Question

A random sample of 700 home owners in a particular city found 112 home owners who had a swimming pool in their backyard. Find a 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard. Express your results to the nearest hundredth of a percent.

Answer 1

Answer: $(13.28\%,\ 18.72\%)$

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 : $\hat{p}=\dfrac{x}{n}=\dfrac{112}{700}=0.16$

z-value for 95% confidence interval : $z_c=1.960$

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 :-

$\hat{p}\pm z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$=0.16\pm (1.96)\sqrt{\dfrac{0.16(1-0.16)}{700}}$

$=0.16\pm (1.96)(0.013856)\\\\\approx 0.16\pm0.0272\\\\ =(0.16-0.0272,\ 0.16+0.0272)=(0.1328,\ 0.1872)=(13.28\%,\ 18.72\%)$

Hence, 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard : $(13.28\%,\ 18.72\%)$