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Unformatted text preview: 1 Solutions to Midterm Fall 07 1. a. Ho: P_roc = P_que; Ha: P_c > P_q. pbar = (0.33*250+0.41*750)/1000 = .39 z = (.41  .33 ) / sqrt((0.39*0.61)*(1/250+1/750)) = .08 / .03562 = 2.246 n.b. sqrt((0.33*0.67)/250+(0.41*0.59)/750) = .0347, with z = .08 / .0347 = 2.30Reject null H if z > 2.326 Conclude the level of support is not higher in ROC than in Quebec b. .39 +/ 2.576 * sqrt(0.39*0.61)/1000)) = .39 +/ 2.576 * .0154 = .39 +/ .0397 = .39 +/ .04 = (.35 to .43) Only 1.5 marks if use .74 instead of .39 Only 1 mark if use .41.33 instead of .39 c. n = (2.576/0.01)^2*0.39*0.61 = 15787 assuming p = .39 n = (2.576/0.01)^2*0.5*0.5 = 16590 assuming p = .50 2. a. the data shows two extreme outliers, making it hard to assume the data come from a normal distribution the most appropriate test cannot be a 1sample ttest, but must be a Wilcoxon test of median. b. Ho: Median = 44; Ha: median < 44 pvalue is .001 from output since the pvalue is < .05, we reject the null H and conclude that 44 is too high for the average since the pvalue is < ....
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This test prep was uploaded on 04/17/2008 for the course MANAGEMENT ADM 2304 taught by Professor Phansalker during the Winter '05 term at University of Ottawa.
 Winter '05
 Phansalker

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