In a comparative study of two new drugs, A and B, 225 patients were treated with drug A, and 250 patients were treated with drug B. (The two treatment groups were randomly and independently chosen.) It was found that 180 patients were cured using drug $\mathrm{A}$ and 190 patients were cured using drug $\mathrm{B}$. Let $p_{1}$ be the proportion of the population of all patients who are cured using drug $A$, and let $p_{2}$ be the proportion of the population of all patients who are cured using drug B. Find a $95 \%$ confidence interval for $p_{1}-p_{2}$. Then find the lower limit and upper limit of the $95 \%$ confidence interval.
Carry your intermediate computations to at least three decimal places. Round your responses to at least three decimal places. (If necessary, consult a list of formulas.)
Answer
Therefore, the 95% confidence interval for the difference in proportions is \(\boxed{(-0.046, 0.126)}\)
Step 1 ：Calculate the sample proportions for each drug. For drug A, the sample proportion (p1) is the number of patients cured divided by the total number of patients treated with drug A. \(p1 = \frac{180}{225} = 0.8\)
Step 2 ：For drug B, the sample proportion (p2) is the number of patients cured divided by the total number of patients treated with drug B. \(p2 = \frac{190}{250} = 0.76\)
Step 3 ：Calculate the difference in proportions, which is \(p1 - p2 = 0.8 - 0.76 = 0.04\)
Step 4 ：Calculate the standard error for the difference in proportions using the formula: \(SE = \sqrt{(p1*(1-p1)/n1) + (p2*(1-p2)/n2)}\) where n1 and n2 are the sample sizes for drug A and B respectively. \(SE = \sqrt{(0.8*(1-0.8)/225) + (0.76*(1-0.76)/250)} = 0.044\)
Step 5 ：Calculate a 95% confidence interval for the difference in proportions as \((p1 - p2) ± Z*SE\), where Z is the Z-score for a 95% confidence interval, which is approximately 1.96.
Step 6 ：Calculate the lower limit of the confidence interval: \((p1 - p2) - Z*SE = 0.04 - 1.96*0.044 = -0.046\)
Step 7 ：Calculate the upper limit of the confidence interval: \((p1 - p2) + Z*SE = 0.04 + 1.96*0.044 = 0.126\)
Step 8 ：Therefore, the 95% confidence interval for the difference in proportions is \(\boxed{(-0.046, 0.126)}\)
