# chap11 - Chapter 11 Tests of Homogeneity and Independence...

This preview shows pages 1–3. Sign up to view the full content.

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: Chapter 11 Tests of Homogeneity and Independence 11.1 Study Suggestions Chapter 11 presents the chi-squared test of homo- geneity (or of independence) and gives a few exam- ples of its use. Following the approach of Chapter 7, emphasis is placed on the type of study, and it is argued that the chi-squared test of homogeneity is appropriate in three different situations. You should understand the rationale behind the formula for the expected count for any cell of the table, namely the product of the row total and the column total divided by the sample size. You also should understand why ( O- E ) 2 E is a reasonable way to measure the amount of evi- dence in a particular cell in support of the alterna- tive hypothesis (the argument appeared in Chapter 10). The computation of the test statistic for the chi- squared test of homogeneity is too tedious to perform by hand. I strongly recommend the use of a com- puter! Make sure you can read computer output of the type given in the text for the test of homogeneity, and that, given the value of the test statistic, you can find the P-value for the test of homogeneity from the table of the chi-squared distributions. If the study has three or more treatments or popu- lations, and the test of homogeneity yields a P-value less than or equal to 0.05, then you must under- stand that you only can conclude that the treatments (populations) are not identical. You cannot conclude which treatments (populations) differ without further testing. The text provides one strategy for testing, but many statisticians believe this problem requires a more sophisticated approach than I advocate. 11.2 Solutions to Odd-Numbered Exercises 5. (c) E = 132(178) / 449 = 52 . 3 . (d) ( O- E ) 2 E = (71- 52 . 3) 2 52 . 3 = 6 . 69 . (f) For a 3 × 2 contingency table, the number of degrees of freedom of the reference chi- squared curve is 2(1) = 2 . The approx- imate P-value is the area to the right of 35.063 under the chi-squared curve with two degrees of freedom. This area is smaller than 0.01. The data are highly statistically significant, and the researcher should reject the null hypothesis that the three populations are identical. Next, the researcher will want to investigate where the differences lie and can do this by mak- ing pairwise comparisons of the popula- tions. (g) First, I will compare the population corre- sponding to no injections with the popu- lation corresponding to an average of 1– 100 injections per month. The standard- ized value of the test statistic for Fisher’s test is z = √ 316[9(144)- 66(98)] p 75(242)(107)(210) =- 4 . 56 . 77 78 CHAPTER 11. TESTS OF HOMOGENEITY AND INDEPENDENCE The approximate P-value for the third al- ternative is twice the area to the right of 4.56 under the standard normal curve....
View Full Document

## This note was uploaded on 10/23/2009 for the course STAT STATS 371 taught by Professor Professorwardrop during the Fall '09 term at University of Wisconsin.

### Page1 / 8

chap11 - Chapter 11 Tests of Homogeneity and Independence...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online