Question

NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. In an earlier study, the population proportion was estimated to be 0.3 . How large a sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 80% confidence level with an error of at most 0.03

Answer 1

A sample of 383 would be required.

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)}`

In an earlier study, the population proportion was estimated to be 0.3.

This means that
`\pi = 0.3`

80% confidence level

So
`\alpha = 0.2`, z is the value of Z that has a pvalue of
`1 - (0.2)/(2) = 0.9`, so
`Z = 1.28`.

How large a sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 80% confidence level with an error of at most 0.03?

A sample of n is needed.

n is found when M = 0.03. So

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

`0.03 = 1.28\sqrt{(0.3*0.7)/(n)}`

`0.03√(n) = 1.28√(0.3*0.7)`

`√(n) = (1.28√(0.3*0.7))/(0.03)`

`(√(n))^2 = ((1.28√(0.3*0.7))/(0.03))^2`

`n = 382.3`

Rounding up:

A sample of 383 would be required.