Chapter 22, Problem 22.12AYK

(a)

To determine

To explain:

The conditions are very similar of the given data.

Explanation of Solution

Given:

  $\begin{array}{l}{x}_1=177,x_2=86\\ n_1=434,n_2=222\end{array}$

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.

To met chi square test, the population distribution is normal.

  $p=\frac{x}{n}$

(b)

To determine

To find:

The value of the statistic $p$ of the given sample.

Answer to Problem 22.12AYK

  $p-$ Value $=0.6136$

Explanation of Solution

Given:

  $\begin{array}{l}{x}_1=177,x_2=86\\ n_1=434,n_2=222\end{array}$

Concept used:

Formula

Standard of error

  $SE=\sqrt{p\left(1-p\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}$

Significance level

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

Calculation:

Parameter $p=$ true proportion of individuals in population

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

  $X$ Is a random variable

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_1:p_1>p_2\end{array}$

  $\begin{array}{l}{\hat{p}}_1=\frac{177}{434}\\ {\hat{p}}_1=0.4078\\ {\hat{p}}_2=\frac{86}{222}\\ {\hat{p}}_2=0.3874\end{array}$

The value of $p$ is

  $\hat{p}\sim N\left(p,\frac{pq}{n}\right)$

  $p$ Comes from null hypothesis and alternative hypothesis

  $z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}}$

Pooled proportion

  $\begin{array}{l}p=\frac{177+86}{434+222}\\ p=0.4009146\end{array}$

Standard of error

  $\begin{array}{l}SE=\sqrt{p\left(1-p\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\\ SE=\sqrt{0.4009146\left(1-0.4009146\right)\left(\frac{1}{434}+\frac{1}{222}\right)}\\ SE=0.0404\end{array}$

  $\begin{array}{l}z=\frac{{\hat{p}}_1-{\hat{p}}_2}{SE}\\ z=\frac{0.4078-0.3874}{0.0404}\\ z=0.5050\end{array}$

For two-tail test

  $p-$ Value

  $\begin{array}{l}=2P\left(z>z_0\right)\\ =2P\left(z>0.5050\right)\\ =0.6136\end{array}$

(c)

To determine

To find:

The value of the chi-square ${\chi }^2$ of the given sample.

Answer to Problem 22.12AYK

  ${\chi }^2=0.5050$

Explanation of Solution

Given:

  $\begin{array}{l}{x}_1=177,x_2=86\\ n_1=434,n_2=222\end{array}$

Concept used:

Formula

Standard of error

  $SE=\sqrt{p\left(1-p\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}$

Significance level

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

Calculation:

Parameter $p=$ true proportion of individuals in population

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

  $X$ Is a random variable

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_1:p_1>p_2\end{array}$

  $\begin{array}{l}{\hat{p}}_1=\frac{177}{434}\\ {\hat{p}}_1=0.4078\\ {\hat{p}}_2=\frac{86}{222}\\ {\hat{p}}_2=0.3874\end{array}$

The value of $p$ is

  $\hat{p}\sim N\left(p,\frac{pq}{n}\right)$

  $p$ Comes from null hypothesis and alternative hypothesis

  $z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}}$

Pooled proportion

  $\begin{array}{l}p=\frac{177+86}{434+222}\\ p=0.4009146\end{array}$

Standard of error

  $\begin{array}{l}SE=\sqrt{p\left(1-p\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\\ SE=\sqrt{0.4009146\left(1-0.4009146\right)\left(\frac{1}{434}+\frac{1}{222}\right)}\\ SE=0.0404\end{array}$

  $\begin{array}{l}{\chi }^2=\frac{{\hat{p}}_1-{\hat{p}}_2}{SE}\\ {\chi }^2=\frac{0.4078-0.3874}{0.0404}\\ {\chi }^2=0.5050\end{array}$

  $p-$ Value $=0.6136$

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

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

(d)

To determine

To explain:

The effectiveness of supplementing $\text{t-PA}$ with endovascular therapy in patients with an acute ischemic strokeof the given data.

Explanation of Solution

Given:

  $\begin{array}{l}{x}_1=177,x_2=86\\ n_1=434,n_2=222\end{array}$

Concept used:

Formula

Standard of error

  $SE=\sqrt{p\left(1-p\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}$

Significance level

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

Calculation:

Parameter $p=$ true proportion of individuals in population

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

  $X$ Is a random variable

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_1:p_1>p_2\end{array}$

  $\begin{array}{l}{\hat{p}}_1=\frac{177}{434}\\ {\hat{p}}_1=0.4078\\ {\hat{p}}_2=\frac{86}{222}\\ {\hat{p}}_2=0.3874\end{array}$

The value of $p$ is

  $\hat{p}\sim N\left(p,\frac{pq}{n}\right)$

  $p$ Comes from null hypothesis and alternative hypothesis

  $z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}}$

Pooled proportion

  $\begin{array}{l}p=\frac{177+86}{434+222}\\ p=0.4009146\end{array}$

Standard of error

  $\begin{array}{l}SE=\sqrt{p\left(1-p\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\\ SE=\sqrt{0.4009146\left(1-0.4009146\right)\left(\frac{1}{434}+\frac{1}{222}\right)}\\ SE=0.0404\end{array}$

  $\begin{array}{l}z=\frac{{\hat{p}}_1-{\hat{p}}_2}{SE}\\ z=\frac{0.4078-0.3874}{0.0404}\\ z=0.5050\end{array}$

For two-tail test

  $p-$ Value

  $\begin{array}{l}=2P\left(z>z_0\right)\\ =2P\left(z>0.5050\right)\\ =0.6136\end{array}$

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 $H_0$

Supplementing $\text{t-PA}$ with endovascular therapy does not significantly increases the proportion of patients with an acute ischemic stroke

So, the regain functional independence.