Question

Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 133 with 61.7% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

Answer 1

Answer:

The 99% confidence interval is (0.508, 0.726).

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:

$n = 133, p = 0.617$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.617 - 2.575\sqrt{\frac{0.617*0.383}{133}} = 0.508$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.617 + 2.575\sqrt{\frac{0.617*0.383}{133}} = 0.726$

The 99% confidence interval is (0.508, 0.726).