Answer: (54.13%, 53.83%)
Step-by-step explanation:
Confidence interval for population proportion is given by :-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
[tex]\hat{p}[/tex] = 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
[tex]\hat{p}=\dfrac{165}{258}\approx0.64[/tex]
For 99.9% confidence , significance level α= 0.001
Critical z-value for 99.9% confidence interval =[tex]z_{\alpha/2}=z_{0.001/2}=z_{0.0005}=3.29[/tex] [By z-table]
Then, the 99.9% confidence interval for the support level percentage in the entire city will be :
[tex]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\%)[/tex]
Hence, the 99.9% confidence interval for the support level percentage in the entire city is (54.13%, 53.83%) .
