Answer: [tex](20.63\%,\ 36.97\% )[/tex]
Step-by-step explanation:
The confidence interval for population proportion(p) is given by :-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex] , where
n= Sample size
z*= Critical z-value.
[tex]\hat{p}[/tex] = sample proportion.
Let p be the true proportion of all adults that have high blood pressure.
As per given , we have
n= 118
Number of adults found to have high blood pressure =34
Then, [tex]\hat{p}=\dfrac{34}{118}\approx0.288[/tex]
Critical z-value for 95% confidence interval : z* = 1.96
Now , the 95% confidence interval for population proportion will be :
[tex]0.288\pm (1.96)\sqrt{\dfrac{0.288(1-0.288)}{118}}[/tex]
[tex]0.288\pm (1.96)\sqrt{0.0017377627}[/tex]
[tex]0.288\pm (1.96)(0.04168648)[/tex]
[tex]0.288\pm0.0817[/tex]
[tex]=(0.288-0.0817,\ 0.288+0.0817) =(0.2063,\ 0.3697 )[/tex]
In percentage , this would be [tex](0.2063,\ 0.3697 )=(20.63\%,\ 36.97\% )[/tex]
Hence, the 95% confidence interval for the true percentage of all adults that have high blood pressure = [tex](20.63\%,\ 36.97\% )[/tex]
