Solutions to Exam 3, Blue Version, Ma381, F05

Solutions to Exam 3, Blue Version, Ma381, F05 - dr W’...

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Unformatted text preview: dr W’ 10”” test this claim you 0 the stated conditions. K m (SMALL) For your sample, 36 = 19.3 mpg and s = 4.2 mpg. Test the car companie’s claim at significance level a = 0.05 by answering the questions below. a. State the hypotheses Ho I a :.- 2, l HA2/441-l b. To continue working this problem the entire population of mpg values for this luxury sedan must (choose one) . . . i. follow some t density curve ii. follow some normal densit curve (”945— '\ (S .5 Maw) in. no 3 (no assumption about the mpg values is needed) c. Sketch, as appropriate, either a t density curve or a standard normal density curve and provide the following labels: i. H0 for the region(s) that we’ll conclude H0 is true and H A for the region(s) that we’ll conclude H A ’is true. _ ii. Numerical value(s) along the horizontal axis that clearly show how the Ho and HA regions are separated. / AW; V ,tq‘KQM-I) d: .05 W -2131 (1. Compute the test statistic. That is, compute xl—J/i' (assuming the null hypothesis true). 5 n VE: f—W: (9.3»21 2:95 e. Which hypothesis is concluded to be correct? H9 (((4654,“(' Ho“) ,u '7 1.5 / M 2. (25 pts.) An automotive batte manufacturer guarantees that the ”of certain battery is 01W 0 test this claim, you rando u y se ec a samp e 0 batteries and find the mean reserve capacity to be 1.57 hours with a standard deviation of 0.34 hours. 14‘ Conduct the appropriate test at significance level a = 0.05 by answering the questions below. (/4 ’32 I) a. State the hypotheses . H 0 : ,0 '-- /-5 HA : AA 7 /.5 b. To continue working this problem the entire population of reserve capacity values for all of these batteries must (choose one) . . . i. follow some normal density curve ii. follow some t density curve iii. nothing (no assump 1011 aout reserve capacity values is need . 5" [:5 44 4:: [e 14" c. Sketch, as appropriate, either a t density curve or a standard normal density curve and provide the following labels: , '—“' i. Ho for the region(s) that we’ll conclude H0 is true and H A for the region(s) that we’ll conclude H A is true; ii. Numerical value(s) along the horizontal axis that clearly show how the Ho and H re ions are se arated. A g P “kitty-é [v arm-J— n '; fins-A - ' 05 / l (I Gr Mo WI .étJ/V/ 3-H s/J; d. Compute the test statistic. That is, compute (assuming the null hypothesis true). e. Which hypothesis is concluded to be correct? H.» C “46% Ho") 3. (25 pts.) Red Dog is a card game in which one can place repeated $1 bets. For purposes of this exam, we need not go into the rulesl, but the payoff table for a single $1 bet is given below: NetWinnin (in dollars) p—e g—A From this probability mass function for a single $1 bet, it can be shown that the mean and standard deviation for net winnings are p E -$ 0.07968 O'E 31.41603 a. The histogram below shows the total net winnings in repeated, simulated sequences of 225 $1 bets. The shape of the histogram below is predicted by a theorem. Name this theorem. C Emnm. LIN m" "714 gonna”! C Cur) Net Winnings in 225 Consecutive $1 Bets in “Red Dog" 0.02 E 2 a, 0.01 o 0.00 — -50 o 50 “" Net Winnings ' If you’re curious about the rules contact your instructor or see, for example, Katz, Nikki (2004), The Everything Card Games Book: A Complete Guide to Over 50 Games to Please Any Crowd, Adams Media, Avon, MA, pp. 203-205. The payout table above is correct for a player who doesn’t ever raise his or her bet and when the game is played with 6 decks. b. Estimate, using the theorem of part a, the chance of having net winnings of _$_20 or more in 225 consecutive $1 bets in Red Dog. (Note: While the histogram on the previous page may help you confirm your calculation, the histogram itself should not be used to answer this question.) P(Z( 4- X‘LJ-WJ- Z215 Z J20) ._ P< (Zmuw- 211;)—|AM 7 ’20—’225C—,o'l‘768) — fig. ’ J22; 1.41503 T 5; P(’<&z 1.762) lay a. : 1— New”) c. Find the 70th percentile of net winnings in 225 consecutive $1 bets in Red Dog! (Note: While the histogram on the previous page may help you confirm your calculation, the histogram itself should not be used to answer this question.) / Nok: 7(—. 7 ‘41" Mk I“; D P ,c-ms = P('2-s .52) : P (Eh-4* gang)_wu ‘ 7<~ 225(-.O7?68’) k a- ‘/2'2.5 l-‘HéO3 4. (25 pts.) The figure below shows a histogram of biacramial diameter values (see the cover page of the exam) for a sample of adult men . The 247 men sampled had an average biacromial diameter of 41.2 cm with a standard deviation of 2.1 cm. Furthermore, note that the histogram of biacromial diameter values looks as if it is well approximated by some normal: —- 0.25 X 2 112' 9’“ 020 I; -; 3:40»qu (5 ’- 7'" z 6" g Q15 DFaMJ'ev- E 0.10 DJ; MAM!” 0.05 Se W om 4L» ”Mk so 40 5o Biacromial Diameter (cm), for the Men a. Estimate the proportion of adult men with biacromial diameter values between 38.5 cm and 44.5 cm. Wax; < 3 < +43) 31%315’71/1 4 By“ 4 +4.5»rr1) 2." 6‘ 2" : {QC-(KL? 4 2 < ['57) = fits?) , $01-27) = IQW" ‘0‘7‘5’5 ‘@ b. Estimate the 20th percentile of adult male biacromial diameter values. If 'lC = 20"“ Wk/ M— 10: FCfif’X) ‘1 P< g f lefl-z') 3? 7“ *‘ < 2.“ .L) .. .. ,fm. I .— “Discussed in Heinz. Grete, Peterson, Louis. Johnson, Roger, and Kerk, Carter (2003), “Exploring relationships in body dimensions", Journal of Statistics Education, vol. 11, no. 2. ...
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