finalanswersspring08

finalanswersspring08 - STA 103 Final Exam Spring 2008 I. H....

Info iconThis preview shows pages 1–4. 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

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: STA 103 Final Exam Spring 2008 I. H. Dinwoodie NAME Calculator and formula sheet allowed. STA103 Spring 2008 Final 1 1. (6 points) Two stocks had daily returns below in a 5 day week: day s1 s2 1 0.07 0.04 2 0.17 0.14 3-0.12 -0.08 4-0.02 -0.08 5 0.07 0.05 (a) What is the sample correlation? r=0.945 (b) Test the null hypothesis that the correlation is 0, against the alternative H a : > 0, at level = . 01. Regression output from fitting the line s2=a + b*s1 was Parameter Estimates Estimate Std Error t Ratio Prob > |t| Intercept-0.01373 0.01680-0.817 0.4737 s1 0.81559 0.16243 5.021 0.0152 Be sure to give the p-value for the test and the test conclusion. Testing H : = 0 versus H a : > 0 is the same as testing for positive slope with H : = 0 , H a : > 0 (recall problem 15.3 and our discussion with JMP output). The output above is the way JMP presents regression, and the p-value .0152 for the slope is two-sided, although the notation is a bit misleading. Then the one-sided p-value is .0076, so at level .01, we reject = 0 (.0076 < .01). STA103 Spring 2008 Final 2 Choose the best answer. (3 points each) 2. Women in 1975 in a certain city had normally distributed incomes that averaged W , with variance 2 . A sample of 100 was taken randomly and it was found that x = 19 . 0 and s x = 5 . 0 (in $1,000). Men in 1975 in that city had normally distributed incomes that average M , and a sample of 100 gave y = 21 . 5 and s y = 6 . 0. The standard error of the estimate 21 . 5- 19 . 0 for M- W is estimated by (a) 5.52 (b) 0.78 X (c) 0.50 (d) 0.55 (e) 0.05 3. A ski lift is designed to hold 20,000 pounds, and claims a capacity of 100 persons. Suppose the weights of all people using the lift have a mean of 190 pounds and with a standard deviation of 45 pounds. What is the probability that a random group of 100 people will total more than the weight limit of 20,000 pounds?...
View Full Document

Page1 / 11

finalanswersspring08 - STA 103 Final Exam Spring 2008 I. H....

This preview shows document pages 1 - 4. Sign up to view the full document.

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