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.

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 , and a confidence interval , we have the following confidence interval of proportions.

In which

Z is the zscore that has a pvalue of .

For this problem, we have that:

In a survey of 4722 American youngsters aged 6–19, 15% were seriously overweight, so .

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

So , z is the value of Z that has a pvalue of , so Z = 2.575.

The lower limit of this interval is:

The upper limit of this interval is:

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.

40 is the final : )