Answer: [tex](0.099,\ 0.157)[/tex]
Step-by-step explanation:
We know that the confidence interval for population standard deviation is given by :-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where , n= sample size.
[tex]\hat{p}[/tex] = sample proprotion.
z* = critical z-value.
Given : A random sample of 500 registrations are selected from a Department of Motor Vehicles database, and 64 are classified as sports utility vehicles.
i.e. n= 500
[tex]\hat{p}=\dfrac{64}{500}=0.128[/tex]
We know that critical z-value for 95% confidence = z*=1.96
Then, the 95% confidence interval to estimate the proportion of sports utility vehicles in California will be :-
[tex]0.128\pm (1.95)\sqrt{\dfrac{0.128(1-0.128)}{500}}[/tex]
[tex]0.128\pm (1.95)\sqrt{0.000223232}[/tex]
[tex]0.128\pm (1.95)(0.0149409504383)[/tex]
[tex]0.128\pm 0.0291348533547\approx0.128\pm0.029=(0.128-0.029,\ 0.128+0.029)=(0.099,\ 0.157)[/tex]
Hence, the 95% confidence interval to estimate the proportion of sports utility vehicles in California. = [tex](0.099,\ 0.157)[/tex]
