Exam_1.xlsx

half-time of a sporting event. The half-time is 25 minutes long. Beginning ea had decided to practice the performance on 50 independent occasions which and she has made a list of how long it has taken her to play the sonata from of these 50 trial performances. From the collected data, the observed samp 22 minutes, and the observed sample standard deviation from the sample da (a) (2 points) Construct a 90% confidence interval for the true mean perfor Remember to show work/formulas/etc. In particular, if you need t not contained in the problem statement, such as c, be sure to show 22 +/- Lower bound of CI 21.534765 Upper bound of CI 22.465235 have decided to practice the performance (and record the corres n 106.158 107 1. (3 pts, total) A pianist is performing Rachmaninoff's 2 nd Piano Sonata tom (b) (1 point) Experimental design: In hindsight, taking the above sample as a a 99% confidence interval for the mean performance time that h

arlier this year, the pianist h she has already done, start to finish in each ple mean performance time of the 50 performances is ata is 2 minutes. rmance time. to find values w work for how you obtain values. 0.465234861470669 B 90 C 1.644854 N 50 Xbar 22 s 2 sponding data) so that she would have been able to construct B 99 C 2.575829 S 2 L 0.5 morrow, during the a given pilot sample, how many times should the pianist has a tolerance level of L = 30 seconds, i.e. if she wanted her 99% CI to look like x-bar plus/minus 30 sec

Further suppose that a 95% confidence interval for the mean Iowan household in Justificatio The are th 2. (2 points, total) Suppose that annual household income has been recorded for a random sample o a) (1 pt) Would a 99% confidence interval be wider or narrower than the one above? i) The 99% confidence interval would be wider. ii) The 99% confidence interval would be narrower. b) (1 pt) The 95% confidence interval can be correctly interpreted as which of the follo i) If the true mean Iowan household income is outside of the inte ii) There is a 95% chance that the true mean Iowan household inco iii) There are 5% of Iowan households that have household income

ncome has been constructed (based on the above sample data) and is (44358, 65892). on: hat could contain mean is larger 2.326>1.645 99%>95% of 100 households in the state of Iowa. Highlight your selection, and provide a brief justification of your answer. owing? (Highlight your selection, no justification needed): erval (44358, 65892), then a particular event has occurred which had only a 5% chance of occurring. ome is between 44358 and 65892. es outside of the interval (44358, 65892).

mean of 5 hours and a standard deviation of 30 minutes. P(X>6) 1-norm .) r day) around the full perimeter of Iowa’s border. ur answer should be an exact answer with an Excel formula coded into your cell.) 15 trips (inclusive) have travel times in excess of 6 hours?

Suppose that each ticket has a 1/600 chance of winning, and that all tickets are independe changing the individual probability of obtaining a winning ticket. Right now, Person A and (Your answer should be an exact answer with an Excel formula coded into your cell (Hint: First determine how many wins are required for Person A to have $10,400, a Prob 0.1008468 0.001667 Expected value of X 3 n*p Standard deviation of X 1.730607 Expected value of Y 6 n*p Standard deviation of Y 2.447448 (Note that you are not required to actually calculate the expected values, only determine E(A) E(B) Justification: 8400>7800 therfore A is larger 4. (5 pts, total) Suppose there exists a mini-lotto game, where each ticket costs $2, and a winning ticket a to buy 1,800 tickets and person B decides that he is going to buy 3,600 tickets. Answer th a) (2 pts) What is the probability that, after all tickets are bought, all tickets are checked f b) (1 pt) Let X be the random variable that counts the number of winning tickets of perso c) (1 pt) Let Y be the random variable that counts the number of winning tickets of person d) (1 pt) Let A represent the final amount that Person A has in her pocket after all tickets are bou pocket after all tickets are bought, checked, and cashed. Which is larger, E(A) or E(B)? H

Cost Spent Total left Won tickets wone tickets bought $2 3600 5400 5000 5 1,800 $2 7200 1800 8600 8 2,600 0.001667 8400 7800 Total cash

131 (215-X)/(215-103) = .75 84 131 5. (2 pts, total) Let Y be a continuous random variable that is uniformly distributed between 103 and 2 (Your answer cell should show work or you can provide written justification of answer.

and the data in column O pertains to a sample of 955 homes sold in West Des Moines. Use the data to answer parts (a) and (b) sale prices in Des Moines (population 1) and the mean home sale prices West Des Moin Remember to show work/formulas/etc. In particular, if you need to find values not contained in the problem statement, such as x1-bar, n1, etc, be sure to make a note of these values somewhere. Des Moines-West Des Moine Lower bound -100491.46234 Upper bound -79835.796107 -100491.462343145 -90163.629225131 +/- -79835.7961071167 Your answer cell can just be a number, but you must obtain this number by creating a pivot ta When Excel asks you where you'd like to place your pivot table, select A30. Put 'beds' down rows and ' 128953.760107817 Average of Sale price Condition Beds above normal below normal normal Total Result 1 108747.34375 122866.666667 153767.878787879 130541.28378 2 121459.28631 138779.116022 131889.588336192 126139.71051 3 122969.15254 121499.753086 128953.760107817 124899.28093 4 122646.51811 106433.488372 133251.461988304 124594.65969 5 127535.45455 132111.111111 103500 125417.66667 6 95400 95400 7 149000 96700 105416.66667 Total Result 122078.2967 129330.542169 131332.90978399 125473.198 6) (4 points, total) The data in columns J, K and L pertain to a sample of 5000 homes sold in Des Moines, a) (3 points) Construct a 95% confidence interval for the difference of means between the mean home b) (1 point) What is the average sale price of a "normal" condition Des Moines home with 3 beds?

Des Moines Data Sale price Condition 120000 normal 106000 above normal nes (population 2). 130000 normal 100000 above normal 119900 above normal 117500 above normal Des Moines 102000 normal XBar 1 125473.198 76000 normal N1 5000 98500 normal S1 91985.7237544377 58750 above normal C 1.95996398454005 55000 above normal West Des Moines 145000 below normal XBar 2 215636.827225131 48000 normal N2 955 10000 normal S2 157800.356304339 110000 above normal 46000 above normal 10327.8331180142 155000 normal 97000 normal 134000 normal 122400 above normal 137000 normal able 129000 normal 'condition' across columns. 148000 above normal 97000 above normal 122000 above normal 90500 above normal 110000 above normal 94000 above normal 209500 normal 276000 above normal 260000 normal 30000 above normal 70000 normal 185000 above normal 205000 above normal 205000 above normal 180000 above normal 215000 above normal 360000 above normal 114000 normal 25500 above normal ,

138900 above normal 130000 below normal 94200 above normal 97900 normal 109500 normal 133000 above normal 80000 above normal 91000 above normal 115000 below normal 114000 below normal 30670 normal 69000 normal 129000 above normal 125900 above normal 49900 below normal 136000 normal 92900 below normal 109500 above normal 123500 normal 131500 below normal 233500 above normal 225000 normal 240000 below normal 48000 below normal 89000 above normal 150000 above normal 133500 normal 183000 above normal 70000 above normal 89900 normal 146500 above normal 119000 below normal 139000 above normal 109560 above normal 49970 above normal 142500 above normal 118500 normal 112000 normal 50000 normal 124900 below normal 140000 normal 173000 normal 165000 above normal 133730 above normal 150000 normal

132500 above normal 87670 above normal 176000 above normal 55000 normal 22500 above normal 59000 normal 137000 above normal 161500 normal 152500 above normal 80000 normal 144500 normal 145000 normal 147500 normal 138500 normal 162500 normal 110500 above normal 140000 above normal 134000 above normal 70570 above normal 149000 above normal 147000 below normal 161000 above normal 119900 above normal 150000 above normal 187000 above normal 121500 normal 153000 above normal 155000 above normal 95600 above normal 163000 normal 140000 normal 162500 above normal 165000 above normal 175000 above normal 128000 normal 60000 above normal 129900 above normal 149500 above normal 134900 above normal 185500 above normal 116000 above normal 140000 above normal 110000 above normal 95000 above normal 140000 above normal

41400 above normal 45540 above normal 60000 above normal 80500 above normal 130000 above normal 110000 above normal 93000 normal 88500 above normal 31000 below normal 63000 above normal 50000 above normal 150000 above normal 142000 above normal 98850 above normal 117500 above normal 225000 above normal 109000 normal 135000 below normal 150000 above normal 80000 above normal 76650 above normal 97000 above normal 71100 above normal 113500 above normal 60000 normal 62010 above normal 123900 below normal 35000 below normal 109500 above normal 87500 above normal 156000 normal 71000 below normal 89900 above normal 188000 normal 176000 normal 55000 above normal 118000 normal 35000 below normal 65000 above normal 121000 above normal 114000 normal 137500 above normal 90000 normal 118000 normal 96000 above normal

129000 normal 67000 above normal 106000 above normal 160000 below normal 60000 normal 103000 above normal 16000 above normal 148500 normal 117000 above normal 133000 normal 54250 above normal 54250 normal 65000 below normal 97500 above normal 101000 above normal 80000 normal 60000 normal 125000 above normal 89900 above normal 67500 above normal 80000 above normal 118500 above normal 138500 above normal 97000 normal 159000 above normal 119000 above normal 78900 normal 25000 normal 7500 normal 13500 below normal 154000 below normal 90500 above normal 127000 above normal 7000 below normal 105500 below normal 61500 above normal 148000 normal 110100 normal 66250 normal 122000 above normal 65700 above normal 130000 above normal 130000 normal 123000 normal 55000 normal