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

Answer:

A sample of 383 would be required.

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:

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

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 $1 - \frac{0.2}{2} = 0.9$, 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

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

$0.03 = 1.28\sqrt{\frac{0.3*0.7}{n}}$

$0.03\sqrt{n} = 1.28\sqrt{0.3*0.7}$

$\sqrt{n} = \frac{1.28\sqrt{0.3*0.7}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.28\sqrt{0.3*0.7}}{0.03})^2$

$n = 382.3$

Rounding up:

A sample of 383 would be required.