Question

Please show me the R outputs with the answer

3. (10 pts) Elderly patients who had suffered hip fractures were randomly assigned to receive either a placebo (n=1,062) or zolendronic acid (n=1,065) in a double blind clinical trial. During the trial 139 placebo patients and 92 zolendronic acid patients had new fractures. Let pl and p2 represent the probabilities of fracture on placebo and zolendronic acid, respectively. (a) Construct the 95% confidence interval for (pl-p2) in R. Show your code and output. Hint: using prop.test R command (b) Interpret your 95% CI in words.

Answer 1

ans a )

> res <- prop.test(x = c(139,92), n = c(1062,1065) ,conf.level = 0.95, correct = TRUE)
> res

   2-sample test for equality of proportions with continuity correction

data: c(139, 92) out of c(1062, 1065)
X-squared = 10.422, df = 1, p-value = 0.001245
alternative hypothesis: two.sided
95 percent confidence interval:
0.01717529 0.07182500
sample estimates:
prop 1 prop 2
0.13088512 0.08638498

b ) we are 95 % confident that the difference of proportion of recieved placebo and zolendronic lies in this ( 0.01717529 ,0.07182500) interval