Chapter 12, Problem 3CS

To determine

To test: the null hypothesis and show that it is rejected at the $\alpha =0.01$ level of significance.

Explanation of Solution

Given information :

Department	A	B	C	D	E	F	Total
Male	825	560	325	417	191	373	2691
Female	108	25	593	375	393	341	1835
Total	933	585	918	792	584	714	4526

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.

  $E=\frac{R\times C}{G}$ , where E is expected frequency, R is row total, C is column total and G is grand total for a cell

  ${\chi }^2$ 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:

Finding the expected frequency for the cell corresponding to:	The expected frequency
Number of male applicantsin department A The row total is 2691, the column total is 933, and the grand total is 4526.	$\frac{2691\cdot 933}{4526}\approx 554.73$
Number of male applicantsin department B The row total is 2691, the column total is 585, and the grand total is 4526.	$\frac{2691\cdot 585}{4526}\approx 347.82$
Number of male applicantsin department C The row total is 2691, the column total is 918, and the grand total is 4526.	$\frac{2691\cdot 918}{4526}\approx 545.81$
Number of male applicantsin department D The row total is 2691, the column total is 792, and the grand total is 4526.	$\frac{2691\cdot 792}{4526}\approx 470.90$
Number of male applicantsin department E The row total is 2691, the column total is 584, and the grand total is 4526.	$\frac{2691\cdot 584}{4526}\approx 347.23$
Number of male applicantsin department F The row total is 2691, the column total is 714, and the grand total is 4526.	$\frac{2691\cdot 714}{4526}\approx 424.52$
Number of female applicantsin department A The row total is 1835, the column total is 933, and the grand total is 4526.	$\frac{1835\cdot 933}{4526}\approx 378.27$
Number of female applicantsin department B The row total is 1835, the column total is 585, and the grand total is 4526.	$\frac{1835\cdot 585}{4526}\approx 237.18$
Number of female applicantsin department C The row total is 1835, the column total is 918, and the grand total is 4526.	$\frac{1835\cdot 918}{4526}\approx 372.19$
Number of female applicantsin department D The row total is 1835, the column total is 792, and the grand total is 4526.	$\frac{1835\cdot 792}{4526}\approx 321.10$
Number of female applicantsin department E The row total is 1835, the column total is 584, and the grand total is 4526.	$\frac{1835\cdot 584}{4526}\approx 236.77$
Number of female applicantsin department F The row total is 1835, the column total is 714, and the grand total is 4526.	$\frac{1835\cdot 714}{4526}\approx 289.48$

All the expected frequencies are at least 5. From the results of previous part we have the below table:

Finding the value of the chi-square corresponding to:	$\frac{{\left(O-E\right)}^2}{E}$
Number of male applicantsin department A The observed frequency is 825 and expected frequency is 554.73	$\frac{{\left(825-554.73\right)}^2}{554.73}\approx 131.68$
Number of male applicantsin department B The observed frequency is 560 and expected frequency is 347.82	$\frac{{\left(560-347.82\right)}^2}{347.82}\approx 129.44$
Number of male applicantsin department C The observed frequency is 325 and expected frequency is 545.81	$\frac{{\left(325-545.81\right)}^2}{545.81}\approx 89.33$
Number of male applicantsin department D The observed frequency is 417 and expected frequency is 470.90	$\frac{{\left(417-470.90\right)}^2}{470.90}\approx 6.17$
Number of male applicantsin department E The observed frequency is 191 and expected frequency is 347.23	$\frac{{\left(191-347.23\right)}^2}{347.23}\approx 70.29$
Number of male applicantsin department F The observed frequency is 373 and expected frequency is 424.52	$\frac{{\left(373-424.52\right)}^2}{424.52}\approx 6.25$
Number of female applicantsin department A The observed frequency is 108 and expected frequency is 378.27	$\frac{{\left(108-378.27\right)}^2}{378.27}\approx 193.11$
Number of female applicantsin department B The observed frequency is 25 and expected frequency is 237.18	$\frac{{\left(25-237.18\right)}^2}{237.18}\approx 189.81$
Number of female applicantsin department C The observed frequency is 593 and expected frequency is 372.19	$\frac{{\left(593-372.19\right)}^2}{372.19}\approx 131.00$
Number of female applicantsin department D The observed frequency is 375 and expected frequency is 321.10	$\frac{{\left(375-321.10\right)}^2}{321.10}\approx 9.05$
Number of female applicantsin department E The observed frequency is 393 and expected frequency is 236.77	$\frac{{\left(393-236.77\right)}^2}{236.77}\approx 103.08$
Number of female applicantsin department F The observed frequency is 341 and expected frequency is 289.48	$\frac{{\left(341-289.48\right)}^2}{289.48}\approx 9.17$

To compute the test statistics, we use the observed frequencies and expected frequency:

  $\begin{array}{l}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\}=\begin{array}{c}\begin{array}{l}131.68+129.44+89.33+6.17+70.29+6.25\\ +193.11+189.81+131+9.05+103.08+9.17\end{array}\end{array}\\ {\chi }^2\approx 1068.37\end{array}$

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

Given $r=2$ , $c=6$ so the number of degree of freedom is $\left(2-1\right)\left(6-1\right)=\left(1\right)\left(5\right)=5$

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

Conclusion:

Test statistic: 1068.37; Critical value: 15.086. ${\chi }^2>CriticalValue$ ,we Reject H₀ (null hypothesis states that gender is independent of department)and weconclude that gender and department are not independent.