Question

The Pew Forum on Religion and Public Life reported on Dec. 9, 2009, that in a survey of 2003 American adults, 25% said they believed in astrology. a. Calculate and interpret a confidence interval at the 99% confidence level for the proportion of all adult Americans who believe in astrology. b. W

Answer 1

Answer:

The confidence interval at the 99% confidence level for the proportion of all adult Americans who believe in astrology is (0.2251, 0.2749). This means that we are 99% sure that the true proportion of all adult Americans who believe in astrology is in this interval.

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:

$p = 0.25, n = 2003$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.25 - 2.575\sqrt{\frac{0.25*0.75}{2003}} = 0.2251$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.25 + 2.575\sqrt{\frac{0.25*0.75}{2003}} = 0.2749$

The confidence interval at the 99% confidence level for the proportion of all adult Americans who believe in astrology is (0.2251, 0.2749). This means that we are 99% sure that the true proportion of all adult Americans who believe in astrology is in this interval.