QMB 3250 Spring 2010 Exam 1 Form A Feb 10, 2010
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* There are 18 questions on this exam. Most are worth 5 points, but there is a 2-point and a 3-point question. The maximum score on the exam is 85 points.
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SIMPLIFIED T-TABLE
Use this for any t-values you need to look up on the exam.
(If your calculator does t-values, they may be different)
tail area: α or α/2df / 0.1 / 0.05 / 0.025 / 0.01 / 0.005
1 to 5 / 1.638 / 2.353 / 3.182 / 4.541 / 5.841
6 to 10 / 1.397 / 1.860 / 2.306 / 2.896 / 3.355
11 to 15 / 1.363 / 1.796 / 2.201 / 2.718 / 3.106
16 to 20 / 1.337 / 1.746 / 2.120 / 2.583 / 2.921
21 to 25 / 1.323 / 1.721 / 2.080 / 2.518 / 2.831
26 to 30 / 1.315 / 1.706 / 2.056 / 2.479 / 2.779
31 to 40 / 1.309 / 1.696 / 2.041 / 2.454 / 2.749
41 to 60 / 1.302 / 1.683 / 2.020 / 2.422 / 2.703
61 to 100 / 1.295 / 1.670 / 2.000 / 2.389 / 2.659
100 - up / 1.282 / 1.645 / 1.960 / 2.326 / 2.576
1.Suppose you do an outlier check on the CARS scores for students that are Marketingmajors. What is the upperfenceyou would use?
IQR=q3-q1=64-57=7 Upper fence = Q3+1.5IQR = 64 + 10.5 = 74.5
73.5 76.5 79.5 82.5
Choose one answer: <------+------+------+------+------>
A B C D E
2.In this outlier check for Marketingmajors, how many points would be considered potential outliers?
Lower fence is 57-10.5=46.5 so three high outlier (75 77 80), one low (40)
A. 0 B. 1 C. 2 D. 3 E. 4 or more
3.Suppose we are constructing a histogram for the CARS scores of all 91 students in the analysis regardless of major.According to the guidelines in our textbook which of the following would be a correct way to set up the categories (or “Bins”)?
Span has to be from 32 to 96 (64 total). This would require K=7 intervals of at least w=9.143; K=8 intervals of at least w=8.0
Choice / # Bins / Bin Width / First Bin / Second Bin / Last Bin / Why wrongA / 7 / 9 / 30.0 to 38.9 / 39.0 to 47.9 / 84.0 to 92.9 / Miss the 96
B / 7 / 9 / 32.0 to 40.9 / 41.0 to 49.9 / 86.0 to 94.9 / Miss the 96
C / 7 / 10 / 30.0 to 39.9 / 40.0 to 49.9 / 90.0 to 99.9 / WORKS OK!
D / 8 / 10 / 30.0 to 39.9 / 40.0 to 49.9 / 100.0 to 100.9 / Last bin empty
E / 8 / 8 / 30.0 to 37.9 / 38.0 to 45.9 / 86.0 to 93.9 / Miss the 96
4.Suppose you are using the sample results for Marketing majors to make an approximately 95% interval estimate for the population median CARS score. Using the “.4n-2” rule developed by Walsh, what is the lowerendpoint of this interval?
R=.4(26)-2 = 8.4 so use the 8th smallest (58) and 8th largest (64)
53.5 55.5 57.5 59.5
Choose one answer: <------+------+------+------+------>
A B C D E
Our text book lists many confidence interval estimation procedures as:
(ESTIMATE) ME where the margin of error (ME) has the form se
with coming from some statistical table and
se being the standard error of the estimate.
5.If you are going to construct a 95% interval estimate for the average CARS score for all Marketingmajors, what is the margin of error ME?
With n=26 use 25 df. The multiplier is t=2.080
ME = 2.080*S/sqrt(26) = 2.080*8.021/5.099 = 3.27
3.15 3.45 3.75 4.05
Choose one: <------+------+------+------+------>
A B C D E
6.Now compare the interval for the average CARS score constructed in question 5 to the interval for the median score from question 4. We will say one interval is “substantially wider” if its span is at least .50 more than the other interval. It is “somewhat wider” if the difference is between .10 and .49. Finally, call them “about even” if the difference is no more than .09 either way.
Mean interval has width 6.54. Median interval has width 6. Difference is 0.54 so that is substantially wider for the mean interval.
What can we say about the intervals for Marketing majors?
A. The interval for the median is substantially wider.
B. The interval for the median is somewhat wider.
C. The two intervals are about even in width.
D. The interval for the average is somewhat wider.
E. The interval for the average is substantially wider.
7.Now assume you want to compare the average CARS score for Marketingmajors to those of Managementmajors. You may assume the variances in scores for these two majors are about equal. Under this assumption, what is the pooled standard deviation for the two samples?
9.30 9.65 10.00 10.35
Choose one: <------+------+------+------+------>
A B C D E
S2pool = {(22-1)*11.6492 + (26-1)*8.0212 }/(22 + 26 -2) = 4458.09/46 = 96.915
Square root of this is 9.8445
8.Again assume you are comparing the average CARS score for Marketing majors to those of Management majors. You want a 95% interval estimate and are willing to believe the variances in scores for thesemajors are about equal. REGARDLESS OF WHAT YOU OBTAINED IN QUESTION 7, assume the pooled standard deviation is 10.00. If so, what is the margin of error (ME) for the interval?
5.1 5.4 5.7 6.0
Choose one: <------+------+------+------+------>
A B C D E
There are 46 df so use t=2.020
ME = 2.02*sqrt{ 10*10/26 + 10*10/22 } = 2.02(2.897) = 5.8516
As we discussed in class, it is true that for finite populations many estimates might be more precise if we use the population size in computing their standard errors. For our three business majors these are the appropriate population sizes:
Tinman University Business MajorMarketing / Management / Finance / Total
Population Size N / 377 / 358 / 512 / 1247
9.If you now incorporate the population size into a 95% interval estimate for the average CARS score for Marketingmajors,what is the standard error se?
Se = (s/sqrt(26))*sqrt((377-26)/(377-1)) = 1.573*.9662 = 1.5199
1.25 1.55 1.85 2.15
Choose one answer: <------+------+------+------+------>
A B C D E
10.If we assume the information on the last page of the exam came from a stratified sample of all students with these three majors, what would be the stratified sample estimate of the overall average CARS score?
XbarST = (377*61.5 + 358*63.091 + 512*55.349)/1247 = 59.43
58.0 59.0 60.0 61.0
Choose one: <------+------+------+------+------>
A B C D E
11.Now assume you are going to use the information in the table above to help plan for a future stratified sample of CARS scores for these three majors. Assume the total size of this sample is going to be n=135. How many Marketingmajors would be included in the sample if they were going to use a proportionalallocation plan?
nMkt = (377/1247)*135 = 40.8 which we would round to 41
39.5 45.5 51.5 57.5
Choose one: <------+------+------+------+------>
A B C D E
12.Again assume you are going to use the information in the table above to help plan for a future stratified sample, and the total size of this sample is going to be n=135. How many Marketingmajors would be included in the sample if they were going to use an optimalallocation plan?
Compute sum of N*S = 377*8.021 + 358*11.649 + 512*9.897 = 12261.523
nFin = (377*8.021/12261.523)135 = 33.29 round to 33
34.5 39.5 44.5 49.5
Choose one: <------+------+------+------+------>
A B C D E
13.(This counts 3 points) What kind of computer do you use for class work?
My setup
A. PC under Windows XP B. PC under Vista C. PC under Windows 7 (soon)
D. Apple E. something else
14.(This counts 2 points) What version of Excel do you use?
A. Excel 2003 B. Excel 2007 C. Excel 2008 D. Something else
15.A CARS scores of 65 or above may show a relatively high degree of anxiety. Suppose we wish to estimate the proportion of all Marketing majors that have this level of computer anxiety. If you make a 95% interval estimate for this proportion, what is the interval’s margin of error (ME)?
P-hat=6/26=.23077 so ME = 1.96*sqrt(.23077(.76923)/26)= 1.96*.0826=.16196
0.105 0.130 0.155 0.180
Choose one: <------+------+------+------+------>
A B C D E
16.Now suppose you are going to compare the proportion of all Marketing majors that have this relatively high anxiety level to the proportion of Management majors that have it. If you make a 95% interval estimate for this comparison, what is the interval’s margin of error (ME)?
For management p-hat = 9/22 =.40909
SE = sqrt{ .23077(.72923)/26 + .40909(.59091)/22}= SQRT{.00683 + .01099}
ME = 1.96*sqrt{.01782} =.2616
0.192 0.217 0.242 0.267
Choose one: <------+------+------+------+------>
A B C D E
Now suppose we want to test the hypothesis the average CARS score is at least 65, versus the alternative that it is smaller. The following hypothesesare formulated:
Ho: µ 65 (Level at least 65) vs. H1: µ < 65 (Average is less)
To test this hypothesis, we will compute the standardized test statistic (T or Z) which has the form:
where SD is either the population or sample std. dev.
17.If we test using the sample of Marketing majors, what is the standardized value of the statistic (T or Z value) used to test the above hypothesis?
Tcalc = (61.5-65)/(8.021/sqrt(26)) = -3.5/1.573 = -2.225
-4.0 -2.0 -1.0 0.0
Choose one answer: <------+------+------+------+------>
A B C D E
18.What is the p-value for the hypothesis test outlined above?
With 25 df, the T-value above is between 2.080 (tail area .025) and 2.518 (tail area .01). The p-value we need is between .025 and .01.
1.0 .10 .05 .025 .01 0.0
Choose one answer: +------+------+------+------+------+
A B C D E