Question

A certain forum reported that in a survey of 2004 American adults, 23% said they believed in astrology. (a) Calculate a confidence interval at the 99% confidence level for the proportion of all adult Americans who believe in astrology. (Round your answers to three decimal places.)

Answer 1

Answer:

The 99% confidence level for the proportion of all adult Americans who believe in astrology is (0.206, 0.254).

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:

$n = 2004, \pi = 0.23$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.23 - 2.575\sqrt{\frac{0.23*0.77}{2004}} = 0.206$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.23 + 2.575\sqrt{\frac{0.23*0.77}{2004}} = 0.254$

The 99% confidence level for the proportion of all adult Americans who believe in astrology is (0.206, 0.254).

Answer 2

The confidence interval at the 99% confidence level for the proportion of all adult Americans who believe in astrology is;

CI = (0.206, 0.254)

We want to find confidence interval at confidence level of 99%. The formula to find confidence level when given proportions like we are given in this question is;

CI = p^ ± (z_α/2)√(p^(1 - p^)/n)

We are given;

p^ = 23% = 0.23

n = 2004

Now,the critical value z_α/2 for confidence level of 99% is gotten from tables to be 2.576.

Thus;

CI = 0.23 ± 2.576√(0.23(1 - 0.23)/2004)

CI = 0.23 ± 0.024

CI = (0.23 - 0.024), (0.23 + 0.024)

CI = (0.206, 0.254)

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