Answer: b. (.54, .71)
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]
,where [tex]\hat{p}[/tex] = sample proportion
z= Critical z-value
n= sample size.
Let p be the proportion of people who support additional handgun control.
As per given , we have
n= 80
[tex]\hat{p}=\dfrac{50}{80}=0.625[/tex]
Critical z-value for 90% confidence interval is 1.645
Now , a 90% confidence interval for the population proportion of those who support additional handgun control will become:
[tex]0.625\pm (1.645)\sqrt{\dfrac{0.625(1-0.625)}{80}}[/tex]
[tex]=0.625\pm (1.645)\sqrt{0.0029296875}[/tex]
[tex]=0.625\pm 0.089\\\\=(0.625-0.089, 0.625+0.089)\\\\=(0.536,\ 0.714)\approx(0.54,\ 0.71)[/tex]
So the correct answer is : b. (.54, .71)
