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 0.32.

1. 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 85% confidence level with an error of at most 0.03? Round your answer up to the next integer.

Answer 1

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.

Explanation:

The (1 - α)% confidence interval for population proportion p is:

`CI=\hat p\pm z_(\alpha/2)* \sqrt{(\hat p(1-\hat p))/(n)}`

The formula to calculate the margin of error is,

`MOE=z_(\alpha/2)* \sqrt{(\hat p(1-\hat p))/(n)}`

The given information is,
`\hat p` = 0.32 and margin of error = 0.03.

The z-value for 85% confidence level is,

`z_(\alpha/2)=z_(0.15/2)=z_(0.075)=1.44`

*Use a z-table for the z-value.

Then the sample size n is given by,

`MOE=z_(\alpha/2)* \sqrt{(\hat p(1-\hat p))/(n)}`

`n=[(z_(\alpha/2)* √(\hat p(1-\hat p)))/(MOE) ]^(2)`

`=[(1.44* √(0.32(1-0.32)))/(0.03)]^(2)`

`=(22.392)^(2)\\=501.402\\\approx502`

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.