D.24 Answers: CIs for odds ratios

Answers to exercises in Sect. 25.9.

Answer to Exercise 25.1: 1. \(99/62 = 1.596774\); about 1.60. 2. \(216/115 = 1.878261\); about 1.88. 3. \(1.596774/1.878261 = 0.850\), as in the output. 4. A few ways; for example: For every 100 men with a smooth scar, about 85 women with a smooth scar. 5. (Graph not shown, but use a stacked or side-by-side bar chart.) 6. Table D.4. 7. Exact 95% CI for the OR, from the output: 0.576 to 1.255. 8. If study repeated study many times (with the same numbers of men and women), about 95% of the CIs would contain population OR. In practice: population OR is probably between 0.576 and 1.255.

TABLE D.4: The odds and percentage of having smooth scars, for women and men

	Odds with smooth scars	Percentage with smooth scars	Sample size
Women:	1.60	61.5%	161
Men:	1.88	65.3%	331
Odds ratio:	0.850

Answer to 25.2: The output can be interpreted in one of two ways (Sect. 25.2):

• Odds are the odds of swimming at the beach; OR compares these odds between those without an ear infection, to those with an ear infection.
• Odds are the odds of not having an ear infection; OR compares these odds for beach swimmers to non-beach swimmers.

Answer to 25.3: 1. Table D.5. 2. Table D.6. 3. OR: Odds of a 1800-hr turbine getting a fissure is 0.352 times the odds of a 3000-hr turbine getting a fissure. 4. CI from 0.13 to 1.14. Plausible values for the population OR that may have produced the sample OR likely to be between these values.

TABLE D.5: The number of fissures for two sets of turbines, run for different numbers of hours

	Fissures	No fissures	Total
About 1800 hours	7	73	80
About 3000 hours	9	33	42
Total	16	106	122

TABLE D.6: The numerical summary for the fissures data

	Odds with fissures	Percentage with fissures	Sample size
About 1800 hours	0.0959	8.75%	80
About 3000 hours	0.2727	21.43%	42
Odds ratio	0.352

Answer to Exercise 25.4: Odds of no rainfall (non-positive SOI): \(14/40 = 0.35\). Odds of no rainfall (negative SOI): \(7/53 = 0.1320755\). Required OR is \(0.35/0.1320755 = 2.65\), as in output. 95% CI from 0.979 to 7.174.

Answer to Exercise 25.5: The 95% CI is from 0.151 to 0.408. The OR of not wearing a hat, comparing males to females (malesless likely to be not wearing a hat; rewording, males more likely to be wearing a hat).