Answer:
[tex]0.104 - 2.58\sqrt{\frac{0.104(1-0.104)}{250}}=0.054[/tex]
[tex]0.104 + 2.58\sqrt{\frac{0.104(1-0.104)}{250}}=0.154[/tex]
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:
[tex]\hat p = \frac{X}{n}= \frac{26}{250}= 0.104[/tex]
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 [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2 =0.005[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
If we replace the values obtained we got:
[tex]0.104 - 2.58\sqrt{\frac{0.104(1-0.104)}{250}}=0.054[/tex]
[tex]0.104 + 2.58\sqrt{\frac{0.104(1-0.104)}{250}}=0.154[/tex]
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
