Answer: [tex](0.0886,\ 0.1514)[/tex]
Step-by-step explanation:
Given : Significance level : [tex]\alpha=1-0.95=0.05[/tex]
Critical value : [tex]z_{\alpha/2}=1.96[/tex]
Sample size : [tex]n=411[/tex]
The number of adults between the ages of 55 and 64 said that they had used online dating : 54
Now, the proportion of adults between the ages of 55 and 64 said that they had used online dating : [tex]p=\dfrac{50}{411}\approx0.12[/tex]
Now, the confidence interval for proportion is given by :-
[tex]p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}\\\\=0.12\pm(1.96)\sqrt{\dfrac{0.12(1-0.12)}{411}}\\\\\approx0.12\pm 0.0314\\\\=(0.12-0.0314,\ 0.12+0.0314)\\\\=(0.0886,\ 0.1514)[/tex]
Hence, a 95% confidence interval for the proportion of all US adults ages 55 to 64 to use online dating is [tex](0.0886,\ 0.1514)[/tex].
