Session 9

1. Session 9 Tests of Association in two-way tables

2. Learning Objectives By the end of this session, you will be able to • conduct and interpret results from a chi-square test for comparing several proportions or (equivalently) testing the association between two categorical variables • explain how results above can be extended to the study of associations in general r x c tables • state assumptions underlying the above test and actions to take if assumptions fail

3. An example Below is a 5x2 table of observed frequenciesshowing animals who did or did not get diseasedafter inoculation with one of five vaccines. Question: Is there an association between occurrence of disease and type of vaccine?

4. Null and alternative hypotheses To answer the question, need to test H0: disease occurrence is independent of type of vaccine (i.e. proportions diseases are the same for all vaccines) H1: the two variables are associated If H0 is true, best estimate of proportion is diseased total divided by grand total = 225/1350 = 0.167 We use this to compute expected values in each cell of the table under the null hypothesis.

5. Computation of expected values Expected values in the first row: Expected value in cell 1 = (225 / 1350)*280 = (225*280) / 1350 = 46.67 Expected value in cell 2 = (1125 / 1350)*280 = (1125*280) / 1350 = 233.33 Can you calculate expected values in the next row? Check that your 2 numbers add to 250.

6. Table of expected values Note:

7. Chi-square test Now compute the chi-square test-statistic given by If H0 is true, X2 follows a 2 distribution with 4 d.f. Note: d.f.=(r-1)(c-1) where r=number of rows and c=number of columns in the table. Comparing 16.56 to we get a p-value of 0.0024, a highly significant result. We may conclude there is strong evidence of an association between disease occurrence and type of vaccine.

8. Extensions to r × c tables Survey results are often expressed in terms of2-way tables. In general, such tables may containr rows and c columns. Questions of interest in such tables centre on whether these is an association between the two variables that have been tabulated. For example if the table tabulates education level of HH head (none, primary, secondary, tertiary) by poverty levels (not poor, poor, very poor), the question “is poverty related to education” may be asked.

9. Chi-square test for an r × c table To answer the above question, the null hypothesis is that the two variables are NOT related, against the alternative that they are. Under the null hypothesis, comparison of expected values with observed values leads to a chi-square test. The d.f. associated with this test = (r-1)(c-1). In the above example, the d.f.=(4-1)(3-1)=6

10. Assumption underlying the test • The chi-square test is approximate • Validity relies on “large” samples • Small samples of unbalanced data (large and small counts together) may invalidate the approximation • Rules of thumb for validity involve the expected values, E • Need large expected values under H0 • Say, most E5 and none less than 1 • If rule of thumb is not satisfied, may have an unreliable p-value

11. Actions when assumptions fail (a) Simple approaches: • Collect more data if this is possible • Collapse rows or columns if the table has more than two rows/columns. But need to recognise that • this leads to loss of information • with some types of variables, there may be no natural way of combining rows/columns

12. Actions when assumptions fail (b) Use a continuity correction This method is often called Yate’s correction and is applicable just to 2x2 tables. First we show the standard chi-square value corresponding to a table with cell counts a, b, c, d as below. (Verify later that this is correct)

13. Actions when assumptions fail (b) Continuity correction (continued)… The approximation of X2 to the chi-square is improved by reducing the absolute value of O-E by ½ before calculating X2. This results in the X2 taking the value below. Note: The equivalent when comparing two proportions using a z-test is to reduce by ½, the r value for the first p=r/n and increase by ½ the r value for the second proportion.

14. Example of use of continuity corrn Above is the example on Bus data used during the practical sessions. Question of interest is whether the proportion of smokers are different across job types.

15. Example of use of continuity corrn The usual chi-square test leads to X2=1.937 Applying the continuity correction, we get X2 = 1.412 Here, there is little difference because the sample sizes are reasonably large. More important to apply the continuity correction for small sample sizes.

16. Actions when assumptions fail • (c) Use an Exact Test • When actions suggested in (a) or (b) are not possible, consider using an Exact Test. • Details of such tests are beyond the scope of this module. • Some software packages (e.g. Stata) have the facility to perform Fisher’s exact test. SPSS does this only for 2x2 tables. Special software also exist for such tests, e.g. StatXact.

17. Limitations • Chi-square tests are limited, in that only two factors are examined at a time. • This may cause erroneous inferences to be made (see Practical 15 for an example). • The inter-relations between more than two factors can be investigating using more sophisticated statistical techniques, e.g. log-linear modelling.

18. References • Altman, D.G., Machin, D., Bryant, T.N., and Gardner, M.J. (2000) Statistics with confidence. (2nd Edition). BMJ Books, Bristol, UK. pp 240. • Armitage, P., Matthews J.N.S. and Berry G. (2002). Statistical Methods in Medical Research. 4th edn. Blackwell.