Question

Jonah wants to construct a confidence interval using 90% confidence to estimate what proportion of silicon wafers at his factory is defective. He wants the margin of error to be no more than 3%. A previous study suggests that about 6% of these wafers are defective. If we assume p=0.06, what is the smallest sample size required to obtain the desired margin of error? 234 416 170 936

Answer 1

Answer:

170

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 of:

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

Assume:

$\pi = 0.06$

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 $Z = 1.645$.

What is the smallest sample size required to obtain the desired margin of error?

This is n for which M = 0.03. So

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

$0.03 = 1.645\sqrt{\frac{0.06*0.94}{n}}$

$0.03\sqrt{n} = 1.645\sqrt{0.06*0.94}$

$\sqrt{n} = \frac{1.645\sqrt{0.06*0.94}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.645\sqrt{0.06*0.94}}{0.03})^2$

$n = 169.6$

Rounding up, 170.