D.22 Answers: CIs for paired data
Answers to exercises in Sect. 23.12.
Answer to Exercise 23.1: Mean of the differences: 5.2; standard error 3.6. Approximate 95% CI: \(5.2 \pm (2\times 3.06)\), or \(5.2\pm 6.12\), from -0.92 to 11.22. Mean taste preference between preferring it better with dip by up to 11.2mm on the 100mm visual analog scale, or preferring it without dip by a little (up to -0.9mm on the 100mm visual analog scale. (Understanding how the differences are defined is needed to understand where this came from.)
A useful summary might be like Table D.2.| Mean | Standard deviation | Standard error | |
|---|---|---|---|
| Raw | 56 | 26.6 | 2.64679892595857 |
| With dip | 61.2 | 28.7 | 2.85575673590267 |
| Differences | 5.2 | 3.06 |
Answer to Exercise 23.2:
1. Computing differences as
Before minus the After measurements seems sensible:
the average blood pressure decrease, the purpose of the drug.
2. The differences (when defined asreductions): 9, 4, 21, 3, 20, 31, 17, 26, and so on.
3. Mean difference: 18.933; standard deviation: 9.027;
standard error: \(9.027/\sqrt{15} = 2.331\).
Approximate 95% CI: \(14.271\) to \(23.56\) mm Hg.
4. Exact 95% CI: 13.934 to 23.93 mm Hg from output.
5. The first uses approximate multipiers.
The second uses exact multipliers.
Answer to Exercise 23.3:
1. Approximate 95% CI for reduction: \(0.66 \pm(2\times 0.37)\), or -0.08 to 1.4:
average could be an increase of up to 0.08 to a reduction of up to 1.4
on the given scale for women.
2. Sample size is not larger than 25, but close:
probably reasonably statistically valid.