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Andreas93 [3]
3 years ago
14

The Associated Press (October 9, 2002) reported that in a survey of 4722 American youngsters aged 6–19, 15% were seriously overw

eight (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.
Mathematics
1 answer:
Degger [83]3 years ago
3 0

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 \pi, and a confidence interval 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:

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

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

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)}{4722}} = 0.15 - 2.575\sqrt{\frac{0.15*0.85}{4722}} = 0.1366

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

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.

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