Chapter 12, Problem 4CS

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 :

  

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 $H_{\mathrm{a}}$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:

For department A:

Department A
Gender	Accept	Reject	Total
Male	512	313	825
Female	89	19	108
Total	601	332	933

Finding the expected frequency for the cell corresponding to:	The expected frequency
Number of male applicants acceptedin department A The row total is 825, the column total is 601, and the grand total is 933.	$\frac{825\cdot 601}{933}\approx 531.43$
Number of male applicants rejectedin department A The row total is 825, the column total is 332, and the grand total is 933.	$\frac{825\cdot 332}{933}\approx 293.57$
Number of female applicantsaccepted in department A The row total is 108, the column total is 601, and the grand total is 933.	$\frac{108\cdot 601}{933}\approx 69.57$
Number of female applicantsrejected in department A The row total is 108, the column total is 332, and the grand total is 933.	$\frac{108\cdot 332}{933}\approx 38.43$

Finding the value of the chi-square corresponding to:	$\frac{{\left(O-E\right)}^2}{E}$
Number of male applicants acceptedin department A The observed frequency is 512 and expected frequency is 531.43	$\frac{{\left(512-531.43\right)}^2}{531.43}\approx 0.71$
Number of male applicants rejectedin department A The observed frequency is 313 and expected frequency is 293.57	$\frac{{\left(313-293.57\right)}^2}{293.57}\approx 1.29$
Number of female applicantsaccepted in department A The observed frequency is 89 and expected frequency is 69.57	$\frac{{\left(89-69.57\right)}^2}{69.57}\approx 5.43$
Number of female applicantsrejected in department A The observed frequency is 19 and expected frequency is 38.43	$\frac{{\left(19-38.43\right)}^2}{38.43}\approx 9.82$

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

  ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\}=\begin{array}{c}0.71+1.29+5.43+9.82\end{array}\Rightarrow {\chi }^2\approx 17.25$

In department A, % of male accepted = $\frac{512}{625}=62.1\%$ and % of female accepted = $\frac{89}{108}=82.4\%$

For department B:

Department B
Gender	Accept	Reject	Total
Male	353	207	560
Female	17	8	25
Total	370	215	585

Finding the expected frequency for the cell corresponding to:	The expected frequency
Number of male applicants acceptedin department B The row total is 560, the column total is 370, and the grand total is 585.	$\frac{560\cdot 370}{585}\approx 354.19$
Number of male applicants rejectedin department B The row total is 560, the column total is 215, and the grand total is 585.	$\frac{560\cdot 215}{585}\approx 205.81$
Number of female applicantsaccepted in department B The row total is 25, the column total is 370, and the grand total is 585.	$\frac{25\cdot 370}{585}\approx 15.81$
Number of female applicantsrejected in department B The row total is 25, the column total is 215, and the grand total is 585.	$\frac{25\cdot 215}{585}\approx 9.19$

Finding the value of the chi-square corresponding to:	$\frac{{\left(O-E\right)}^2}{E}$
Number of male applicants acceptedin department B The observed frequency is 353 and expected frequency is 354.19	$\frac{{\left(353-354.19\right)}^2}{354.19}\approx 0.00$
Number of male applicants rejectedin department B The observed frequency is 207 and expected frequency is 205.81	$\frac{{\left(207-205.81\right)}^2}{205.81}\approx 0.01$
Number of female applicantsaccepted in department B The observed frequency is 17 and expected frequency is 15.81	$\frac{{\left(17-15.81\right)}^2}{15.81}\approx 0.09$
Number of female applicantsrejected in department B The observed frequency is 8 and expected frequency is 9.19	$\frac{{\left(8-9.19\right)}^2}{9.19}\approx 0.15$

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

  $\begin{array}{l}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\}=0.00+0.01+0.09+0.15\\ {\chi }^2\approx 0.25\end{array}$

In department B, % of male accepted = $\frac{353}{560}=63.0\%$ and % of female accepted = $\frac{17}{25}=68.0\%$

For department C:

Department C
Gender	Accept	Reject	Total
Male	120	205	325
Female	202	391	593
Total	322	596	918

Finding the expected frequency for the cell corresponding to:	The expected frequency
Number of male applicants acceptedin department C The row total is 325, the column total is 322, and the grand total is 918.	$\frac{325\cdot 322}{918}\approx 114.00$
Number of male applicants rejectedin department C The row total is 325, the column total is 596, and the grand total is918.	$\frac{325\cdot 596}{918}\approx 211.00$
Number of female applicantsaccepted in department C The row total is 593, the column total is 322, and the grand total is 918.	$\frac{593\cdot 322}{918}\approx 208.00$
Number of female applicantsrejected in department C The row total is 593, the column total is 596, and the grand total is 918.	$\frac{593\cdot 596}{918}\approx 385.00$

Finding the value of the chi-square corresponding to:	$\frac{{\left(O-E\right)}^2}{E}$
Number of male applicants acceptedin department C The observed frequency is 120 and expected frequency is 114	$\frac{{\left(120-114\right)}^2}{114}\approx 0.32$
Number of male applicants rejectedin department C The observed frequency is 205 and expected frequency is 211	$\frac{{\left(205-211\right)}^2}{211}\approx 0.17$
Number of female applicantsaccepted in department C The observed frequency is 202 and expected frequency is 208	$\frac{{\left(202-208\right)}^2}{208}\approx 0.17$
Number of female applicantsrejected in department C The observed frequency is 391 and expected frequency is 385	$\frac{{\left(391-385\right)}^2}{385}\approx 0.09$

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

  $\begin{array}{l}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\}=0.32+0.17+0.17+0.09\\ {\chi }^2\approx 0.75\end{array}$

In department C, % of male accepted = $\frac{120}{325}=36.9\%$ and % of female accepted = $\frac{202}{593}=34.1\%$

For department D:

Department D
Gender	Accept	Reject	Total
Male	138	279	417
Female	131	244	375
Total	269	523	792

Finding the expected frequency for the cell corresponding to:	The expected frequency
Number of male applicants acceptedin department D The row total is 417, the column total is 269, and the grand total is 792.	$\frac{417\cdot 269}{792}\approx 141.63$
Number of male applicants rejectedin department D The row total is 417, the column total is 523, and the grand total is792.	$\frac{417\cdot 523}{792}\approx 275.37$
Number of female applicantsaccepted in department D The row total is 375, the column total is 269, and the grand total is 792.	$\frac{375\cdot 269}{792}\approx 127.37$
Number of female applicantsrejected in department D The row total is 375, the column total is 523, and the grand total is 792.	$\frac{375\cdot 523}{792}\approx 247.63$

Finding the value of the chi-square corresponding to:	$\frac{{\left(O-E\right)}^2}{E}$
Number of male applicants acceptedin department D The observed frequency is 138 and expected frequency is 141.63	$\frac{{\left(138-141.63\right)}^2}{141.63}\approx 0.09$
Number of male applicants rejectedin department D The observed frequency is 279 and expected frequency is 275.37	$\frac{{\left(279-275.37\right)}^2}{275.37}\approx 0.05$
Number of female applicantsaccepted in department D The observed frequency is 131 and expected frequency is 127.37	$\frac{{\left(131-127.37\right)}^2}{127.37}\approx 0.10$
Number of female applicantsrejected in department D The observed frequency is 244 and expected frequency is 247.63	$\frac{{\left(244-247.63\right)}^2}{247.63}\approx 0.05$

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

  $\begin{array}{l}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\}=0.09+0.05+0.10+0.05\\ {\chi }^2\approx 0.30\end{array}$

In department D, % of male accepted = $\frac{138}{417}=33.1\%$ and % of female accepted = $\frac{131}{375}=34.9\%$

For department E:

Department E
Gender	Accept	Reject	Total
Male	53	138	191
Female	94	299	393
Total	147	437	584

Finding the expected frequency for the cell corresponding to:	The expected frequency
Number of male applicants acceptedin department E The row total is 191, the column total is 147, and the grand total is 584.	$\frac{191\cdot 147}{584}\approx 48.08$
Number of male applicants rejectedin department E The row total is 191, the column total is 437, and the grand total is584.	$\frac{191\cdot 437}{584}\approx 142.92$
Number of female applicantsaccepted in department E The row total is 393, the column total is 147, and the grand total is 584.	$\frac{393\cdot 147}{584}\approx 98.92$
Number of female applicantsrejected in department E The row total is 393, the column total is 437, and the grand total is 584.	$\frac{393\cdot 437}{584}\approx 294.08$

Finding the value of the chi-square corresponding to:	$\frac{{\left(O-E\right)}^2}{E}$
Number of male applicants acceptedin department E The observed frequency is 53 and expected frequency is 48.08	$\frac{{\left(53-48.08\right)}^2}{48.08}\approx 0.50$
Number of male applicants rejectedin department E The observed frequency is 138 and expected frequency is 142.92	$\frac{{\left(138-142.92\right)}^2}{142.92}\approx 0.17$
Number of female applicantsaccepted in department E The observed frequency is 94 and expected frequency is 98.92	$\frac{{\left(94-98.92\right)}^2}{98.92}\approx 0.24$
Number of female applicantsrejected in department E The observed frequency is 299 and expected frequency is 294.08	$\frac{{\left(299-294.08\right)}^2}{294.08}\approx 0.08$

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

  $\begin{array}{l}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\}=0.50+0.17+0.25+0.08\\ {\chi }^2\approx 1.00\end{array}$

In department E, % of male accepted = $\frac{53}{191}=27.7\%$ and % of female accepted = $\frac{94}{393}=23.9\%$

For department F:

Department F
Gender	Accept	Reject	Total
Male	22	351	373
Female	24	317	341
Total	46	668	714

Finding the expected frequency for the cell corresponding to:	The expected frequency
Number of male applicants acceptedin department F The row total is 373, the column total is 46, and the grand total is 714.	$\frac{373\cdot 46}{714}\approx 24.03$
Number of male applicants rejectedin department F The row total is 373, the column total is 668, and the grand total is714.	$\frac{373\cdot 668}{714}\approx 348.97$
Number of female applicantsaccepted in department F The row total is 341, the column total is 46, and the grand total is 714.	$\frac{341\cdot 46}{714}\approx 21.97$
Number of female applicantsrejected in department F The row total is 341, the column total is 668, and the grand total is 714.	$\frac{341\cdot 668}{714}\approx 319.03$

Finding the value of the chi-square corresponding to:	$\frac{{\left(O-E\right)}^2}{E}$
Number of male applicants acceptedin department F The observed frequency is 22 and expected frequency is 24.03	$\frac{{\left(22-24.03\right)}^2}{24.03}\approx 0.17$
Number of male applicants rejectedin department F The observed frequency is 351 and expected frequency is 348.97	$\frac{{\left(351-348.97\right)}^2}{348.97}\approx 0.01$
Number of female applicantsaccepted in department F The observed frequency is 24 and expected frequency is 21.97	$\frac{{\left(24-21.97\right)}^2}{21.97}\approx 0.19$
Number of female applicantsrejected in department F The observed frequency is 317 and expected frequency is 319.03	$\frac{{\left(317-319.03\right)}^2}{319.03}\approx 0.01$

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

  $\begin{array}{l}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\}=0.17+0.01+0.19+0.01\\ {\chi }^2\approx 0.39\end{array}$

In department E, % of male accepted = $\frac{22}{373}=5.9\%$ and % of female accepted = $\frac{24}{341}=7.0\%$

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

For all the contingency table $r=2$ , $c=2$ so the number of degree of freedom is $\left(2-1\right)\left(2-1\right)=\left(1\right)\left(1\right)=1$

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

The critical value is same for all the contingency table.

Conclusion:

For department A:

Test statistic: 17.25; Critical value: 6.635. ${\chi }^2>CriticalValue$ ,we Reject H₀

For department B:

Test statistic: 0.25; Critical value: 6.635. ${\chi }^2<CriticalValue$ , we do notreject H₀

For department C:

Test statistic: 0.75; Critical value: 6.635. ${\chi }^2<CriticalValue$ , we do notreject H₀

For department D:

Test statistic: 0.30; Critical value: 6.635. ${\chi }^2<CriticalValue$ , we do notreject H₀

For department E:

Test statistic: 1.00; Critical value: 6.635. ${\chi }^2<CriticalValue$ , we do notreject H₀

For department F:

Test statistic: 0.39; Critical value: 6.635. ${\chi }^2<CriticalValue$ , we do notreject H₀

In departmentA, 82.4% of the women were accepted, but only 62.1% of themen were accepted.