COURSE: DSCI 3710
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Exam 1 Version A
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Spring 2015
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INSTRUCTIONS:
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Please print your name and student ID number on this exam. Also, put your
signature on this exam.
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On your scantron PRINT your name and exam version.

This exam has 25 questions. You have 75 minutes to complete this exam. The
exam is open book and open notes. You may use a laptop computer or any type
of hand calculator but please show all your work on the exam and mark all
answers on the scantron. Usage of cell phones, digital cameras and other
communication devices is prohibited.

Please DO NOT pull this exam apart. When you have completed the exam, please
turn your scantron and exam booklet into your instructor, at the front desk.

Good luck and we wish you well on the exam.
Note: Whenever question(s) are connected you may be asked to assume a result (given
a value) as an answer for the previous question but this result (value) may or may not
be correct. The procedure is set in place to prevent you from losing points on a
subsequent question because you made a mistake on some previous question/s.
Use the information given in the following paragraph to answer the first four questions.
260 people are randomly chosen at a shopping mall to taste-test a new brand of fruit drink. They
are asked to rate the drink on a scale from 1 to 7, with 1 being very bad and 7 being very good.
The results of the survey reveal that the average rating is 5.22 with a standard deviation of 2.31.
The marketing division of the fruit drink distributor is only interested in selling this drink if the
true mean rating is more than 4.75.
1. What is the alternative hypothesis for testing whether the fruit drink distributor should sell
this drink?
A. Ha: µ ≠ 4.75
B. Ha: µ = 2.25
C. Ha: µ > 4.75
D. Ha: µ < 4.75
E. Ha: µ > 2.25
2. What is the calculated value of the test statistic?
A. 3.28
B. 2.86
C. 4.24
D. 2.86
E. 1.65
3. What is the rejection region for testing at the 0.10 level of significance whether the fruit
drink distributor should sell this drink?
A. Z > 1.645
B. Z > 1.96
C. Z < -1.28
D. Z < -1.96
E. Z > 1.28
4. Assuming the calculated value of the test statistic is 1.78, what is the conclusion of testing at
the .10 level of significance whether the fruit drink distributor should sell this drink?
A. Based on the sample data, there isn't sufficient evidence to conclude that the average
rating is more than 4.75.
B. Based on the sample data, there isn't sufficient evidence to conclude that the average
rating is no more than 4.75.
C. Based on the sample data, there is sufficient evidence to conclude that the average
rating is no more than 4.75.
D. Based on the sample data, there is sufficient evidence to conclude that the average
rating is more than 4.75.
E. None of the above.
Use the information given in the next paragraph to answer the next four questions.
The table below shows the test scores of 10 employees who participated in a new corporate
training program that lasted one week. The scores shown are those before and after completing
the program, for each employee. Test whether there is significant improvement in the scores after
the training. Excel analysis at the 5% significance level is shown.
Employee
1
2
3
4
5
6
7
8
9
10
Before
(1)
98
75
64
86
xx
52
61
74
63
51
After
(2)
95
81
75
89
yy
59
68
79
64
70
t-Test: Paired Two Sample for Means
Mean
Variance
Observations
Pearson Correlation
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Before
After
69.7
75.6
215.57 121.82
10
10
0.931
0
9
-3.1246
0.0061
1.8331
0.0122
2.2622
5. What is the alternative hypothesis for testing the belief that there was an increase in the
mean test scores after the training program?
A. Ha: µ2 > 0
B. Ha: µ1 > µ2
C. Ha: µ2 > µ1
D. Ha: µ2 = µ1
E. Ha: µ1 ≠ µ2
6. What is the table value (at the 5 % level) of the appropriate critical test statistic to test the
belief that there is an increase in the mean test scores of employees who go through the
training program?
A. 1.833
B. 0.000
C. -3.1246
D. 2.2622
E. 0.0122
7. What is the p-value for testing the hypothesis that there is an increase in the mean test scores
of employees who participated in the training program?
A. -3.1246
B. 0.0061
C. 0.000
D. 2.2622
E. 0.0122
8. Based on the excel output and the significance level of 5%. What would be your
conclusion about whether the training program increased the mean test scores?
A. Conclude that there is insufficient evidence that the training program increased the
mean test scores since p is less than the significance level
B. Conclude that there is sufficient evidence that the training program increased the
mean test scores since p-value is less than the significance level.
C. Conclude that there is insufficient evidence that the training program increased the
mean test scores since p is more than the significance level.
D. Conclude that there is sufficient evidence that the training program increased the
mean test scores since p-value is more than the significance level.
E. none of the above
Please use the information given in the paragraph below to answer the four questions that
follow.
A university librarian at a university with a large multinational student body is interested in the
proportion of students who would like the library search functions available in a language other
than English. If over half of the student population favors the Multilanguage search option, the
university administration will approve the cost of the new service. A random sample of 350
students was obtained and 186 students indicated they would like the new services. A hypothesis
test was conducted using the KPK macro-based MS Excel output provided below to determine if
a majority of students favor the new hours. A significance level of 5% was used.
Z Test for One Proportion
Sample Proportion
Number of Observations
Ho:XXXX
Z*
P[Z  Z*]
Z Critical,  = 0.05
95% CI for Pop. Proportion
0.531429
350
Ha:XXXX
1.175949
XXXX
XXXX
0.479075
to
0.583782
9. What are the null and alternative hypotheses for this test?
A. Ho: p = 0.5
Ha: p ≠ 0.5
B. Ho: p < 0.5
Ha: p > 0.5
C. Ho:  < 0.5
Ha:  > 0.5
D. Ho: p > 0.5
Ha: p < 0.5
E. Ho:  < 0.05
Ha:  > 0.05
10. What is the critical value of the test statistic?
A. -1.645
B. 2.33
C. 0.479
D. 1.645
E. -2.33
11. What is the p-value associated with this test?
A. 0.1986
B. 0.2180
C. 0.1198
D. 0.2396
E. 0.0599
12. Assume the calculated test statistic is 1.36. What conclusion can be drawn?
A. Fail to reject Ho. There is evidence that supports the cost of the new service.
B. Reject Ho. There is evidence that supports the cost of the new service.
C. Reject Ho. The increased service is not supported.
D. Reject Ha. There is evidence that supports the cost of the new service.
E. Fail to reject Ho. There is insufficient evidence to support the cost of the new service.
Use the following information to answer the next four questions.
A tax preparation firm is interested in comparing the quality of work at its two different office
locations. To test if there is a difference, the firm randomly selected samples of tax returns that
were prepared at each office and checked them for accuracy. In the Denton (Variable 1) office
they found that 40 of 450 returns contained some error(s) and at the Flower Mound (Variable 2)
office they found that 60 of 540 returns contained some error(s). MS Excel output provided
below to determine if there is a difference between the qualities of work (as gauged by the mean
proportion of errors) at these two offices. The significance level has been used for this test is
0.01.
Z Test for Two Proportions
Sample Proportion
Number of Observations
Ho:XXXX
Z*
2 * P[Z Z*] two tail
 Z Critical ,  = 0.01
99% CI for p1 - p2
Variable 1
Variable 2
0.088889
450
Ha:XXXX
-1.155349
XXXX
YYYYY
-0.071340
0.111111
540
to
0.026895
13. What is the alternate hypothesis?
A. p1 < p2
B. p1 = p2
C. p1 ≠ p2
D. p1 < p2
E. p1 > p2
14. What is the critical (Table) value of the test statistic?
A. ±1.645 B. ±1.96
C. ±1.28
D. ±2.33
E. ±2.58
15. What is the p-value associated with the calculated value of the test statistic?
A. 0.0012
B. 0.2366
C. 0.2479
D. 0.1240
E. 0.0620
16. What conclusion can be drawn?
A. Reject Ha and conclude that the two offices make about the same number of errors.
B. Reject Ho and conclude that the two offices make a different number of errors.
C. Fail to reject Ho and conclude that the two offices do not make a significantly different
number of errors.
D. Reject Ho and conclude that the two offices make about the same number of errors.
E. None of the above.
Use the information given in the paragraph below to answer the next five questions.
The manager of a Pizza Delivery service recorded the delivery times (in minutes) for 11
randomly selected orders for each of four drivers. The manager claims there are significant
differences in the true mean delivery times among the 4 drivers. Test his belief at the 0.05
significance level, based on the results presented in the ANOVA Table shown below.
Anova: Single Factor
SUMMARY
Groups
Driver1
Driver2
Driver3
Driver4
Count
11
11
11
11
Sum
231.000
229.000
205.000
223.000
Average
21.000
20.818
18.636
20.273
Variance
70.600
82.964
42.255
47.618
ANOVA
Source of
Variation
Between Groups
Within Groups
SS
38.182
2434.364
df
3
40
MS
12.727
60.859
F
XXXX
Total
2472.545
43
TUKEY MULTIPLE COMPARISON
TEST
Critical Q
Distance Alpha
3.79
8.915
0.05
Means joined by a double line are not significantly different.
Driver3
18.636
Driver4
20.273
Driver2
20.818
Driver1
21.000
P-value
0.889
F crit
XXXX
17. Which of the following is the correct alternative hypothesis?
A. Ha: the mean time to deliver orders is the same for all four drivers.
B. Ha: not all the drivers have the same mean time to deliver.
C. Ha: the mean time to deliver orders for each driver is different from the others.
D. Ha: the mean time to deliver orders is the same for drivers 1 and 2.
E. Ha: the mean time to deliver orders is the same among order destinations.
18. What is the calculated value of the test statistic to test the claim that there is difference in the
mean delivery times for the four drivers?
A. 0.209
B. 8.369
C. 0.169
D. 1.245
E. 2.365
19. What is the critical (table) value of the test statistic to test the manager’s hypothesis?
A. 0.18
B. 2.11
C. 2.96
D. 3.28
E. 2.84
20. Assuming the calculated value of the test statistic is 2.68, what is the p-value of the test?
A. p > 0.1
B. 0.05 < p < 0.10
C. 0.025 < p < 0.05
D. 0.01 <p < 0.025
E. p < 0.01
21. Based on the excel output given above, what are the statistical decision and reason for the
decision concerning the manager’s hypothesis?
A. Fail to reject Ho because the p-value of the test is larger than α.
B. Reject Ho because the p-value of the test is larger than the F critical (table) value.
C. Reject Ho because the p-value of the test is larger than α.
D. Fail to reject Ho because the p-value of the test is smaller than α.
E. Reject Ho because the p-value of the test is smaller than α.
Use the information given on the next page to answer the next four questions.
The personnel manager of Star Inc. wishes to determine if the number of sick days reported by
each employee in 2013 at its Dallas plant was more than the number of sick days reported in
2013 at its Austin plant. Random samples of employees in Dallas and Austin are selected, and
the following results are obtained with Excel assuming equal variances and a significance level
of 0.01.
employees Dallas
Austin
1
9.2
8.4
2
8.5
8.2
3
7.6
7.6
4
6.6
5.1
5
x.y
y.x
6
5.9
5.7
7
8.4
7.5
8
7.6
8.9
9
6.3
9.1
10
7.1
5.3
11
5.4
6.1
12
5.8
7.9
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Dallas Austin
7.292 7.442
1.730 2.341
12
12
2.035
0
22
-0.258
XXXX
2.508
XXXX
2.819
22. What is the null hypothesis for this test?
A. H0: dallas - austin > 0.15
A. H0: dallas > austin
B. H0: dallas = austin
C. H0: dallas < austin
D. H0: dallas  austin
23. What is the (absolute) critical value of the test statistic?
A. -0.934
B. 3.268
C. 2.508
D. 0.00
E. 1.689
24. Which one of the following would best describe the p-value of the test?
B. 0.001 < p < 0.01
C. 0.01 < p < 0.025
D. 0.025 < p < 0.05
E. 0.05 < p < 0.1
F. p > 0.1
25. Assuming the variance in the number of sick days reported for all employees at the Dallas
plant is equal to that for all employees at the Austin plant, what is the calculated value of the
test statistic for testing whether the number of sick days reported by each employee in 2013
at its Dallas plant was more than the number of sick days reported in 2013 at its Austin plant?
A. -0.258
B. -2.508
C. 2.508
D. 0.00
E. 2.819
Thank you