A quality control expert wants to estimate the proportion of

A quality control expert wants to estimate the proportion of defective components that are being manufactured by his company to within 2.5%. A sample of 300 components showed that 20 were defective. How large a sample is needed to estimate the true proportion of defective components with 99% confidence?

Solution

2.32=z for 99%. So margin of error = sqrt(p*(1-p)/n)*2.32 sqrt(20/300*280/300/n)*2.32=.025 540=n