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Week 2 Exercises - #9.62 H0 p = 0.95 H1 p> 0.95 Decision...

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#9.54 (a) H0: mu = 10 H1: mu > 10 one-tailed t-test with a = 0.01; df 35-1 = 34, critical value t = 2.326 Test statistic: T = (14.44-10)/[4.45/sqrt(35)] = 5.9028 Since T > 2.326 weject H0 and conclude that there is statistical evidence that the true mean is > 10 (b) p-value = 0.00000057579 ----------- #9.56 (a) phat = 38/60 = 0.6333 H0: p = 0.5 H1: p > 0.5 a = 0.10 Test statistic Z = (phat - p)/sqrt[p(1-p)/n] = (0.6333 - 0.5)/sqrt[0.5(1-0.5)/60] = 2.065 Decision rule: Reject H0 if Z > 1.2816 Since 2.065 > 1.2816 we reject H0 and conclude that the coin is biased. (b) p-value = 0.019434 Since the p-value < 0.10 we reject the null hypothesis and conclude that the coin is biased. -------
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Unformatted text preview: #9.62 H0: p = 0.95 H1: p > 0.95 Decision rule: Reject H0 if z > 1.96. Test statistic Z = (phat-p)/sqrt[p*(1-p)/n] = (485/500-0.95)/sqrt(0.95*(1-0.95)/500) = 2.0519 Since 2.01519 > 1.96 we reject H0 and conclude that they are exceeding their goal.--------#9.64 H0: u = 250 H1: u > 250 (one-sided) a = 0.05 Test statistic t = (xbar - u)/(s/sqrt(n)) df = 25-1 = 24 Rejection region: Reject H0 when T > t(0.95, 24) = 1.711 Value of test statistic: t = (275.66 - 250)/(78.11/sqrt(25)) = 1.643 Since 1.643 < 1.711 we fail to reject H0 and conclude that the average patient does not pay more than $250. This was a close decision, the p-value is 0.0567!...
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