Question

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

a. Define the random variables$\mathrm{X}$and ${\mathrm{P}}^{\text{'}}$in words.

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

c. Construct a confidence interval for the population proportion of people over $50$who 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 $97\%$this study.

Short answer

a. $X$is

Variable at random The number of people over the age of fifty who ran and died in the same eight-year period.

$P^{\text{'}}$is the percentage of people over fifty who ran and died in the same eight-year period

b. Distribution is $\mathcal{N}\left(0.015,\sqrt{\frac{0.015(1-0.015)}{451}}\right)$

c. Confident level is $0.0026\le \mathrm{p}\le 0.0274$,

error bound is $0.0124$.

d. True population interval is is localid="1649859297863" $0.0026\le p\le 0.0274$

and the graph is

Step-by-step solution

Step 1

a.

$\hat{p}$is the percentage of people over $50$who ran and died in the same eight-year period. So, the point estimate for the true population proportion is

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

$=0.015$

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

Therefore,

$\mathrm{x}=451\times 0.015=6.765.$

Step 2

b)

A proportion, given $\hat{\mathrm{p}}=0.015$and $\mathrm{n}=451$,

The distribution we should use is,

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

Step 3

c).

The population that falls into this category is,

$\hat{\mathrm{p}}-{\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}\sqrt{\frac{\hat{\mathrm{p}}(1-\hat{\mathrm{p}})}{\mathrm{n}}}\le \mathrm{p}\le \hat{\mathrm{p}}+{\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}\sqrt{\frac{\hat{\mathrm{p}}(1-\hat{\mathrm{p}})}{\mathrm{n}}}$

Where,

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

and

$z_{\frac{\alpha }{2}}=2.17..............2$

From equation $1$and$2$

$0.015-2.17\sqrt{\frac{0.015(1-0.015)}{451}}\le p\le 0.015+2.17\sqrt{\frac{0.015(1-0.015)}{451}}$

$0.015-0.0124\le P\le 0.015+0.0124$

$90\%$CI is ,

$0.0026\le p\le 0.0274$

Step 4

c.

Graph is,

EBM$=0.0124$

Introduction

The genuine proportion statistics exam score appears in$97\%$of all confidence intervals computed this manner.

For example, we would anticipate $97$of these confidence ranges to reflect the genuine population proportion if we generated $100$of them.

The true population proportion is between $0.0026$and $0.0274$, according to ninety seven percent confidence.