{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Week 2 Exercises

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

This preview shows pages 1–3. Sign up to view the full content.

#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. -------

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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!...
View Full Document

{[ snackBarMessage ]}