Question

A simple random sample of size n equals 450 n=450 individuals who are currently employed is asked if they work at home at least once per week. Of the 450450 employed individuals​ surveyed, 3434 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: $(0.044,\ 0.108)$

Step-by-step explanation:

The confidence interval for population proportion formula :-

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

where n= sample size.

$\hat{p}$= sample proportion.

$z_{\alpha/2}$ is the two-tailed z-value as per confidence level or significance level $(\alpha)$ .

To find : 99% confidence interval for the population proportion of employed individuals who work at home at least once per week.

Given : Of the 450 employed individuals​ surveyed, 34 responded that they did work at home at least once per week.

As per given , we have

n= 450

$\hat{p}=\dfrac{34}{450}\approx0.076$

Two-tailed z value for 99% confidence level i.e. significance level of $\alpha=0.01$

$z_{\alpha/2}=z_{0.005}=2.576$

Then , 99% confidence interval for the population proportion of employed individuals who work at home at least once per week will be :-

$0.076\pm (2.576)\sqrt{\dfrac{0.076(1-0.076)}{450}}\\\\=0.076\pm0.032=(0.076-0.032,\ 0.076+0.032)\\\\=(0.044,\ 0.108)$

Hence, a 99% confidence interval for the population proportion of employed individuals who work at home at least once per week = $(0.044,\ 0.108)$