Question

In a sample of 258 individuals selected randomly from a city of 750,339 people, 165 were found to be supportive of a new public works project. Find the 99.9% confidence interval for the support level percentage in the entire city

Answer 1

Answer: (54.13%, 53.83%)

Step-by-step explanation:

Confidence interval for population proportion is given by :-

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$\hat{p}$ = sample proportion

n= sample size.

z* = critical z-value.

Let p be the proportion of individuals supportive of a new public works project.

As per given , we have

n= 258

$\hat{p}=\dfrac{165}{258}\approx0.64$

For 99.9% confidence , significance level α= 0.001

Critical z-value for 99.9% confidence interval =$z_{\alpha/2}=z_{0.001/2}=z_{0.0005}=3.29$ [By z-table]

Then, the 99.9% confidence interval for the support level percentage in the entire city will be :

$0.64\pm (3.29)\sqrt{\dfrac{0.64(1-0.64)}{258}}\\\\\approx 0.64\pm0.0983\\\\=(0.64-0.0983,\ 0.64+0.0983) = (0.5417,\ 0.7383= (54.13\%,\ 53.83\%)$

Hence, the 99.9% confidence interval for the support level percentage in the entire city is (54.13%, 53.83%) .