Chapter 21, Problem 21.1AYK

(a)

To determine

To explain:

Thechi-square and p-value of the given data.

Answer to Problem 21.1AYK

P-value $=$

  $0.000318$

Explanation of Solution

Given:

Two titled downward $=20,40\text{degree}$

Concept used:

Formula

Degree of freedom

  $=n-1$

Significance level

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

Calculation:

  $\begin{array}{l}{H}_0:p_1=p_2=p_3\\ H_a:\text{one}\text{of}\text{probability}\text{is}\text{differ}\end{array}$

Since the sample size are large.

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

Applying chi-square test of independence

  $E_i=\frac{\text{row}\text{total}\text{×column}\text{total}}{\text{grand}\text{total}}$

Draw the table

	Observed counts	$E$	$\frac{{\left(O-E\right)}^2}{E}$
Vertical	$31$	$17.66667$	$10.06289$
$20-\text{degree}$	$14$	$17.66667$	$0.761006$
$40-\text{degree}$	$8$	$17.66667$	$5.289308$

Chi-square $=\frac{{\left(O_i-E_i\right)}^2}{E_i}$

Chi-square $=16.1132$

Degree of freedom

  $\begin{array}{l}=n-1\\ =3-1\\ =2\end{array}$

P-value $=$

  $0.000318$

The p-value of the test is $0.000318$

Since p-value is greater than $0.05$

The result significant at $p<0.05$

So, the null hypothesis is fail to reject.

(b)

To determine

To explain:

Thenull and alternative hypothesis of the given data.

Answer to Problem 21.1AYK

P-value $=$

  $0.000318$

Explanation of Solution

Given:

Two titled downward $=20,40\text{degree}$

Concept used:

Formula

Degree of freedom

  $=n-1$

Significance level

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

Calculation:

  $\begin{array}{l}{H}_0:p_1=p_2=p_3\\ H_a:\text{one}\text{of}\text{probability}\text{is}\text{differ}\end{array}$

Since the sample size are large.

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

Applying chi-square test of independence

  $E_i=\frac{\text{row}\text{total}\text{×column}\text{total}}{\text{grand}\text{total}}$

Null hypothesis $H_0$ :-No relationship between expected ratio and observed ratio.

Alternative hypothesis $H_1$ :-A significant relationship between expected ratio and observed ratio 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 $H_0$

Since p-value is greater than $0.05$

The result significant at $p<0.05$

So, the null hypothesis is fail to reject.