Question

It was reported that in a survey of 4794 American youngsters aged 6 to 19, 15% were seriously overweight (a body mass index of at least 30; this index is a measure of weight relative to height). Calculate a confidence interval using a 99% confidence level for the proportion of all American youngsters who are seriously overweight. (Round your answers to three decimal places.)

Answer 1

The 99% confidence level for the proportion of all American youngsters who are seriously overweight is (0.137, 0.163)

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 = 4794, \pi = 0.15`

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a pvalue of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.575`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.15 - 2.575\sqrt{(0.15*0.85)/(4794)} = 0.137`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.15 + 2.575\sqrt{(0.15*0.85)/(4794)} = 0.163`

The 99% confidence level for the proportion of all American youngsters who are seriously overweight is (0.137, 0.163)