# Hw5a - df = 16 t =-2.33 p = 0.016 or 0.018 if they are...

This preview shows page 1. Sign up to view the full content.

STAT 350 Homework #5 Solution Addendum 7.46 – Test for equal variances: p – value = 0.027. Therefore it is NOT appropriate to assume equal variances 7.48 - Test for equal variances: p – value = 0.015. Therefore it is NOT appropriate to assume equal variances 7.52 – this problem should be analyzed using a paired t-test, so the issue of equal variances is irrelevant. 8.26 – Test for equal variances: p -value = 0.205, therefore it IS appropriate to use the test with equal variances. Therefore the following would also be acceptable. Then the following would also be acceptable:
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: df = 16, t = -2.33 p = 0.016 ( or 0.018 if they are using the t table from the text) Fail to reject null hypothesis at α = 0.01. Two-Sample T-Test and CI Sample N Mean StDev SE Mean 1 8 1.710 0.530 0.19 2 10 2.530 0.870 0.28 Difference = mu (1) - mu (2) Estimate for difference: -0.820 95% upper bound for difference: -0.207 T-Test of difference = 0 (vs <): T-Value = -2.33 P-Value = 0.016 DF = 16 Both use Pooled StDev = 0.7407 8.40- this problem should be analyzed using a paired t-test, so the issue of equal variances is irrelevant....
View Full Document

## This note was uploaded on 03/09/2010 for the course STAT 350 taught by Professor Staff during the Spring '08 term at Purdue.

Ask a homework question - tutors are online