Answer: [tex](0.164,\ 0.276)[/tex]
Step-by-step explanation:
The confidence interval of population proportion is given by :-
[tex]\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
Given : Sample size : [tex]n=150[/tex]
The proportion of individuals lease a car :[tex]hat{p}=\dfrac{33}{150}=0.22[/tex]
Significance level : [tex]\alpha: 1-0.9=0.1[/tex]
Critical value : [tex]z_{\alpha/2}=1.645[/tex]
Then, the confidence interval of population proportion is given by :-
[tex]0.22\pm (1.645)\sqrt{\dfrac{0.22(1-0.22)}{150}}\\\\\approx0.22\pm0.056=(0.164,\ 0.276)[/tex]
Hence, the confidence interval for the population proportion with a 90% confidence level= [tex](0.164,\ 0.276)[/tex]