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Assignment 4

Mean 1845925 st dev 654370 b 95 ci 1539674 2152176 c

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Mean = $18,459.25 St. Dev = $6,543.70 b) 95% CI: (15396.74, 21521.76) c) No, the 95% CI for µ is an interval that has a 95% chance of containing the mean winnings of the entire population, NOT the winnings of 95% of similar winners. d) I. H o : μ = μ o H a : μ μ o μ o = 34064 This is a two-sided alternative because the question asks whether or not the population mean differs from Jennings’ average (alternative hypothesis has a “not equal” sign), which means that μ o can either be larger or smaller than μ . II. t = = -10.6647 P-value = 2 x Pr (t >|t-stat|) < 0.01. P-value = 2 x < 0.01 P-value < 0.02. At the 5% significance level, 0.02 < alpha, therefore we have significant evidence to prove that the mean winnings of the population are different from Ken Jennings’ mean winnings. III. The 95% CI from part b) does support the conclusion of our hypothesis test. The CI we found was (15396.74, 21521.76). Ken Jennings’ mean winnings are not inside this interval, which means that we reject the null hypothesis. e) In this situation, Type I error would mean that Ken Jennings’ average of $34,064 actually represented the population mean winnings, but this was rejected in the hypothesis test. Type II
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error would mean that the previous statement was actually false, but we accepted the hypothesis. The only error that could have happened in part d) was the Type I error.
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Mean 1845925 St Dev 654370 b 95 CI 1539674 2152176 c No the...

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