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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 -since the p-value is < ....
View Full Document

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.

Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online