QMB 3250 Fall 2009 Exam 1 Form A Sept 29, 2009 Page 1 of 6

* If you correctly bubble in your name, UF ID number and any test location codes, you will have one incorrect answer "forgiven". DON'T LOSE THIS FREEBIE BY BEING CARELESS!

* There are 18 questions on this exam. Each question is worth 5 points. The maximum score on the exam is 90 points.

* If you wish to file a protest about anything concerning this exam, check the syllabus for procedures.

When you are done with this exam and are ready to turn it in, please do the following:

1. Make sure you get the Form Code from the header and bubble that in on the scantron. 2. Sign your name, IN INK, in the box below. 3. Put your UF ID number in the space below the box. Record the room code. 4. Take your exam and a picture ID to one of the exam proctors when you go up to turn in your bubble sheet. The proctor will compare the signature on the ID with the one in the box.

UF ID: ______- ______

CODE FOR ROOM LOCATION: __ __

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/2 df 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 QMB 3250 Fall 2009 Exam 1 Form A Sept 29, 2009 Page 2 of 6

------Use this information for problems 1-4 below and 5-6 (next page) ------

Each summer, some of Florida’s finest high school basketball players take their talent to the Gator Basketball Camp Invitational, a week-long clinic committed to improving the skills of participants. While the players work out, coaches analyze the play of the athletes as well as scout for future collegiate talent. The height of a random sample of some of the participants in several past summer camps was recorded and appears below.

Heights (inches) of Camp Participants

52 53 62 63 63 64 64 66 66 67 67 68 69 69 70 70 71 71 72 72 72 73 73 73 73 73 74 74 75 75 75 76 77 78 81 82

1. Austin Longlegs (a former camp participant who retired from basketball and became a QMB 3250 TA) analyzed this data in PHStat and found the first quartile was 66 and the third quartile was 74. If we were to do an outlier check on this data, what would be the upper fence we would compute?

80.0 82.5 85.0 87.5 Choose one answer: <------+------+------+------+------> A B C D E IQR = 74-66=8 Upper Fence = 74 + 1.5(8) = 86

2. Assuming the same quartiles as in question #1, how many of the camp participants would be considered an outlier in terms of height?

A. None B. 1 C. 2 D. 3 E. 4 or more Lower Fence = 66-12 = 54 so two low outliers. There are none above.

3. According to the guidelines in our QMB3250 book, which one of the following sets of intervals would be appropriate to use in a frequency distribution of this data?

A. 7 intervals of width 5, the first two being (50.1 to 55) and (55.1 to 60) B. 7 intervals of width 4, the first two being (48.1 to 52) and (52.1 to 56) C. 7 intervals of width 6, the first two being (48.1 to 54) and (54.1 to 60) D. 7 intervals of width 5, the first two being (55.1 to 60) and (60.1 to 65) E. none of the above are appropriate

With n=36 use 5-7 intervals. Span needs to be 82-52 = 30 at minimum A Last category is (80.1 to 85) so this scheme works. B Is not wide enough. C has the last category of (84.1 to 90) which is empty. D misses the first data values

4. Basketball is a game dominated by height; accordingly, college coaches don’t put much promise in any players they see who are shorter than 5’6’’ (66 inches). Use the above data to make an estimate for the proportion of all camp participants who are shorter than 66 inches. What is the standard error of this estimate?

.061 .064 .067 .070 Choose one answer: <------+------+------+------+------> A B C D E p-hat = 7/36 = .1944 SE = sqrt[.1944(1-.1944)/36] = .06596

------NOTE: QUESTIONS 5 and 6 (NEXT PAGE) ALSO USE THIS DATA ------QMB 3250 Fall 2009 Exam 1 Form A Sept 29, 2009 Page 3 of 6

5. Suppose we were going to make an approximately 95% confidence interval estimate of the median height of all camp participants. What is the lower endpoint of this interval?

61.5 64.5 67.5 70.5 Choose one answer: <------+------+------+------+------> A B C D E Compute r=.4n-2 = .4(36)-2 = 12.4 12th smallest is 68

6. What is the upper endpoint of the interval for the median that is specified in question #5?

61.5 64.5 67.5 70.5 Choose one answer: <------+------+------+------+------> A B C D E 12th smallest is 73

-- THE TWO QUESTIONS BELOW USE THE TRAFFIC FINE DATA (NOT BASKETBALL PLAYER HEIGHT) --

What costs you more, running a red light or failing to stop at a stop sign? We sampled 9 cities across the nation, and recorded the fines levied for these violations. Listed are the fines plus any court costs, assuming there was no injury or property damage caused by the violation.

City 1 2 3 4 5 6 7 8 9 Mean S.D. RedLight 166 135 170 150 170 146 138 158 170 155.89 14.11 StopSign 142 144 145 149 149 153 158 158 164 151.33 7.45 Diff. 24 -9 25 1 21 -7 -20 0 6 4.56 15.90

Compare the average cost by computing a confidence interval with the general form:

(XBAR1 – XBAR2)   se where  comes from some statistical table and se is the standard error for the difference in means.

7. What is the value of the standard error se?

1.35 2.35 3.35 4.35 Choose one: <------+------+------+------+------> A B C D E SE = 15.90/sqrt(9) = 5.30

8. Suppose you are doing the cost comparison by computing a 90% confidence interval for the difference in means. REGARDLESS OF WHAT YOU OBTAINED IN QUESTION 7, assume that se = 4.00. If so, what is the best conclusion reached about the fines for the two types of traffic violations? Multiplier = t(for 90% CI with 8 df) = 1.860 from page 1 ME = 1.860(4.00) = 7.44 is smaller than the diff of 4.56 so there is not a significant difference.

A. It will cost you significantly more to run a red light than a stop sign. B. It will cost you significantly more to run a stop sign than a red light. C. We cannot conclude either type of fine really costs more. D. It does not matter in Gainesville, everybody does it. E. Some other conclusion not listed in A, B, C or D. QMB 3250 Fall 2009 Exam 1 Form A Sept 29, 2009 Page 4 of 6

-- USE THE INFORMATION BELOW FOR QUESTIONS 9-11 (and 12-15 on the next page) --

A consumer group is comparing electric usage in three cities in Florida. To control for home and family size, their study looked at samples of single-family, detached homes of 1400 to 1600 square feet, occupied by four people.

Over a 3-month period, the group records usage in units of 1000 kilowatt hours (for example, the Gainesville average of 5.373 means 5373 kilowatt hour). For the three cities, here are the number of homes in the sample, the average usage, and the standard deviation in usage:

City n Average S.D. Gainesville 40 5.373 1.101 Orlando 60 6.320 1.302 Tampa 70 5.896 0.984

Our text book lists many of the 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.

9. If you are going to use these results to construct a 95% interval estimate for the average electric usage in Gainesville what is the margin of error ME?

.230 .270 .310 .350 Choose one answer: <------+------+------+------+------> A B C D E n=40 use 39 df. For 95%, t=2.041 se = 1.101/sqrt(40) = .17408 and ME = 2.041(.17408) = .3553

10. Now assume you are comparing the average usage in Gainesville to that in Orlando. If the variance in usage is assumed about equal in these cities, what is the pooled standard deviation for the two samples?

1.10 1.18 1.26 1.34 Choose one: <------+------+------+------+------> A B C D E Pooled variance = [ (40-1)*1.101*1.101 + (60-1)*1.302*1.302 ]/(40 + 60 – 2) = 1.50299 Pooled SD = sqrt(1.50299) = 1.22596

11. Again assume you are comparing Gainesville to Orlando and believe the variance in usage in these cities is about equal. REGARDLESS OF WHAT YOU OBTAINED IN QUESTION 10, assume the pooled standard deviation is 1.800. If so, what is the best conclusion about the comparison of average electric usage between these two cities?

A. We really need to factor in the third city to tell. B. Average electric usage is significantly lower in Gainesville than Orlando. C. Average electric usage is significantly higher in Gainesville than Orlando. D. We cannot say there really is a difference in the average electric usage. E. The correct conclusion is something other than A, B, C or D.

SE = sqrt[ 1.8*1.8/40 + 1.8*1.8/60 ] = sqrt[.135] = .36742 For interval, use t = 2.00 for the “standard” 95% interval ME = 2.00(.36742) = .7348 QMB 3250 Fall 2009 Exam 1 Form A Sept 29, 2009 Page 5 of 6 Diff = 5.373 – 6.320 = -.947. The ME is smaller than this so it indicates a significant difference (Gainesville usage is less). QMB 3250 Fall 2009 Exam 1 Form A Sept 29, 2009 Page 6 of 6

---- USE THE INFORMATION BELOW AND ON PAGE 4 FOR QUESTIONS 12-15 ----

One of the employees of the consumer group from the previous page is Jeri Ashley, an intern who was a teaching assistant in QMB 3250 at the University of Florida. She pointed out that if they were able to use the population sizes (number of homes) for the three cities, many of their estimates might be more precise. With the number of applicable homes (N = population size for that city) added to the table, we have:

City N n Avg SD Gainesville 2500 40 5.373 1.101 Orlando 9000 60 6.320 1.302 Tampa 8000 70 5.896 0.984 Totals 19500 170

12. If you now incorporate the population size into the 95% interval estimate for the average usage in Gainesville, what is the standard error se?

.305 .325 .345 .365 Choose one answer: <------+------+------+------+------> A B C D E se = [(1.101)/sqrt(40)][sqrt((2500-40)/(2500-1))] = .1171 OOPS. I had meant to ask about the ME again, so on all versions the answers fall into A.

13. If you assume the information in the table above came from a stratified sample of electric customers in the three cities, what would be the stratified sample estimate of the overall average usage?

5.5 5.7 5.9 6.1 Choose one: <------+------+------+------+------> A B C D E XbarST = (2500*5.373 + 9000*6.320 + 8000*5.896)/19500 = 6.0246

14. Now assume you are going to use the information in the table above to help plan for a future stratified sample of electric customers in the three cities. Assume the total size of this sample is going to be n=350. How many Orlando homes would be included in the sample if they were going to use a proportional allocation plan?

125 135 150 165 Choose one: <------+------+------+------+------> A B C D E N(Orlando) = (9000/19500)(350) = 161.5

15. 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=350. How many Orlando homes would be included in the sample if they were going to use an optimal allocation plan?

125 135 150 165 Choose one: <------+------+------+------+------> A B C D E Sum of N*s = 2500(1.101)+9000(1.302)+8000(.984) = 22342.5 N(Orlando) = (9000(1.302)/22342.5)(350) = 183.6 QMB 3250 Fall 2009 Exam 1 Form A Sept 29, 2009 Page 7 of 6

In a factory, small amplifiers that average 20 watts of power are produced. An engineer has designed a testing procedure that allows workers to cull out any amplifiers that are on the low side of the power distribution. If so, it is likely that the amplifiers that actually get used will average more than 20 watts and will have a standard deviation that is unknown. To determine if this is true, the following hypothesis test is performed:

Ho: µ ≤ 20 (No improvement) versus H1: µ > 20 (Output increase)

To test this hypothesis, we will compute the standardized test statistic (T or Z) which has the form:

X  20 T or Z  where SD is either the population or sample standard deviation. SD n

16. Assume we are using a sample of size 12 to test this hypothesis. If we test at a significance level of  = .05, what is the decision rule?

A. Something other than those listed in B, C, D or E B. Reject H0 if TCALC > 1.796 C. Reject H0 if TCALC > 1.363 or TCALC < -1.363 D. Reject H0 if ZCALC > 1.645 E. Reject H0 if ZCALC > 1.282 or ZCALC < -1.282 It would be upper-tailed and a T-distribution with df=11 (you don’t know the population SD). T.05 = 1.796

17. A sample of 21 amplifiers produces average power of 20.8 watts with a standard deviation of 1.4 watts. Given this information, what is the standardized value of the statistic (T or Z value) used to test the above hypothesis?

1.5 2.0 2.5 3.0 Choose one answer: <------+------+------+------+------> A B C D E TCALC = (20.8-20)/(1.4/Sqrt(21)) = .8/.3055 = 2.6186

18. Regardless of what you actually got for question 17, assume you calculated that the standardized value of the test statistic was 2.43. Assuming the sample size was still n=21, what is the p-value for the test?

1.0 .10 .05 .025 .01 0.0 Choose one answer: +------+------+------+------+------+ A B C D E

If n=21, P(T20 > 2.43) is between .025 (T=2.120) and .01 (T=2.583). Since this is a one-sided test the p-value is between .025 and .01.