Answer: [tex](0.081,\ 0.163)[/tex]
Step-by-step explanation:
Let [tex]\hat{p}[/tex] be the sample proportion.
As per given , we have
n= 411
[tex]\hat{p}=\dfrac{50}{411}=0.121654501217\approx0.122[/tex]
Standard error : se=0.016
Critical value for 99% confidence : [tex]z_{\alpha/2}=2.576[/tex]
Confidence interval for population proportion :-
[tex]\hat{p}\pm z_{\alpha/2} (se)\\\\=0.122\pm (2.576)(0.016)\\\\=0.122\pm0.041216\\\\=(0.122-0.041216,\ 0.122+0.041216)\\\\=(0.080784,\ 0.163216)\approx(0.081,\ 0.163)[/tex]
Hence, the 99% confidence interval for the proportion of all US adults ages 55 to 64 to use online dating : [tex](0.081,\ 0.163)[/tex]
