In a (Gallup poll, 26.6% of 5000 people reported a Body Mass Index (BMl) 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. Choose the correct interpretation for the 98% confidence interval for the proportion of the population who arc obese? A. The proportion of adults in the United States, aged 18 or older, who are obese is between 0.2515 and 0.2805. B. We are 98% confident the proportion of adults polled who are obese is between 0.2515 and 0.2805. C. In 98% of all samples, the proportion of adults polled who are obese will be between 0.2515 and 0.2805. D. We are 98% confident that the proportion of adults in the United States, aged 18 or older, who are obese is between 0.2515 and 0.2805.

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

26.6% of 5000 people reported a Body Mass Index (BMl) greater than 30

This means that [tex]\pi = 0.266, n = 5000[/tex]

98% confidence level

So [tex]\alpha = 0.02[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.327[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.266 - 2.327\sqrt{\frac{0.266*0.734}{5000}} = 0.2515[/tex]

The upper limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.266 + 2.327\sqrt{\frac{0.266*0.734}{5000}} = 0.2805[/tex]

The correct interpreation is that we are 98% confident that the proportion of adults in the United States, aged 18 or older, who are obese is between 0.2515 and 0.2805, which means that the correct answer is given by option D.