally divided among the three ideological groups.
What would our E column be if this is true? (1)
b) Put together an O column and do a chi-square test of the hypothesis that the political expert has
mentioned. (3)
c) Do a confidence interval for the difference between the proportion of Democrats that are conservatives
and the proportion of Republicans that are conservatives.
a) I got E by multiplying 613 by 1 .
3
b) H 0 : Uniformity
Row
O
p
1
2
3
227
228
158
613
0.333333
0.333333
0.333333
1.00000
E
204.333
204.333
204.333
613.000
OE
22.6667
23.6667
-46.3333
0.0000
O  E 2
E
2.5144
2.7412
10.5063
15.7618
O2
E
252.181
254.408
122.173
628.762
b) H 0 : Uniformity
O  E 2
O2
 n . Both of these two formulas are shown above. There is no reason
E
E
O  E 2  15.7618 or
to do both. DF  r  1  2 . So we have  2 
E
2
O
2 
 n  628 .762  613  15.762 . Since  .205  5.9915 we reject the null hypothesis.
E
2 

or  2 



c) The Formula Table says the following.
Interval for
Confidence
Hypotheses
Interval
Difference
H 0 : p  p 0
p  p  z 2 s p
between
H 1 : p  p 0
p  p1  p 2
proportions
p 0  p 01  p 02
p1 q1 p 2 q 2
q  1 p
s p 

or p 0  0
n1
n2
Test Ratio
z
p  p 0
 p
If p  0
 p 
p 01q 01 p 02 q 02

n1
n2
Or use
n1  399 , x1  100 , p1 
s p
Critical Value
pcv  p0  z 2  p
If p  0
 1
1

 n1 n 2
 p  p 0 q 0 
p0 



n1 p1  n 2 p 2
n1  n 2
pq
.2506 .7494 
100
 .2506 and 1 1 
 .00047068
399
n1
399
n 2  214 , x 2  127 , p 2 
p q
.5935 .4065 
127
 .00112737
 .5935 and 2 2 
n2
214
214
If p  p1  p 2 a 95% confidence interval would be p  ( p1  p 2 )  z.025 s p
 .2506  .5935   1.960 .00047068  .00112737  .3429   1.960 .00159805  .3429  1.960 0.03997 
 .3429  .0783 . Certainly significant.
d) Make an 87% confidence interval for the same difference. Do not use the t table. Solution: This is the
easiest problem on the exam. All you need to know is z .065  1.51 , which you found on page 1.
8
252y0622 11/8/06
15. (Abramovic) Two independent samples, one of ten randomly selected women and the second of
ten randomly selected men are shown below. (Compare with question on 252y0621)
Female Male
1
467
514
2
470
512
3
470
470
4
465
409
5
466
507
6
461
505
7
460
502
8
459
501
9
449
495
10
446
497
If we wish to show that male results exceed female results and suspect that the underlying
distribution is not Normal. We should use: (2)
a) Sign Test
b) *Wilcoxon-Mann-Whitney Test for independent samples
c) Wilcoxon signed rank test
d) t-test for paired data
e) t-test for independent samples with equal variances.
f) F-test.
g) None of the above. (Give the correct one!)
16. Turn in your computer output for the first problem only. To get full credit it must be noted what
hypotheses were tested and what were the results. Use a 5% significance level. (4)
[41]
9
252y0622 11/8/06
ECO252 QBA2
SECOND EXAM
November 7-8
TAKE HOME SECTION
Name: ____KEY________________
Student Number: _________________________
Class days and hour: _________________________
III. Do at least 3 problems (at least 7 each) (or do sections adding to at least 20 points - Anything extra you
do helps, and grades wrap around) . Note: Look at 252thngs (252thngs) on the syllabus supplement part of
the website before you start (and before you take exams). Show your work! State H 0 and H 1 where
appropriate. You have not done a hypothesis test unless you have stated your hypotheses, run the
numbers and stated your conclusion.. (Use a 95% confidence level unless another level is specified.)
Answers without reasons usually are not acceptable. Neatness and clarity of explanation are
expected. This must be turned in when you take the in-class exam. Note that from now on neatness
means paper neatly trimmed on the left side if it has been torn, multiple pages stapled and paper
written on only one side.
1.
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A bank is testing various ways to serve walk-in customers. It tries 10 different new ways to
organize this service and gets the following results. Each result can be considered as a random
sample of 14 times that it takes a teller to serve ten walk-in customers. The data is presented
below. Method 1 is the method currently in use. You are asked to compare method 1 with one
other method by comparing the mean or median time it takes to serve the customers and to find out
if there is a significant difference between the methods. On the basis of your results you can
conclude that i) the old method is better, ii) the new method is better or iii) more information is
needed to make a decision. The second method to compare to method 1 is determined by the
second to last digit of your student number: if it is 0 use method 10; if it is 1 use method 11. If it is
2, 3, 4 etc use the method with the corresponding number. Label this problem clearly with the
number of the method that you are comparing with Method 1.
Method
1
2.9
3.9
2.6
3.1
3.9
2.6
3.3
3.0
3.5
3.1
3.2
2.4
4.0
4.3
2
2.8
2.6
2.6
2.9
2.9
2.8
2.3
2.4
2.0
2.5
2.4
2.0
4.1
4.3
3
2.6
2.7
3.2
2.8
3.6
2.1
2.3
2.6
2.6
2.9
2.4
2.3
0.5
4.7
4
7.7
2.9
4.3
2.7
3.4
4.4
5.5
3.4
3.4
3.5
4.0
3.4
4.1
3.8
5
2.4
13.4
5.8
1.5
9.8
2.7
2.7
4.5
2.3
5.8
4.8
4.2
5.8
6.1
6
6.6
3.7
9.7
1.9
10.1
4.5
2.9
9.9
3.0
31.5
3.5
5.3
9.8
5.3
7
3.5
8.4
4.3
3.3
11.9
3.7
3.0
2.9
3.6
5.4
4.4
3.0
4.3
5.4
8
3.4
8.3
4.2
3.2
11.0
3.6
2.9
2.8
3.5
5.3
4.3
2.9
4.2
5.3
9
3.5
3.4
3.8
3.4
3.6
3.5
4.8
3.5
5.3
3.7
3.4
3.6
3.8
3.8
10
2.3
6.9
3.3
5.3
3.0
3.3
6.1
3.1
2.6
4.4
15.0
6.9
2.1
10.4
11
3.4
4.4
3.1
3.6
4.4
3.1
3.8
3.5
4.0
3.6
3.7
2.9
4.5
4.8
Minitab gives us the following computations for Method 1. n  14, x  3.271, s x  0.155,
s  0.580, Minimum  2.400 Q1  2.825, x.50  3.150, Q3  3.900, Maximum  4.300. For
other information, see the end of this document. You can assume that the data for method 1 has an underlying Normal
distribution. Use a 95% confidence level.
a) Compute a standard deviation for the method that you are comparing with Method 1. Carry at
least as many significant figures as you see in the Minitab computations above. (1)
b) Test for a significant difference between the methods on the assumption that your method
represents data taken from the Normal distribution and the times have approximately equal variances. Use a
test ratio, critical value or a confidence interval (3) or all three (6).
c) Assume that the data represents a Normally distributed population but the variances are not
equal (4 – extra credit)
10
252y0622 11/8/06
d) Assume that the Normal distribution does not apply. (4)
e) Test to see if the data from your method is Normally distributed (3)
f) Test to see if the standard deviations of the two methods are equal (1)
g) On the basis of the sections of this problem that you have been able to do write a
recommendation. (1)
[11]
[16]
2.
A researcher asked a group of artists and non- artists whether they believed in extra-sensory
perception. The data assembled is as below.
Of 114 artists, 58.77% believed in extrasensory perception. 35.96% more or less
believed in extrasensory perception and the remainder did not believe.
Of 344 non artists, 37.50% believed in extrasensory perception, 53.20% more or less
believed and the remainder did not believe.
a) Is there any association between being an artist and believing in ESP? (6)
b) Do a confidence interval for the difference in proportions who do not believe in ESP among
artists and non artists. Is the difference significant? (2)
[24]
3.
A researcher wants to see if the type of vehicle driven affects safety related behavior. The
researcher observes the following behavior at a stop sign on a rural intersection.
Behavior
Sedan Wagon Van
SUV
Stopped
180
24
30
14
Rolled through 107
18
10
20
Ran sign
60
8
11
16
Personalize the data as follows: To the number 180, add the last digit of your student number.
a) Is there an association between the vehicle driven and behavior? (5)
b) Cut down the number of rows in the problem by adding together the people who stopped and
rolled through. Assume that you were testing the hypothesis that the proportion of individuals who
who ran the stop sign was independent of the type of vehicle driven and that you had rejected your
null hypothesis. Use a Marascuilo procedure to find the 6 possible differences between
proportions and to find out if there are any pairs of vehicle types where the difference between the
proportions is insignificant. (4)
[33]
There will be extra credit offered for verifying your results using Minitab. A chi squared test can
be done by putting columns of data in without any totals and following the procedure used in the first part
of 252chisqx2. To compare two columns you can use the following instructions.
MTB >
SUBC>
MTB >
SUBC>
MTB >
MTB >
SUBC>
MTB >
SUBC>
normtest c2;
kstest.
VarTest c1 c2;
Unstacked.
TwoSample c1 c2.
TwoSample c1 c2;
Pooled.
Mann-Whitney 95.0 c1
Alternative 0.
#To test for normality.
#To test for equal variances.
#To compare means
#To compare means with equal variances assumed.
c2;
#To compare medians.
You will not be familiar with all the procedures used, but to get credit, you should be able to say in a
general way what was tested and what the results were.
11
252y0622 11/8/06
Additional computations for the material in Problem 1. The data for methods 2-11 were stacked with
method 1 as shown below.
Method
Row 2
3
4
5
6
7
8
9
10
11
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
2.9
3.9
2.6
3.1
3.9
2.6
3.3
3.0
3.5
3.1
3.2
2.4
4.0
4.3
2.8
2.6
2.6
2.9
2.9
2.8
2.3
2.4
2.0
2.5
2.4
2.0
4.1
4.3
2.9
3.9
2.6
3.1
3.9
2.6
3.3
3.0
3.5
3.1
3.2
2.4
4.0
4.3
2.6
2.7
3.2
2.8
3.6
2.1
2.3
2.6
2.6
2.9
2.4
2.3
0.5
4.7
2.9
3.9
2.6
3.1
3.9
2.6
3.3
3.0
3.5
3.1
3.2
2.4
4.0
4.3
7.7
2.9
4.3
2.7
3.4
4.4
5.5
3.4
3.4
3.5
4.0
3.4
4.1
3.8
2.9
3.9
2.6
3.1
3.9
2.6
3.3
3.0
3.5
3.1
3.2
2.4
4.0
4.3
2.4
13.4
5.8
1.5
9.8
2.7
2.7
4.5
2.3
5.8
4.8
4.2
5.8
6.1
2.9
3.9
2.6
3.1
3.9
2.6
3.3
3.0
3.5
3.1
3.2
2.4
4.0
4.3
6.6
3.7
9.7
1.9
10.1
4.5
2.9
9.9
3.0
31.5
3.5
5.3
9.8
5.3
2.9
3.9
2.6
3.1
3.9
2.6
3.3
3.0
3.5
3.1
3.2
2.4
4.0
4.3
3.5
8.4
4.3
3.3
11.9
3.7
3.0
2.9
3.6
5.4
4.4
3.0
4.3
5.4
2.9
3.9
2.6
3.1
3.9
2.6
3.3
3.0
3.5
3.1
3.2
2.4
4.0
4.3
3.4
8.3
4.2
3.2
11.0
3.6
2.9
2.8
3.5
5.3
4.3
2.9
4.2
5.3
2.9
3.9
2.6
3.1
3.9
2.6
3.3
3.0
3.5
3.1
3.2
2.4
4.0
4.3
3.5
3.4
3.8
3.4
3.6
3.5
4.8
3.5
5.3
3.7
3.4
3.6
3.8
3.8
2.9
3.9
2.6
3.1
3.9
2.6
3.3
3.0
3.5
3.1
3.2
2.4
4.0
4.3
2.3
6.9
3.3
5.3
3.0
3.3
6.1
3.1
2.6
4.4
15.0
6.9
2.1
10.4
The numbers in each column above were ranked as follows.
Method
Row 2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
15.0
23.5
9.5
18.5
23.5
9.5
21.0
17.0
22.0
18.5
20.0
5.0
25.0
27.5
12.5
9.5
9.5
15.0
15.0
12.5
3.0
5.0
1.5
7.0
5.0
1.5
26.0
27.5
14.5
24.5
9.0
17.5
24.5
9.0
21.0
16.0
22.0
17.5
19.5
5.5
26.0
27.0
9.0
12.0
19.5
13.0
23.0
2.0
3.5
9.0
9.0
14.5
5.5
3.5
1.0
28.0
5.5
19.5
2.5
8.5
19.5
2.5
11.0
7.0
16.5
8.5
10.0
1.0
21.5
24.5
28.0
5.5
24.5
4.0
13.5
26.0
27.0
13.5
13.5
16.5
21.5
13.5
23.0
18.0
9.0
16.5
5.5
11.5
16.5
5.5
14.0
10.0
15.0
11.5
13.0
3.5
18.0
20.0
3.5
28.0
24.0
1.0
27.0
7.5
7.5
21.0
2.0
24.0
22.0
19.0
24.0
26.0
5.5
16.5
3.5
9.5
16.5
3.5
12.0
7.5
13.5
9.5
11.0
2.0
18.0
19.0
23.0
15.0
24.0
1.0
27.0
20.0
5.5
26.0
7.5
28.0
13.5
21.5
25.0
21.5
4.5
18.5
2.5
9.5
18.5
2.5
12.5
7.0
14.5
9.5
11.0
1.0
20.0
22.0
14.5
27.0
22.0
12.5
28.0
17.0
7.0
4.5
16.0
25.5
24.0
7.0
22.0
25.5
6.0
18.5
2.5
9.5
18.5
2.5
13.0
8.0
15.5
9.5
11.5
1.0
20.0
23.5
14.0
27.0
21.5
11.5
28.0
17.0
6.0
4.0
15.5
25.5
23.5
6.0
21.5
25.5
4.0
23.5
2.5
6.5
23.5
2.5
9.0
5.0
14.5
6.5
8.0
1.0
25.0
26.0
14.5
11.0
21.0
11.0
17.5
14.5
27.0
14.5
28.0
19.0
11.0
17.5
21.0
21.0
2.9
3.9
2.6
3.1
3.9
2.6
3.3
3.0
3.5
3.1
3.2
2.4
4.0
4.3
3.4
4.4
3.1
3.6
4.4
3.1
3.8
3.5
4.0
3.6
3.7
2.9
4.5
4.8
10
7.0
18.5
5.0
11.0
18.5
5.0
15.0
8.5
17.0
11.0
13.0
3.0
20.0
21.0
2.0
25.5
15.0
23.0
8.5
15.0
24.0
11.0
5.0
22.0
28.0
25.5
1.0
27.0
11
4.5
20.5
2.5
8.5
20.5
2.5
12.0
6.0
14.5
8.5
11.0
1.0
22.5
24.0
13.0
25.5
8.5
16.5
25.5
8.5
19.0
14.5
22.5
16.5
18.0
4.5
27.0
28.0
12
252y0622 11/8/06
1.
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A bank is testing various ways to serve walk-in customers. It tries 10 different new ways to
organize this service and gets the following results. Each result can be considered as a random
sample of 14 times that it takes a teller to serve ten walk-in customers. The data is presented
below. Method 1 is the method currently in use. You are asked to compare method 1 with one
other method by comparing the mean or median time it takes to serve the customers and to find out
if there is a significant difference between the methods. On the basis of your results you can
conclude that i) the old method is better, ii) the new method is better or iii) more information is
needed to make a decision. The second method to compare to method 1 is determined by the
second to last digit of your student number: if it is 0 use method 10; if it is 1 use method 11. If it is
2, 3, 4 etc use the method with the corresponding number. Label this problem clearly with the
number of the method that you are comparing with Method 1.
Service Method
1
2.9
3.9
2.6
3.1
3.9
2.6
3.3
3.0
3.5
3.1
3.2
2.4
4.0
4.3
2
2.8
2.6
2.6
2.9
2.9
2.8
2.3
2.4
2.0
2.5
2.4
2.0
4.1
4.3
3
2.6
2.7
3.2
2.8
3.6
2.1
2.3
2.6
2.6
2.9
2.4
2.3
0.5
4.7
4
7.7
2.9
4.3
2.7
3.4
4.4
5.5
3.4
3.4
3.5
4.0
3.4
4.1
3.8
5
2.4
13.4
5.8
1.5
9.8
2.7
2.7
4.5
2.3
5.8
4.8
4.2
5.8
6.1
6
6.6
3.7
9.7
1.9
10.1
4.5
2.9
9.9
3.0
31.5
3.5
5.3
9.8
5.3
7
3.5
8.4
4.3
3.3
11.9
3.7
3.0
2.9
3.6
5.4
4.4
3.0
4.3
5.4
8
3.4
8.3
4.2
3.2
11.0
3.6
2.9
2.8
3.5
5.3
4.3
2.9
4.2
5.3
9
3.5
3.4
3.8
3.4
3.6
3.5
4.8
3.5
5.3
3.7
3.4
3.6
3.8
3.8
10
2.3
6.9
3.3
5.3
3.0
3.3
6.1
3.1
2.6
4.4
15.0
6.9
2.1
10.4
11
3.4
4.4
3.1
3.6
4.4
3.1
3.8
3.5
4.0
3.6
3.7
2.9
4.5
4.8
Minitab gives us the following computations for Method 1. n  14, x  3.271, s x  0.155,
s  0.580, Minimum  2.400 Q1  2.825, x.50  3.150, Q3  3.900, Maximum  4.300. For
other information, see the end of this document. You can assume that the data for method 1 has an underlying Normal
distribution. Use a 95% confidence level.
a) Compute a standard deviation for the method that you are comparing with Method 1. Carry at
least as many significant figures as you see in the Minitab computations above. (1)
Solution: The mean and standard deviation computed by Minitab will have to do.
Variable
tel 1
tel 2
tel 3
tel 4
tel 5
tel 6
tel 7
tel 8
tel 9
tel 10
tel 11
N
14
14
14
14
14
14
14
14
14
14
14
N*
0
0
0
0
0
0
0
0
0
0
0
Mean
3.271
2.757
2.664
4.036
5.129
7.69
4.793
4.636
3.793
5.336
3.771
SE Mean
0.155
0.181
0.243
0.338
0.859
1.99
0.669
0.623
0.150
0.970
0.155
StDev
0.580
0.677
0.909
1.265
3.214
7.45
2.503
2.331
0.561
3.630
0.580
Minimum
2.400
2.000
0.500
2.700
1.500
1.90
2.900
2.800
3.400
2.100
2.900
Q1
2.825
2.375
2.300
3.400
2.625
3.38
3.225
3.125
3.475
2.900
3.325
Median
3.150
2.600
2.600
3.650
4.650
5.30
4.000
3.900
3.600
3.850
3.650
Q3
3.900
2.900
2.975
4.325
5.875
9.83
5.400
5.300
3.800
6.900
4.400
Maximum
4.300
4.300
4.700
7.700
13.400
31.50
11.900
11.000
5.300
15.000
4.800
Since methods 6 and 9 seem to be the extremes, here are their computations. I have used both the
computational and definitional methods, though the definitional method is a waste of time.
13
252y0622 11/8/06
Computational
Method
Definitional
x 9  x 9   x 9  x 9 2
2
Row x 9
x9
1 3.5 12.25
2 3.4 11.56
3 3.8 14.44
4 3.4 11.56
5 3.6 12.96
6 3.5 12.25
7 4.8 23.04
8 3.5 12.25
9 5.3 28.09
10 3.7 13.69
11 3.4 11.56
12 3.6 12.96
13 3.8 14.44
14 3.8 14.44
53.1 205.49
x
-0.29286
-0.39286
0.00714
-0.39286
-0.19286
-0.29286
1.00714
-0.29286
1.50714
-0.09286
-0.39286
-0.19286
0.00714
0.00714
0.00000
0.08577
0.15434
0.00005
0.15434
0.03719
0.08577
1.01434
0.08577
2.27148
0.00862
0.15434
0.03719
0.00005
0.00005
4.08929

x 92  nx 92 205 .49  14 3.79286 2 4.08929
53 .1
 3.79286, s 92 


 0.324560
n
13
n 1
13
13
The formula above is the computational formula. The definitional formula is
x9 
s 92
9

 x

2
9
 x9

2
4.08929
 0.324560
13
n 1
Method
Computational
Definitional

2
Row
x6
x6
1
6.6
43.56
2
3.7
13.69
3
9.7
94.09
4
1.9
3.61
5 10.1 102.01
6
4.5
20.25
7
2.9
8.41
8
9.9
98.01
9
3.0
9.00
10 31.5 992.25
11
3.5
12.25
12
5.3
28.09
13
9.8
96.04
14
5.3
28.09
107.7 1549.35
x6 
x
6
n
s 62 

x 6  x 6   x 6  x 6 2
-1.0929
-3.9929
2.0071
-5.7929
2.4071
-3.1929
-4.7929
2.2071
-4.6929
23.8071
-4.1929
-2.3929
2.1071
-2.3929
0.0000
107 .7
 7.69286, s 62 
13

x 62
 x6
n 1

s 9  0.324560  0.560857
1.194
15.943
4.029
33.557
5.794
10.194
22.971
4.871
22.023
566.780
17.580
5.726
4.440
5.726
720.829
x
2
6
 nx 62
n 1

1549 .35  14 7.69286 2 720 .829

 55.4484
13
13
2

720 .829
 55.4484
13
s 6  55 .4484  7.44637
The results for all 10 versions of this problem appear in 252y0621s1. A careful reading of
the explanation of Minitab output in this document should enable you to follow the
computer output for any of the 10 versions.
14
252y0622 11/8/06
b) Test for a significant difference between the methods on the assumption that your method
represents data taken from the Normal distribution and the times have approximately equal variances. Use a
test ratio, critical value or a confidence interval (3) or all three (6).
Solution: The formula table gives us the formulas below. (Method D2 – Comparison of two means with
samples coming from populations with similar variances.)
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Difference
H 0 : D  D0 *
D  d  t 2 s d
d cv  D0  t  2 s d
d  D0
t
between Two
H 1 : D  D0 ,
sd
1 1
Means (
sd  s p

D  1   2
n1  1s12  n2  1s22
n n
2
unknown,
1
variances
assumed equal)
2
sˆ p 
n1  n2  2
DF  n1  n2  2
Let’s use serving Method 4 this time. We have from the printout in part a)
n1  14, x1  3.271 and s1  0.580 . n 4  14, x 4  4.036 and s 4  1.265. I will replace 2 with 4 in
H :    4
the formulas above. DF  n1  1  n 4  1  13  13  26  n1  n 4  2. Our hypotheses are  0 1
 H 1 : 1   4
H :    4  0
H : D  0
or  0 1
or if D  1   4 ,  0
. d  x1  x4  3.271 4.036  0.764286. Sorry
 H 1 : 1   4  0
H 1 : D  0
about the extra digits, but I’m using the computer solution. Obviously, if you had rounded to three digits
you would have gotten -0.765 but the rounding should not affect the answer by much.
n  1s12  n4  1s 22 . But, because the sample sizes are identical we can do some cancelling.
sˆ 2p  1
n1  n 4  2
13 0.580 2  13 1.265 2
0.580 2  1.265 2
0.3364  1.600

 0.9682 . This is the pooled
2
2
variance. sˆ p  0.9840 . The computer got .9841.   .05 so t.26
025  2.056 .
sˆ 2p  

26


 1
1
1
1
1 
1
 2
  0.9682     0.9682    0.13831  0.3719 . Recall

 sˆ 2p  
n1 n 2
 14 14 
 14 
 n1 n 2 
that our alternate hypothesis is H 1 : D  0 so this is a two-sided test.
s d  sˆ p
Test Ratio: t 
x  x 2   10   20 
d  D0
or t  1
.
sd
sd
If this test ratio lies between  t.26
025  2.056 , do not reject H 0 .
t
d  D0
 0.764286  0

 2.055 .
sd
0.3719
Make a diagram with zero in the middle showing shaded ‘reject’ regions below -2.056 and above 2.056.
Since –2.055 does not fall in the 'reject' region, do not reject H 0 . Or you can say that, since -2.055 falls
26
 1.706 and t.26
between t .05
025  2.056 , for a one-sided test, .025  p  value  .05 . But this is a two-sided
test so that .05  p  value  .10 Since the p-value is above   .05 , do not reject H 0 .
Critical Value: d CV  D0  t  2 s d or x1  x 2 CV  10   20   t  2 s d . For a two-sided test we want two
critical values above and below D0  0. If d  x1  x 2 is between the critical values, do not reject H 0 .
d CV  D0  t  2 s d  0  2.056 0.3719   0.7646 . Make a diagram with 0 in the middle showing shaded
‘reject’ regions below -0.7646 and above 0.7646. Since d  0.7643 does not fall in the 'reject' region, do
not reject H 0 .
15
252y0622 11/8/06
Confidence Interval: D  d  t  2 s d . We already know that t  s d  2.0560.3719  0.7646 so we can say
2
D  0.7643  0.7646 or 1.529  D  0.0003 . Make a diagram with -0.7643 in the middle. Represent
the confidence interval by shading the area between -1.529 and 0.0003. Since zero is in this area, do not
reject H 0 . Or simply note that the error part of the confidence interval is larger that the sample mean
difference so the interval must include zero.
This is probably the closest of the tests; you may or may not end up rejecting the null hypothesis,
depending on how carefully you do the calculations.
A summary of the results for the other serving methods follows. With most of these the confidence interval
does not include zero and the p-value is below   .05 so we reject the null hypothesis.
Serving
Confidence interval for D
d
t ratio
Method
(0.024746, 1.003825)
2
0.514286
2.16
(0.015054, 1.199232)
3
0.607143
2.11
(-1.528864, 0.000293)
4
-0.764286
-2.05
(-3.65142, -0.06286)
5
-1.85714
-2.13
(-8.52457, -0.31829)
6
-4.42123
-2.21
(-2.93275,
-0.11011)
7
-1.52143
-2.22
(-2.68400, -0.04458)
8
-1.36429
-2.12
(-0.964545, -0.078312)
9
-0.521429
-2.42
(-4.08381, -0.04476)
10
-2.06429
-2.10
(-0.950373, -0.049627)
11
-0.50000
-2.28
The Minitab output for comparing Service methods 1 and 4 appears below.
MTB > TwoSample c1 c4;
SUBC>
Pooled.
p-value
0.040
0.045
0.050
0.043
0.036
0.036
0.043
0.023
0.045
0.031
ŝ p
0.6301
0.7621
0.9841
2.3095
5.2813
1.8166
1.6987
0.5704
2.5994
0.5707
#This is method D2.
Two-Sample T-Test and CI: tel 1, tel 4
Two-sample T for tel 1 vs tel 4
N
Mean StDev SE Mean
tel 1 14 3.271 0.580
0.15
tel 4 14
4.04
1.27
0.34
#Sample means, std. deviations etc
Difference = mu (tel 1) - mu (tel 4)
#Printout of d
Estimate for difference: -0.764286
95% CI for difference: (-1.528864, 0.000293)
#95% Confidence interval
T-Test of difference = 0 (vs not =): T-Value = -2.05 P-Value = 0.050 DF =26
# H0 ,
H 0 , value of t -ratio. P-value
# seems to be about .0501, if the confidence
# interval is correct, since it includes zero.
Both use Pooled StDev = 0.9841
#Value of
sˆ 2p
16
252y0622 11/8/06
c) Assume that the data represents a Normally distributed population but the variances are not
equal (4 – extra credit)
Solution: If the variances are not equal and we are assuming a Normally distributed population, we can use
method D3, the Satterthwaite approximation. The formula table has the following formulas.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Difference
H 0 : D  D0 *
D  d  t 2 s d
d cv  D0  t  2 s d
d  D0
t
between Two
H 1 : D  D0 ,
sd
s2 s2
Means(
sd  1  2
D  1   2
unknown,
n1
variances
assumed
unequal)
DF 
n2
 s12 s22 
  
n

 1 n2 
2
   
s12
2
n1
n1  1
s 22
2
n2
n2  1
Let’s continue to use serving Method 4 . We have from the printout in part a)
n1  14, x1  3.271 and s1  0.580 . n 4  14, x 4  4.036 and s 4  1.265. I will replace 2 with 4 in
 H 0 : 1   4
 H 0 : 1   4  0
the formulas above. Our hypotheses are 
or 
or if D  1   4 ,
 H 1 : 1   4
 H 1 : 1   4  0
2
H 0 : D  0
s2  s 
. d  x1  x4  3.271 4.036  0.764286. We could take stuff like 4   4  from the

n1  n 4 
H 1 : D  0


SEmean (Standard error of the mean) column in 1a), but I think that using a table would look better.
s12
n1

s 42
n4

0.580 2
= 0.0240285
14
1.265 2
=0.1143017
14
 s12 s 42 



 n1 n 4 


Thus s d 
DF 
s12 s 42

 0.1383302  0.3719276
n1 n 4
 s12 s 22 



 n1 n 2 


2
2
2
 
 
 
 
 n1 
 n2 
 
 

n1  1
n2 1
s12
=0.1383302
s 22

0.1383302 2
0.0240285 2  0.1143017 2
13

0.0191352
0.0191352

0.00004413  0.001004991
0.001049404
13
=18.23. So we use 18 degrees of freedom.   .05 so t .18
025  2.101 . You should only do one of the
following.
x  x 2   10   20 
d  D0
or t  1
Test Ratio: t 
.
sd
sd
If this test ratio lies between  t .18
025  2.101 , do not reject H 0 .
t
d  D0
 0.764286  0

 2.055 .
sd
0.3719
Make a diagram with zero in the middle showing shaded ‘reject’ regions below -2.101 and above 2.101.
17
252y0622 11/8/06
Since –2.055 does not fall in the 'reject' region, do not reject H 0 . Or you can say that, since -2.055 falls
18
between t.18
05  1.734 and t .025  2.101 , for a one-sided test, .025  p  value  .05 . But this is a two-sided
test so that .05  p  value  .10 Since the p-value is above   .05 , do not reject H 0 .
Critical Value: d CV  D0  t  2 s d or x1  x 2 CV  10   20   t  2 s d . For a two-sided test we want two
critical values above and below D0  0. If d  x1  x 2 is between the critical values, do not reject H 0 .
d CV  D0  t  2 s d  0  2.1010.3719   0.7814 . Make a diagram with 0 in the middle showing shaded
‘reject’ regions below -0.7814 and above 0.7814. Since d  0.7643 does not fall in the 'reject' region, do
not reject H 0 .
Confidence Interval: D  d  t  2 s d . We already know that t  s d  2.1010.3719  0.7814 so we can say
2
D  0.7643  0.7814 or 1.546  D  0.017 . Make a diagram with -0.7643 in the middle. Represent the
confidence interval by shading the area between -1.546 and 0.017. Since zero is in this area, do not reject
H 0 . Or simply note that the error part of the confidence interval is larger that the sample mean difference
so the interval must include zero.
Since the Computer printout is fairly well labeled, you should be able to check your results against it.
MTB > TwoSample c1 c4.
#This is method D3. This is the default method.
Two-Sample T-Test and CI: tel 1, tel 4
Two-sample T for tel 1 vs tel 4
N
Mean StDev SE Mean
tel 1 14 3.271 0.580
0.15
tel 4 14
4.04
1.27
0.34
#Sample means, std. deviations etc
Difference = mu (tel 1) - mu (tel 4) #Printout of d
Estimate for difference: -0.764286
95% CI for difference: (-1.545748, 0.017177)
# 95% Confidence interval
T-Test of difference = 0 (vs not =): T-Value = -2.05 P-Value = 0.055 DF =18
# H0 ,
H 0 , value of t -ratio. p-value
# is above 5%, so we cannot reject the null hypothesis.
d) Assume that the Normal distribution does not apply. (4)
[11]
Solution: Because the data are assumed to be independent nonnormal random samples and thus not paired,
H 0 : 1   2
use the Wilcoxon-Mann-Whitney Rank Sum Test. 
or the null hypothesis is simply 'similar
H 1 : 1   2
distributions.' n1  14 and n2  14 . In the table below, r1 and r2 represent bottom to top ranking.
row
x1
r1
x4
r4
1
2
3
4
5
6
7
8
9
10
11
12
13
Sum
2.9
3.9
2.6
3.1
3.9
2.6
3.3
3.0
3.5
3.1
3.2
2.4
4.0
5.5
19.5
2.5
8.5
19.5
2.5
11.0
7.0
16.5
8.5
10.0
1.0
21.5
158
7.7
2.9
4.3
2.7
3.4
4.4
5.5
3.4
3.4
3.5
4.0
3.4
4.1
28.0
5.5
24.5
4.0
13.5
26.0
27.0
13.5
13.5
16.5
21.5
13.5
23.0
248
18
252y0622 11/8/06
Recall that n1  n 4  14 and that the total number of numbers that we have ranked is
nn  1 2829 

 406 and that the two rank
2
2
sums add to 158 + 248 = 406. The outline says that the smaller of SR1 and SR2 is called W and is compared
with Table 5 or 6. W  158 . For values of n1 and n 2 that are too large for the tables, W has the normal
distribution with mean W  1 2 n1 n1  n 2  1  1 2 14 29   203 and variance
n  n1  n 2  28. Note that the sum of the first 28 numbers is
 W2  16 n2 W  16 14 203   473 .6667 . If the significance level is 5% and the test is two-sided, we
W  W
reject our null hypothesis if z 
does not lie between z.025  1.960 . In this case
W
W  W
158  203
 45
z


 2.068 . Since this is not between -1.960 and 1.960, we reject H 0 .
W
21
.7639
473 .6667
To get a p-value for this result, use 2Pz  2.068   2Pz  2.07   2.5  .4803   2.0197   .0394 .
This is a pretty good approximation. The Minitab printout is as follows.
MTB > Mann-Whitney 95.0 c1
SUBC>
Alternative 0.
c4;
Mann-Whitney Test and CI: tel 1, tel 4
N Median
tel 1 14
3.150
tel 4 14
3.650
Point estimate for ETA1-ETA2 is -0.500
95.4 Percent CI for ETA1-ETA2 is (-1.100,-0.000)
W = 158.0
Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.0409
The test is significant at 0.0404 (adjusted for ties)
The explanation is that the authors of Minitab use the Greek letter eta   as their symbol for the median.
Minitab thus gives us an almost 95% confidence interval for the difference between the medians. It gets the
same value of W that we do, but gives us two more exact p-values. The first one fails to adjust for ties.
(We have not adjusted for ties either!) The second one takes the ties into account and is thus more accurate.
In any case, all the p-values we have are below our 5% significance level so that we reject the null
hypothesis of equal medians.
e) Test to see if the data from your method is Normally distributed (3)
Solution: Assume   .05 H0 : Normal The most practical method is the Lilliefors method, which is a
version of the K-S method with the tables adjusted for the loss of degrees of freedom from estimating the
mean and standard deviation.
The numbers must be in order before we begin computing cumulative probabilities! The x column
below is x 4 in order. Checking the data in 1a) we find that x 4  4.036 and s 4  1.265 . A computer-aided
xx
. (This is really a t .) Fe is the cumulative distribution, which
s
we have usually gotten from the Normal table by adding or subtracting 0.5. In the interest of getting this
typed rapidly, I have let Minitab compute these. Thus Pz  1.05613   .145455 . O is our observed
distribution - there is one of every number. cumO was entered by hand and is simply the cumulative
solution is presented. We compute z 
distribution gotten by adding down the O column. Fo is the relative cumulative distribution and is gotten
by dividing the cumO column by 14. Finally D  Fo  Fe . We pick the value of D with the largest
absolute value, which is 0.2439.
For   .05 and n  14 the critical value from the Lilliefors table is 0.227. Since the largest deviation here
is .2439 which is larger than .227, we reject H 0 .
19
252y0622 11/08/06
x  x 4.4  4.036

 0.28775 .
s
1.265
Using the z table Px  4.4  Pz  0.28775   Pz  0.29   Pz  0  P0  z  0.29 
To illustrate the computation here, pick line 12. x  4.4. z 
 .5  .1141  .6141 . This is close enough to the correct value to give us max D  .6141 .85714
 0.2430 . This is still large enough to give us a rejection.
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
x
2.7
2.9
3.4
3.4
3.4
3.4
3.5
3.8
4.0
4.1
4.3
4.4
5.5
7.7
z
-1.05613
-0.89802
-0.50277
-0.50277
-0.50277
-0.50277
-0.42372
-0.18656
-0.02846
0.05059
0.20870
0.28775
1.15731
2.89644
O cumO
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
Fo
Fe
D
0.07143
0.14286
0.21429
0.28571
0.35714
0.42857
0.50000
0.57143
0.64286
0.71429
0.78571
0.85714
0.92857
1.00000
0.145455
0.184586
0.307564
0.307564
0.307564
0.307564
0.335887
0.426002
0.488648
0.520175
0.582657
0.613230
0.876428
0.998113
-0.074027
-0.041729
-0.093278
-0.021850
0.049579
0.121007
0.164113
0.145426
0.154209
0.194111
0.203057
0.243913
0.052144
0.001887
The Minitab output follows. The output for other service methods is with the computer output.
MTB > normtest c4;
SUBC> kstest.
Probability Plot of tel 4
Note that the value of KS on the graph is the value of the maximum deviation in the computer aided output
above. The associated p-value of 0.032 is below 5% and thus gives a rejection of the null hypothesis. This
invalidates tests that assume the Normal distribution.
f) Test to see if the standard deviations of the two methods are equal (1)
Solution: Let’s continue to use serving Method 4. We have from the printout in part a)
n1  14, x1  3.271 and s1  0.580. n 4  14, x 4  4.036 and s 4  1.265.
Our Hypotheses are H 0 :  12   22 and H1 :  12   22 . DF1  n1  1  13 and DF2  n 2  1  13 , Since
the table is set up for one sided tests, if we wish to test H 0 :  12   22 , we must do two separate onesided tests. First test
s12
s 22

.580 2
1.265 2
13,13 and then test
 0.2102 against F.025
s 22
s12

1
 4.757
0.2102
20
252y0622 11/08/06
13,13 . If either test is failed, we reject the null hypothesis. The table does not contain F 13,13 .
against F.025
.025
12,13  3.15 and F 14,13  3.08 . So F 13,13 must be between them. Since 4.757 is
But we can see that F.025
.025
.025
above the table F, we reject the null hypothesis.
The Minitab output follows.
MTB > VarTest c1 c4;
SUBC>
Unstacked.
Test for Equal Variances: tel 1, tel 4
95% Bonferroni confidence intervals for standard deviations
N
Lower
StDev
Upper
tel 1 14 0.402354 0.57969 1.00741
tel 4 14 0.878207 1.26528 2.19886
F-Test (normal distribution)
Test statistic = 0.21, p-value = 0.008
Levene's Test (any continuous distribution)
Test statistic = 1.30, p-value = 0.264
Test for Equal Variances for tel 1, tel 4
A graph that isn’t really needed has been left out and the Bonferroni intervals are for later. The main point
here is that the F test, which assumes a Normal distribution, gives a very low p-value, which rules out
using the test (MethodD2) that compares means assuming equal variances.
g) On the basis of the sections of this problem that you have been able to do write a
recommendation. (1)
[16]
Solution: Every service method will give us a different set of answers. But in comparing service
methods 1 and 4 we have found that the only valid comparison is using the Wilcoxon-Mann-Whitney
method. It shows that the median time needed to serve 10 customers differs significantly between the two
service methods. Since the median time for service method 4 is higher, we don’t want to adopt it.
If you rejected Normality use Wilcoxon -Mann-Whitney (Method 5a).
If you rejected equal variances but not Normality, use (Satterthwaite) comparison of means with no
assumption of equal variances (Method 3).
If you did not reject either, use comparison of means with equal variances (Method 2). If you
rejected equality of means or medians, use the service method with the lowest mean or median.
21
252y0622 11/08/06
2.
A researcher asked a group of artists and non- artists whether they believed in extra-sensory
perception. The data assembled is as below.
Of 114 artists, 58.77% believed in extrasensory perception. 35.96% more or less
believed in extrasensory perception and the remainder did not believe.
Of 344 non artists, 37.50% believed in extrasensory perception, 53.20% more or less
believed and the remainder did not believe.
a) Is there any association between being an artist and believing in ESP? (6)
Solution: There are three different categories for both artists and non artists, so this is a test of
H : Occupations Homogeneous
(independence or) homogeneity. So let’s say  0
. It really does not
H 1 : not Homogeneous
matter whether you put level of belief in the rows or columns. If you choose rows, we start with this
O
Occupation
Level of belief
Artists Non artists Total
pr
?
Believed
? 
?
?
diagram.


Believed somewhat
? 
?
?
?


Did not believe
?
?
?
?


Total
114 344
458 1.0000
For the artists there are .5877(114) = 67 believers, .3596(114) = 41 somewhat believers and 114 – 67 –
41 = 6 non believers. For the non artists there are .3750(344) = 129 believers , .5320(344) = 183
somewhat believers and 344 – 129 – 183 = 32 nonbelievers. We can put these in, add them up and then
compute the proportion of each group in the total.
O
Occupation
Level of belief
Artists Non artists Total
pr


Believed
67
129
196
.42795
The O is the hard part. To get the E just


Believed somewhat
183 
224
.49608
 41
 6
Did not believe
32 
38
.08297


Total
114
344
458 1.00000
apply the row proportions to the column totals.
E
Occupation
Level of belief
Believed
Believed somewhat
Did not believe
Total
Artists Non artists
Total
pr
 48.786

147.214
196
.42795
The table of computations follows.


55.755
168.245
224
.48908


 9.459
28.541 
38
.08297


114.000
344.000
458 1.00000
OE
Row
O
E
1
2
3
4
5
6
Sum
67
41
6
129
183
32
458
48.786
55.755
9.459
147.214
168.245
28.541
458.000
O  E 2
E
18.214 6.80010
-14.755 3.90476
-3.459 1.26490
-18.214 2.25352
14.755 1.29401
3.459 0.41921
0.000 15.93650
O  E 2
O2
E
92.014
30.150
3.806
113.040
199.049
35.878
473.937
O2
 n . Both of these two
E
E
formulas are shown above. There is no reason to do both.. DF  r  1c  1  3  12  1  2 . So we have
The formula for the chi-squared statistic is  2 
2 

O  E 2
E
 15 .93650 or  2 


or  2 

O2
 n  473 .937  458  15.937 . If we compare our results
E
22
252y0622 11/08/06
2
with  2 .05  5.9915, we notice that our computed value exceeds the table value Since our computed value
of chi-squared is less than the table value, we reject our null hypothesis.
b) Do a confidence interval for the difference in proportions who do not believe in ESP among
artists and non artists. Is the difference significant? (2)
[24]
The Formula Table says the following.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
pcv  p0  z 2  p
Difference
p  p 0
H 0 : p  p 0
p  p  z 2 s p
z
between
If
p  0
 p
H 1 : p  p 0
p  p1  p 2
proportions
 1
1 
If p  0
 p  p 0 q 0   
p 0  p 01  p 02
p1 q1 p 2 q 2
q  1 p
 n1 n 2 
s p 

p 01q 01 p 02 q 02
 p 

or p 0  0
n1
n2
n1
n2
n p  n2 p 2
p0  1 1
n1  n 2
Or use s p
Though you do not need a hypothesis to do a confidence interval, if it is a 2-sided interval, the
H : p  p 2
H : p  p 2  0
 H : p  0
hypotheses  0 1
or  0 1
or if p  p1  p 2 ,  0
are implied.
 H 1 : p1  p 2
H 1 : p1  p 2  0
 H 1 : p  0
O
Occupation
Level of belief
Artists Non artists Total
pr


Believed
67
129
196
.42795
We had
. We can say that for artists


Believed somewhat
41
183
224
.49608


 6
Did not believe
32 
38
.08297


Total
114
344
458 1.00000
32
6
 .0930 .
 .0526 and for non artists p 2 
344
114
n1  114 , n 2  344 , q1  1  p1  1  .0526  .9474 and q 2  1  p 2  1  .0930  .9070 .
p1 
p  p1  p 2  .0526  .0930  .0404 . z 2  z.025  1.960 . p  p  z 2 sp
s p 
.
p1 q1 p 2 q 2
.0526 .9474  .0930 .9070 



 .000437  .000245  .000682  0.026122 .
n1
n2
114
344
So p  .0404  1.960 0.026122   .0404  .0512 or -0.0916 to .0108. Note that this interval includes
zero and thus prevents us from rejecting the null hypothesis.
23
252y0622 11/08/06
It’s happy Minitab time!!! To check our work on the last two pages we put the columns of O in columns 10
and 11 on the spreadsheet and go to work.
MTB > ChiSquare c10 c11.
Chi-Square Test: O1, O2
Expected counts are printed below observed counts
Chi-Square contributions are printed below expected counts
O1
67
48.79
6.800
O2
129
147.21
2.254
Total
196
2
41
55.76
3.905
183
168.24
1.294
224
3
6
9.46
1.265
32
28.54
0.419
38
Total
114
344
458
1
Chi-Sq = 15.936, DF = 2, P-Value = 0.000
MTB > PTwo 114 6 344 32.
Test and CI for Two Proportions
Sample
X
N Sample p
1
6 114 0.052632
2
32 344 0.093023
Difference = p (1) - p (2)
Estimate for difference: -0.0403917
95% CI for difference: (-0.0916005, 0.0108172)
Test for difference = 0 (vs not = 0): Z = -1.55
For the chi-squared test, just look at the computation of  2 

P-Value = 0.122
O  E 2
two pages back. For
E
once I got it right on the first try and the p-value of zero confirms my rejection of the null hypothesis. You
O  E 2 column in this chi squared printout page.
should be able to find the elements of O, E and the
E
For the two proportions test, the 95% confidence interval Minitab gets
 .09160  p1  p 2  .01082  is essentially the same as the interval on the last page. The p-value of 0.122,
since it is above our significance level of 5%, confirms our inability to reject the null hypothesis that equal
proportions did not believe. (Oh ye of little faith!)
24
252y0622 11/08/06
3.
A researcher wants to see if the type of vehicle driven affects safety related behavior. The
researcher observes the following behavior at a stop sign on a rural intersection.
Behavior
Sedan Wagon Van
SUV
Stopped
180
24
30
14
Rolled through 107
18
10
20
Ran sign
60
8
11
16
Personalize the data as follows: To the number 180, add the last digit of your student number.
a) Is there an association between the vehicle driven and behavior? (5)
Solution: Our null hypothesis seems to be H 0 : Independence , but homogeneity would do as well.
  .05 . Let’s start by putting this all in the O table. This is Version 0; other versions appear in
252y0621s1. As usual we use the row totals to tell us what proportion of the data is in each row and list it as
pr .
O
Vehicle type
Behavior
Sedan Wagon Van SUV
Total
pr
 180
Stopped
24
30
14 
248
.49799


Rolled through  107
18
10
20  155
.31125


Ran sign
60
8
11
16
95
.19076


Total
347
50
51
50
498 1.00000
Now we use the row proportions and the column sums to compute the numbers in the E table.
E
Vehicle type
Behavior
Sedan

Stopped
172 .8025

Rolled through  108 .0038
 66 .1937
Ran sign

Total
347 .0000
The table of computations follows.
Row
O
E
1
2
3
4
5
6
7
8
9
10
11
12
180
107
60
24
18
8
30
10
11
14
20
16
498
172.803
108.004
66.194
24.900
15.563
9.538
25.398
15.874
9.729
24.900
15.563
9.538
498.000
Wagon Van
24 .8995 25 .3975
15 .5625 15 .8738
9.5380
50 .0000
OE
SUV
Total
pr

24 .8995
248
.49799

15 .5625  155
.31125
9.7288
9.5380 
95
.19076

51 .0001 50 .0000
498 1.00000
O  E 2
E
7.1975 0.29979
-1.0038 0.00933
-6.1937 0.57954
-0.8995 0.03249
2.4375 0.38178
-1.5380 0.24800
4.6025 0.83406
-5.8738 2.17349
1.2712 0.16610
-10.8995 4.77114
4.4375 1.26531
6.4620 4.37801
-0.0001 15.1390
O2
E
187.497
106.006
54.386
23.133
20.819
6.710
35.437
6.300
12.437
7.872
25.703
26.840
513.139
O  E 2
O2
 n . Both of these two
E
E
formulas are shown above. There is no reason to do both.. DF  r  1c  1  3  14  1  6 . So we have
The formula for the chi-squared statistic is  2 
2 

2 6 

O  E 2
E
 15 .1390 or  2 


or  2 

O2
 n  513 .139  498  15.139 . If we compare our results with
E
 12.5916 , we notice that our computed value exceeds the table value Since our computed value of
chi-squared is less than the table value, we reject our null hypothesis.
.05
25
252y0622 11/08/06
b) Cut down the number of rows in the problem by adding together the people who stopped and
rolled through. Assume that you were testing the hypothesis that the proportion of individuals who ran the
stop sign was independent of the type of vehicle driven and that you had rejected your null hypothesis. Use
a Marascuilo procedure to find the 6 possible differences between proportions and to find out if there are
any pairs of vehicle types where the difference between the proportions is insignificant. (4)
[33]
Solution: The outline says that the Marascuilo procedure can be used 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 (columns), the chi - squared has c  1
degrees of freedom and the standard deviation is s p 
p a q a pb qb
,

na
nb
you can say that you have a significant difference between p a and p b . This is equivalent to using a
c 1  p a q a
confidence interval of p a  p b   p a  p b    2 
Set up the O table. For the first column p1 
later.
O
Sedan
Didn’t run
Ran
Sum ni 
Proportion
that ran
 pi 
Proportion
that didn’t
run qi 
pq
n
Wagon

 n
 a

pb qb
nb




pq
60
 .172911 . q1  1  p1 .
is also computed for use
347
n
Van
SUV
Total
403
95
5452
287
60
347
42
8
50
40
11
51
34
16
50
.172911
.160000
.215686
.320000
.827089
.840000
.784314
.680000
.00041214
.0026880
.0033170
.0043520
pr
We have the following possible combinations of column proportions. Chi-square has three degrees of
3
freedom so use  2 .05  7.8147
1 and 2 .172911  .160000   7.8142 .0041214  .0026880
1 and 3
1 and 4
2 and 3
2 and 4
3 and 4
  .01291   .23069
.172911  .215686   7.8142 .0041214  .0033170   .04278   .24109
.172911  ..320000   7.8142 .0041214  .0043520   .14709   .25732
.160000  .215686   7.8142 .0026880  .0033170   .05569   .21648
.160000  .320000   7.8142 .0026880  .0043520   .16000   .36108
.215686  .320000   7.8142 .0033170  .0043520   .10431   .24464
Note that the error part of these intervals is always larger than the difference between the sample
proportions. This surprised me until I looked at the chi-squared test based on the 2 4 O above. I seems
that once we mush together those the null hypothesis is not rejected, so it seems there is no significant
difference between the proportions that run the sign between vehicles. Just to see where the differences are
coming from, I went back to the original O and computed the proportion of each group that actually
stopped.
26
252y0622 11/08/06
O
Behavior
Stopped
Rolled through
Ran sign
Vehicle type
Sedan Wagon Van SUV
Total
pr
 180

24
30
14
248
.49799


107
18
10
20
155
.31125


 60

8
11
16
95
.19076


Total
347
50
51
50
498 1.00000
ps
.51873 .48000 .58824 .28000
This is shown on the line marked p s below. It looks like there is a substantial difference especially
between the proportion of van drivers and SUV drivers that actually stopped. I computed
pq
for these two
n
types and got .004749 and .004032. This means that the interval is going to be
.58824  .28000  
7.8142 .004749  .004032   .3082   .2619 . This interval does not include zero.
The Minitab output follows.
MTB > Print c1-c4
Data Display
Row
1
2
3
O11
180
107
60
O21
24
18
8
O31
30
10
11
# This is the O table in C1-C4.
O41
14
20
16
MTB > ChiSquare C1 C2 C3 C4
# As explained before, this table contains O, E and the
#
O  E 2
E
Chi-Square Test: O11, O21, O31, O41
column.
Expected counts are printed below observed counts
Chi-Square contributions are printed below expected counts
O11
O21
O31
O41 Total
1
180
24
30
14
248
172.80 24.90 25.40 24.90
0.300 0.033 0.834 4.771
2
107
108.00
0.009
18
15.56
0.382
10
15.87
2.173
20
15.56
1.265
155
3
60
66.19
0.580
8
9.54
0.248
11
9.73
0.166
16
9.54
4.378
95
Total
347
50
51
50
498
Chi-Sq = 15.139, DF = 6, P-Value = 0.019
#The low p-value means a rejection of the
#null hypothesis at the 5% level.
MTB > print c6-c9
Data Display
Row
1
2
012
287
60
O22
42
8
O32
40
11
O42
34
16
# This is a new O table in C6-C9. The top two rows
# have been combined.
27
252y0622 11/08/06
MTB > ChiSquare C6 C7 C8 C9
Chi-Square Test: 012, O22, O32, O42
Expected counts are printed below observed counts
Chi-Square contributions are printed below expected counts
012
O22
O32
O42 Total
1
287
42
40
34
403
280.81 40.46 41.27 40.46
0.137 0.058 0.039 1.032
2
60
66.19
0.580
8
9.54
0.248
11
9.73
0.166
16
9.54
4.378
95
Total
347
50
51
50
498
Chi-Sq = 6.638, DF = 3, P-Value = 0.084
#The p-value is now above 5%, so we
# cannot reject the null hypothesis at the 5%
#level.
MTB > print c11-c14
Data Display
Row
1
2
P1
0.172911
0.827089
P2
0.16
0.84
P3
0.215686
0.784314
P4
0.32
0.68
# The first line here is the p and the
#second is the q for the Marascuilo
#procedure
Note that in 1b) it says to assume that data have equal variances and in 3b) it says to assume that
the null hypothesis had been rejected. You do not get credit for testing assumptions if it is not
offered.
28