Question

The Associated Press (October 9, 2002) reported that in a survey of 4722 American youngsters aged 6–19, 15% were seriously overweight (a body mass index of at least 30; this index is a measure of weight relative to height). Calculate and inter- pret a confidence interval using a 99% confidence level for the proportion of all American youngsters who are seriously overweight.

Answer 1

Answer:

We are 99% sure that the proportion of all American youngsters who are seriously overweight is between 0.1366 and 0.1634.

Step-by-step explanation:

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

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

In a survey of 4722 American youngsters aged 6–19, 15% were seriously overweight, so $n = 4722, \pi = 15$.

Calculate and inter- pret a confidence interval using a 99% confidence level for the proportion of all American youngsters who are seriously overweight.

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so Z = 2.575.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{4722}} = 0.15 - 2.575\sqrt{\frac{0.15*0.85}{4722}} = 0.1366$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{4722}} = 0.15 + 2.575\sqrt{\frac{0.15*0.85}{4722}} = 0.1634$

The 99% confidence level for the proportion of all American youngsters who are seriously overweight is (0.1366, 0.1634).

We are 99% sure that the proportion of all American youngsters who are seriously overweight is between 0.1366 and 0.1634.