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

The minimum sample size needed is 125.

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\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 - (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{(\pi(1-\pi))/(n)}`

`0.1 = 2.575\sqrt{(0.25*0.75)/(n)}`

`0.1√(n) = 2.575{0.25*0.75}`

`√(n) = \frac{2.575{0.25*0.75}}{0.1}`

`(√(n))^(2) = (\frac{2.575{0.25*0.75}}{0.1})^(2)`

`n = 124.32`

Rounding up

The minimum sample size needed is 125.