Question

stat crunch A simple random sample of size n=250 individuals who are currently employed is asked if they work at home at least once per week. Of the 250 employed individuals​ surveyed, 41 responded that they did work at home at least once per week. Construct a​ 99% confidence interval for the population proportion of employed individuals who work at home at least once per week.

Answer 1

Answer:

The 99% confidence interval for the population proportion of employed individuals who work at home at least once per week is between (0.104, 0.224).

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:

$n = 250, \pi = \frac{41}{250} = 0.164$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.164 - 2.575\sqrt{\frac{0.164*0.836}{250}} = 0.104$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.164 + 2.575\sqrt{\frac{0.164*0.836}{250}} = 0.224$

The 99% confidence interval for the population proportion of employed individuals who work at home at least once per week is between (0.104, 0.224).