Question

Suppose a sample of 518518 people is drawn. Of these people, 150150 passed out. Using the data, construct the 85%85% confidence interval for the population proportion of people who black out at G forces greater than 66. Round your answers to three decimal places.

Answer 1

The 85% confidence interval for the population proportion of people who black out at G forces greater than 66 is (0.261, 0.319)

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

`n = 518, p = (150)/(518) = 0.29`

85% confidence level

So
`\alpha = 0.15`, z is the value of Z that has a pvalue of
`1 - (0.15)/(2) = 0.9250`, so
`z = 1.44`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.29 - 1.44\sqrt{(0.29*0.71)/(518)} = 0.261`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.29 + 1.44\sqrt{(0.29*0.71)/(518)} = 0.319`

The 85% confidence interval for the population proportion of people who black out at G forces greater than 66 is (0.261, 0.319)