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27. Bob selects independent random samples from Subgp A Subgp B two populations and obtains the yalues Gp S F Tot | Gp S F

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27. Bob selects independent random samples from Subgp A Subgp B two populations and obtains the yalues Gp S F Tot | Gp S F Tot 0.700 andj = 0.500. He constructs the 95% 3 17 20 40 40 80 confidence interval for py and gets: 2 40 60 Tot 60 Tot 140 0.200 + 1.96(0.048) = 0.200 0.094. Note that 0.048 is called the estimated standardn observational study yields the following error of - py (the ESE of the estimate). "collapsed table." Tom wants to estimate the mean of the success Group S F Total rates: Pi + p2 43 57 100 2 2 36 64 100 (a) Calculate Tom's point estimate. Total 79 121 200 (b) Given that the estimated standard erBelow are two (partial) component tables for ror of(p1 + p2)/2 is 0.024, calculate these data. Explain why Simpson's Paradox the 95% confidence interval estimate ofcannooccur for these data. (pi + p2)/2. Hint: The answer has our usual form: Subgp A Subgp B Pt. est.+ 1.96 x ESE of the estimate GP S F Tot| Gp S F Tot 3 17 20 40 40 80 28. Carl selects one random sample from a popu- 4 40 2 60 lation and calculates three confidence intervalslot 60 Tot 140 forp. His intervals are below. 31. An observational study yields the following A B C "collapsed table." p +0.080 p +0.040 p0.072 Group S F Total Match each confidence interval to its level, with 45 55 100 levels chosen from: 80%, 90%, 95%, 98%, and 38 62 100 99%. Note: Clearly, two of these levels will Total 83 117 200 not be used. You dotneed to explain your reasoning. Below are two (partial) component tables for these data. Complete these tables so that Simp- 29. An observational study yields the following's Paradooccurssee Course Notes). Note "collapsed table." that there might be more than one possible cor- Group S F Total rect answer. 43 57 100 39 61 100 Subgp A Subgp B 82 118 200 Gp S F Tot | Gp S F Total Tot 5 15 20 40 40 80 Below are two (partial) component tables for 2 60 40 these data. Complete these tables so that Simp- Tot 80 Tot 120 son's Paradox is occurring (see Course Notes). Note that there is more than one possible cafresh observational study yields the following answer. "collapsed table."Table & Table 9 Current Current Table 4 Table 5 Prev. S F Total| Prev. S F Total Current Current 46 43 89 36 53 89 Prev. S F Total Prev. S F Total 44 66 110 F 54 56 110 47 42 89 52 37 89 Total 90 109 199 Total 90 199 42 28 70 F 37 33 70 Total 89 70 159 Total 89 70 159 Table 10 Table 11 Current Current Table 6 Table 7 Prev. S F Total| Prev. S F Total Current Current 45 45 90 35 55 90 Prev. S F Total Prev. S F Total F 44 65 109 54 55 109 54 36 90 18 42 90 Total 89 110 199 Total 89 110 199 F 36 33 69 F 42 27 69 Total 90 69 159 Total 69 159 18. Abby and Dana each perform 160 dichotomous trials. A success is the desirable outcome; it re- quires more skill than does a failure. You are Table 8 Table 9 given the following information. Current Current Prev. S F Total| Prev. S F Total . Each of the women achieves exactly 90 53 36 89 47 42 89 successes. F 37 33 70 F 43 27 70 Abby exhibited evidence of improving Total 90 69 159 Total 90 69 159 skill over time; and Dana exhibited no ev- Table 10 Table 11 idence of changing skill over time Current Current Abby had failures on her first and last tri-Prev. S F Total| Prev. S F Total als; Dana had a failure on her first trial and S 48 42 90 53 37 90 a success on her last trial. F 41 28 69 36 33 69 . Abby performed better after a success Total 89 70 159 Total 89 70 159 than after a failure; and Dana performed better after a failure than after a success. A box contains 14 red cards and six blue cards for a total of 20 cards. Walt is going to select For each woman, identify her two tables from=10 cards at randomith replacementm the tables below. Hint: For each woman, choose one from Tables 1-3 and one from Tables 4-1 1the box. Let denote the number of red cards (Hint: If there is more than one table that satiat Walt obtains. Betlenote the number of isfies the conditions stated above, just give melue cards that Walt obtains. Yale is going to one of them. If no table satisfies the conditionselect = 10 cards at randomithout replace- say it is impossible.) menfrom the box. Letdenote the number of red cards that Yale obtains. Finallydeet Table 1 Table 2 note the number of blue cards that Yale obtains. Half S F Total Half F Total You may use the fact that the probability his- Ist 35 45 80 Ist 50 30 30 2nd 55 25 tograms of the sampling distributions\\of 80 2nd 40 40 80 Total 90 70 160 Total 90 70 Y and are pictured below. The number above 160 each rectangle is its height which also equals Table 3 its area. Note that 0 means zero, whereas .000 Half S F Total means smaller than .0005, but not zero. Ist 45 35 80 2nd 45 35 80 (a) Place and next to the probability his- Total 90 160 togram of the sampling distribution of 6Chapters 5-7 . Alex had successes on his first and last tri- als; Bruce had a success on his first trial 15. On each of four days next week (Monday thru and a failure on his last trial. Thursday), Earl will shoot six free throws. As- sume that Earl's shots satisfy the assumptions . Alex performed better after a failure than of Bernoulli trials with 0.37. after a success; and Bruce performed bet- (a) Compute the probability that on any par- ter after a success than after a failure. ticular day Earl obtains exactly two suc- cesses. For future reference, if Earl obfor each man, identify his two tables from the tains exactly two successes on any partictables below. Hint: For each man, choose one ular day, then we say that the event "Bradfrom Tables 1-3 and one from Tables 4-11. has occurred. (Hint: If there is more than one table that sat- isfies the conditions stated above, just give me (b) Refer to part (a). Compute the probabilityne of them.) that: next week Brad will occur on Mon- day and Thursday and will not occur on Tuesday and Wednesday. (Note: You are being asked to compute one probability.) Table 1 Table 2 16. On each of four days next week (Monday theyHalf S F Total | Half S F Total Thursday), Dan will shoot five free throws. AS- 2nd Ist 35 65 100 1st 45 55 100 55 45 100 2nd 45 55 100 sume that Dan's shots satisfy the assumptionstotal 90 110 200 Total 90 110 200 of Bernoulli trials with 0.74. Table 3 (a) Compute the probability that on any par-Half S F Total ticular day Dan obtains exactly three suc- Ist 70 30 100 cesses. For future reference, if Dan ob- 2nd 20 80 100 tains exactly three successes on any partic-Total 90 110 20q ular day, then we say that the event "Mel" has occurred. (b) Refer to part (a). Compute the probabil- ity that: next week Mel will occur exactly Table 4 Table 5 once and that one occurrence will be on prey. Current Current Monday. (Note: You are being asked to- S F Total | Prev. S F Total 43 46 89 30 59 89 compute one probability.) 46 64 110 F 59 51 110 Total 89 110 199 Total 89 110 199 17. Alex and Bruce each perform 200 dichotomous trials. A success is the desirable outcome; it re- Table 6 Table 7 quires more skill than does a failure. You are Current Current given the following information. Prev. S F Total| Prev. S F Total 43 47 90 33 57 90 . Each of the men achieves exactly 90 suc- F 47 62 109 57 52 109 cesses. Total 90 109 199 Total 90 109 199 . Alex exhibited evidence of improving skill over time; and Bruce exhibited evi- dence of declining skill over time. 5Set 1: 0.2450 0.4688 0.9233 12. An unbalanced CRD is performed with a to- Set 2: 0.1445 0.2890 0.9625 tal of 800 subjects. Three hundred subjects are placed on the first treatment and 500 are placed (a) Which set contains the correct P-values: on the second treatment. There is a total of 356 or 2? (No explanation is needed.) successes, with 126 of the successes on the first (b) For the set you selected in part (a), matcheatment. Use the standard normal curve to ob- each P-value to its alternative. (No explatain the approximate P-value for the third alter- nation is needed.) Note: Even if you picknativep # p. the wrong set in part (a), you can still getA sample space has three possible outcomes, B, full credit for part (b). C, and D. It is known that') = P(D). The operation of the chance mechanism is simulated 9. A comparative study is performed; you arco,000 times (runs). The sorted frequencies of given the following information. the three outcomes (B, C, and D) are: . The total number of subjects equals 29. 2322, 2360, and 5318. . The observed value of the test statistic is greater than 0. (a) What is your approximationaf To receive credit you must explain your an- I used the website to obtain the exact P-value for swer. Fisher's test for each of the three possible alter(b) What is the best approximationof natives. These three P-values are below along To receive credit you must explain your with three bogus P-values. answer. Set 1: 0.1445 0.2890 0.9622 14. A sample space has four possible outcomes, A, Set 2: 0.0762 0.1297 0.9868 B, C, and D. It is known then) + P(B) = 0.60 andP(C)

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