Answer: [tex](0.367,\ 0.473)[/tex]
Step-by-step explanation:
The confidence interval for population mean is given by :-
[tex]\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where [tex]\hat{p}[/tex] is the sample proportion, n is the sample size , [tex]z_{\alpha/2}[/tex] is the critical z-value.
Given : Significance level : [tex]\alpha:1-0.99=0.01[/tex]
Sample size : n= 85
Critical value : [tex]z_{\alpha/2}=2.576[/tex]
Sample proportion: [tex]\hat{p}=\dfrac{36}{85}\approx0.42[/tex]
Now, the 99% confidence level will be :
[tex]\hat{p}\pmz_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.42\pm(2.576)\sqrt{\dfrac{0.42(1-0.42)}{85}}\\\\\approx0.42\pm0.053\\\\=(0.42-0.053,\ 0.42+0.053)=(0.367,\ 0.473)[/tex]
Hence, the 99% confidence interval estimate of the true proportion of families who own at least one DVD player is [tex](0.367,\ 0.473)[/tex]
