252x0763 11/03/07
Corrected
ECO252 QBA2
SECOND EXAM
Nov 1-5 2007
Name
Class________________________
Student Number_______________
Show your work! Make Diagrams! Exam is normed on 50 points. Answers without reasons are not
usually acceptable.
x ~ N 22, 18  - If you are not using the supplement table, make sure that I know it.
1. P0  x  20.1
2. P42  x  42 
3. Px  24 
4.
x.045 (Do not try to use the t table to get this.)
252x0763 11/03/07
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
33
43
23
29
33
37
26
Women
x2
Difference
d
33
33
26
41
33
48
44
0
10
-3
-12
0
-11
-18
Minitab computes the following.
Variable
x1
x2
d
N
7
7
7
Mean
32.00
36.86
-4.86
SE Mean
2.55
2.91
StDev
6.76
7.69
Minimum
23.00
26.00
-18.00
Q1
26.00
33.00
-12.00
Median
33.00
33.00
-3.00
Q3
37.00
44.00
0.00
Maximum
43.00
48.00
10.00
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 a 91% 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]
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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.
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 13 non-accounting students and 10 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)
c)
d)
e)
f)
g)
h)
s12
s 22
s12
s 22
s12
s 22
s 22
s12
s 22
s12
s 22
s12
s 22
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
12,9   _______
2. F.05
250,150 is, at most, ______. If you did not get this from the Supplementary Tables, you must explain how
F.05
you found this.
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Exhibit 1:
Sample size
Married
25
Unmarried
30
Standard error 24.2534
Mean Std Deviation
268.90
77.25
455.10 102.40
d  D 0  186 .20

 7.6773
sd
24 .2534
Difference between means -186.20
(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.
4. If the researcher is trying to show that unmarried cardholders will increase their spending more than
married 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
5. If the researcher in exhibit 1 is attempting to show that unmarried cardholders will spend significantly
more than married 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
6. If the researcher does not believe that the population standard deviation 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 .)
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7. I am testing the hypothesis H 0 :   300 . I get a value of x  162 , which results in a p-value of .076.
What are the p-values for H 0 :   300 and 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]
8. A 2007 survey (Baltic Surveys for the International Republican Institute) of a sample of 1062 opinions in
Moldova asked what countries are the greatest social and economic threats to Moldova. The data presented
said that 23% (244 people) said that Russia was one of those countries and that 25% (265) that the US was
one.
a) If 15% (159 people) said that both of the countries are in this threatening group, and country 1 is Russia
and country 2 is the US, and we wish to test the statement that the US is significantly less popular than
Russia in Moldova, what would our null and alternative hypotheses be? Answer in terms of 1 ,  2 and
D  1   2 , or  1 and  2 or p1 , p 2 and p  p1  p 2 as appropriate. (Example if  1 ,  2 and F 
1
2
were a reasonable answer, you might get two points for saying H 0 :  1   2 , H 1 :  1   2 and wrongly
saying that this becomes H 1 : F  0 and the corresponding H 0 : F  0 ) (3)
b) Test this hypothesis using the data above (4) (Note that it is possible to get a) right and still choose the
wrong method, if this happens there will be partial credit.)
[21]
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9. A 2007 survey (Baltic Surveys for the International Republican Institute) of a sample of 1062 opinions in
Moldova asked if a family member was working abroad. 62% said yes. A similar survey in 2004 found
64% saying yes to the same question. If we assume that the earlier survey included 1000 people and that
the two surveys are independent random samples, can we say that there has been a significant change in the
number working abroad? (State and test your null and alternate hypotheses.) (5)
[28]
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10. Computer question.
a. Turn in your first computer output. Only do b, c and d if you did. (3)
b. (Groebner) A rental car company has just stared to rent SUVs to individuals. It has long
experience with automobiles and knows that highway miles per gallon exceeds mpg in cities. To
verify this it takes a random sample of 25 SUVs driven on the highway (sample 1) and a second
sample of 25 SUVs driven in the city (sample 2). Actually this was done by giving each renter 14
gallons and having them drive until they ran out of gas. At this point they were picked up and
given a coupon good for another week’s rental. Excel computation from these data said that for
highway driving, the sample mean was 19.6468 mpg and the sample standard deviation was
18.4637. Corresponding figures for city driving were 16.6064 and 29.6064. 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 highway mpg was higher than city mpg? (2)
e. On the basis of run C, could we have made things easier by assuming equal variances? Why?
(1)
[36]
MTB > TwoT 25 19.6468 18.4637 25 16.6064 29.6064;
SUBC>
Alternative -1.
A) Two-Sample T-Test and CI
Sample
N Mean StDev SE Mean
1
25 19.6
18.5
3.7
2
25 16.6
29.6
5.9
Difference = mu (1) - mu (2)
Estimate for difference: 3.04040
95% upper bound for difference: 14.79096
T-Test of difference = 0 (vs <): T-Value = 0.44
P-Value = 0.667
DF = 40
P-Value = 0.333
DF = 40
MTB > TwoT 25 19.6468 18.4637 25 16.6064 29.6064;
SUBC>
Alternative 1.
B) Two-Sample T-Test and CI
Sample
N Mean StDev SE Mean
1
25 19.6
18.5
3.7
2
25 16.6
29.6
5.9
Difference = mu (1) - mu (2)
Estimate for difference: 3.04040
95% lower bound for difference: -8.71016
T-Test of difference = 0 (vs >): T-Value = 0.44
MTB > VarTest 15 18.4637 15 29.6064.
C) Test for Equal Variances
95% Bonferroni confidence intervals for standard deviations
Sample
N
Lower
StDev
Upper
1 15 3.01576 4.29694 7.28309
2 15 3.81883 5.44118 9.22251
F-Test (normal distribution)
Test statistic = 0.62, p-value = 0.388
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Blank
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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 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
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)
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]
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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
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.
252x0763 11/03/07
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.
252x0763 11/03/07
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.