The procurement manager for a large company wishes to estimate the proportion of parts from a supplier that are defective. She has selected a random sample of n = 200 incoming parts and has found 11 to be defective. Based on a 95 percent confidence level, the upper and lower limits for the confidence interval estimate are approximately 0.0234 to 0.0866.

Question

Quizplus · Accepted Answer

The confidence interval for the proportion of defective parts can be calculated using the formula for the confidence interval of a proportion: $p \pm Z\sqrt{\frac{p(1-p)}{n}}$, where $p$ is the sample proportion of defective parts (11/200 = 0.055), $n$ is the sample size (200), and $Z$ is the Z-score corresponding to the desired confidence level (approximately 1.96 for 95%). Plugging these values into the formula gives a confidence interval approximately from 0.0234 to 0.0866, confirming the statement.

The Procurement Manager for a Large Company Wishes to Estimate