A quality manager in a small manufacturing company wants to estimate the proportion of items produced by a very specialized process that fail to meet a customer's specification.Because it is very expensive to determine if an item produced by the process meets the specification,only a very small number of items can be tested.A random sample of 15 items was selected,and in the sample three of them failed.At the 98% confidence level,what is the margin of error in the plus four estimate of p?
A)0.198
B)0.023
C)0.235
D)0.261
E)Not within ± 0.01 of any of the above

Question

Quizplus · Accepted Answer

First, we need to calculate the point estimate for the proportion of items that fail to meet the customer's specification:

$\hat{p} = \frac{3}{15} = 0.2$

Next, we can use the formula for confidence intervals for proportions to calculate the margin of error:

$ME = z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

where $z_{\alpha/2}$ is the critical value from the standard normal distribution for the given level of confidence, $\hat{p}$ is the point estimate, and $n$ is the sample size.

At the 98% confidence level, the critical value is $z_{\alpha/2} = 2.33$.

Substituting in the values, we get:

$ME = 2.33 \sqrt{\frac{0.2(0.8)}{15}} \approx 0.235$

Therefore, the margin of error is 0.235, which corresponds to choice C.

A Quality Manager in a Small Manufacturing Company Wants to Estimate