MidtSoln_W08 - Solutions to 2304 Midterm Winter 2008 1 Test...

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Solutions to 2304 Midterm, Winter 2008 1. Test and CI for One Proportion Test of p = 0.06 vs p > 0.06 95% Lower Sample X N Sample p Bound Z-Value P-Value 1 22 250 0.088000 0.058529 1.86 0.031 (a) -Ho: p=.06; Ha: p > .06 -z = (22/250 - .06 )/sqrt(.06*.94/250) = 1.86, -decision rule is reject if z > 1.645 at .05 level of significance -reject Ho and conclude proportion of “defective” worse than .06 (b) -p-value = Prob(z > 1.86 ) = .5 - .4686 = .031 -Would reject Ho since p-value < .05 (c) .088 +/- 1.96 * sqrt(.088*.912/250) (1 mark for stderr) = .088 +/- 1.96 * .018 (1 mark for rest of calc., including crit.val) = .088 +/- .035 = (.05, .12) (d) n = (.088*.912) *(1.96/.02)^2=771 , based on best estimate of p only 1 mark if they assume p=q=.5 and find n = 2401 2. (a) -matched pairs because each pair of before-after tests correspond to the same mgr (b) -referring to the boxplot of differences, this is relatively symmetric and there are no outliers; therefore, it is reasonable to assume the differences came from a
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This note was uploaded on 10/16/2009 for the course MANAGEMENT ADM 2304 taught by Professor Phansalker during the Winter '05 term at University of Ottawa.

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MidtSoln_W08 - Solutions to 2304 Midterm Winter 2008 1 Test...

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