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

The minimum sample size needed is 2401.

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)}`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (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{(\pi(1-\pi))/(n)}`

`0.02 = 1.96\sqrt{(0.5*0.5)/(n)}`

`0.02√(n) = 1.96*0.5`

`√(n) = (1.96*0.5)/(0.02)`

`(√(n))^(2) = ((1.96*0.5)/(0.02))^(2)`

`n = 2401`

The minimum sample size needed is 2401.