1. Categorical Data

2. To identify any association between two categorical data. Example: 1,073 subjects of both genders were recruited for a study where the onset of severe chest pain is recorded for each subject. Variables: - Onset of severe chest pain (+ve / –ve) -Gender (male / female) Categorical Data Analysis

3. Commonly denoted as  2 Useful in testing for independence between categorical variables (e.g. genetic association between cases / controls) Comparison of observed, against what is expected under the null hypothesis. Assumptions Sufficiently large data in each cell in the cross-tabulation table. Chi-square tests

4. In general, require (a) Smallest expected count is 1 or more (b) At least 80% of the cells have an expected count of 5 or more Yate’s Continuity Correction Provides a better approximation of the test statistic when the data is dichotomous (2  2) Small Cell Counts

5. Null hypothesis of a hypothesized distribution for the data. Expected frequencies calculated under the hypothesized distribution. For example: The number of outbreaks of flu epidemics is charted over the period 1500 to 1931, and the number of outbreaks each year is tabulated. The variable of interest counts the number of outbreaks occurring in each year of that 432 year period. E.g. there were 223 years with no flu outbreaks. Goodness-of-fit test

6. Hypotheses: H 0 : Data follows a Poisson distribution with mean 0.692 H 1 : Data does not follow a Poisson distribution with mean 0.692 Note: Mean 0.692 is obtained from the sample mean. Expected frequency for X = 0 =432  P(X = 0), where X ~ Poisson(0.692) Test Statistic, with df = (6 – 1). This yields a p-value of 0.99, indicating that we will almost certainly be wrong if we reject the null hypothesis. Goodness-of-fit test Sample mean = (0 x 223 + 1 x 142 + 2 x 48 + 3 x 15 + 4 x 4 + 5 x 0) / 432 = 0.692

7. Test of independence Most common usage of the Pearson’s chi-square test. H0: The two categorical variables are independent H1: The two categorical variables are associated (i.e. not independent) Under the independence assumption, if outcome A is independent to outcome B, then P(A and B happen jointly) = P(A happen) x P(B happen)

8. Calculating expected frequencies P(Chest pain +) = 83/1073P(Chest pain -) = 990/1073 P(Males) = 520/1073P(Females) = 553/1073 P(Males with chest pain +) = 83/1073 x 520/1073 = 0.0375 Expected(Males with chest pain +) = 1073 x P(.) = 1073 x 0.0375 = 40.224 Observed(Males with chest pain +) = 46

9. Expected frequencies calculated by: Degrees of freedom = (r – 1)  (c – 1) Test of independence

10. Chi-square test

11. Looking at the validity of the assumption of sufficiently large sample sizes!

12.  2 -test identifies whether there is significant association between the two categorical variables. But does not quantify the strength and direction of the association. Need odds ratio to do this. Odds ratio defines “how many times more likely” it is to be in one category compared to the other: Example: For the previous example on severe chest pain, males are about 1.4 times more likely to experience severe chest pains than females. Quantification of the effect Always know what is the outcome/event of interest, and what is the baseline reference! Otherwise OR can be interpreted both ways!

13. Pos. outcomeNeg. outcome Exposure (+)ab Exposure (-)cd Odds ratio and relative risk Calculation of odds ratio is pretty straightforward. - Use the leading diagonal divided by the antidiagonal. Relative risk is more tricky though, since it’s not symmetric! While it’s commonly used interchangeable with OR, the interpretation and calculation are very different!

14. Case-Control Study Compare affected and unaffected individuals Usually retrospective in nature Temporal sequence cannot be established (timing for the onset of the disease) No information on population incidence of the disease Cohort Study Usually random sampling of subjects within the population Prospective, retrospective or both Long follow-up; loss to follow-up Costly to conduct Temporal sequence can be established Provides information on population incidence of the disease Exegesis on epidemiology Odds ratio is the right metric here! Relative risk is the appropriate metric here!

15. Not straightforward to obtain confidence intervals of odds ratio (due to complexity in obtaining the variance) Straightforward to obtain the variance of the logarithm of odds ratio. Odds ratio is always reported together with the p- values (obtained from Pearson’s Chi-square test), and the corresponding confidence intervals. Confidence interval of odds ratio

16. Ca (+ve)Ca (-ve) Smoking (+)1,3011,205 Smoking (-)56152 Odds and Odds Ratio Odds Ratio (OR)=(1301/56)/(1205/152)=2.93 Pearson’s Chi-square= 47.985, on df = 1  p-value= 0 Var[log(OR)]= = 0.026 95% Confidence interval= = (2.14, 4.02) Case study on smoking and lung cancer

17. Beyond 2 x 2 tables

18. Nominal or ordinal For categorical variables with two possible outcomes: - Does not matter whether the variable is nominal or ordinal For categorical variables with more than 2 outcomes: - Important to note whether the variable is nominal or ordinal - Test to use is very different, and thus conclusion reached can be very different. Example: Consider the same dataset on severe chest pain, suppose we have the smoking status of every individual, classified into: - Non-smoker - Daily smoker - Excessive smoker Smoking intensity

19. OR smoker = 1.52 (0.88, 2.63), p = 0.180 OR ex-smoker = 2.11 (1.00, 4.51), p = 0.081 with non-smoker as reference category. Chi-square test for trend

20. Linear-by-linear association Adopts a correlational approach by calculating the Pearson correlation coefficient between the rows and the columns, allowing for ordinal outcomes in either. Recode rows as: yes = 0, no = 1. Recode columns as: non-smoker = 0, daily smoker = 1, ex-smoker = 2

21. Linear-by-linear association 53 observations 19 observations Pearson Correlation = -0.0706 Consider the test statistic: T = (N – 1) r 2 ~ Chi-square(1) = (1050 – 1)  (-0.0706) 2 = 5.2356

22. Nominal vs. Ordinal Importance of recognising the kind of variables we have in order to identify the right test!

23. Summarise data using cross-tabulation tables, with percentages Recognise whether any of the variables are ordinal Perform a chi-square of independence to test for association between the two categorical variables, or the linear-by-linear test if there is at least one ordinal variable out of the two variables Check the validity of the assumption on the sample size Quantify any significant association using odds ratios Always report odds ratios with corresponding 95% confidence interval Procedure for Categorical Data Analysis

24. Categorical Data Analysis in SPSS

25. Example: Let’s consider the lung cancer and smoking example: 1. Establish the relationship between the onset of lung cancer and smoking status. Quantify this relationship if it is statistically significant. Ca (+ve)Ca (-ve) Smoking (+)1,3011,205 Smoking (-)56152

26. Data entry Slightly counter-intuitive, event of interest and outcome of interest should be coded as 0, and the baseline reference outcome/event coded as 1.

27. Define what 0 and 1 corresponds to:

31. Definition of 0s and 1s converted to what you specified under “Values”. Percentages are much easier and more meaningful to interpret than absolute numbers!

32. Highly significant, P < 0.001 Odds ratio of getting lung cancer with corresponding 95% CI, with non-smoker as baseline Relative risk of getting lung cancer with corresponding 95% CI, with non-smoker as baseline

33. understand the use of a chi-square test for testing independence between two categorical outcomes understand the assumptions on sample sizes for the use of a chi-square test know how to quantify the association using odds ratio/relative risk, with corresponding 95% confidence intervals differentiate between the tests to be used for nominal categorical and ordinal categorical variables. perform the appropriate analyses in SPSS and RExcel Students should be able to