252y0762 10/30/07
ECO252 QBA2
SECOND EXAM
Nov 1-5 2007
Name
KEY
Class________________________
Student Number_______________
Show your work! Make Diagrams! Exam is normed on 50 points. Answers without reasons are not
usually acceptable.
x ~ N 12, 9 - If you are not using the supplement table, make sure that I know it.
21  12 
  21  12
z
1. P21  x  21  P 
 P3.67  z  1.00   P3.67  z  0  P0  z  1.00 
9
9 

 .4999  .3413  .8412
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 -3.67 and 1.00. Because this is on one side of zero we must add the area between
-3.67 and zero to the area between zero and 1.00. If you wish, make a completely separate diagram for x .
Draw a Normal curve with a mean at 12. Indicate the mean by a vertical line! Shade the area between 21 and 21. This area is on both sides of the mean (12) so we add to get our answer.
14  12 

2. Px  14   P  z 
 Pz  0.22   Pz  0  P0  z  0.22   .5  .0871  .4129
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 above 0.22. Because this is on one side of zero we must subtract the area between zero and
0.22 from the area above zero. If you wish, make a completely separate diagram for x . Draw a Normal
curve with a mean at 12. Indicate the mean by a vertical line! Shade the above 14. This area is on one
side of the mean (12) so we subtract to get our answer.
10 .05  12 
 0  12
z
3. P0  x  10 .05   P 
  P1.33  z  0.22 
9
 9

 P1.33  z  0  P0.22  z  0  .4082  .0871  .3211
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 -1.33 and -0.22. Because this is on one side of zero we must subtract the area
between -0.22 and zero from the larger area between -0.22 and zero. If you wish, make a completely
separate diagram for x . Draw a Normal curve with a mean at 12. Indicate the mean by a vertical line!
Shade the area between zero and 10.05. This area is on one side of the mean (12) so we subtract to get our
answer.
x.055 (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 .055 is the value of z with 5.5% of the distribution above it. Since 100 – 5.5 = 94.5, it is also
the 94.5th percentile. Since 50% of the standardized Normal distribution is below zero, your diagram should
show that the probability between z .055 and zero is 94.5% - 50% = 44.5% or P0  z  z.055   .4450 . The
closest we can come to this is P0  z  1.60   .4452 . (1.59 is also acceptable here.) So z .055  1.60 . To
4.
get from z .055 to x.055 , use the formula x    z , which is the opposite of z 
x
.

x  12  1.609  26.40 . If you wish, make a completely separate diagram for x . Draw a Normal curve
with a mean at 12. Show that 50% of the distribution is below the mean (12). If 5.5% of the distribution is
above x.055 , it must be above the mean and have 44.5% of the distribution between it and the mean.
Check:
26 .40  12 

Px  26.40   P  z 
  Pz  1.60   Pz  0  P0  z  1.60   .5  .4452  .0548  .055
9


252y0762 10/30/07
II. (5+ points) Do all the following. Look them over first – There is a section III in the in-class exam and
the computer problem is at the end. Show your work where appropriate. There is a penalty for not
doing Problem 1a.
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.
6. Make sure that you state your null and alternative hypothesis, that I know what method you are using and what the
conclusion is when you do a statistical test.
1. (Groebner) We wish to compare the amount of time men and women spend in the supermarket. The two
columns below, x1 and x 2 represent two independent samples with 7 shoppers in each sample. You may
assume that the parent distributions are Normal. d  x1  x 2
Row
1
2
3
4
5
6
7
Men
x1
32
42
22
28
32
36
25
Women
x2
Difference
d
33
33
26
41
33
48
44
-1
9
-4
-13
-1
-12
-19
Minitab computes the following.
Variable
x1
x2
d
N
7
7
7
Mean
31.00
36.86
-5.86
SE Mean
2.55
2.91
StDev
6.76
7.69
Minimum
22.00
26.00
-19.00
Q1
25.00
33.00
-13.00
Median
32.00
33.00
-4.00
Q3
36.00
44.00
-1.00
Maximum
42.00
48.00
9.00
a. Compute the sample variance for the d column – Show your work! (2)
b. Is there a significant difference between the variances for men and women? State your hypotheses and
your conclusion clearly! (2)
c. Test to see there is a difference between the average amount of time men and women shop. (3)
d. Using the sample means and standard deviations computed above and changing each sample size from 7
to 100, find an 89% 2-sided confidence interval for the difference between the amount of time men and
women shop. Does it indicate a significant difference between men’s and women’s times? Why? (3) [10]
Solution: a. Compute the sample variance for the d column – Show your work! (2)
d 2  862
d  48 ,
So n  7 ,
Row d
d2
1
2
3
4
5
6
7
-2
8
-5
-14
-2
-13
-20
-48
4
64
25
196
4
169
400
862


First
d
 d   48  6.85714  x
n
7
1
 x 2  31 .00  36 .86  5.86
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The formula for the sample variance is s 2 
s d2
d

2
 nd 2
n 1

773  7 5.85714 2
6
x
2
 nx 2
. For the difference this becomes
n 1
773  240 .14286


6
 d  d 
2

n 1
532 .85714
 88 .80952 .
6
s d  88 .80952  9.4239 .
b. Is there a significant difference between the variances for men and women? State your hypotheses and
your conclusion clearly! (2)
Solution: From the formula table we have the following.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
D1:Ratio of
 22 s22 DF1 , DF2
s2
H0 : 12   22
 2 F.5  .5  
F DF1 , DF2  12
Variances
2
2
1
s1
s
H : 2   2
1 , DF2
F1DF


2
1
1
DF1  n1  1
FDF1 , DF2
DF2  n 2  1
2
1
2
2
and
F DF2 , DF1 
 2

.5  .5   2    or
1  
2

We are testing
H0 : 12   22
s 22
s12
by using a variance ratio. We have been given n1  n 2  7 , s1  6.76 and
H1 : 12   22
s 2  7.69 . To do a 5% two sided test we must compare
s12
s 22
and
s 22
s12
6,6   5.82 . The larger of
against F.025
2
s 22
 7.69 
6,6 

  1.137 . Since this is not larger than F.025  5.82 , we cannot reject the
2
s1  6.76 
null hypothesis of equal variances. Since the other ratio is below 1 and there are no numbers below 1 on the
F table we do not bother to compute or compare it. There is no significant difference between variances.
the two ratios is
c. Test to see there is a difference between the average amount of time men and women shop. (3)
Here is Table 3. Because we have shown that there is no significant difference between the variances, we
will use method D2.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
D2: Difference
H 0 : D  D0 *
d cv  D0  t  2 s d
D  d  t 2 s d
d  D0
t
between Two
H 1 : D  D0 ,
s
1
1
d
Means (
sd  s p

D





n

1s12  n2  1s22
1
2
n1 n2
unknown,
sˆ 2p  1
n1  n2  2
variances
assumed equal)
DF  n1  n2  2
D3: Difference
between Two
Means(
unknown,
variances
assumed
unequal)
H 0 : D  D0 *
D  d  t 2 s d
s12 s22

n1 n2
sd 
DF 
 s12 s22 
  
n

 1 n2 
2
   
s12
2
n1
n1  1
s 22
2
n2
n2  1
H 1 : D  D0 ,
D  1   2
t
d  D0
sd
d cv  D0  t  2 s d
252y0762 10/30/07
2
2
n1  7, x1  31.00, s12  6.76  , n 2  7, x 2  38 .86 and s 22  7.69  .
 H 0 : 1   2
H :    2  0
or  0 1
DF  n1  1  n 2  1  6  6  12  n1  n 2  2. Our hypotheses are 
 H 1 : 1   2
 H 1 : 1   2  0
H : D  0
or if D  1   2 ,  0
. Our D0  0 . This is a two-sided test. d  x1  x 2  5.86 .
H 1 : D  0
n1  1s12  n2  1s 22
66.76 2  67.69 2
6.76 2  7.69 2
45 .6976  59 .1361

2
n1  n 4  2
12
2
 52.41685 . This is the pooled variance. sˆ p  7.23995 . The computer got 7.2391.   .05 so
sˆ 2p 




12
t.12
025  2.179 and t .05  1.782
 1
1
1
1 
1 1
2
  52 .4185     52 .4185    14.9767  3.8700 . Recall

 sˆ 2p  
7
7
7
n1 n 2
n
n


 
2 
 1
that our alternate hypothesis is H 1 : D  0 so this is a two-sided test.
Use only one of the following methods!
x  x 2   10   20 
d  D0
Test Ratio: t 
.
or t  1
sd
sd
s d  sˆ p
12
If this test ratio lies between  t.12
025  2.179 and t .025  2.179 , do not reject H 0 .
t
d  D0
 5.86  0

 1.514 .
sd
3.8700
Make a diagram with zero in the middle showing shaded ‘reject’ regions below  t.12
025  2.179 and
above t .12
025  2.179 . Since -1.514 does not fall in the 'reject' region, do not reject H 0 .
Here is the t table for 12 degrees of freedom.
Significance Level
df .45 .40 .35 .30 .25 .20 .15 .10 .05 .025 .01 .005 .001
12 0.128 0.259 0.395 0.539 0.695 0.873 1.083 1.356 1.782 2.179 2.681 3.055 3.930
12
Or you can say that, since 1.514 falls between t.12
05  1.782 and t .10  1.356 , for a one-sided test,
.05  p  value  .10 . For this 2-sided test, .10  p  value  .20 . (The computer gets .156) Since the pvalue is above   .05 , do not reject H 0 .
Critical Value: The formula for a 2-sided critical value for the sample difference between means is
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 on
either side of D0  0. If d  x1  x 2 is above the critical value, reject H 0 .
d CV  D0  t  2 s d  0  2.179 3.8700   8.433 . Make a diagram with 0 in the middle showing two
shaded ‘reject’ regions below -8.433 and above 8.433. Since d  5.86 falls between -8.433 and 8.433
and is not in the 'reject' region, do not reject H 0 .
Confidence Interval: The formula for a 2 sided confidence interval is D  d  t  2 s d . Since our alternate
hypothesis is two-sided, this is what we will use. We already know that t  s d  2.1793.8700  8.433 .
2
So we can say D  5.86  8.433 . Make a diagram with d  6.86 in the middle. Represent the
confidence interval by shading the area between -5.86 - 8.433 = -14.29 and -5.86 + 8.433 = 2.57 (The
confidence interval will always include d .) Since D0  0 is in this area, do not reject H 0 .
There is no significant difference between the time men and women shop.
The Minitab run follows.
252y0762 10/30/07
Two-Sample T-Test and CI: x1, x2
Two-sample T for x1 vs x2
N
Mean StDev SE Mean
x1 7 31.00
6.76
2.6
x2 7 36.86
7.69
2.9
Difference = mu (x1) - mu (x2)
Estimate for difference: -5.85714
95% CI for difference: (-14.28799, 2.57371)
T-Test of difference = 0 (vs not =): T-Value = -1.51
DF = 12
Both use Pooled StDev = 7.2391
P-Value = 0.156
d. Using the sample means and standard deviations computed above and changing each sample size from 7
to 100, find an 89% 2-sided confidence interval for the difference between the amount of time men and
women shop. Does it indicate a significant difference between men’s and women’s times? Why? (3) [10]
If we use the values that we found before, we now have n1  100 , x1  31.00. s12  6.76 2 , n 2  100 ,
x 2  38 .86 , s 22  7.69 2 and d  x1  x 2  5.86 . Because of the large sample size, we will use Method
D1 with sample variances substituting for population variances.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
D1. Difference
H 0 : D  D0 *
d cv  D0  z d
D  d z 2  d
d  D0
z
between Two
H
:
D

D
,

1
0
d
Means (
 12  22
D







1
2
known)
d
n
n
1
2
d  x1  x 2
sd 
s12 s 22

n1 n 2

6.76 2 7.69 2

 0.45698 0.59136  0.45698  0.6760 . The formula for a 2100
100
sided confidence interval is D  d  z  2 s d . The confidence level is 1    .89 , so   .11 and
 .13 2  .055 . On the first page of the exam we found z .055  1.60 . So D  5.86  1.600.6760 
 5.86  1.08 . This interval is -6.95 to -4.78 and does not include zero, so there is a significant difference
between the means.

2
252y0762 10/30/07
III. (18+ points) Do as many of the following as you can. (2points each unless noted otherwise). Look
them over first – the computer problem is at the end. Show your work where appropriate.

 .10

Note that if you have a table like this 
 .90 , and if you know one number on the inside of

.20 .80 1.00
the table, you can get the rest by subtracting.
1. A professor wishes to see if the variability of scores for people taking the introductory accounting course
is different. He takes a sample of the scores of 10 non-accounting students and 13 accounting students and
gets the following results: n1  13, n 2  10, s12  210.2 and s 22  36.5 . Though this is a 2-sided test with
a 95% confidence level, he can actually do the entire test by comparing
s12
9,12
a)
against F.05
2
s2
b)
s12
s 22
s12
c)
s 22
s12
d)
s 22
s 22
e)
s12
s 22
f)
s12
s 22
g)
s12
s 22
h)
s12
Explanation: s12
9,12
against F.025
12,9 
against F.05
12,9 
against F.025
9,12
against F.05
9,12
against F.025
12,9 
against F.05
12,9 
against F.025
 210.2 has n1  1  13  1  12 degrees of freedom and s 22  36.5 has n 2  1  10  1  9
degrees of freedom. In theory, if   .05 , we should test
s12
s 22
12,9  and
against F.025
s 22
s12
9,12 and
against F.025
reject the null hypothesis of equality if either computed ratio exceeds the table value. But
s 22
s12
is obviously
below 1, and there are no numbers on the F table below 1, so the only test that we really have to do is
9,12
against F.025
9,12  3.44
2. F.025
s12
s 22
150, 250 is, at most, ______. If you did not get this from the Supplementary Tables, you must explain how
F.025
you found this. Explanation: The first available numerator degrees of freedom below 150 on the table is
100 and the first available denominator degrees of freedom below 250 is 200, since the table Fs go up as we
100, 200  1.39 .
move to the Northwest on the table use F.025
252y0762 10/30/07
150, 250 .
If we are more ambitious, we might notice the values that bracket F.025
100, 200  1.39 F 200, 200  1.32
F.025
.025
100,400  1.35 F 200,400  1.27
F.025
.025
150, 200  1.355 and F 150,400  1.31 . The difference between these two
From these we might guess that F.05
.05
is .045. If we go three quarters of the way between them we might guess 1.34 or 1.35.
Exhibit 1:
Sample size
Married
25
Unmarried
30
Standard error 24.2534
Mean Std Deviation
268.90
77.25
455.10 102.40
Difference between means -186.20
d  D 0  186 .20

 7.6773
sd
24 .2534
(Groebner et. al.) Bank managers want to find out if an incentive interest rate will cause more of an
increase in spending by married cardholders than by unmarried cardholders. Let x1 represent the increase
of spending by a random sample of 25 married cardholders and x 2 represent increase of spending by a
random sample of 30 unmarried cardholders. Sample data is above.
3. If the bank finds that the difference between married and unmarried couples is 186 .20  110 .03 .
a. *The difference is statistically significant because 186.20 is larger than 110.03
b. The difference is statistically significant because the confidence interval supports a null
hypothesis.
c. The difference is statistically insignificant because 110.03 is smaller than 186.20
d. The difference is statistically insignificant because the confidence interval would lead us to
reject a null hypothesis.
Explanation: Do the math! 186 .20  110 .03 is below zero but 186 .20  110 .03 is also below zero, so
the interval does not include zero and we can say that the difference is significantly different from zero.
Note that ‘No significant difference’ is your null hypothesis, so that b and d could never be true.
4. If the researcher is trying to show that married cardholders will increase their spending more than
unmarried cardholders, and, assuming that the population mean is appropriate to compare salaries,
D  1   2 , her null hypothesis should be:
a. D  0
b. D  0
c. D  0
d. * D  0
e. D  0
f. D  0
g. None of the above
Explanation: Let x1 represent the increase of spending by a random sample of 25 married cardholders and
x 2 represent increase of spending by a random sample of 30 unmarried cardholders. Then she is trying to
show that 1   2 or D  1   2  0 . This does not contain an equality, so it must be an alternative
hypothesis. The opposite of this is D  1   2  0
252y0762 10/30/07
5. If the researcher in exhibit 1 is attempting to show that married cardholders will spend significantly more
than unmarried cardholders the appropriate critical value for the difference between the sample means is
(assuming that t or z  is chosen correctly):
a. * 0  t or z 24 .2534
b. 0  t or z 24 .2534
c. 0  t or z 24 .2534
d. 186 .20  t or z 24.2534
e. 186 .20  t or z 24.2534
f. 186 .20  t or z 24.2534
We have H 1 : D  1   2  0 . We need a critical value for d , the difference between sample means that
is above zero. The formula for a critical value for a difference between means, when the alternate
hypothesis is H 1 : D  D0 is d CV  D0  t  2 s d must become d CV  D0  t  2 s d  0  t  s d
2
6. If the researcher believes that the population standard deviations for men and women are both the same,
the appropriate degrees of freedom for the test (in problem 5) are:
a. 53
b. Gotten by the formula
DF 
 s12 s22 
  
n

 1 n2 
2
   
s12
2
n1
n1  1
s 22
2
n2
n2  1
c. 25 (The smaller of 25 and 30)
d. 55
e. None of the above. (Use z instead of t .)
Explanation: Let x1 represent the increase of spending by a random sample of 25 married cardholders and
x 2 represent increase of spending by a random sample of 30 unmarried cardholders. Then we have 25 – 1
= 24 degrees of freedom for the first sample and 30 – 1 = 29 degrees of freedom in the second sample for a
total of 53. This is, of course because we are using Method D2.
7. I am testing the hypothesis H 0 :   300 . I get a value of x  345 , which results in a p-value of .076.
What are the p-values for H 0 :   300 and (Note Error!) H 0 :   300 ?
a. Both are .076
b. Both are .038
c. The first is .038 and the second is .962
d. *The first is .962 and the second is .038
e. Both are .962
f. Not enough information
[14]
Explanation: Make a diagram. Make an almost Normal curve with a mean of 300. x  345 is above 300.
Since it is above 300, p-value is defined as 2Px  345  . Since half of .076 is .038, Px  345   .038 . Its
opposite is Px  345   1  .038  .962 . H 0 :   300 has the alternate hypothesis H 1 :   300 . We will
reject the null hypothesis if the sample mean is too far above 300 and we have a right-sided test. The pvalue is defined as Px  345  , when the population mean is 300, so it must be .038 and the p-value for
H 0 :   300 must be Px  345   .962 .
252y0762 10/30/07
8. Surveys taken for the International Republican Institute in the last two years say that in 2007 41% (618
out of 1507) of Turks believe that genocide had been committed against the Armenians in the early 20 th
century. In 2006 39% (which would be 588 out of 1507) held a similar belief. Can we say that there has
been a significant increase in the proportion that believed that genocide had occurred?
a) If year 1 is 1006 and year 2 is 2007, what would our null and alternative hypotheses be? Answer in terms
of 1 ,  2 and D  1   2 , or  1 and  2 or p1 and p 2 and p  p1  p 2 as appropriate. (Example: If  1 ,
 2 and F 
1
were a reasonable answer, you might get two points for saying H 0 :  1   2 ,
2
H 1 :  1   2 and wrongly saying that this becomes H 1 : F  0 and the corresponding H 0 : F  0 ) (3)
Solution: We are asking if the proportion that believe in the genocide in the second survey is larger than
the proportion that believed it in the first survey. This is p1  p 2 and must be an alternative hypothesis. If
H 0 : p1  p 2 or p  0
.
p  p1  p 2 , we have 
H 1 : p1  p 2 or p  0
b) If we find that the 2006 and 2007 surveys were surveys of the same people and that 38% (573 people)
believed in the genocide in both periods, test your hypothesis.(5)
[21]
H 0 : p1  p 2 or p  0
Solution: We are testing 
. Since we are worrying about p  p1  p 2 being less
H 1 : p1  p 2 or p  0
than zero, this is a left-sided test. This is a McNemar problem, since we are putting two questions to the
same group. We found that 2007 41% (618 out of 1507) of Turks believed that genocide had been
committed against the Armenians in the early 20th century, in 2006 39% (which would be 588 out of 1507)
held a similar belief and that 38% (573 people) believed in the genocide in both periods. The general design
question 2
question 1
yes no
of the table is
. We can make the following table from the data given.
yes
 x11 x12 
x

no
 21 x 22 
Y 06
N 06
Y 07
573


618
N 07



588
. Here Y06 is “Yes in 2006” and N06 is “no.” Y07 is “Yes in 2007.” If we just
1507
make the table add up we get
Y 06
N 06
Y 07 N 07
573 15  588


. We can finish by filling the remaining blanks.

 919
618 889 1507
Y 07 N 07
Y 06 573
15  588
x  x 21
15  45
30 2


  15  3.873
. We will compare z  12


60
N 06  45 874  919
15  45
x12  x 21
618 889 1507
against  z  . If the significance level is 5%, reject the null hypothesis if our computed value of z is below
z .05  1.645 . In this case we reject the null hypothesis and conclude that belief in the genocide has
increased.
252y0762 10/30/07
9. A survey of 2714 respondents, of whom 55% (1493) were women, conducted by the International
Republican Institute in Iraq in 2005, says that 12% of men and 15% of women believed that Iraqi women
had sufficient rights, opportunities and protections under the new constitution. Does this show a significant
difference between attitudes of men and women? Assume that the men and women are two independent
samples. (State and test your null and alternate hypotheses.) (5) Note error – 12% was men and 15%
women, but it’s not worth correcting.
[28]
 p  x1
 1
n1  H 0 : p1  p 2
H : p  p 2  0
 H : p  0
Solution: If 
or  0 1
. If p  p1  p 2 , then  0
.

 H 1 : p1  p 2
H 1 : p1  p 2  0
 H 1 : p  0
 p2  x2
n2

Since we are comparing proportions of two independent samples, use Method D6a.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
pcv  p0  z 2  p
Difference
p  p 0
p  p  z 2 s p
H 0 : p  p 0
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
s p 

 n1 n 2 
p 01q 01 p 02 q 02
 p 

n1
n2
or p 0  0
n1
n2
n p  n2 p 2
p0  1 1
Or use s p
n1  n 2
So   .05 , z  z.025  1.960, If group 1 is women, n1  1493 , and n 2  2714  1493  1221 , p1  .15
2
q1  .85  ,
p 2  .12 q 2  .88  and p 0 
1493 .15   1221 .12  224  147
371


 .1367 q 0  .8633  .
1493  1221
2714
2714
p  .15  .12  .03 .
1 
 1

  .1367 .8633 0.00149   0.0001757  .013255
1493
1221


 p  .1367 .8633 
 .15.85  .12 .88  
s p  

  .0000854 .0000865  .0001719  .013110
1221 
 1493
Do only one of the following three methods.
.03  0
 2.263 . Make a diagram of a Normal curve with 0 in the middle and
Test ratio: z 
.013255
two ‘reject’ regions, one below z .025  1.960 and one above z .025  1.960 , . Since 2.263 is in
the upper ‘reject region, reject the null hypothesis. Since the alternative hypothesis is
H 1 : p  0 , which is 2-sided, the p-value is the probability that p is as extreme or more
extreme than .03 or that z is as extreme or more extreme than 2.288.
p  value  2Pz  2.288   2(.5  .4890 )  2(.0110 )  .0220 . . Because this is below .05, we reject
the null hypothesis.
Critical value: Since the alternative hypothesis is H 1 : p  0 , there must be two critical values,
one above and one below zero. p cv  0  1.960 .013255   .0260 . Make a diagram of a
Normal curve with 0 in the middle and a ‘reject’ regions below -.0260 and above .0260. Since
p  .03 is in the ‘reject region, reject the null hypothesis.
Confidence interval: p  p  z s p  .03  1.960.013110  .03  .026 . Make a diagram of
2
a Normal curve with .03 in the middle. Represent the confidence interval by shading the entire
region between .004 and .056. Since p 0  0 is not in the confidence interval, reject the null
hypothesis.
252y0762 10/30/07
10. Computer question.
a. Turn in your first computer output. Only do b, c and d if you did. (3)
b. (Groebner) A study was done in North Carolina to compare the incentive to invest under plan 1
(tax sheltered annuities) and plan 2 (401k). Two independent samples of 15 individuals were
selected. Each group was eligible for only one of the two plans. The first sample gave a sample
mean investment of $2119.70 with a sample standard deviation of $709.70. The second sample
had a sample mean of $1777.70 with a sample standard deviation of $593.90. We start out by
wanting to test our belief that people in plan 1 will invest less than those in plan 2 (!!!!). What are
our null and alternative hypotheses? (1)
c. Is the appropriate computer run A or B below? Why? (1)
d. What is our conclusion – do we reject the null hypothesis using a 5% significance level? Why?
Can we say that people in plan 1 will invest more than those in plan 2? (2)
e. On the basis of run C, could we have made things easier by assuming equal variances? Why?
(1)
[36]
Solution: b. We start out by wanting to test our belief that people in plan 1 will invest less than those in
plan 2 (!!!!). What are our null and alternative hypotheses? (1)
If we assume that the underlying distribution is Normal, so that the mean is relevant, the alternate
hypothesis is H 1 : 1   2 . this means that the null hypothesis is H 0 : 1   2 .
c. Is the appropriate computer run A or B below? Why? (1) A says the following.
T-Test of difference = 0 (vs <): T-Value = 1.43
P-Value = 0.918
DF = 27
“T-Test of difference = 0 (vs <)” means that the alternate hypothesis is H 1 : 1   2 . , so A is
our test.
d. What is our conclusion – do we reject the null hypothesis using a 5% significance level? Why? Can we
say that people in plan 1 will invest more than those in plan 2? (2) The 91.8% p-value means that we
cannot reject the null hypothesis at the 5% level, since the p-value is above the 5% significance level. We
certainly have no reason to believe that people in plan 1 will invest more than people in plan 2!
e. On the basis of run C, could we have made things easier by assuming equal variances? Why? We assume
that the null hypothesis in the F test below is ‘equal variances.’ The p-value of .514 for this null hypothesis
means that we would not reject the null hypothesis of equality if we use confidence levels in the .005 to .10
confidence level range. This means that we could use Method D2 instead of D3, which is used here.
MTB > TwoT 15 2119.7 709.7 15 1777.7 593.9;
SUBC>
Alternative -1.
A) Two-Sample T-Test and CI
#Implies H 1 : 1   2 . H 0 : 1   2 .
Sample
N Mean StDev SE Mean
1
15 2120
710
183
2
15 1778
594
153
Difference = mu (1) - mu (2)
Estimate for difference: 342.000
95% upper bound for difference: 748.985
T-Test of difference = 0 (vs <): T-Value = 1.43
P-Value = 0.918
DF = 27
MTB > TwoT 15 2119.7 709.7 15 1777.7 593.9;
SUBC>
Alternative 1.
B) Two-Sample T-Test and CI
Sample
1
2
N
15
15
Mean
2120
1778
StDev
710
594
#Implies H 1 : 1   2 . H 0 : 1   2 .
SE Mean
183
153
Difference = mu (1) - mu (2)
Estimate for difference: 342.000
95% lower bound for difference: -64.985
T-Test of difference = 0 (vs >): T-Value = 1.43
P-Value = 0.082
DF = 27
252y0762 10/30/07
MTB > VarTest 15 503674 15 352717.
C) Test for Equal Variances
95% Bonferroni confidence intervals for standard deviations
Sample
N
Lower
StDev
Upper
1 15 498.095 709.700 1202.90
2 15 416.822 593.900 1006.63
F-Test (normal distribution)
Test statistic = 1.43, p-value = 0.514
252y0762 10/30/07
ECO252 QBA2
SECOND EXAM
March 23, 2007
TAKE HOME SECTION
Name: _________________________
Student Number: _________________________
Class hours registered and attended (if different):_________________________
IV. Neatness Counts! Show your work! Always state your hypotheses and conclusions clearly. (19+ points). In each section
state clearly what number you are using to personalize data. There is a penalty for failing to include your student number on this
page, not stating version number in each section and not including class hour somewhere. Please write on only one side of the
paper. You must do 3a (penalty).
1. (Groebner, et. al.) Your company produces hair driers for retailers to sell as house brands. A design change will produce
considerable savings, but the new design will not be adopted unless it is more reliable. For a sample of 250 hair driers with the old
design, 75 failed in a simulated 1-year period. For a sample of 250 driers with the new design 50  a fail in a simulated one year
period, where a is the second-to-last digit of your student number. Use a 90% confidence level. Make sure that I know what value
you are using for a .
a) Can we say that the proportion of the redesigned driers that fail is significantly lower than that of the driers with the current (old)
design? (3)
b) Do a 95% two-sided confidence interval for the difference between the two proportions. (1)
c) After you have implemented your decision on using the new design, a newly-hired engineer recommends another design change
(the newest design) that she claims will decrease the proportion that fail even further. For a sample of 100 driers, 18 fail in a simulated
one-year period. Do a test of the equality of the three proportions, again using a 90% confidence level. (4)
d) Follow your results in c) with a Marascuilo procedure for finding pairwise differences between the proportions for the three
designs. Assuming that there is no cost-saving in going to the newest design, would you recommend going to it? Write a paragraph
long report on your conclusions from the two hypothesis tests and what decisions these implied. (4)
[12]
2. (Groebner et al) A ripsaw is cutting lumber into narrow strips and should be set to produce a product whose width differs from the
width specified by an amount given by a Normal distribution with a mean of zero and a standard deviation of 0.01 inch. Because we
have been getting complaints about the uniformness of our product, we wish to verify the Normal distribution specified is correct. We
cut 600  b pieces (where b is the last digit of your student number. Our results are as follows.
Deviation from specified
Number of pieces
width
Below -0.02
0
-0.02 to -0.01
84
-0.01 to 0
266
0 to 0.01
150 + b
0.01 to 0.02
94
0.02 and above
6
a) To use a chi-squared procedure to check the distribution, find the values of E (3)
b) State and test the null hypothesis. (2)
c) We have learned another procedure that can be used to test for a Normal distribution when the parameters are given. Use it now to
verify your results. Can you say that the saw is working as advertised? (4)
[21]
3. (Groebner et al.) Two groups of 16 individuals were asked to do their income taxes using two tax preparation software packages.
The data is below (in number of minutes required) and may be considered two independent random samples. To personalize the data
add the last digit of your student number to every number in the TT00 column. Use 10 if your number ends in 0. Label the column
clearly as TT1, TT2 through TT10 according to the number used. Let d  TTa  TC .
Row TT00 TC
1
65 88
2
51 71
3
74 89
4
89 66
5
88 78
6
96 64
7
37 74
8
66 99
9
86 79
10
54 68
11
60 93
252y0762 10/30/07
12
13
14
15
16
45
42
55
58
38
93
86
86
81
83
Minitab has given us the following results
Variable
N N*
Mean SE Mean
TC
16
0 81.13
2.60
TT1
16
0 63.75
4.76
TT2
16
0 64.75
4.76
TT3
16
0 65.75
4.76
TT4
16
0 66.75
4.76
TT5
16
0 67.75
4.76
TT6
16
0 68.75
4.76
TT7
16
0 69.75
4.76
TT8
16
0 70.75
4.76
TT9
16
0 71.75
4.76
TT10
16
0 72.75
4.76
StDev
10.40
19.05
19.05
19.05
19.05
19.05
19.05
19.05
19.05
19.05
19.05
Minimum
64.00
38.00
39.00
40.00
41.00
42.00
43.00
44.00
45.00
46.00
47.00
Q1
71.75
47.50
48.50
49.50
50.50
51.50
52.50
53.50
54.50
55.50
56.50
Median
82.00
60.00
61.00
62.00
63.00
64.00
65.00
66.00
67.00
68.00
69.00
Q3
88.75
84.00
85.00
86.00
87.00
88.00
89.00
90.00
91.00
92.00
93.00
Maximum
99.00
97.00
98.00
99.00
100.00
101.00
102.00
103.00
104.00
105.00
106.00
a) Find the mean and standard deviation of d . (1) Assume the Normal distribution in b), c), and e).
b) Find out if there is a significant difference between the mean times for the two packages, using a test ratio, a critical value or a
confidence interval.(4) (2 extra points if you use all three methods and get the same results on all three, 3 extra (extra) points if you do
not assume equal variances)
c) Test the variances of the two samples for equality on the assumption that they come from the Normal distribution. (2)
d) Test the d column to see if the data was Normally distributed (5)
e) Actually the data above was taken from only 16 people, who were randomly assigned to use one of the methods first. Would this
mean that what you did above was correct? If not do b) over again. (3)
f) In view of the fact that the data was taken from only 16 people and dropping the assumption of Normality, find out if there is a
significant difference between the medians of the two packages. (3)
g) If you did d) e) and f), can you report on your results, indicating which result was correct? (1) [40]
Be prepared to turn in your Minitab output for the first computer problem and to answer the questions on the problem sheet
about it or a similar problem.
Solution.
1. (Groebner, et. al.) Your company produces hair driers for retailers to sell as house brands. A design
change will produce considerable savings, but the new design will not be adopted unless it is more reliable.
For a sample of 250 hair driers with the old design, 75 failed in a simulated 1-year period. For a sample of
250 driers with the new design 50  a fail in a simulated one year period, where a is the second-to-last
digit of your student number. Use a 90% confidence level. Make sure that I know what value you are using
for a .
What the problem says.   .10
75
 .3000 .
Old design: Out of a sample of n1  250, x1  75 failed. So p1 
250
50  a
New design: Out of a sample of n 2  250, x 2  50  a failed. So p 2 
.
250
a) Can we say that the proportion of the redesigned driers that fail is significantly lower than that of the
driers with the current (old) design? (3)
Solution: From the formula table we have the following.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
pcv  p0  z 2  p
Difference
p  p 0
p  p  z 2 s p
H 0 : p  p 0
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
s p 

 n1 n 2 
p 01q 01 p 02 q 02
 p 

n1
n2
or p 0  0
n1
n2
n p  n2 p 2
p0  1 1
Or use s p
n1  n 2
The problem above asks if p1  p 2 . This is an alternative hypothesis since it is missing an equality.
252y0762 10/30/07
We are testing H 0 : p1  p 2 or p  0 H 1 : p1  p 2 or p  0 where p  p1  p 2 . The proportion of
p  p 0
75  50  a
the pooled sample that failed is p 0 
. q 0  1  p 0 . The test ratio is z 
, where
250  250
 p
p 0  0 and  p 
 1
1 

p 0 q 0  
 n1 n 2 
The confidence interval has the form p  p  z s p , where s p 
2
p1 q1 p 2 q 2

n1
n2
The following are identical for all versions of this problem: x1  75, n1  250, n 2  250,
 1
pq
.30 .70 
1 
75
 .00084 and    =0.008. Using the computer as a desk
 .3000 and 1 1 
250
n1
250
 n1 n 2 
calculator, but working with columns, I got the following.
p1 
Version
x2
0
1
2
3
4
5
6
7
8
9
10
50
51
52
53
54
55
56
57
58
59
60
p0
0.250
0.252
0.254
0.256
0.258
0.260
0.262
0.264
0.266
0.268
0.270
p2
p1  p 2
p2 q2
n2
 p 2
s p 2
 p
0.200
0.204
0.208
0.212
0.216
0.220
0.224
0.228
0.232
0.236
0.240
0.100
0.096
0.092
0.088
0.084
0.080
0.076
0.072
0.068
0.064
0.060
0.0006400
0.0006495
0.0006589
0.0006682
0.0006774
0.0006864
0.0006953
0.0007041
0.0007127
0.0007212
0.0007296
0.0015000
0.0015080
0.0015159
0.0015237
0.0015315
0.0015392
0.0015468
0.0015544
0.0015620
0.0015694
0.0015768
0.0014800
0.0014895
0.0014989
0.0015082
0.0015174
0.0015264
0.0015353
0.0015441
0.0015527
0.0015612
0.0015696
0.0387298
0.0388326
0.0389342
0.0390348
0.0391342
0.0392326
0.0393300
0.0394263
0.0395215
0.0396158
0.0397089
s p
0.0384708
0.0385945
0.0387162
0.0388359
0.0389535
0.0390692
0.0391829
0.0392946
0.0394044
0.0395122
0.0396182
Since our alternate hypothesis is H 1 : p1  p 2 or p  0 , this is a right-sided test. You will get credit for
one of the methods below.
p  p 0
0.100  0

 2.582 . For Version 10
Test Ratio Method: For Version 0 z 
.0387298
 p
0..060  0
 1.511 We compare this with z  z.10  1.282 . The rejection zone is the area above
.0397089
1.282. (Remember that for t, zero is never in the rejection zone.) Of course in both versions this value of z
is in the rejection zone and we reject the null hypothesis. For Version 0, the p-value is
Pz  2.582   .5  .4951  .0049 Pz  1.511   .5  .4345  .0655 . Both of these are below   .10 ,
implying rejection.
Critical value method: The two-sided formula for a critical value is pcv  p0  z  p , but we need one
z
2
critical value that, in light of our alternate hypothesis, H 1 : p  0 , must be above zero. So, for Version 0,
we have pcv  p0  z p  0  1.282 .0387298   .0497. Since p  p1  p 2  .1 is above the critical
value, we reject the null hypothesis. In version 10, the critical value is  0  1.282 .0397089   .0509. This
time p  .060, and we still reject the null hypothesis.
Confidence interval method: Again the usual formula, p  p  z s p , must be modified to become, in
2
Version 0, p  p  z s p  .100 1.282.0384708  .0507 . Since it is impossible for p  .0503 and
H 0 : p  0 to both to be true, we reject the null hypothesis. For Version 10,
p  p  z s p  .060  .1.282.0396182  .0092 and the null hypothesis is still contradicted.
The Minitab output for these two versions appears below. Note that the differences between the
means, the lower limit of the confidence interval, the values of z and the p-values are as above.
252y0762 10/30/07
MTB > PTwo 250 75 250 50;
SUBC>
Confidence 90.0;
SUBC>
Alternative 1;
SUBC>
Pooled.
Test and CI for Two Proportions
Sample
X
N Sample p
1
75 250 0.300000
2
50 250 0.200000
Difference = p (1) - p (2)
Estimate for difference: 0.1
90% lower bound for difference: 0.0506977
Test for difference = 0 (vs > 0): Z = 2.58
P-Value = 0.005
MTB > PTwo 250 75 250 60;
SUBC>
Confidence 90.0;
SUBC>
Alternative 1;
SUBC>
Pooled.
Test and CI for Two Proportions
Sample
X
N Sample p
1
75 250 0.300000
2
60 250 0.240000
Difference = p (1) - p (2)
Estimate for difference: 0.06
90% lower bound for difference: 0.00922726
Test for difference = 0 (vs > 0): Z = 1.51
P-Value = 0.065
b) Do a 95% two-sided confidence interval for the difference between the two proportions. (1)
In Version 1, p  p  z s p  .100  1.960.0384708  .100  .075 or .065 to .175. For Version 10,
2
p  .060  .1.960 .0396182   .060  .078 or -0.072 to .138.
c) After you have implemented your decision on using the new design, a newly-hired engineer recommends
another design change (the newest design) that she claims will decrease the proportion that fail even
further. For a sample of 100 driers, 18 fail in a simulated one-year period. Do a test of the equality of the
three proportions, again using a 90% confidence level. (4)
What the problem says.   .10
75
 .3000 .
Old design: Out of a sample of n1  250, x1  75 failed. So p1 
250
50  a
New design: Out of a sample of n 2  250, x 2  50  a failed. So p 2 
(which is between .20 and
250
.24).
18
 .18 .
Newest design: Out of a sample of n3  100, x 3  18 failed. So p 3 
100
O
Design
failed
didn' t
Total
old new nwst Total
pr
 75 50 18  143
.2383 To get the E just apply the row proportions to the column


.7617
175 200 82  457
250 250 100
600 1.0000
E
Design
old
new
nwst
Total
pr
totals.
failed  59 .575 59 .575 23 .830  142 .98 .2383


didn' t 190 .425 190 .425 76 .170  457 .02 .7617
Total 250 .000 250 .000 100 .000 600 .00 1.0000
The table of computations follows.
252y0762 10/30/07
Row
1
2
3
4
5
6
Totals
O
75
175
50
200
18
82
600
OE
E
59.575
190.425
59.575
190.425
23.830
76.170
600.000
15.425
-15.425
-9.575
9.575
-5.830
5.830
0.000
O  E 2
O  E 2
237.931
237.931
91.681
91.681
33.989
33.989
O  E 2
E
3.99380
1.24947
1.53891
0.48145
1.42631
0.44622
9.13617
O2
E
94.419
160.824
41.964
210.056
13.596
88.276
609.136
O2
 n . Both of these two
E
E
formulas are used above. There is no reason to do both. DF  r  1c  1  2  13  1  2 . So we have
The formula for the chi-squared statistic is  2 
2 
2
 2 .10

O  E 2
E
 9.13617 or  2 


or  2 

O2
 n  609 .136  600  9.136 . If we compare our results with
E
 4.6052 , we notice that our computed value exceeds the table value. Since our computed value of
chi-squared is larger than the table value, we do reject our null hypothesis.
The Minitab solution to Version 0 follows. The numbers are identical to those above.
MTB > ChiSquare c1 c2 c3.
Chi-Square Test: old, new, newer
Expected counts are printed below observed counts
Chi-Square contributions are printed below expected counts
old
new newer Total
1
75
50
18
143
59.58
59.58 23.83
3.989
1.541 1.428
2
175
200
82
457
190.42 190.42 76.17
1.248
0.482 0.447
Total
250
250
100
600
Chi-Sq = 9.135, DF = 2, P-Value = 0.010
The Minitab solution to Version 10 follows. Once again we reject the null hypothesis at the 10% level.
However, notice how much the p-value rises.
MTB > ChiSquare c1 c2 c3.
Chi-Square Test: old, new, newer
Expected counts are printed below observed counts
Chi-Square contributions are printed below expected counts
old
new newer Total
1
75
60
18
153
63.75
63.75 25.50
1.985
0.221 2.206
2
175
190
82
447
186.25 186.25 74.50
0.680
0.076 0.755
Total
250
250
100
600
Chi-Sq = 5.922, DF = 2, P-Value = 0.052
d) Follow your results in c) with a Marascuilo procedure for finding pairwise differences between the
proportions for the three designs. Assuming that there is no cost-saving in going to the newest design,
252y0762 10/30/07
would you recommend going to it? Write a paragraph long report on your conclusions from the two
hypothesis tests and what decisions these implied. (4)
[12]
I have repeated the O table for both Version 0 and Version 10. This was to give you the values of
pq
n
that are needed for the procedure.
Version 0
O
old

failed
75

didn' t  175
Sum
250
p
.30
pq
.00084
n
Design
new nwst
Total

50
18
143

200
82  457
250
100 600
.20
.18
.00064 .00148
Version 10
O
old

failed
75

didn' t  175
Sum
250
p
.30
pq
.00084
n
Design
new nwst
Total

60
18
143

190
82  457
250
100 600
.24
.18
.00073 .00148
c 1  p a q a
The general formula for the procedure is p a  p b   p a  p b    2 

 n
 a

pb qb
nb




2
We have already found  2 .10  4.6052 . There are three possible contrasts for version 0.
Old-new
Old-newest
New-newest
.30  .20  
.30  .18  
.20  .18  
4.6052 .00084  .00064   .10  .0826
4.6052 .00084  .00148   .18  .1033
4.6052 .00064  .00148   .02  .0988
The first two differences are significant since the differences between estimated proportions are larger than
the error terms. This seems to be telling us that there is no significant benefit in jumping from the new
model to the newest model.
There are three possible contrasts for version 10.
Old-new
Old-newest
New-newest
.30  .24  
.30  .18  
.24  .18  
4.6052 .00084  .00073   .06  .0850
4.6052 .00084  .00148   .18  .1033
4.6052 .00073  .00148   .06  .1008
Only the second difference is significant since the difference between estimated proportions is larger than
the error term. This seems to be telling us that there is no significant benefit in jumping from the old model
to the new model or in jumping from the new model to the newest model. If we go back to the first part of
this problem, note that the difference between old and new was significant at the 10% level, but not at the
5% level. Perhaps the problem is that we were using too low a significance level and should not have
moved to the new model, but waited until something like the newest model was available.
2. (Groebner et al) A ripsaw is cutting lumber into narrow strips and should be set to produce a product
whose width differs from the width specified by an amount given by a Normal distribution with a mean of
zero and a standard deviation of 0.01 inch. Because we have been getting complaints about the uniformness
252y0762 10/30/07
of our product, we wish to verify the Normal distribution specified is correct. We cut 600  b pieces (where
b is the last digit of your student number. Our results are as follows.
Deviation from
Number of pieces
specified width
Below -0.02
0
-0.02 to -0.01
84
-0.01 to 0
266
0 to 0.01
150 + b
0.01 to 0.02
94
0.02 and above
6
a) To use a chi-squared procedure to check the distribution, find the values of E (3)
We are going to need both the cumulative distribution and the frequency distribution for the given values of
x
x . Consider the interval -0.02 to -0.01. These numbers can be transformed by z 
into

0.02  0
0.01  0
z
 2 and z 
 1 . The cumulative distribution would then be F 2  Pz  2
0.01
0.01
 Pz  0  P2  z  0  .5  .4772  .0228 and F 1  Pz  1  Pz  0  P1  z  0
.5  .3413  .1587 . So the frequency for the -0.02 to -0.01 group is P2  z  1  F 1  F 2
 .1587  .0668  .0919 . To get E , use the formula E  fn , where f is the frequency and n is the sum of
the O column.
x interval
 to-0.02
-0.02 to -0.01
-0.01 to 0
0 to 0.01
0.01 to 0.02
0.02 and above
E 0  f 600  E10  f 610 
f
.0228
13.68
13.908
.1359
81.54
82.899
.3413
204.78
208.193
.3413
204.78
208.193
.1359
81.54
82.899
.0228
3.68
13.908
1.0000
600.00
610.000
F z 
.0228
.1587
.5000
.8413
.9772
1
z
-2.0
-1.0
0
1.0
2.0

b) State and test the null hypothesis. (2) H 0 : x is Normal . The alternate, of course is that x is not
Normal. No significance level is specified, so use 5%. Version 0 follows.
Row
1
2
3
4
5
6
O
E
0
84
266
150
94
6
600
13.68
81.54
204.78
204.78
81.54
13.68
600.00
OE
-13.68
2.46
61.22
-54.78
12.46
-7.68
0.00
O  E 2
E
13.6800
0.0742
18.3020
14.6540
1.9040
4.3116
52.9258
O2
E
0.000
85.116
339.858
122.963
106.588
2.588
652.926
O  E 2
O2
 n . Both of these two
E
E
formulas are used above. There is no reason to do both. DF  r  1  6  1  5 . So we have
The formula for the chi-squared statistic is  2 
2 

O  E 2
E
 52.9258 or  2 


or  2 

O2
 n  652 .926  600  52 .926 . If we compare our results
E
2
with  2 .05  11.0705 , we notice that our computed value exceeds the table value. Since our computed
value of chi-squared is larger than the table value, we reject our null hypothesis.
Version 10 would seem to be closer to Normal, since the data is more concentrated toward the center.
252y0762 10/30/07
Row
1
2
3
4
5
6
O
0
84
266
160
94
6
610
However,  2 

O  E 2
OE
E
13.908
82.899
208.193
208.193
82.899
13.908
610.000
O  E 2
-13.908
1.101
57.807
-48.193
11.101
-7.908
0.000
E
13.9080
0.0146
16.0507
11.1558
1.4865
4.4964
47.1122
 47.1122 or  2 
E

O2
E
0.000
85.116
339.858
122.963
106.588
2.588
657.112
O2
 n  657 .112  600  57 .112 , so that, if we
E
2
compare our results with  2 .05  11.0705 , we still reject our null hypothesis.
c) We have learned another procedure that can be used to test for a Normal distribution when the
parameters are given. Use it now to verify your results. Can you say that the saw is working as advertised?
(4) The Kolmogorov-Smirnov method can be used to test for any distribution with known parameters.
The cumulative Normal distribution was used to get the E in 2a. It is repeated here as Fe . To get the
relative frequencies O
in the observed data, divide O by its sum n and then sum down the O column
n
n
to get Fo . The absolute difference between Fo and Fe is D . Maximum values are starred.
Version 0
O
D  Fo  Fe
Fo
Fe
O
n
Row
1
2
3
4
5
6
0
84
266
150
94
6
600
0.000000
0.140000
0.443333
0.250000
0.156667
0.010000
0.00000
0.14000
0.58333
0.83333
0.99000
1.00000
0.0228
0.1587
0.5000
0.8413
0.9772
1.0000
0.0228000
0.0187000
*0.0833333
0.0079667
0.0128000
0.0000000
Version 10
Row
1
2
3
4
5
6
O
O
0
84
266
160
94
6
610
Fo
n
0.000000
0.137705
0.436066
0.262295
0.154098
0.009836
D  Fo  Fe
Fe
0.00000
0.13770
0.57377
0.83607
0.99016
1.00000
0.0228
0.1587
0.5000
0.8413
0.9772
1.0000
0.0228000
0.0209951
*0.0737705
0.0052344
0.0129639
0.0000000
The last few lines of the Kolmogorov-Smirnov table appear below.
n
  .20
  .10
  .05
  .02
  .01
37
38
39
40
Over
40
.172
.170
.168
.165
.196
.194
.191
.189
.218
.215
.213
.210
.244
.241
.238
.235
.262
.258
.255
.252
1.07
1.22
1.36
1.52
1.63
n
n
n
n
n
1.36
 .0555 . The maximum value in the column D for Version 1 is
600
0.0833333, which exceeds the critical value, so we reject the null hypothesis of Normality. The 5% value
The 5% critical value for n  600 is
252y0762 10/30/07
1.36
 .0551 and the maximum value in D is 0.0737705, so, we again reject the
610
hypothesis of Normality,
for n  610 is
[21]
3. (Groebner et al.) Two groups of 16 individuals were asked to do their income taxes using two tax
preparation software packages. The data is below (in number of minutes required) and may be considered
two independent random samples. To personalize the data add the last digit of your student number to
every number in the TT00 column. Use 10 if your number ends in 0. Label the column clearly as TT1, TT2
through TT10 according to the number used. Let d  TTa  TC .
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
TT00
65
51
74
89
88
96
37
66
86
54
60
45
42
55
58
38
TC
88
71
89
66
78
64
74
99
79
68
93
93
86
86
81
83
Minitab has given us the following results
Variable
TC
TT1
TT2
TT3
TT4
TT5
TT6
TT7
TT8
TT9
TT10
N
16
16
16
16
16
16
16
16
16
16
16
N*
0
0
0
0
0
0
0
0
0
0
0
Mean
81.13
63.75
64.75
65.75
66.75
67.75
68.75
69.75
70.75
71.75
72.75
SE Mean
2.60
4.76
4.76
4.76
4.76
4.76
4.76
4.76
4.76
4.76
4.76
StDev
10.40
19.05
19.05
19.05
19.05
19.05
19.05
19.05
19.05
19.05
19.05
Minimum
64.00
38.00
39.00
40.00
41.00
42.00
43.00
44.00
45.00
46.00
47.00
Q1
71.75
47.50
48.50
49.50
50.50
51.50
52.50
53.50
54.50
55.50
56.50
Median
82.00
60.00
61.00
62.00
63.00
64.00
65.00
66.00
67.00
68.00
69.00
Q3
88.75
84.00
85.00
86.00
87.00
88.00
89.00
90.00
91.00
92.00
93.00
Maximum
99.00
97.00
98.00
99.00
100.00
101.00
102.00
103.00
104.00
105.00
106.00
252y0762 10/30/07
a) Find the mean and standard deviation of d . (1) Assume the Normal distribution in b), c), and e).
Believe it or not here are all of them. Signs are reversed.
Row
d1
d2
d3
d4
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
-22
-19
-14
24
11
33
-36
-32
8
-13
-32
-47
-43
-30
-22
-44
-278
-21
-18
-13
25
12
34
-35
-31
9
-12
-31
-46
-42
-29
-21
-43
-262
-20
-17
-12
26
13
35
-34
-30
10
-11
-30
-45
-41
-28
-20
-42
-246
-19
-16
-11
27
14
36
-33
-29
11
-10
-29
-44
-40
-27
-19
-41
-230
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
d12
d 22
d 32
d 42
d6
d7
-18
-15
-10
28
15
37
-32
-28
12
-9
-28
-43
-39
-26
-18
-40
-214
d5
-17
-14
-9
29
16
38
-31
-27
13
-8
-27
-42
-38
-25
-17
-39
-198
-16
-13
-8
30
17
39
-30
-26
14
-7
-26
-41
-37
-24
-16
-38
-182
-15
-12
-7
31
18
40
-29
-25
15
-6
-25
-40
-36
-23
-15
-37
-166
d8
-14
-11
-6
32
19
41
-28
-24
16
-5
-24
-39
-35
-22
-14
-36
-150
d9
d 52
d 62
d 72
d 82
d 92
d 10
-13
-10
-5
33
20
42
-27
-23
17
-4
-23
-38
-34
-21
-13
-35
-134
2
d10
484
441
400
361
324
289
256
225
196
169
361
324
289
256
225
196
169
144
121
100
196
169
144
121
100
81
64
49
36
25
576
625
676
729
784
841
900
961 1024 1089
121
144
169
196
225
256
289
324
361
400
1089 1156 1225 1296 1369 1444 1521 1600 1681 1764
1296 1225 1156 1089 1024
961
900
841
784
729
1024
961
900
841
784
729
676
625
576
529
64
81
100
121
144
169
196
225
256
289
169
144
121
100
81
64
49
36
25
16
1024
961
900
841
784
729
676
625
576
529
2209 2116 2025 1936 1849 1764 1681 1600 1521 1444
1849 1764 1681 1600 1521 1444 1369 1296 1225 1156
900
841
784
729
676
625
576
529
484
441
484
441
400
361
324
289
256
225
196
169
1936 1849 1764 1681 1600 1521 1444 1369 1296 1225
13782 13242 12734 12258 11814 11402 11022 10674 10358 10074
The sample means are
Version
1
2
3
4
5
6
7
8
9
10
-17.375 -16.375 -15.375 -14.375 -13.375 -12.375 -11.375 -10.375 -9.375 -8.375
Regardless of which version you used n  16 and
s d2 
d
2
 nd 2
n 1

 
 n d1
2
2
2
 nd 2  8951 .75 ,
 596 .783 and s d  596 .783  24 .4291 .
For example for d 1 ,
d12
 d  d    d
d
1
d
 -278,
2
1
 13782, d 1 
 13782  16 17 .375   8951.75,
2
s d21
278
 -17.375,
16
d

2
1
 2
 n d1
n 1

8951 .75
 596.783,
15
b) Find out if there is a significant difference between the mean times for the two packages, using a test
ratio, a critical value or a confidence interval.(4) (2 extra points if you use all three methods and get the
same results on all three, 3 extra (extra) points if you do not assume equal variances)
Here is Table 3. If we assume equal variances, we will use D2. If we do not assume equal variances, we
will use method D3.
252y0762 10/30/07
Interval for
Confidence
Interval
Hypotheses
D2: Difference
between Two
Means (
unknown,
variances
assumed equal)
D  d  t 2 s d
H 0 : D  D0 *
D3: Difference
between Two
Means(
unknown,
variances
assumed
unequal)
D  d  t 2 s d
sd  s p
1
1

n1 n2
Test Ratio
t
H 1 : D  D0 ,
D  1   2
sˆ 2p 
d  D0
sd
Critical Value
d cv  D0  t  2 s d
n1  1s12  n2  1s22
n1  n2  2
DF  n1  n2  2
s12 s22

n1 n2
sd 
DF 
 s12 s22 
  
n

 1 n2 
t
H 1 : D  D0 ,
D  1   2
d  D0
sd
d cv  D0  t  2 s d
2
   
s12
2
n1
n1  1
x1  TC
n1  16
In Version 1
x 2  TT1
n 2  16
In Version 10
x 2  TT10 n 2  16
H 0 : D  D0 *
s 22
2
n2
n2  1
We will assume   .05 .
x1  81.13
s1  10.40
x 2  63.75
s 2  19.05
d  x1  x 2  17.38
x 2  72.75
s 2  19.05
d  x1  x 2  8.38
 H 0 : 1   2
 H 0 : 1   2  0
Regardless of our assumption about the variance, we will use 
or 
or if
 H 1 : 1   2
 H 1 : 1   2  0
H 0 : D  0
. This is a two-sided test.
D  1   2 , 
H 1 : D  0
If we assume that  12   22 we will do the following DF  n1  1  n 2  1  15  15  30  n1  n 2  2.
sˆ 2p 
n1  1s12  n2  1s 22
n1  n 4  2

. sˆ 2p  

15 10.40 2  15 19.05 2
the pooled variance. sˆ p  15.347 .   .05 so
30
t.30
025

108 .16  362 .9025
 235 .53125 . This is
2
 2.042 .
 1
1
1
1
1 
1
 2
  235 .53125     235 .53125    29.4414  5.4260 .

 sˆ 2p  
n1 n 2
 16 16 
 16 
 n1 n 2 
Recall 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
30
If this test ratio lies between  t.30
025  2.042 and t .025  2.042 , do not reject H 0 .
For Version 1 t 
d  D0
d  D0
17 .38  0
8.38  0

 1.544

 3.203 .and for Version 10 t 
sd
5.4260
sd
5.4260
Make a diagram with zero in the middle showing shaded ‘reject’ regions below  t.30
025  2.042 and
above t.30
025  2.042 . For Version 1, 3.203 falls in the upper 'reject' region, so reject H 0 . For Version 10,
1.544 does not fall in the upper 'reject' region, so do not reject H 0 .
Here is the t table for 30 degrees of freedom. {ttable}
252y0762 10/30/07
Significance Level
df .45
.40
.35
.30
.25
.20
.15
.10
.05 .025
.01 .005 .001
30 0.127 0.256 0.389 0.530 0.683 0.854 1.055 1.310 1.697 2.042 2.457 2.750 3.385
30
For Version 1, t  3.203 , which is between t .30
005  2.750 and t .001  3.385 because we double p-values for
a two-sided test, .002  p  value  .01 . Since the p-value is below   .05 , we reject the null hypothesis .
30
For Version 10, t  1.544 which is between t.30
10  1.310 and t .05  1.697. So .10  p  value  .20 and we
do not reject the null hypothesis because the p-value is above   .05 ’
Critical Value for the observed difference: The formula for a 2-sided critical value for the sample
difference between means is 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 on either side of D0  0. t .30
025  2.042 and s d  5.4260 . If d  x1  x 2 is not
between the the critical values, reject H 0 . d CV  D0  t  s d  0  2.042 5.4260   11.080 . Make a
2
diagram with 0 in the middle showing a lower shaded ‘reject’ region below -11.080 and an upper ‘reject
region above 11.080. For Version 1 d  17.38 falls in the upper ‘reject region so we reject H 0 . For
Version 10, d  8.38 does not fall in a 'reject' region, so do not reject H 0 . Sample means for the d
columns appear below, so we can see that there will be a rejection for Versions 1-7 only.
Version
1
2
3
4
5
6
7
8
Mean
17.375 16.375 15.375 14.375 13.375 12.375 11.375 10.375
9
9.375
10
8.375
Confidence Interval: The formula for a 2 sided confidence interval is D  d  t  2 s d . For Version 1, this
would be D  17.38  11.080 . Because 11.080 is less in absolute value than 17.38, this interval does not
include D0  0 and we reject the null hypothesis of equal means. For Version 10, the interval
is D  8.38  11 .080 . This interval includes zero and thus does not conflict with the null hypothesis
H 0: D  0 or 1   2 . Make a diagram with d in the middle. Represent the confidence interval by
shading the area between d  11 .80 and d  11 .80 . (The confidence interval will always include d .) If
zero is not in this area, reject H 0 .
To work without an assumption of equal variances, recall the following.   .05 , x1  TC, n1  16,
x1  81.13 and s1  10.40. In Version 1, x 2  TT1, n 2  16, x 2  63.75, s 2  19.05 and
d  x1  x 2  17.38. In Version 10, x 2  TT10, n 2  16, x 2  72.75, s 2  19.05 and d  x1  x 2  8.38.
 H 0 : 1   2
 H 0 : 1   2  0
H 0 : D  0
Our hypotheses are 
or 
or if D  1   2 , 
. This is a twoH
:



H
:




0
2
2
 1 1
 1 1
H 1 : D  0
sided test.
We can start by computing the degrees of freedom and the standard error for the difference between the
means.
s12
10 .40 2

= 6.7600
16
n1
s 22
n2
 s12 s 22 



 n1 n 2 



19 .05 2
16
= 22.6814
= 29.4414
252y0762 10/30/07
Thus s d 
DF 
s12 s 42

 29.4414  5.4260
n1 n 4
 s12 s 22 



 n1 n 2 


2
2
2

29 .4414 2
6.7600 2  22 .6814 2

866 .7960
866 .7960

=23.21.
3.0465  34 .2964
37 .3429
 s12 
 s 22 
 
 
15
15
 n1 
 n2 
 
 

n1  1
n2 1
So we, rounding down to be conservative, use 23 degrees of freedom. Here is the t table for 23 degrees of
freedom.
Significance Level
df .45 .40 .35 .30 .25 .20 .15 .10 .05 .025 .01 .005 .001
23 0.127 0.256 0.390 0.532 0.685 0.858 1.060 1.319 1.714 2.069 2.500 2.807 3.485
  .05 , so t.23
05  1.714 . x1  TC, n1  16, x1  81.13 and s1  10.40. In Version 1, x 2  TT1,
n 2  16, x 2  63.75, s 2  19.05 and d  x1  x 2  17.38. In Version 10, x 2  TT10, n 2  16,
x 2  72.75, s 2  19.05 and d  x1  x 2  8.38. s d  5.4260
Test Ratio: t 
x  x 2   10   20 
d  D0
or t  1
. If this test ratio lies between  t.23
025  2.069 and
sd
sd
t.23
025  2.069 , do not reject H 0 . For Version 1, t 
t
d  D0
17 .38  0

 3.203 . For Version 10,
sd
5.4260
d  D0
8.38  0

 1.544 . Make a diagram with zero in the middle showing shaded ‘reject’ regions
sd
5.4260
23
below  t.23
025  2.069 and above t .025  2.069 . If you are doing Version 1, 3.203 falls in the upper 'reject'
region, so reject H 0 . If you are doing Version 10, 1.544 does not fall in a reject region, so do not reject
23
 2.807 and t.23
H 0 . For Version 1, t  3.203 falls between t .005
001  3.485, so you can say that, for a one-
sided test, .001  p  value  .005 , or for a 2-sided test, .002  p  value  .01 . Since the p-value is below
23
23
  .05 , reject H 0 . For Version 10, t  1.544 falls between t .10
 1.319 and t .05
 1.714 so you can say
that, for a one-sided test, .05  p  value  .10 or for a 2-sided test, .10  p  value  .20 . Since the p-value
is above   .05 , do not reject H 0 .
Critical Value for the observed difference: The formula for a 2-sided critical value for the sample
difference between means is 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 on either side of D0  0. t .23
025  2.069 and s d  5.4260 . If d  x1  x 2 is not
between the critical values, reject H 0 . d CV  D0  t  s d  0  2.069 5.4260   11.226 . Make a
2
diagram with 0 in the middle showing a lower shaded ‘reject’ region below -11.226 and an upper ‘reject
region above 11.226. For Version 1 d  17.38 falls in the upper ‘reject region so we reject H 0 . For
Version 10, d  8.38 does not fall in a 'reject' region, so do not reject H 0 . Versions 1-7 reject H 0
Version
1
2
3
4
5
6
7
8
Mean
17.375 16.375 15.375 14.375 13.375 12.375 11.375 10.375
9
9.375
10
8.375
252y0762 10/30/07
Confidence Interval: The formula for a 2 sided confidence interval is D  d  t  2 s d . For Version 1, this
would be D  17.38  11.226 . Because 11.226 is less in absolute value than 17.38, this interval does not
include D0  0 and we reject the null hypothesis of equal means. For Version 10, the interval
is D  8.38  11 .080 . This interval includes zero and thus does not conflict with the null hypothesis
H 0: D  0 or 1   2 . Make a diagram with d in the middle. Represent the confidence interval by
shading the area between d  11.226 and d  11.226 . (The confidence interval will always include d .) If
zero is not in this area, reject H 0 .
The Minitab output follows.
MTB > TwoSample c4 c2.
Two-Sample T-Test and CI: TT1, TC
Two-sample T for TT1 vs TC
N Mean StDev SE Mean
TT1 16 63.8
19.0
4.8
TC
16 81.1
10.4
2.6
Difference = mu (TT1) - mu (TC)
Estimate for difference: -17.3750
95% CI for difference: (-28.5986, -6.1514)
T-Test of difference = 0 (vs not =): T-Value = -3.20
DF = 23
P-Value = 0.004
MTB > TwoSample c13 c2.
Two-Sample T-Test and CI: TT0, TC
Two-sample T for TT0 vs TC
N Mean StDev SE Mean
TT0 16 72.8
19.0
4.8
TC
16 81.1
10.4
2.6
Difference = mu (TT0) - mu (TC)
Estimate for difference: -8.37500
95% CI for difference: (-19.59858, 2.84858)
T-Test of difference = 0 (vs not =): T-Value = -1.54
DF = 23
P-Value = 0.136
c) Test the variances of the two samples for equality on the assumption that they come from the Normal
distribution. (2)
From the formula table we have the following.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Ratio of Variances  22 s22 DF , DF
s2
H0 : 12   22
 2 F.5  1.5  2 
F DF1 , DF2  12
2
2
1
s1
s
H : 2   2
1 , DF2
F1DF


2
1
DF1 , DF2
F
2
DF1  n1  1
DF2  n 2  1
 2

.5  .5   2    or
1  
2

1
1
2
2
and
F DF2 , DF1 
s 22
s12
 H 0 :  12   22
We are testing 
by using a variance ratio.   .05 . x1  TC, n1  16, so n1  1  15 and
 H 1 :  12   22
s1  10.40. In Versions 1-10, x 2  TT1, n 2  16, so n2  1  15 and s 2  19.05.
252y0762 10/30/07
To do a 5% two sided test we must compare
s12
s 22
and
15,15 , but it must be between F 14,15  2.89 and
F.025
.025
s 22
s12
s 22
15,15 . Our F table does not have
against F.025
s12
16,15
F.025
 2.84 . The larger of the two ratios is
2
 19 .05 
15,15

  3.36 . Since this is larger than F.025 , we reject the null hypothesis of equal variances
10
.
40


d) Test the d column to see if the data was Normally distributed (5)
 H 0 : Normal

 H 1 : not Normal
Fortunately for me, all of your versions of d have identical standard deviations and, when the means are
subtracted are identical. If you subtract the mean in any version you get the d1  d1 column shown below.
I used the d 1 column, first putting it in order as d1 ord . I then subtracted a mean of -17.88 and divided by
the standard deviation of 24.43 to get the column marked z unrounded . Because I wanted to look up all
the values of z in the Normal table, I rounded my values of z unrounded to only reveal two places to the
right of the decimal point. (In the past I have used Minitab to compute the cumulative distributions, but I
wanted to see how long it took using the table. It went pretty fast, but the results are less accurate than
Minitab computations.) This is tabulated as " z" - your values of z should be substantially identical. I now
computed the expected cumulative distribution Fe , by getting Pz " z" using the Normal table. For
example, Pz  1.21  Pz  0  P1.21  z  0  .5  .3869  .1131 or Pz  2.06 
 Pz  0  P0  z  2.06   .5  .4803  .9803 . Since we have the equivalent of 16 groups, each with 1
member, each group has a frequency of 116 , so I put the numbers 1 through 16 in the Fo column and
divided by 16. The absolute value of the difference between the two cumulative columns is in the D
column. The largest value is starred.
Row
d1  d1
d1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
-4.62
-1.62
3.38
41.38
28.38
50.38
-18.62
-14.62
25.38
4.38
-14.62
-29.62
-25.62
-12.62
-4.62
-26.62
-22
-19
-14
24
11
33
-36
-32
8
-13
-32
-47
-43
-30
-22
-44
d1 ord z unrounded " z"
-47
-44
-43
-36
-32
-32
-30
-22
-22
-19
-14
-13
8
11
24
33
-1.21244
-1.08964
-1.04871
-0.76218
-0.59844
-0.59844
-0.51658
-0.18911
-0.18911
-0.06631
0.13835
0.17929
1.03889
1.16169
1.69382
2.06222
-1.21
-1.09
-1.05
-0.76
-0.60
-0.60
-0.52
-0.19
-0.19
-0.07
0.14
0.18
1.04
1.16
1.69
2.06
Fe
0.1131
0.1379
0.1469
0.2236
0.2743
0.2743
0.3015
0.4247
0.4247
0.4721
0.5557
0.5714
0.8508
0.8770
0.9545
0.9803
Fo
0.0625
0.1250
0.1875
0.2500
0.3125
0.3750
0.4375
0.5000
0.5625
0.6250
0.6875
0.7500
0.8125
0.8750
0.9375
1.0000
D  Fo  Fe
0.0506
0.0129
0.0406
0.0264
0.0382
0.1007
0.1360
0.0753
0.1378
0.1529
0.1318
*0.1786
0.0383
0.0020
0.0170
0.0197
An excerpt from the Lilliefors table is shown below. The maximum is .1786, which is below the
5% critical value of .213, so we cannot reject the null hypothesis of Normality.
n
16
  .20
.173
  .15
.182
  .10
  .05
  .01
.195
.213
.250
252y0762 10/30/07
e) Actually the data above was taken from only 16 people, who were randomly assigned to use one of the
methods first. Would this mean that what you did above was correct? If not do b) over again. (3)
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
d  x1  x 2
Means (paired
D




1
2
data.)
s
df  n  1 where
sd  d
n1  n 2  n
n
We are testing
H 0 : 1   2
H 1 : 1   2
or
H0 : D  0
H1 : D  0
where D  1   2 .
Regardless of which version you used n  16 and s d  596 .783  24 .4291 ,
sd 
sd

n
s d2
596 .783

 37 .2989  6.1073 . Values of d appear below.
n
16
Version
1
2
3
4
5
6
7
8
Mean
17.375 16.375 15.375 14.375 13.375 12.375 11.375 10.375
9
9.375
10
8.375
15
If we assume that the significance level is   .05 , use t n1  t .025
 2.131 . We can use one of the three
2
methods.
Test Ratio: t 
x  x 2   10   20 
d  D0
15
or t  1
. If this test ratio lies between  t .025
 2.131 , do
sd
sd
not reject H 0 .
For Version 1 t 
d  D0
8.375  0
17 .375  0

 2.8450 . For Version 10 t 
6.1073
sd
6.1073
=1.3713. Make a diagram with zero in the middle showing shaded ‘reject’ regions below -2.131 and
above 2.131. For version 1, the computed t falls in the rejection zone and we reject H 0 , while for Version
10 the computed t does not fall in the 'reject' region and we do not reject H 0 . Or you can look at the 15
line of the t-table, which reads as below. {ttable}
df .45 .40 .35 .30 .25 .20 .15 .10 .05 .025 .01 .005 .001
15 0.128 0.258 0.393 0.536 0.691 0.866 1.074 1.341 1.753 2.131 2.602 2.947 3.733
15
15
 2.947 , for a one 2.602 and t .005
For version 1, we can say that, since t  2.8450 falls between t .01
sided test, ..005  p  value  .01 . But this is a two-sided test so that .01  p  value  .02 Since the pvalue is below   .05 , reject H 0 . For version 10, we can say that, since t  1.3713 falls between
15
15
t .10
 1.341 and t .05
 1.753 , for a two-sided test, .20  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  2.1316.1073   13 .015 . Make a diagram with 0 in the
middle showing shaded ‘reject’ regions below -13.015 and above 13.015.
Version
1
2
3
4
5
6
7
8
Mean
17.375 16.375 15.375 14.375 13.375 12.375 11.375 10.375
d falls in the 'reject' region for Versions 1-5. Reject H 0 for these versions.
9
9.375
10
8.375
252y0762 10/30/07
Confidence Interval: D  d  t  2 s d . We already know that t  s d  2.1316.1073  13.015 and d  0.5
2
so we can say D  d  13.015 or Make a diagram with d in the middle. Represent the confidence
interval by shading the area between D  d  13.015 and D  d  13.015 . The intervals for Versions 1-5
will not include zero and we will not reject H 0 .
f) In view of the fact that the data was taken from only 16 people and dropping the assumption of
Normality, find out if there is a significant difference between the medians of the two packages. (3) This is
 H :   2
a two sided Wilcoxon Rank Sum test.  0 1
n  11 . I have copied out four of the d s. Of course,
 H 1 : 1   2
signs are still reversed, but this should not create problems. d 1 is the original data, d 1 is the absolute
value of the original column, r1 is a ranking from 1 to 16, signed r1 is the ranking revised to give the
averaged rank to tied items and to restore the signs.
Row
signed r5
d1
r1
signed r1
d5
r5
d5
d1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
-22
-19
-14
24
11
33
-36
-32
8
-13
-32
-47
-43
-30
-22
-44
22
19
14
24
11
33
36
32
8
13
32
47
43
30
22
44
6.0
5.0
4.0
8.0
2.0
12.0
13.0
10.0
1.0
3.0
11.0
16.0
14.0
9.0
7.0
15.0
-6.5
-5.0
-4.0
8.0
2.0
12.0
-13.0
-10.5
1.0
-3.0
-10.5
-16.0
-14.0
-9.0
-6.5
-15.0
Version 1 totals T   23 T   113
Row
d7
d7
r7
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
-16
-13
-8
30
17
39
-30
-26
14
-7
-26
-41
-37
-24
-16
-38
16
13
8
30
17
39
30
26
14
7
26
41
37
24
16
38
5.0
3.0
2.0
11.0
7.0
15.0
12.0
9.0
4.0
1.0
10.0
16.0
13.0
8.0
6.0
14.0
-18
-15
-10
28
15
37
-32
-28
12
-9
-28
-43
-39
-26
-18
-40
18
15
10
28
15
37
32
28
12
9
28
43
39
26
18
40
6.0
4.0
2.0
10.0
5.0
13.0
12.0
10.0
3.0
1.0
10.0
16.0
14.0
8.0
7.0
15.0
-6.5
-4.5
-2.0
10.0
4.5
13.0
-12.0
-10.0
3.0
-1.0
-10.0
-16.0
-14.0
-8.0
-6.5
-15.0
Version 5 totals T   30.5 T   105 .5
signed r7
d0
-5.5
-3.0
-2.0
11.5
7.0
15.0
-11.5
-9.5
4.0
-1.0
-9.5
-16.0
-13.0
-8.0
-5.5
-14.0
-13
-10
-5
33
20
42
-27
-23
17
-4
-23
-38
-34
-21
-13
-35
Version 7 totals T   37.5 T   98.5
d0
13
10
5
33
20
42
27
23
17
4
23
38
34
21
13
35
r0
4.0
3.0
2.0
12.0
7.0
16.0
11.0
9.0
6.0
1.0
10.0
15.0
13.0
8.0
5.0
14.0
signed r0
-4.5
-3.0
-2.0
12.0
7.0
16.0
-11.0
-9.5
6.0
-1.0
-9.5
-15.0
-13.0
-8.0
-4.5
-14.0
Version 10 totals T   41 T   95
If we add together the numbers in signed r1 with a + sign, we get T  . If we do the same for numbers with
a – sign, we get T  . To check this, note that these two numbers must sum to the sum of the first n
nn  1 16 17 

 136 , and that all T    T  do add to 136.
numbers, and that this is
2
2
252y0762 10/30/07
Version 1 totals T   23 T   113
Version 5 totals T   30.5 T   105 .5
Version 7 totals T   37.5 T   98.5
Version 10 totals T   41 T   95
We check T  , which is the smaller of the two rank sums in every case against the numbers in table 7.
{wsignedr} For a two-sided 5% test, we use the   .025 column. For n  16 , the critical value is 30, and
we reject the null hypothesis only if our test statistic is below this critical value. Since our test statistic in
Version 1 is 23, we reject the null hypothesis of equal medians in that case alone (though I suspect that
Versions 2 through 4 will also yield rejections.)
g) If you did d) e) and f), can you report on your results, indicating which result was correct? (1) [40]
Because we now know that the data was paired and we cannot reject the hypothesis of Normality, section
d) of this question gives the correct result. If you did Versions 1-5 there is a clear difference and the TT
package is a winner. For Versions 6-10, there is no significant difference between the packages.
Be prepared to turn in your Minitab output for the first computer problem and to answer the
questions on the problem sheet about it or a similar problem.
252y0762 10/30/07
4. (Extra Credit) Check your work on Minitab. Remind me that you did extra credit on your front page.
For a Chi-squared test of Independence or Homogeneity, put your observed data in adjoining columns.
Use the Stat pull-down menu. Choose Tables and then Chi Squared Test. Your output will show O and E
as a single table. You will be given a p-value for the hypothesis of Independence or Homogeneity.
For a test of Normality, when sample mean and variance are to be computed from the sample, put your
complete set of numbers in one column. . Use the Stat pull-down menu. Choose Basic Statistics and then
Normality test. Check Kolmogorov-Smirnov to get a Lilliefors test. You will be given a p-value for the
hypothesis of Normality.
For a Chi-squared test of goodness of fit, put your observed data in C1 and your expected data or
frequencies in C2. The expected data may be proportions adding to 1 or counts adding to n . Use the Stat
pull-down menu. Choose Tables and then Chi Squared Test of Goodness of Fit. Pick specific proportions or
historic counts. Observed counts is C1 and the other column requested will be C2. The computed degrees
of freedom will have to be reduced if you computed any statistics from the data before setting up the
expected count or frequency. You are warned not to use expected counts below 5.
For a test of Two Proportions, Use the Stat pull-down menu. Choose Basic Statistics and then Two
Proportions. Check Summarized Data and then enter n1 , x1 , n 2 and x 2 . Use Options to set the inequality
in the alternate hypotheses and check Pooled Estimate unless you are doing a confidence interval.
To fake computation of a sample variance or standard deviation of the data in column c1 using
column c2 for the squares,
MTB
MTB
MTB
MTB
MTB
MTB
>
>
>
>
>
>
let C2 = C1*C1
name k1 'sum'
name k2 'sumsq'
let k1 = sum(c1)
let k2 = sum(c2)
print k1 k2
Data Display
sum
sumsq
MTB
MTB
MTB
MTB
>
>
>
>
3047.24
468657
* performs multiplication
** would do a power, but multiplication
is more accurate.
This is equivalent to let k2 = ssq(c1)
This is a progress report for my data
set.
name k1 'meanx'
let k1 = k1/count(c1)
/means division. Count gives n.
let k2 = k2 - (count(c1))*k1*k1
print k1 k2
Data Display
meanx
sumsq
152.362
4372.53
MTB > name k2 'varx'
MTB > let k2 = k2/((count(c1))-1)
MTB > print k1 k2
Data Display
meanx
varx
152.362
230.133
MTB > name k2 'stdevx'
MTB > let k2 = sqrt(k2)
MTB > print k1 k2
Data Display
meanx
stdevx
152.362
15.1701
Print C1, C2
Sqrt gives a square root.
252y0762 10/30/07
To check your mean and standard deviation, use
`
MTB > describe C1
To check for equal variances for data in C1 and C2, use
MTB > VarTest c1 c2;
SUBC>
Unstacked.
Both an F test and a Levine test will be run.
To put a items in column C1 in order in column C2, use
MTB > Sort c1 c2;
SUBC>
By c1.
Commands like Count C1, Sum C1 and SSq C1 can be used alone without Let if the values don’t need to
be stored. In the above I have continuously named and renamed the constants k1 and k2. There are many
constants in Minitab on an invisible worksheet. (k1 …….k100 at least), so you can preserve your results by
using separate locations for subsequent computations.