Question

In a Gallup poll, 26.6% of 5000 people reported a Body Mass Index (BMI) greater than 30, which is classified as obese. The 5000 people polled are a random sample of adults, aged 18 and older, from the United States population. What is the 99% confidence interval for the proportion of the population who are obese

Answer 1

The 99% confidence interval for the proportion of the population who are obese is (0.25, 0.282)

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 = 5000, \pi = 0.266`

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)} = 0.266 - 2.575\sqrt{(0.266*0.734)/(5000)} = 0.25`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.266 + 2.575\sqrt{(0.266*0.734)/(5000)} = 0.282`

The 99% confidence interval for the proportion of the population who are obese is (0.25, 0.282)