Question

A machine that produces a major part for an airplane engine is monitored closely. In the past, 10% of the parts produced would be defective. With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is _________.

Answer 1

Answer:

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.

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:

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

For this problem, we have that:

$p = 0.1$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is

We need a sample size of at least n, in which n is found M = 0.04.

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

$0.04 = 1.96\sqrt{\frac{0.1*0.9}{n}}$

$0.04\sqrt{n} = 0.588$

$\sqrt{n} = \frac{0.588}{0.04}$

$\sqrt{n} = 14.7$

$(\sqrt{n})^{2} = (14.7)^{2}$

$n = 216$

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.