Question

The dwarves of the Grey Mountains wish to conduct a survey of their pick-axes in order to construct a 99% confidence interval about the proportion of pick-axes in need of repair. What minimum sample size would be necessary in order ensure a margin of error of 10 percentage points (or less) if they use the prior estimate that 25 percent of the pick-axes are in need of repair?

Answer 1

Answer:

The minimum sample size needed is 125.

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:

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

For this problem, we have that:

$\pi = 0.25$

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 $Z = 2.575$.

What minimum sample size would be necessary in order ensure a margin of error of 10 percentage points (or less) if they use the prior estimate that 25 percent of the pick-axes are in need of repair?

This minimum sample size is n.

n is found when $M = 0.1$

So

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

$0.1 = 2.575\sqrt{\frac{0.25*0.75}{n}}$

$0.1\sqrt{n} = 2.575{0.25*0.75}$

$\sqrt{n} = \frac{2.575{0.25*0.75}}{0.1}$

$(\sqrt{n})^{2} = (\frac{2.575{0.25*0.75}}{0.1})^{2}$

$n = 124.32$

Rounding up

The minimum sample size needed is 125.