It was reported that in a survey of 4794 American youngsters aged 6 to 19, 15% were seriously overweight (a body mass index of at least 30; this index is a measure of weight relative to height). Calculate a confidence interval using a 99% confidence level for the proportion of all American youngsters who are seriously overweight. (Round your answers to three decimal places.)

The 99% confidence level for the proportion of all American youngsters who are seriously overweight is (0.137, 0.163)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , 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:

99% confidence level

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

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 (0.137, 0.163)

answer: 42

use a calculator to find that 845/73 = 11.575

the whole part of the answer is 11. multiply this by 73 to get 11*73 = 803.

subtract that result from 845 to get 845 - 803 = 42

answer: i guess 5 units?

step-by-step explanation: