Chapter 12, Problem 1CQ

A contingency table containing observed values has four rows and five columns. The value of the chi-square statistic for testing independence is 22.87. Is $H_0$ rejected at the $\alpha =0.05$ level?

To determine

To find : whether $H_0$ rejected at the $\alpha =0.05$ level.

Answer to Problem 1CQ

Yes. The null hypothesis $H_0$ is rejected because ${\chi }^2=22.87\left(given\right)$ is greater than the critical value 21.026.

Explanation of Solution

Given information : The value of chi-square statistics for testing independence is 22.87. A contingency table containing observed values has four rows and five columns.

Concept involved:

In order to decide whether the presumed hypothesis for data sample stands accurate for the entire population or not we use the hypothesis testing.

  $H_0$ represents null hypothesis test and $H_a$ represents alternative hypothesis test.

  ${\chi }^2$ represent value of chi square test statistics found using the formula ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$ where O is observed frequency and E is expected frequency.

The value of test statistics and the critical value identified from the table help us to decide whether to reject or do not reject null hypothesis.

The critical value from Table A.4, using degrees of freedom of contingency table of any given study is provided.

If ${\chi }^2\ge CriticalValue$ , then reject $H_0$ otherwise reject $H_0$ .

The values of two qualitative variables are connected and denoted in a contingency table. This table consists of rows and column. The variables in each row and each column of the table represent a category. The number of rows of contingency table is represented by letter ‘r’ and number of column of contingency table is represented by letter ‘c’.

The formula to find the number of degree of freedom of contingency table is $\left(r-1\right)\left(c-1\right)$ .

Calculation:

Here r represents the number of rows and c represents the number of columns.

Given $r=4$ , $c=5$ so the number of degree of freedom is $\left(4-1\right)\left(5-1\right)=\left(3\right)\left(4\right)=12$

Degrees of freedom	Table A.4 Critical Values for the chi-square Distribution
	0.995	0.99	0.975	0.95	0.90	0.10	0.05	0.025	0.01	0.005
1	0.000	0.000	0.001	0.004	0.016	2.706	3.841	5.024	6.635	7.879
2	0.010	0.020	0.051	0.103	0.211	4.605	5.991	7.378	9.210	10.597
3	0.072	0.115	0.216	0.352	0.584	6.251	7.815	9.348	11.345	12.838
4	0.207	0.297	0.484	0.711	1.064	7.779	9.488	11.143	13.277	14.860
5	0.412	0.554	0.831	1.145	1.610	9.236	11.070	12.833	15.086	16.750
6	0.676	0.872	1.237	1.635	2.204	10.645	12.592	14.449	16.812	18.548
7	0.989	1.239	1.690	2.167	2.833	12.017	14.067	16.013	18.475	20.278
8	1.344	1.646	2.180	2.733	3.490	13.362	15.507	17.535	20.090	21.955
9	1.735	2.088	2.700	3.325	4.168	14.684	16.919	19.023	21.666	23.589
10	2.156	2.558	3.247	3.940	4.865	15.987	18.307	20.483	23.209	25.188
11	2.603	3.053	3.816	4.575	5.578	17.275	19.675	21.920	24.725	26.757
12	3.074	3.571	4.404	5.226	6.304	18.549	21.026	23.337	26.217	28.300
13	3.565	4.107	5.009	5.892	7.042	19.812	22.362	24.736	27.688	29.819
14	4.075	4.660	5.629	6.571	7.790	21.064	23.685	26.119	29.141	31.319
15	4.601	5.229	6.262	7.261	8.547	22.307	24.996	27.488	30.578	32.801

Here degree of freedom is 12. The test statistics ${\chi }^2=22.87\left(given\right)$ is greater than the critical value 21.026. So reject $H_0$