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 (\(\chi^2=20.923\); \(P<0.001\)).
| Answer call | Respond to text | Reply to email | |
|---|---|---|---|
| Low exposure | 275.7 | 275.7 | 272.61 |
| High exposure | 81.3 | 81.3 | 80.39 |