91Ó°ÊÓ

The given is the data statistics about whether running is good for men and women after the age of fifty by Stanford University

The objective is to find the random variable, distribution kind, and confidence level

a) During the study's first eight years. The $50$-Plus Fitness Association had $n=451$ members, and $1.5\%$ of them died.

The percentage of adults over $50$ who ran and died in the same eight-year period is called $\hat{p}$. As a result, the true population proportion point estimate is

$\hat{p}=1.5\%=0.015$

Random variable $X$ is the number of people over $50$who ran and died in the same eight-year period. Therefore,

b) Given $\hat{p}=0.015$ and $n=451$, the distribution we should apply for predicting a proportion is

c) An approximate confidence interval on the proportion of the population that belongs to this class is if is the proportion of observations in a random sample of size that belong to a class of interest.

$\hat{p}-z_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}≤p≤\hat{p}+z_{\frac{\mathrm{σ}}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\hat{p}-z_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}≤p≤\hat{p}+z_{\frac{\mathrm{σ}}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

where $z_{\frac{a}{2}}$is the upper $\frac{a}{2}$percentage of the distribution of normal point. For $97\%$two-sided confidence interval;

$\frac{\mathrm{Î\pm }}{2}=\frac{1-0.97}{2}=0.015$

and

$$

z_{\frac{a}{2}}=2.17 \text {. }

$$

From the equations , $95 \%$ two-sided CI for the population proportion is

$$

\begin{aligned}

0.015-2.17 \sqrt{\frac{0.015(1-0.015)}{451}} & \leq p \leq 0.015+2.17 \sqrt{\frac{0.015(1-0.015)}{451}} \\

0.015-0.0124 & \leq p \leq 0.015+0.0124

\end{aligned}

$$

f