EC315_HW7_JamieGossett

EC315_HW7_JamieGossett - Jamie Gossett EC315 Week 7...

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Jamie Gossett EC315 Week 7 Homework 24. H0: μ2 ≤ μ1 H1: μ2 > μ1 a = 0.05 (one-tailed) z-crit = 1.645 test statistic z = (μ2-μ1) / √[(s1)²/n + (s2)²/m] = (351-345) / √[28²/60 + 21²/54] = 1.3 Since 1.3 < 1.645 we fail to reject H0 and conclude that the difference in means is not statistically significant. In other words, we cannot claim that the number of units produced on the afternoon shift is larger. 34. These are independent samples of sufficient size to use the z-test for proportions. n1 = 270 x1 = 56 n2 = 203 x2 = 52 α = 0.05 p1’ = x1/n1 = 56 / 270 = 0.2074 p2’ = x2/n2 = 52 / 203 = 0.2562 pbar = (x1 + x2) / (n1 + n2) = (56 + 52) / (270 + 203) = 0.2283 qbar = 1 – pbar = 1 – 0.2283 = 0.7717 H0: p1 = p2 Ha: p1 ≠ p2 (claim) At α = 0.05 for a two-tailed test, zcrit = ±1.96 ztest = [ (p1’ – p2’) – (p1 – p2) ] / √[(pbar)(qbar)(1/n1 + 1/n2)] ztest = [ (0.2074 – 0.2562) – 0 ] / √[(0.2283)(0.7717)(1/270 + 1/203)] ztest = -1.2515 Since ztest > zcrit, the null hypothesis cannot be rejected. There is not sufficient evidence to support the claim that there is a difference in the
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This note was uploaded on 04/23/2011 for the course EC 315 taught by Professor Barcus during the Spring '10 term at Park.

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EC315_HW7_JamieGossett - Jamie Gossett EC315 Week 7...

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