Stat 312 Homework Problems - Week 10

003 006 at the 5 level of significance we reject the

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

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

Unformatted text preview: > 3.16) = 2(.003) = .006. At the 5% level of significance, we reject the null hypothesis and conclude that the data provides sufficient evidence to conclude that there is a difference in the mean shear strength between the two steel types. 26. Because the standard deviations are so similar, the degrees of freedom for the two test types (pooled versus unpooled) are hardly any different. The alternate hypothesis of Ha: 1 - 2 > 0 is one-sided, and the SAS output shows a positive test statistic, so the p-values reported by SAS are twice the actual size. The p-value is therefore .0004. Since this is much smaller than the significance level of .01, we reject the null hypothesis. The data indicates that the mean potential drop for type alloy connections (type 1) exceeds the mean for EC connections (type 2). Since the null hypothesis was rejected, the only possible error that could have been committed was a Type I error. Beverage 29. The data is summarized in the table. The following analysis assumes that the Strawberry compression strength of both beverages is normally distributed. The hypothesis Cola that cola compressive strength is greater gives t...
View Full Document

Ask a homework question - tutors are online