Chapter 10, Problem 1CQ

To determine

Explain whether the null hypothesis $H_0$ will be rejected at the level of significance of $\alpha =0.05$.

Answer to Problem 1CQ

The null hypothesis will be rejected.

Explanation of Solution

Calculation:

The contingency table with 4 rows and 5 columns is given. The chi-square test statistic value is 22.87.

Contingency table:

A contingency table is obtained as using two qualitative variables. One of the qualitative variable is row variable that has one category for each row of the table another is column variable has one category for each column of the table.

The hypotheses are:

Null Hypothesis:

$H_0:\text{The rows and columns are independent}\text{.}$

Alternate Hypothesis:

$H_1:\text{The The rows and columns are not independent}.$

Chi-Square statistic:

The chi-square statistic is obtained as ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$ where $O_1,O_2,\mathrm{...},O_k$ be the observed frequencies and $E_1,E_2,\mathrm{...},E_k$ be the expected frequencies for k number of categories.

It is known that under the null hypothesis $H_0$ the test statistic ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$ follows chi-square distribution with $\left(r-1\right)\left(c-1\right)$ degrees of freedom for r number of rows and c number of columns, provided that all the expected frequencies are greater than or equal to 5.

In the given question there are 4 rows and 5 columns.

Hence, the degrees of freedom is

$\begin{array}{c}\left(r-1\right)\left(c-1\right)=\left(4-1\right)\left(5-1\right)\\ =\left(3\right)\left(4\right)\\ =12\end{array}$

Thus, the degrees of freedom is 12.

Level of significance:

The level of significance is given as 0.05.

Critical value:

In a test of hypotheses, the critical value is the point by which one can reject or accept the null hypothesis.

Software procedure:

Step-by-step software procedure to obtain critical value using MINITAB software is as follows:

• Select Graph>Probability distribution plot > view probability.
• Select Chi -Square under distribution.
• In Degrees of freedom, enter 12.
• Choose Probability Value and Right Tail for the region of the curve to shade.
• Enter the Probability value as 0.05 under shaded area.
• Select OK.
• Output using MINITAB software is given below:

Hence, the critical value at $\alpha =0.05$ is 21.03.

  Rejection rule:

If the ${\chi }^2$ value is greater than or equal to critical value, that is ${\chi }^2\ge {\chi }_{\alpha ;k-1}^2$, then reject the null hypothesis $H_0$. Otherwise, do not reject $H_0$.

Conclusion:

Here, the ${\chi }^2$ value is greater than the critical value.

That is, ${\chi }^2\left(=22.87\right)>{\chi }_{0.05,12}^2\left(=21.03\right)$

Thus, the decision is “reject the null hypothesis”.

Thus, there is not enough evidence to conclude that the rows and columns are independent.