Question

The percent defective for parts produced by a manufacturing process is targeted at 4%. The process is monitored daily by taking samples of sizes n = 160 units. Suppose that today%u2019s sample contains 14 defectives.
How many units would have to be sampled to be 95% confident that you can estimate the fraction of defective parts within 2% (using the information from today%u2019s sample--that is using the result that ?

Answer 1

Answer:

A sample of 767 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 given by:

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

For this problem, we have that:

14 defectives out of 160, so $\pi = \frac{14}{160} = 0.0875$

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 $Z = 1.96$.

How many units would have to be sampled to be 95% confident that you can estimate the fraction of defective parts within 2% of the percentage?

We would need a sample of n.

n is fround when $M = 0.02$.  So

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

$0.02 = 1.96\sqrt{\frac{0.0875*0.9125}{n}}$

$0.02\sqrt{n} = 1.96\sqrt{0.0875*0.9125}$

$\sqrt{n} = \frac{1.96\sqrt{0.0875*0.9125}}{0.02}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.0875*0.9125}}{0.02})^2$

$n = 766.8$

Rounding up

A sample of 767 is needed.