Answer:
The sample that would be required in order to estimate the fraction of people who black out at 6 or more Gs is of size 502.
Step-by-step explanation:
The (1 - α)% confidence interval for population proportion p is:
[tex]CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The formula to calculate the margin of error is,
[tex]MOE=z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The given information is, [tex]\hat p[/tex] = 0.32 and margin of error = 0.03.
The z-value for 85% confidence level is,
[tex]z_{\alpha/2}=z_{0.15/2}=z_{0.075}=1.44[/tex]
*Use a z-table for the z-value.
Then the sample size n is given by,
[tex]MOE=z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
[tex]n=[\frac{z_{\alpha/2}\times \sqrt{\hat p(1-\hat p)}}{MOE} ]^{2}[/tex]
[tex]=[\frac{1.44\times \sqrt{0.32(1-0.32)}}{0.03}]^{2}[/tex]
[tex]=(22.392)^{2}\\=501.402\\\approx502[/tex]
Thus, the sample that would be required in order to estimate the fraction of people who black out at 6 or more Gs is of size 502.
