Question

122. Stanford University conducted a study of whether running is healthy for men and warner overage $50$. During the first eight years of the study, $1.5\%$of the $451$members of the $50$-Plus Fatness Association died. We are interested in the proportion of people over $50$who ran and tied in the same eight-year period.

a. Define the random variables$XandF$in words.

b. Which distribution should you use for this problem? Explain your choice.

c. Construct a $97\%$confidence interval for the population proportion of people average$50$ we ran and died in the same eight-year period.

i. State the confidence interval.

ii. Sketch the graph.

iii. Calculate the error bound.

d. Explain what a " $97\%$confidence interval" means for this study.

Short answer

a. Variable at random The number of people over the age of 50 who ran and died in the same eight-year period is $X$.

The percentage of adults over $50$who ran and died in the same eight-year period is called $\hat{p}$.

b. The normal distribution $\mathcal{N}(0.015,\sqrt{0.015(1-0.015)})$.

c. The confidence interval of the proportion of population is $0.0026\le p\le 0.0274andEBM=0.0124$

d. The true population proportion is estimated to be between $0.0026$ and $0.0274$.

Step-by-step solution

Introduction

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

Step 1 

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,

$x=451\times 0.015=6.765$

Step 2

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

$N\left(0.015,\sqrt{\frac{0.015(1-0.015)}{451}}\right)$

Step 3

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.

$\hat{p}-z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\le p\le \hat{p}+z_{\frac{\sigma }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\hat{p}-z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\le p\le \hat{p}+z_{\frac{\sigma }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

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

$\frac{\alpha }{2}=\frac{1-0.97}{2}=0.015$

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

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