An article reported that in a study of a particular wafer in

An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 171 of these passed the probe. Assuming a stable process, calculate a 95% (two-sided) confidence interval for the proportion of all dies that pass the probe. (Round your answers to three decimal places.). You may need to use the appropriate table in the Appendix of Tables to answer this question

Solution

p=171/356 = 0.4803371

Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So the lower bound is

p - Z*sqrt(p*(1-p)/n) = 0.4803371 -1.96*sqrt(0.4803371*(1-0.4803371)/356) =0.428

So the upper bound is

p + Z*sqrt(p*(1-p)/n) = 0.4803371 +1.96*sqrt(0.4803371*(1-0.4803371)/356) =0.532