Chapter 22, Problem 22.11AYK

(a)

To determine

To find:

The study provides evidence that among men with a neurodegenerative disorder for the given data.

Answer to Problem 22.11AYK

  ${\chi }^2=39.35$

Explanation of Solution

Given:

Participated in a contact sport $=66\text{men}$

None of the control men $=132$

Concept used:

Formula

  $p=\frac{x}{n}$

Significance level

  $\begin{array}{l}\alpha =1-\text{confidence}\\ \alpha =1-0.95\end{array}$

Calculation:

Parameter $p=$ sample proportion

The best point of estimate of $p$ is sample proportion

Since the sample size are large.

The condition for the two samples $z$ test are met.

The population distribution is normal.

Draw the first table

CTEContact sportsNoContact sports Total

Observed	$21$	$2$	$23$
Expected	$7.67$	$15.33$	$23.00$
$\text{O-E}$	$13.33$	$-13.33$	$0.00$
$\frac{{\left(\text{O-E}\right)}^2}{\text{E}}$	$23.19$	$11.59$	$34.78$

Draw the second table

No CTEContact sportsNoContact sports Total

Observed	$45$	$130$	$175$
Expected	$58.33$	$116.67$	$175.00$
$\text{O-E}$	$-13.33$	$13.33$	$0.00$
$\frac{{\left(\text{O-E}\right)}^2}{\text{E}}$	$3.05$	$1.52$	$4.57$

Draw the third table

TotalContact sportsNoContact sports Total

Observed	$66$	$132$	$198$
Expected	$66.00$	$132.00$	$198.00$
$\text{O-E}$	$0.00$	$0.00$	$0.00$
$\frac{{\left(\text{O-E}\right)}^2}{\text{E}}$	$26.24$	$13.12$	$39.35$

Test statistic

  $X$ Is a random variable

The null hypothesis

  $\begin{array}{l}{\chi }^2={\displaystyle \sum \left\{\frac{{\left(O_i-E_i\right)}^2}{E_i}\right\}}\\ {\chi }^2=39.35\end{array}$

(b)

To determine

To explain:

The chi-square test is nevertheless appropriate for the given data.

Explanation of Solution

Given:

Participated in a contact sport $=66\text{men}$

None of the control men $=132$

Concept used:

Formula

  $p=\frac{x}{n}$

Significance level

  $\begin{array}{l}\alpha =1-\text{confidence}\\ \alpha =1-0.95\end{array}$

Calculation:

Parameter $p=$ sample proportion

The best point of estimate of $p$ is sample proportion

Since the sample size are large.

The condition for the two samples $z$ test are met.

The population distribution is normal.

Draw the first table

CTEContact sportsNoContact sports Total

Observed	$21$	$2$	$23$
Expected	$7.67$	$15.33$	$23.00$
$\text{O-E}$	$13.33$	$-13.33$	$0.00$
$\frac{{\left(\text{O-E}\right)}^2}{\text{E}}$	$23.19$	$11.59$	$34.78$

Draw the second table

No CTEContact sportsNoContact sports Total

Observed	$45$	$130$	$175$
Expected	$58.33$	$116.67$	$175.00$
$\text{O-E}$	$-13.33$	$13.33$	$0.00$
$\frac{{\left(\text{O-E}\right)}^2}{\text{E}}$	$3.05$	$1.52$	$4.57$

Draw the third table

TotalContact sportsNoContact sports Total

Observed	$66$	$132$	$198$
Expected	$66.00$	$132.00$	$198.00$
$\text{O-E}$	$0.00$	$0.00$	$0.00$
$\frac{{\left(\text{O-E}\right)}^2}{\text{E}}$	$26.24$	$13.12$	$39.35$

Test statistic

  $X$ Is a random variable

The null hypothesis

  $\begin{array}{l}{\chi }^2={\displaystyle \sum \left\{\frac{{\left(O_i-E_i\right)}^2}{E_i}\right\}}\\ {\chi }^2=39.35\end{array}$

Significance level

  $\begin{array}{l}\alpha =1-\text{confidence}\\ \alpha =1-0.95\\ \alpha =0.05\end{array}$

Since $p-$ Value is greater than $0.05$ significance level.

So the hypothesis is fail to reject null hypothesis $H_0$