chapter19

# chapter19 - Case Studies 2 × 2 Tables of Counts The...

This preview shows pages 1–8. Sign up to view the full content.

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

View Full Document

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

View Full Document

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

View Full Document

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

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

Unformatted text preview: Case Studies 2 × 2 Tables of Counts The Chi-Squared Test r × c Tables of Counts Chapter 19: More Tools for Tables of Counts STAT 3022 Fall 2011 University of Minnesota November 28, 2011 Case Studies 2 × 2 Tables of Counts The Chi-Squared Test r × c Tables of Counts Introduction Data in Chapter 18 can be displayed as 2 × 2 tables of counts. Cross-classification of row factor (e.g., obese or not obese) column factor (e.g., heart disease or no heart disease) This chapter focuses on sampling schemes leading to tables of counts appropriate wording of hypotheses, based on sampling scheme & question of interest extensions to larger tables Case Studies 2 × 2 Tables of Counts The Chi-Squared Test r × c Tables of Counts Sex Role Stereotypes and Personnel Decisions Example Male bank supervisors were asked to decide whether to promote a hypothetical personnel file. half described a female candidate, half described a male files were identical in all other respects Promoted Yes No Male 21 3 Female 14 10 Was there bias against female applicants? Or can observed difference in promotion rates be attributed to chance? Case Studies 2 × 2 Tables of Counts The Chi-Squared Test r × c Tables of Counts (continued) Example Wish to test H : π m- π f = 0 vs . H a : π m- π f > Sample proportions are ˆ π m = 21 24 = 0 . 875 , ˆ π f = 14 24 = 0 . 583 Difference is ˆ π m- ˆ π f = 0 . 292 Proportion from combined sample is ˆ π c = 21 + 14 24 + 24 = 0 . 729 Case Studies 2 × 2 Tables of Counts The Chi-Squared Test r × c Tables of Counts (continued) Example Check adequacy of Normal approximation: 24( . 729) = 17 . 5 , 24( . 271) = 6 . 5 SE (ˆ π m- ˆ π f ) (test version) is r ( . 729)( . 271) 24 + ( . 729)( . 271) 24 = 0 . 128 z-statistic is z = ˆ π m- ˆ π f SE (ˆ π m- ˆ π f ) = . 292 . 128 = 2 . 281 What is the one-sided p-value? What do we conclude? Case Studies 2 × 2 Tables of Counts The Chi-Squared Test r × c Tables of Counts Hypotheses of Homogeneity Chapter 18 presented tests for identical hypotheses: H : π 1- π 2 = 0 and H : ω 1 /ω 2 = 1 , where π 1 and π 2 are population proportions or probabilities, ω 1 and ω 2 are corresponding odds. Definition A hypothesis of homogeneity is one where interest is focused on whether the distribution of the binary response is homogeneous (the same) across populations. Case Studies 2 × 2 Tables of Counts The Chi-Squared Test r × c Tables of Counts Hypotheses of Independence We could test the hypothesis H : row categorization is independent of column categorization Definition A hypothesis of independence is used to investigate an association between row and column factors, without specifying one of them as a response....
View Full Document

{[ snackBarMessage ]}

### Page1 / 36

chapter19 - Case Studies 2 × 2 Tables of Counts The...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online