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.
|Odds with smooth scars||Percentage with smooth scars||Sample size|
- 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 0.389 times the odds of a 3000-hr turbine getting a fissure. 4. CI from 0.133 to 1.14. Plausible values for the population OR that may have produced the sample OR likely to be between these values.
|About 1800 hours||7||66||73|
|About 3000 hours||9||33||42|
|Odds with fissures||Percentage with fissures||Sample size|
|About 1800 hours||0.1061||9.59%||73|
|About 3000 hours||0.2727||21.43%||42|
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).