Question

post the r code please

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

= FALSE) testresults <- prop.test (x = c (139, 92), n = c (1062, 1065), conf.level = 0.95, correct > testresults 2-sample test for equality of proportions without continuity correction data c (139, 92) out of c (1062, 1065) X-squared = 10.877, df 1, p-value = 0.0009736 alternative hypothesis: two.sided 95 percent confidence interval 0.01811559 0.07088471 sample estimates: prop 2 prop 1 0.13088512 0.08638498

(a) given that, sample one, x1 =139, n1 1062, p1= x1/n1-0.1309 sample two, x2 92, n2 1065, p2= x2/n2=0.0864 CI (p1-p2) t sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 ) where, p1, p2 proportion of both sample observation n1,n2 size of both group a 1 - (confidence Level/100) Za/2 Z-table value CI confidence interval CI [ (0.1309-0.0864) 1.96 * 0.0135] = [ 0.0181,0.0709 ] (b) interpretations: 1) we are 95% sure that the interval [ 0.0181,0.0709]contains the difference between true population proportion P1-P2 2) if a large number of samples are collected, and a confidence interval is created for each sample, 95% of these intervals will contains the difference between true population mean P1-P2