# MidtSoln_F09 - Solutions to Midterm Fall 2009 Note: Many...

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Solutions to Midterm Fall 2009 Note: Many students are using a sheet of notes that may be misleading. It suggests the general decision rule “reject Ho if |statistic| > |critical value|”, but this should be used only for 2-sided tests since it defines a 2-tailed rejection region. However, it can lead to incorrect decisions for 1-sided tests unless you also ensure that the statistic is positive for the “>” alternative and negative for the “<” alternative. It is easier and safer to use the rejection regions “z > z α ” and “z < -z α ”, which clearly define upper and lower tail rejection regions, for the “>” and “<” alternatives, respectively. Question One. This was originally conceived as a comparison of two proportions. Test and CI for Two Proportions Sample X N Sample p 1 149 383 0.389034 2 115 383 0.300261 Difference = p (1) - p (2) Estimate for difference: 0.0887728 95% lower bound for difference: 0.0325302 Test for difference = 0 (vs > 0): Z = 2.58 P-Value = 0.005 a. -Ho: p1-p2=0; Ha: p1-p2>0 or p2-p1<0 -p-hat = (149+115)/(383+383) = .345 is pooled proportion -z = (.39 - .30)/sqrt(.345*(1-.345)*(2/383)) = 2.62 -reject Ho if z > 1.645 -conclude drop in Liberal support b. p-value = Prob(Z > 2.62) = .5-.4956 = .0044 c. at least .09 - 1.645*sqrt(.39*.61/383 +.3*.7/383) = .09 – 1.645 * .034 = .034 (lower limit) d. p-value < .05 and 1-sided interval does not cover p1-p2 = 0, both leading to the same decision. e. p-hat is assumed to be normally distributed—this is approximately true if np 5 and nq 5. It is incorrect to say that the data are normally distributed—a binomial population with

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## This note was uploaded on 01/10/2010 for the course MANAGEMENT ADM 2304 taught by Professor Phansalker during the Fall '05 term at University of Ottawa.

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MidtSoln_F09 - Solutions to Midterm Fall 2009 Note: Many...

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