1 of 2 STA 2023 Homework S This is the ifth homework asignment for STA 2023 Itdu
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1 of 2 STA 2023 Homework S This is the ifth homework asignment for STA 2023 Itdue in class on Friday,November 17 1. Confidence lnterval of You randomly tample N individuals from a population and estimate each sampled individual's height. The average height [u) of the population is unknown, but the standard deviation (al is nown and equals 3.5 inches. You calculate the sample average ) and obtain -66 inches a. Create both a 95% and 99% confidence interval for for: No25 Suppose that a random sample of heights is obtained from, a population having "67 inches and 5 inches The sample average } is computed for 500 ndependent samples of size N. This ais provided in the Excel tle labeled Heights for i, N,25N" 100 "IN" 1000 b. c. For ii,produce a hisbogram of the sample average (R For Hil, compute the percentage of times the 95% condence interval o' doesnot ootan the true mean 0.67 inches). Foe iHI, compute the percentage of times the 99% cotdence interval of doesnot contain the true mean d. 67 inches). 2. t-distribution: You randomly sample Nindividuals from a population ofl students and record each sampled student's GPA The mean GPA u) and the standard deviation a) of the pooulation are both unknown, You cakculate the sample average R) and the sample variance () and obtain R 32 ands 016 a. Estimate 95% and 99% corf dence interval for under a large sample assumption for b.Estinate a 95% and 99% confidence interval for under a smal sample assumption see: Suppose that a random sample of GPAS is obtained from a populaton having -3.2 GPA ande«04 The sample average (X) and sample varlance (s) is computed for 500 independent samples of size N. This data is provided in the Excel fe labeled GPAs for O N-8 iQ N-16 i) N-50 produce a histogram of the sample average GPA (R) compute the percentage of times the 9S%confidence interval of-under the large e. For 00 d. For sample assumption does not contain the true mean 32 For40, compute the percentage of times the 95% confidence interval of under th" sma sample assumption does not contain the true mean 32 e. What can you conclude from your revalts in (d) and (e)? 3. Confidence Interval: You operate a qality control company that evaluates the operational quality of slot machines that are distributed through your partner vendor. For each slot machine that you test, the expected loss Ieain for the casino) is 10 cents per $1 bet. In order to test each machine, you run 1000 bets and calculate the amount that a hypothetical player would receive as a paryout. For each bet, the player either wins $5 or loses a. Compute a 95% confidence interval for the verage loss per bet for the outcomes below which slot machines of anyl would you suspect as being defective? If you suspect a machine to be defective, state your leved of confidence Note A machi e is defective drites s'enficantly from an expected 0 cents per ane L. Payout $1250 [net winnings $250 Payout $1100 (net winnings $100 Payout S7001net winnings-5300, b. Repeat 4ab using a 95% confidence interval of the proportion of winning games. Compare your resultsExplanation / Answer
i) N=25
One-Sample Z
The assumed standard deviation = 3.5
N Mean SE Mean 95% CI
25 66.000 0.700 (64.628, 67.372)
One-Sample Z
The assumed standard deviation = 3.5
N Mean SE Mean 99% CI
25 66.000 0.700 (64.197, 67.803)
ii) N=100
One-Sample Z
The assumed standard deviation = 3.5
N Mean SE Mean 95% CI
100 66.000 0.350 (65.314, 66.686)
One-Sample Z
The assumed standard deviation = 3.5
N Mean SE Mean 99% CI
100 66.000 0.350 (65.098, 66.902)
iii) N=1000
One-Sample Z
The assumed standard deviation = 3.5
N Mean SE Mean 95% CI
1000 66.000 0.111 (65.783, 66.217)
One-Sample Z
The assumed standard deviation = 3.5
N Mean SE Mean 99% CI
1000 66.000 0.111 (65.715, 66.285)
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