A major textbook publisher has a contract with a printing company. Part of the contract stipulates that no more than 5 percent of the pages should have any type of printing error. As a quality control measure, the publisher periodically selects a random sample of n = 100 pages. Then, depending on the proportion of pages with errors, they either say nothing to the printer or they complain that the quality has slipped. Suppose the publisher wants no more than a .10 chance of mistakenly blaming the printer for poor quality, what should the cut-off proportion be?
A) About 0.0279
B) Approximately 0.0779
C) About 0.0221
D) About 0.10

Question

Quizplus · Accepted Answer

The cut-off proportion is determined using the binomial distribution or normal approximation to the binomial distribution. Given that the publisher tolerates up to a 5% error rate and wants to keep the chance of a Type I error (falsely blaming the printer) below 10%, we use the normal approximation to find the z-score corresponding to a 90% confidence level (since 1 - 0.10 = 0.90). The z-score for 90% confidence is approximately 1.28. We then solve for the proportion using the formula for the margin of error in a proportion: $ME = z \sqrt{\frac{p(1-p)}{n}}$, where $p = 0.05$ (the acceptable error rate), $n = 100$ (the sample size), and $z = 1.28$. Solving for the cut-off proportion gives a value closer to option B, which accounts for the possibility of observing a slightly higher proportion of errors by chance without indicating a true increase in the error rate beyond the acceptable 5%.

A Major Textbook Publisher Has a Contract with a Printing