9B Key Even 

9B.2 Telephone survey non-response

(A) Proportions contacted

• Point estimates: p-hat₁ = 200 / 291 = 0.6873 and p-hat₂ = 58 / 100 = 0.5800. Difference = 0.1073 or about 11%. 
• Hypothesis test: The overall p-hat = (200 + 58) / (291 + 100) = 0.6598; SE = 0.05492 and z_stat = 1.95, Two-sided P-value = 0.0512 [not continuity corrected]. This provides good evidence the increase was not random. 
• 95% confidence interval by Wilson score method:  -0.000128 to 0.217489 (0.0% to 21.7%)
• Size of effect / interpretation: Although the observed difference is not likely to be explained by random sampling chance, the data are consistent with a small or negligible change. The lower limit of the confidence interval is about 0%. Therefore, the effects may be quite modest, although further study is warranted.

(B) Proportions completing survey

• Point estimates: p-hat₁ = 134 / 291 = 0.4605; p-hat₂ = 33 / 100 = 0.3300; Difference in proportions = 0.1305 or about 13%
• Hypothesis test: p-hat_pooled = 0.4271; SE = 0.05734; z_stat = 2.28; P = 0.0226; good evidence for a non-random difference
• 95% confidence interval: 95% confidence interval by the plus-four method 0.0196 to 0.2352 or from about a 2% to 24% improvement. 
• Size of effect / interpretation: Although the difference is "significant" the confidence interval indicates that the size of the difference could very well be as small as 2%. At the same time, it could be quite large (24%). Therefore, further study is warranted.

9B.4 The "Father of Medical Statistics".

(A) Mortality proportion in group bleed early₁ = 18 / 41 = 0.4390 (about 44%);  mortality in group bleed late ₂ = 9 / 36 = 0.2500 (25%). 

• Hypotheses: H₀:  p₁ = p₂ vs. H₁:  p₁ ≠  p₂ (one-sided H₁ would be H₁:  p₁ >  p₂)

• Test statistic: p-hat_pooled = (18 + 9) / (41 + 36) = 0.3506; q-hat_pooled  = 1 = 0.3506 = 0.6494; SE = sqrt[(0.3506)(0.6494)(41^-1 + 36^-1) = 0.1090; z_stat = (0.4390 - 0.2500) / 0.1090 = 1.73

• One-sided P = 0.0418; two-sided P = 0.0836

• Interpretation: The difference is significant (one-sided) or marginally significant (two-sided). It is not likely the observed difference is due to sampling chance. 

(B) ₁  - ₂ = 0.1890 (95% C.I. for  p₁ = p₂: -0.0242 to 0.3770). This means the risk difference is about 19% (19% greater risk in those bled early). The confidence interval suggests the risk difference may be anywhere between −2% and 38%, suggesting the result is quite imprecise.

9B.6. Induction of labor

• Proportions: (estimated risks of meconium staining): Induced ₁ = 1 / 111 = 0.0090. Non-induced ₂  =   13 / 117 = 0.1111
• Risk difference: ₁  - ₂  =   0.0090 - 0.1111 = -0.1021 (about a 10% reduction)
• 95% confidence interval for the risk difference p₁ - p₂ = -0.172 to -0.042 (by Wilson score method, which should be similar to the plus-four method) We can say with 95% confidence that risk is reduced between 4% and 17%.

9B.8 Cytarabine and cerebellar toxicity. 

• Risk difference = ₁  - ₂  = (11 / 25)  - (3 / 34) = 0.4400 - 0.0882 = 0.3518
• 95% confidence interval for  p₁ - p₂ by the plus-four method = 0.3333 ± (1.96)(0.10903) = 0.3333 ± 0.2137 = (0.120, 0.547)
• P = 0.002

9B.10 Framingham Heart Study. 

(A) Risk in high cholesterol women ₁  = 30 / 689 = 0.04354; ₂ = 8 / 445 = 0.02556; Risk difference ₁  - ₂  = 0.04354 - 0.01798 = 0.02556 (2.6%); 95% confidence interval for the risk difference by the Wilson score method = 0.004166 to 0.045560 (0.4% to 4.6%).
(B) Risks were much higher for men: For high-cholesterol men and women in the cohort: Six-year risk, men = 0.12028 (12.0%) while six-year risk, women = 0.04354 (4.3%). For low-cholesterol men and women: Six-year risk, men = 0.03524 (3.5%), while six-year risk women = 0.01798 (1.7%). 

9B.12 Scandinavian Simvastatin Survival Study (4S).₁ = 182 / 2221 = 0.0819 (8.2%), ₂ = 256 / 2223 = 0.1152 (11.5%), In testing H₀:  p₁ = p₂ = p, z_stat = 3.67, P = 0.000 [ 2.0E-4 ] (two-tailed).

9B.14 UGDP. ₁ = 26 / 204 = 0.1275 and ₂ = 2 / 64 = 0.03125 ; P = 0.028 (not continuity-corrected) or P = 0.050 (continuity-corrected). The phenformin group had a significantly higher risk  (13% vs. 3%).

9B.16. Drug testing student athletes. ₁ = 5 / 95 = 0.05263; ₂ = 12 / 62 = 0.1935; z_stat = 2.78; P = 0.0054 (two-tailed) [with continuity correction:  z_stat,cc = 2.57, P = 0.010]

9B.18. Prevalence of cigarette use among racial/ethnic populations. 

(A) The sampling distribution of ₁  - ₂ has expected value =  p₁ − p₂ = 0.40 - 0.12 = 0.28. The distribution will be Normal (assuming samples are large) with standard deviation SE = sqrt[(.4)(.6)/1000 + (.12)(.88)/1000] = 0.01859. In shorthand, ₁  - ₂ ~N(0.28, 0.01859). 
(B) To determine this probability, you must standardize a value of 0.26 and then look up its cumulative probability in the z table: Pr(₁  - ₂ <= 0.26) = Pr(Z <= [(0.26 - 0.28) / 0.01859]) = Pr(Z <= -1.08) = 0.1400 (one-tailed) and 0.2800 (two-tailed). This would not be too unusual.
(C) Pr(₁  - ₂ <= 0) = Pr(Z <= [(0.00 - 0.28) / 0.01859]) = Pr(Z <= -15.06) = 0.0000; nearly impossible due to chance.