2007fallsolutions

# 2007fallsolutions - Solutions to Midterm Fall 07 1 a-Ho...

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1 Solutions to Midterm Fall 07 1. a. -Ho: P_roc = P_que; Ha: P_c > P_q. -p-bar = (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 = 2.30 -Reject 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 1-sample t-test, but must be a Wilcoxon test of median. b. -Ho: Median = 44; Ha: median < 44 -p-value is .001 from output -since the p-value is < .05, we reject the null H and conclude that 44 is too high for the average number of photos c. the CI is x_bar +/- t(alpha/2) * s/sqrt(n) or 39.55 +/- 2.093 * 7.95 or 39.55 +/- 16.6 you should be uncomfortable with this CI since the value 2.093 is dependent on the assumption that the data come from a normally distributed population which is clearly questionable.

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