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:

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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

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

This means that [tex]\pi = 0.3[/tex]

80% confidence level

So [tex]\alpha = 0.2[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.2}{2} = 0.9[/tex], so [tex]Z = 1.28[/tex].

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

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.03 = 1.28\sqrt{\frac{0.3*0.7}{n}}[/tex]

[tex]0.03\sqrt{n} = 1.28\sqrt{0.3*0.7}[/tex]

[tex]\sqrt{n} = \frac{1.28\sqrt{0.3*0.7}}{0.03}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.28\sqrt{0.3*0.7}}{0.03})^2[/tex]

[tex]n = 382.3[/tex]

Rounding up:

A sample of 383 would be required.

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