1. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. Relationships Between Categorical Variables Chapter 6

2. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 2 Is there a relationship between the two variables, so that the category into which individuals fall for one variable seems to depend on the category they are in for the other variable?

3. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 3 Example 6.1 Smoking and Divorce Risk Data on smoking habits and divorce history for the 1669 respondents who had ever been married. Among smokers, 49% have been divorced, 51% have not. Among nonsmokers, only 32% have been divorced, 68% have not. The difference between row percents indicates a relationship.

4. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 4 Number in category Total number in group Risk = Relative Risk = Risk in category 1 Risk in category 2 Percent increase in risk = (relative risk – 1) x 100%

5. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 5 Odds = Number in category 1 to Number in category 2 = (Number in category 1/Number in category 2) to 1 Odds Ratio = (Odds for group 1) / (Odds for group 2)

6. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 6 6.3 Misleading Statistics About Risk Questions to Ask: What are the actual risks? What is the baseline risk? What is the population for which the reported risk or relative risk applies? What is the time period for this risk?

7. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 7 Example 6.4 Disaster in the Skies? Case Study 1.2 Revisited Look at risk of controller error per flight: In 1998: 5.5 errors per million flights In 1997: 4.8 errors per million flights “Errors by air traffic controllers climbed from 746 in fiscal 1997 to 878 in fiscal 1998, an 18% increase.” USA Today Risk of error increased but the actual risk is very small.

8. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 8 Example 6.5 Dietary Fat and Breast Cancer Two reasons info is useless: 1.Don’t know how data collected nor what population the women represent. 2.Don’t know ages of women studied, so don’t know baseline rate. “Italian scientists report that a diet rich in animal protein and fat – cheeseburgers, french fries, and ice cream, for example – increases a woman’s risk of breast cancer threefold.” Prevention Magazine’s Giant Book of Health Facts (1991, p. 122).

9. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 9 Example 6.5 Dietary Fat and Breast Cancer (cont) Age is a critical factor. Accumulated lifetime risk of woman developing breast cancer by certain ages: By age 50: 1 in 50 By age 60: 1 in 23 By age 85: 1 in 9 Annual risk 1 in 3700 for women in early 30’s. If Italian study was on very young women, the threefold increase in risk represents a small increase.

10. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 10 6.4 The Effect of a Third Variable and Simpson’s Paradox Example 6.7 Educational Status and Driving after Substance Use 1996 nationwide survey of 11,847 individuals 16 or over. Response was Driving Status with 3 categories: Unimpaired = never drove while impaired Alcohol = drove within 2 hours of alcohol use, but never after drug use Drug = drove within 2 hours of drug use and possibly after alcohol use.

11. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 11 Example 6.7 Educational Status and Driving after Substance Use As amount of education increases, the proportion who drove within two hours of alcohol use also increases. One difference between the educ. groups is age. Not enough information on age so age is a lurking variable.

12. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 12 Example 6.8 Blood Pressure and Oral Contraceptive Use Hypothetical data on 2400 women. Recorded oral contraceptive use and if had high blood pressure. Percent with high blood pressure is about the same among oral contraceptive users and nonusers.

13. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 13 Example 6.8 Blood Pressure and Oral Contraceptive Use (cont) Many factors affect blood pressure. If users and nonusers differ with respect to such a factor, the factor confounds the results. Blood pressure increases with age and users tend to be younger. In each age group, the percentage with high blood pressure is higher for users than for nonusers => Simpson’s Paradox.

14. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 14 6.5 Assessing the Statistical Significance of a 2x2 Table Question: Can a relationship observed in the sample data be inferred to hold in the population represented by the data? A statistically significant relationship or difference is one that is large enough to be unlikely to have occurred in the observed sample if there is no relationship or difference in the population.

15. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 15 Five Steps to Determining Statistical Significance: 1.Determine the null and alternative hypotheses. 2.Verify necessary data conditions, and if met, summarize the data into an appropriate test statistic. 3.Assuming the null hypothesis is true, find the p-value. 4.Decide whether or not the result is statistically significant based on the p-value. 5.Report the conclusion in the context of the situation.

16. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 16 Step 1: Null and Alternative Hypotheses null hypothesis: The two variables are not related. alternative hypotheses: The two variables are related.

17. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 17 Step 2: The Chi-square Statistic Chi-square statistic measures the difference between the observed counts and the counts that would be expected if there were no relationship. Large difference => evidence of a relationship. Compute expected count for each cell: Expected count = (Row total)  (Column total) Total n for table Compute for each cell: (Obs count – Exp count) 2 Exp count Compute test statistic by totaling over all cells: (Obs count – Exp count) 2 Exp count

18. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 18 Step 3: The p-value of the Chi-square Test Q: If there is actually no relationship in the population, what is the likelihood that the chi-square statistic could be as large as it is or larger? A: The p-value Large test statistic => evidence of a relationship. So how large is enough to declare significance? Note: The p-value is generally reported in computer output.

19. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 19 Steps 4 and 5: Making and Reporting a Decision Common rule: p-value  0.05 => say relationship is statistically significant and we reject the null hypothesis p-value > 0.05 => cannot say relationship is statistically significant and we cannot reject the null hypothesis Large test statistic => small p-value => evidence a real relationship exists in the population. Note: For 2  2 tables, a test statistic of 3.84 or larger is significant.

20. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 20 Example 6.10 Randomly Pick S or Q Of 92 college students asked: “Randomly choose one of the letters S or Q”, 66% (61/92) picked S. Of another 98 students asked: “Randomly choose one of the letters Q or S”, 46% (45/98) picked S. Can we conclude order of letters on the form and the response are related? The p-value = 0.005 which is less than 0.05, so the relationship is statistically significant.

21. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 21 Factors that Affect Statistical Significance The strength of the observed relationship Example 6.10 Of those with “S or Q”, 66% picked S. Of those with “Q or S”, 46% picked S. Difference in percentages (66% - 46%) reflects the strength of the observed relationship.

22. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 22 Factors that Affect Statistical Significance: (cont) How many were studied (sample size) Example: I.Treatment A had 8 of 10 patients improve. Treatment B had 5 of 10 patients improve. Strength = 80% - 50% = 30% seems large but study is too small. The p-value is 0.16. II.Treatment A had 80 of 100 patients improve. Treatment B had 50 of 100 patients improve. Strength = 80% - 50% = 30% is again large. The p-value is 0.000000087, which is very significant.

23. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 23 Practical versus Statistical Significance Statistical Significance does not mean the relationship is of practical importance. Example 6.12 Aspirin and Heart Attacks p-value is 0.000 => relationship is statistically significant. Placebo: 189/11034 = 1.71% had attack Aspirin: 104/11037 = 0.94% had attack Difference only 1.71 – 0.94 = 0.77%, or less than 1%. With large sample this important difference was detected.

24. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 24 Interpreting a Nonsignificant Result The sample results are not strong enough to safely conclude that there is a relationship in the population. The observed relationship could have resulted by chance, even if there is no relationship in the population. This is not the same as saying there is no relationship.

25. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 25 Case Study 6.1 Drinking, Driving, and the Supreme Court “Random Roadside Survey” of drivers under 20 years of age. p-value is 0.201 => the observed association could easily have occurred even if there is no relationship in the population. This result was used by Supreme Court to overturn a law that allowed sale of beer to females but not males.