Question

Rosetta wants to estimate the percentage of people who rent their home. She surveys 250 individuals and finds that 48 rent their home. Use a calculator to find the confidence interval for the population proportion with a 90% confidence level. 0.10 0.05 0.025 0.01 0.005 1.282 1.645 1.960 2.326 2.576

Answer 1

Answer:

$0.192 - 1.645\sqrt{\frac{0.192(1-0.192)}{250}}=0.151$

$0.192 + 1.645\sqrt{\frac{0.192(1-0.192)}{250}}=0.233$

Step-by-step explanation:

Information given

$X= 48$ number of people who rent their home

$n= 250$ represent the sample size

$\hat p =\frac{48}{250}= 0.192$ represent the proportion of people who rent their home

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$. And the critical value would be given by:

$z_{\alpha/2}=\pm 1.645$

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:

$0.192 - 1.645\sqrt{\frac{0.192(1-0.192)}{250}}=0.151$

$0.192 + 1.645\sqrt{\frac{0.192(1-0.192)}{250}}=0.233$