1. FPP 28 Chi-square test

2. More types of inference for nominal variables Nominal data is categorical with more than two categories Compare observed frequencies of nominal variable to hypothesized probabilities One categorical variable with more than two categories Chi-squared goodness of fit test Test if two nominal variables are independent Two categorical variables with at least one having more than two categories Chi-squared test of independence

3. Goodness of fit test Do people admit themselves to hospitals more frequently close to their birthday? Data from a random sample of 200 people admitted to hospitals Days from birthday Number of admissions within 711 8-3024 31-9069 91+96

4. Goodness of fit test Assume there is no birthday effect, that is, people admit randomly. Then, Pr (within 7) = =.0411 Pr (8 - 30) = =.1260 Pr (31-90) = =.3288 Pr (91+) = =.5041 So, in a sample of 200 people, we’d expect to be in “within 7” to be in “8 - 30” to be in “31 - 90” to be in “91+”

5. Goodness of fit test If admissions are random, we expect the sample frequencies and hypothesized probabilities to be similar But, as always, the sample frequencies are affected by chance error So, we need to see whether the sample frequencies could have been a plausible result from a chance error if the hypothesized probabilities are true. Let’s build a hypothesis test

6. Goodness of fit test Hypothesis Claim (alternative hyp.) is admission probabilities change according to days since birthday Opposite of claim (null hyp.) is probabilities in accordance with random admissions. H 0 : Pr (within 7) =.0411 Pr (8 - 30) =.1260 Pr (31-90) =.3288 Pr (91+) =.5041 H A : probabilities different than those in H 0.

7. Goodness of fit test: Test statistic Chi-squared test statistic

8. Goodness of fit test: Test statistic CellObsExpDifDif 2 Dif 2 /Exp In 7 8-30 31-90 91+

9. Goodness of fit test: Calculate p- value X 2 has a chi-squared distribution with degrees of freedom equal to number of categories minus 1. In this case, df = 4 – 1 = 3.

10. Goodness of fit test: Calculate p- value To get a p-value, calculate the area under the chi-squared curve to the right of 1.397 Using JMP, this area is 0.703. If the null hypothesis is true, there is a 70% chance of observing a value of X 2 as or more extreme than 1.397 Using the table the p-value is between 0.9 and 0.70

11. Chi-squared table

12. JMP output admissions

13. Goodness of fit test: Judging p- value The.70 is a large p-value, indicating that the difference between the observed and expected counts could well occur by random chance when the null hypothesis is true. Therefore, we cannot reject the null hypothesis. There is not enough evidence to conclude that admissions rates change according to days from birthday.

14. Independence test Is birth order related to delinquency? Nye (1958) randomly sampled 1154 high school girls and asked if they had been “delinquent”. Eldest24450 In Between29312 Youngest35211 Only2370

15. Sample of conditional frequencies % Delinquent for each birth order status Based on conditional frequencies, it appears that youngest are more delinquent Could these sample frequencies have plausibly occurred by chance if there is no relationship between birth order and delinqeuncy Oldest.05 Middle.085 Youngest.14 Only.25

16. Test of independence Hypotheses Want to show that there is some relationship between birth order and delinquency. Opposite is that there is no relationship. H 0 : birth order and delinquency are independent. H A : birth order and delinquency are dependent.

17. Implications of independence Expected counts Under independence, Pr(oldest and delinquent) = Pr(oldest)*Pr(delinquent) Estimate Pr(oldest) as marginal frequency of oldest Estimate Pr(delinquent) as marginal frequency of delinquent Hence, estimate Pr(oldest and delinquent) as The expected number of oldest and delinquent, under independence, equals This is repeated for all the other cells in table

18. Test of independence Expected counts Next we compare the observed counts with the expected to get a test statistic Oldest45.59428.41 In Between32.80308.2 Youngest23.66222.34 Only8.9584.05

19. Use the X 2 statistic as the test statistic:

20. Test of independence: Calculate the p-value X 2 has a chi-squared distribution with degrees of freedom: df = (number rows – 1) * (number columns – 1) In delinquency problem, df = (4 - 1) * (2 - 1) = 3. The area under the chi-squared curve to the right of 42.245 is less than.0001. There is only a very small chance of getting an X 2 as or more extreme than 42.245.

22. JMP output for chi-squared test This is a small p-value. It is unlikely we’d observe data like this if the null hypothesis is true. There does appear to be an association between delinquency and birth order.

23. Chi-squared test details Requires simple random samples. Works best when expected frequencies in each cell are at least 5. Should not have zero counts How one specifies categories can affect results.

24. Chi-squared test items What do I do when expected counts are less than 5? Try to get more data. Barring that, you can collapse categories. Example: Is baldness related to heart disease? (see JMP for data set) Baldness Disease Number of people None Yes 251 None No 331 Little Yes 165 Little No 221 Some Yes 195 Some No 185 Combine “extreme” and “much” categories Much Yes 50 Much or extreme Yes 52 Much No 34 Much or extreme No 35 Extreme Yes 2 Extreme No 1 This changes the question slightly, since we have a new category.

25. Chi-squared test for collapsed data for baldness example Based on p-value, baldness and heart disease are not independent. We see that increasing baldness is associated with increased incidence of heart disease.