Question

A simple random sample of size n=250 individuals who are currently employed is

Answer 1

Answer:

$0.104 - 2.58\sqrt{\frac{0.104(1-0.104)}{250}}=0.054$

$0.104 + 2.58\sqrt{\frac{0.104(1-0.104)}{250}}=0.154$

The 99% confidence interval would be given by (0.054;0.154) . So we are confident at 99% that the true proportion of people that they did work at home at least once per  week is between 0.054 and 0.154

Step-by-step explanation:

For this case we can estimate the population proportion of people that they did work at home at least once per  week with this formula:

$\hat p = \frac{X}{n}= \frac{26}{250}= 0.104$

We need to find the critical value using the normal standard distribution the z distribution. Since our condifence interval is at 99%, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$. And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.104 - 2.58\sqrt{\frac{0.104(1-0.104)}{250}}=0.054$

$0.104 + 2.58\sqrt{\frac{0.104(1-0.104)}{250}}=0.154$

The 99% confidence interval would be given by (0.054;0.154) . So we are confident at 99% that the true proportion of people that they did work at home at least once per  week is between 0.054 and 0.154