Suppose three faulty components are identified in a random sample of 20 products taken from a production line. What statement can be made about the defect rate of all finished products?
The proportion of defective products in the sample is p = {}^{3}/_{20} = 0.15, and the sample size is n = 20. Therefore, the 95% confidence interval for the population mean is given by
p\pm\,z\sqrt{\frac{p(1-p)}{n}}\,=\,0.15\pm\,1.96\sqrt{\frac{0.15(1-0.15)}{20}}= 0.15 ± 1.96 * 0.0798
= 0.15 ± 0.1565
= (-0.0065, 0.3065)
That is, we can say with 95% confidence that the defective rate of finished products lies between 0 and 0.3065.