## D.29 Answers: Tests for odds ratios

Answers to exercises in Sect. 31.13.

Answer to Exercise 31.1: The missing entries: Odds: 1.15; Percentage: 58.1%.

$$\chi^2=4.593$$; approximately $$z=\sqrt{4.593/1}=2.14$$; expect small $$P$$-value. Software gives $$P=0.032$$. Evidence that the difference between the sample proportions is unlikely to be due to sampling variation. The test is statistically valid.

The sample provides moderate evidence ($$\text{chi-square}=4.593$$; two-tailed $$P=0.032$$) that the population odds of finding a male sandfly in eastern Panama is different at 3 feet above ground (odds: 1.15) compared to 35 feet above ground (odds: 1.71; OR: $$0.67$$; 95% CI from $$0.47$$ to $$0.97$$).

Answer to Exercise 31.2: One option: $$H_0$$: The population OR is one; $$H_1$$: The population OR is not one. From software, $$\chi^2=0.667$$; $$P=0.414$$, which is large. No evidence ($$P= 0.414$$) that the odds of having a smooth scar is different for women and men (chi-square: 0.667). The test is statistically valid.
Answer to Exercise 31.4: One option: $$H_0$$: The population OR is one; $$H_1$$: The population OR is not one. From software, $$\chi^2=3.845$$; $$P=0.050$$. Moderate evidence ($$P= 0.05$$) that the odds of having no rainfall is different for non-positive SOI Augusts and negative-SOI Augusts (chi-square: 3.845). The test is statistically valid.
Answer to Exercise 31.5: 1. $$22/366 \times 100 = 6.0$$%. 2. $$79/386 \times 100 = 20.5$$%. 3. $$22/344 = 0.06395349$$, or about 0.0640. 4. $$79/307 = 0.257329$$, or about 0.257. 5. $$0.257/0.0640 = 4.02$$. 6. $$0.0640/0.257= 0.249$$. 7. From $$0.151$$ to $$0.408$$. 8. $$\chi^2 = 33.763$$% (approximately $$z = 5.81$$) and $$P < 0.001$$. 9. Strong evidence ($$P < 0.001$$; $$\chi^2 = 33.763$$; $$n=752$$) that the odds of wearing hat is different for males (odds: 0.257) and females (odds: 0.0640; OR: 0.249, 95% CI from 0.151 to 0.408). 10. Yes.
Answer to Exercise 31.6: 1. Low exposure (in order): 73.7%, 72.5%, 85.6%. High exposure (in order): 26.3%, 27.5%, 14.4%. 2. In order: 2.80, 2.64, 5.92. 3. Various ways; probably the easiest: $$H_0$$: No association between level of exposure and type of interaction (in the population). 4. Table D.8. 5. Approximately $$z=\sqrt{20.923/2} = 3.23$$: expect small $$P$$-value. 6. Very strong evidence in the sample of an association between level of exposure and type of interaction in the population ($$\chi^2=20.923$$; $$P<0.001$$).
TABLE D.8: The expected counts for the phone-use data