Respuesta :
You know that of 122 adults selected randomly from one town, 27 of them smoke.
Then, you need to use this formula for Confidence Interval:
[tex]CI=p\pm z_{\frac{\alpha}{2}}\sqrt[]{\frac{p(1-p)}{n}}[/tex]Where "p" is the proportion, "z" is the value of confidence level, and "n" is the number of elements in the sample.
You need to find the 99% confidence interval for the true percentage of all adults in the town that smoke.
By definition, the value of "z" for a 99% confidence interval is:
[tex]z_{\frac{\alpha}{2}}=2.57583[/tex]In this case:
[tex]\begin{gathered} n=122 \\ \\ p=\frac{27}{122} \end{gathered}[/tex]Therefore, substituting values into the formula and evaluating, you get:
[tex]CI=\frac{27}{122}\pm(2.57583)_{}\sqrt[]{\frac{\frac{27}{122}(1-\frac{27}{122})}{122}}[/tex][tex]CI=\frac{27}{122}\pm(2.57583)_{}\sqrt[]{\frac{\frac{27}{122}(1-\frac{27}{122})}{122}}[/tex]You will two values:
[tex]CI_1=\frac{27}{122}+(2.57583)_{}\sqrt[]{\frac{\frac{27}{122}(1-\frac{27}{122})}{122}}\approx31\text{\%}[/tex][tex]CI_2=\frac{27}{122}-(2.57583)_{}\sqrt[]{\frac{\frac{27}{122}(1-\frac{27}{122})}{122}}\approx13\text{\%}[/tex]Hence, the answer is: First option.
