Question

Use the given data to construct a confidence interval for the population proportion of the requested level. x=52, n=72, confidence level 99.9% Round the answer to at least three decimal places.

Answer 1

Answer:

The 99.9% confidence interval for the population proportion is (0.548, 0.896).

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:

$n = 72, \pi = \frac{x}{n} = \frac{52}{72} = 0.722$

99.9% confidence level

So $\alpha = 0.001$, z is the value of Z that has a pvalue of $1 - \frac{0.001}{2} = 0.9995$, so $Z = 3.29$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.722 - 3.29\sqrt{\frac{0.722*0.278}{72}} = 0.548$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.722 + 3.29\sqrt{\frac{0.722*0.278}{72}} = 0.896$

The 99.9% confidence interval for the population proportion is (0.548, 0.896).