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
, and a confidence interval
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
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:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{4722}} = 0.15 - 2.575\sqrt{\frac{0.15*0.85}{4722}} = 0.1366](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7B4722%7D%7D%20%3D%200.15%20-%202.575%5Csqrt%7B%5Cfrac%7B0.15%2A0.85%7D%7B4722%7D%7D%20%3D%200.1366)
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](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7B4722%7D%7D%20%3D%200.15%20%2B%202.575%5Csqrt%7B%5Cfrac%7B0.15%2A0.85%7D%7B4722%7D%7D%20%3D%200.1634)
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.