Respuesta :
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 [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
In a survey of 4722 American youngsters aged 6–19, 15% were seriously overweight, so [tex]n = 4722, \pi = 15[/tex].
Calculate and inter- pret a confidence interval using a 99% confidence level for the proportion of all American youngsters who are seriously overweight.
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so Z = 2.575.
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{4722}} = 0.15 - 2.575\sqrt{\frac{0.15*0.85}{4722}} = 0.1366[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{4722}} = 0.15 + 2.575\sqrt{\frac{0.15*0.85}{4722}} = 0.1634[/tex]
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.
