Question

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

Answer 1

Answer:

A sample of 1032 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

A sample of 300 components showed that 20 were defective.

This means that $\pi = \frac{20}{300} = 0.0667$

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

How large a sample is needed to estimate the true proportion of defective components to within 2.5 percentage points with 99% confidence?

A sample of n is needed.

n is found when M = 0.025. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.025 = 2.575\sqrt{\frac{0.0667*0.9333}{n}}$

$0.02\sqrt{n} = 2.575\sqrt{0.0667*0.9333}$

$\sqrt{n} = \frac{2.575\sqrt{0.0667*0.9333}}{0.02}$

$(\sqrt{n})^2 = (\frac{2.575\sqrt{0.0667*0.9333}}{0.02})^2$

$n = 1031.9$

Rounding up

A sample of 1032 is needed.