Question

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.)

Answer 1

Answer:

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 $\pi$, and a confidence level of $1-\alpha$, 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:

$n = 4794, \pi = 0.15$

99% confidence level

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.15 - 2.575\sqrt{\frac{0.15*0.85}{4794}} = 0.137$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.15 + 2.575\sqrt{\frac{0.15*0.85}{4794}} = 0.163$

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