Answer: [tex](13.28\%,\ 18.72\%)[/tex]
Step-by-step explanation:
Given : A random sample of 700 home owners in a particular city found 112 home owners who had a swimming pool in their backyard.
i.e. n= 700 and x= 112
Sample proportion : [tex]\hat{p}=\dfrac{x}{n}=\dfrac{112}{700}=0.16[/tex]
z-value for 95% confidence interval : [tex]z_c=1.960[/tex]
Now, the 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard will be :-
[tex]\hat{p}\pm z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
[tex]=0.16\pm (1.96)\sqrt{\dfrac{0.16(1-0.16)}{700}}[/tex]
[tex]=0.16\pm (1.96)(0.013856)\\\\\approx 0.16\pm0.0272\\\\ =(0.16-0.0272,\ 0.16+0.0272)=(0.1328,\ 0.1872)=(13.28\%,\ 18.72\%)[/tex]
Hence, 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard : [tex](13.28\%,\ 18.72\%)[/tex]
