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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

User Ovnia
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Answer:

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)

User Alexismorin
by
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