Question

NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than six. In an earlier study the population proportion was estimated to be 0.45. How large of a sample would be required in order to estimate the fraction of people who black out at six and more gs at the 95% confidence level with an error of at most 0.04? Round your answer out to the next integer

Answer 1

595

Explanation:

Given:

Proportion (p) = 0.45

Margin of error (E) = 0.04

At 95% confidence level the z is:

`\begin{gathered} \alpha=1-95\% \\ \\ \alpha=1-0.95=0.05 \\ \\ \alpha\text{/2}=(0.05)/(2)=0.025 \\ \\ \text{ The corresponding value of Z would be:} \\ \\ Z_{\alpha\text{/2}}=1.96 \end{gathered}`

We can calculate the value of the sample by the following formula:

`\begin{gathered} E=Z_{\alpha\text{/2}}\cdot\sqrt{(p\cdot(1-p))/(n)} \\ \\ \text{ We replacing:} \\ \\ 0.04=1.96\cdot \sqrt{(0.45\cdot \left(1-0.45\right))/(n)} \\ \\ \sqrt{(0.45\left(1-0.45\right))/(n)}=(0.04)/(1.96) \\ \\ (0.45\left(1-0.45\right))/(n)=\left(\:(0.04)/(1.96)\right)^2 \\ \\ (1)/(n)=(\left(\:(0.04)/(1.96)\right)^2)/(0.45\left(1-0.45\right)) \\ \\ n=(0.45\left(1-0.45\right))/(\left(\:(0.04)/(1.96)\right)^2) \\ \\ n=594.2475\cong595 \end{gathered}`

The sample size is 595