Question

A random sample of elementary school children in New York state is to be selected to estimate the proportion p who have received a medical examination during the past year. An interval estimate of the proportion p with a margin of error of 0.06 0.06 and 99% 99 % confidence is required.

Answer 1

The minimum sample size required is 461.

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)}`

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`.

An interval estimate of the proportion p with a margin of error of 0.06. What is the minimum sample size required?

The minimum sample size required is n, which is found when M = 0.06.

We don't have an estimate for the true proportion, which means that we use
`\pi = 0.5`. So

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

`0.06 = 2.575\sqrt{(0.5*0.5)/(n)}`

`0.06√(n) = 2.575*0.5`

`√(n) = (2.575*0.5)/(0.06)`

`(√(n))^2 = ((2.575*0.5)/(0.06))^2`

`n = 460.5`

Rounding up

The minimum sample size required is 461.