solpracfinals06

solpracfinals06 - PSTAT 120C: Solutions to Practice Final...

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: PSTAT 120C: Solutions to Practice Final June 9, 2006 1. (a) The critical value for a 2 distribution with (4- 1)(5- 1) = 12 degrees of freedom is 21.026. Therefore this result is not significant. There is not sufficient evidence to prove that the freshmen or juniors attended different numbers of lectures compared to the seniors or sophomores. (b) The marginal totals are Number of lectures attended 1 2 3 4 Total Freshmen 7 13 23 12 15 70 Sophomores 14 19 20 4 13 70 Juniors 15 15 17 3 10 60 Seniors 16 10 12 7 5 50 Total 52 57 72 26 43 250 To check that the expected values are all greater than 5 it is only necessary to check the smallest row and column 26 50 250 = 5 . 2 > 5 so that the 2 approximation is appropriate. 2. The expected values under the null hypothesis are Event A B C D Observed 34 30 12 24 Expected 32 32 18 18 The chi-squared statistic is 2 = 4 32 + 4 32 + 36 18 + 36 18 = 4 . 25 This 2 has two degrees of freedom because we estimated two means. The value 4.15 is not significant when compare to the critical value 2 . 95 , 2 = 5 . 991. Therefore, we accept the991....
View Full Document

Page1 / 4

solpracfinals06 - PSTAT 120C: Solutions to Practice Final...

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