Chapter 12, Problem 2CS

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
Accept	601	370	322	269	147	46	1755
Reject	332	215	596	523	437	668	2771
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 applicants who were accepted by department A The row total is 1755, the column total is 933, and the grand total is 4526.	$\frac{1755\cdot 933}{4526}\approx 361.78$
Number of applicants who were accepted by department B The row total is 1755, the column total is 585, and the grand total is 4526.	$\frac{1755\cdot 585}{4526}\approx 226.84$
Number of applicants who were accepted by department C The row total is 1755, the column total is 918, and the grand total is 4526.	$\frac{1755\cdot 918}{4526}\approx 355.96$
Number of applicants who were accepted by department D The row total is 1755, the column total is 792, and the grand total is 4526.	$\frac{1755\cdot 792}{4526}\approx 307.11$
Number of applicants who were accepted by department E The row total is 1755, the column total is 584, and the grand total is 4526.	$\frac{1755\cdot 584}{4526}\approx 226.45$
Number of applicants who were accepted by department F The row total is 1755, the column total is 714, and the grand total is 4526.	$\frac{1755\cdot 714}{4526}\approx 276.86$
Number of applicants who were rejected by department A The row total is 2771, the column total is 933, and the grand total is 4526.	$\frac{2771\cdot 933}{4526}\approx 571.22$
Number of applicants who were rejected by department B The row total is 2771, the column total is 585, and the grand total is 4526.	$\frac{2771\cdot 585}{4526}\approx 358.16$
Number of applicants who were rejected by department C The row total is 2771, the column total is 918, and the grand total is 4526.	$\frac{2771\cdot 918}{4526}\approx 562.04$
Number of applicants who were rejected by department D The row total is 2771, the column total is 792, and the grand total is 4526.	$\frac{2771\cdot 792}{4526}\approx 484.89$
Number of applicants who were rejected by department E The row total is 2771, the column total is 584, and the grand total is 4526.	$\frac{2771\cdot 584}{4526}\approx 357.55$
Number of applicants who were rejected by department F The row total is 2771, the column total is 714, and the grand total is 4526.	$\frac{2771\cdot 714}{4526}\approx 437.14$

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 applicants who were accepted by department A The observed frequency is 601 and expected frequency is 361.78	$\frac{{\left(601-361.78\right)}^2}{361.78}\approx 158.18$
Number of applicants who were accepted by department B The observed frequency is 370 and expected frequency is 226.84	$\frac{{\left(370-226.84\right)}^2}{226.84}\approx 90.35$
Number of applicants who were accepted by department C The observed frequency is 322 and expected frequency is 355.96	$\frac{{\left(322-355.96\right)}^2}{355.96}\approx 3.24$
Number of applicants who were accepted by department D The observed frequency is 269 and expected frequency is 307.11	$\frac{{\left(269-307.11\right)}^2}{307.11}\approx 4.73$
Number of applicants who were accepted by department E The observed frequency is 147 and expected frequency is 226.45	$\frac{{\left(147-226.45\right)}^2}{226.45}\approx 27.88$
Number of applicants who were accepted by department F The observed frequency is 46 and expected frequency is 276.86	$\frac{{\left(46-276.86\right)}^2}{276.86}\approx 192.50$
Number of applicants who were rejectedby department A The observed frequency is 332 and expected frequency is 571.22	$\frac{{\left(332-571.22\right)}^2}{571.22}\approx 100.18$
Number of applicants who were rejected by department B The observed frequency is 215 and expected frequency is 358.16	$\frac{{\left(215-358.16\right)}^2}{358.16}\approx 57.22$
Number of applicants who were rejectedby department C The observed frequency is 596 and expected frequency is 562.04	$\frac{{\left(596-562.04\right)}^2}{562.04}\approx 2.05$
Number of applicants who were rejectedby department D The observed frequency is 523 and expected frequency is 484.89	$\frac{{\left(523-484.89\right)}^2}{484.89}\approx 2.99$
Number of applicants who were rejectedby department E The observed frequency is 437 and expected frequency is 357.55	$\frac{{\left(437-357.55\right)}^2}{357.55}\approx 17.66$
Number of applicants who were rejectedby department F The observed frequency is 668 and expected frequency is 437.14	$\frac{{\left(668-437.14\right)}^2}{437.14}\approx 121.92$

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

  $\begin{array}{l}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\}=\begin{array}{c}\begin{array}{l}158.18+90.35+3.24+4.73+27.88+192.50\\ +100.18+57.22+2.05+2.99+17.66+121.92\end{array}\end{array}\\ {\chi }^2\approx 778.91\end{array}$

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: 778.91; Critical value: 15.086. ${\chi }^2>CriticalValue$ , so we Reject H₀ (null hypothesis states that acceptance to graduate school is independent of department). Weconclude that acceptance rates differ among departments.