Question

The dwarves of the Grey Mountains wish to conduct a survey of their pick-axes in order to construct a 95% 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 2 percentage points (or less) if they do not have a prior estimate of the percentage of pick-axes that are in need of repair?

Answer 1

Answer:

The minimum sample size needed is 2401.

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:

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

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 $Z = 1.96$.

What minimum sample size would be necessary in order ensure a margin of error of 2 percentage points (or less) if they do not have a prior estimate of the percentage of pick-axes that are in need of repair?

The minimum sample size needed is n.

n is found when $M = 0.02$

We do not have a prior estimate of the percentage of pick-axes that are in need of repair, so we use $\pi = 0.5$

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

$0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.02\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.02}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.02})^{2}$

$n = 2401$

The minimum sample size needed is 2401.