Question

Suppose a production facility purchases a particular component part in large lots from a supplier. The production manager wants to estimate the proportion of defective parts received from this supplier. She believes the proportion defective is no more than 0.22 and wants to be within 0.02 of the true proportion of defective parts with a 90% level of confidence. How large a sample should she take?

Answer 1

Answer:

We need a sample size of at least 1161.

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 for the interval is:

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

For this problem, we have that:

$\pi = 0.22$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

How large a sample should she take?

We need a sample size of at least n.

n is found when $M = 0.02$

So

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

$0.02 = 1.645\sqrt{\frac{0.22*0.78}{n}}$

$0.02\sqrt{n} = 1.645\sqrt{0.22*0.78}$

$\sqrt{n} = \frac{1.645\sqrt{0.22*0.78}}{0.02}$

$(\sqrt{n})^{2} = (\frac{1.645\sqrt{0.22*0.78}}{0.02})^{2}$

$n = 1160.88$

Rounding up

We need a sample size of at least 1161.