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

	Answer call	Respond to text	Reply to email
Low exposure	275.7	275.7	272.61
High exposure	81.3	81.3	80.39