Chapter 10.7, Problem 13E

To determine

To find:

To show that there is a relationship between grades and the combination of subject and gender

Answer to Problem 13E

Solution:

There is a relationship between grades and the combination of particular subject and gender for student in the state.

Explanation of Solution

The following table represents the random sample of relationship between grades and the combination of particular subject and gender for student in the state is,

	Observed Sample of 440 Students
	A	B	C	D	F	Total
Math, Female	12	30	44	15	9	110
Math, Male	5	24	37	36	8	110
English, Female	16	41	39	10	4	110
English, Male	10	40	35	19	6	110
Total	43	135	155	80	27	440

Step 1: To state the null and alternative hypotheses.

Let the null hypothesis be that grades and combination of subjects and gender are independent of one another.

Null hypothesis: $H_0$: Grades and combination of subjects and gender are independent.

Alternative hypothesis: $H_a$: Grades and combination of subjects and gender are dependent.

Step 2: To determine which distribution to use for the test statistic, and state the level of significance.

To determine there is a relationship between grades and the combination of particular subject and gender for student in the state. Since we assume that the necessary condition that have been met. So, use the chi-square test for this association and the level of siginificance is $\alpha =0.005$.

Step 3: To calculate the test statistic.

First calculate the expected value for each cell in the contingency table.

Formula for calculating expected value of a frequency in a contingency table:

The expected value of the frequency for the i^th possible outcome in a contingency table is given by

$E_i=\frac{\left(rowtotal\right)\left(columntotal\right)}{n}$

Where n is the sample size.

From the above table represent the value of n is 146.

The expected values of for the Grade A and the combination of subjects and gender is given below,

$\begin{array}{rll}{E}_{\text{Grade A and Math, Female }}& =& \frac{43\times 110}{440}\\ & =& \frac{4730}{440}\\ & =& 10.75\\ E_{\text{Grade A and Math, Male }}& =& \frac{43\times 110}{440}\\ & =& \frac{4730}{440}\\ & =& 10.75\\ E_{\text{Grade A and English, Female }}& =& \frac{43\times 110}{440}\\ & =& \frac{4730}{440}\\ & =& 10.75\\ E_{\text{Grade A and English, Male }}& =& \frac{43\times 110}{440}\\ & =& \frac{4730}{440}\\ & =& 10.75\end{array}$

The expected values of for the Grade B and the combination of subjects and gender is given below,

$\begin{array}{rll}{E}_{\text{Grade B and Math, Female }}& =& \frac{135\times 110}{440}\\ & =& \frac{14850}{440}\\ & =& 33.75\\ E_{\text{Grade B and Math, Male }}& =& \frac{135\times 110}{440}\\ & =& \frac{14850}{440}\\ & =& 33.75\end{array}$

$\begin{array}{rll}{E}_{\text{Grade B and English, Female }}& =& \frac{135\times 110}{440}\\ & =& \frac{14850}{440}\\ & =& 33.75\\ E_{\text{Grade B and English, Male }}& =& \frac{135\times 110}{440}\\ & =& \frac{14850}{440}\\ & =& 33.75\end{array}$

The expected values of for the Grade C and the combination of subjects and gender is given below,

$\begin{array}{rll}{E}_{\text{Grade C and Math, Female }}& =& \frac{155\times 110}{440}\\ & =& \frac{17050}{440}\\ & =& 38.75\\ E_{\text{Grade C and Math, Male }}& =& \frac{155\times 110}{440}\\ & =& \frac{17050}{440}\\ & =& 38.75\end{array}$

$\begin{array}{rll}{E}_{\text{Grade C and English, Female }}& =& \frac{155\times 110}{440}\\ & =& \frac{17050}{440}\\ & =& 38.75\\ E_{\text{Grade C and English, Male }}& =& \frac{155\times 110}{440}\\ & =& \frac{17050}{440}\\ & =& 38.75\end{array}$

The expected values of for the Grade D and the combination of subjects and gender is given below,

$\begin{array}{rll}{E}_{\text{Grade D and Math, Female }}& =& \frac{80\times 110}{440}\\ & =& \frac{8800}{440}\\ & =& 20\\ E_{\text{Grade D and Math, Male }}& =& \frac{80\times 110}{440}\\ & =& \frac{8800}{440}\\ & =& 20\\ E_{\text{Grade D and English, Female }}& =& \frac{80\times 110}{440}\\ & =& \frac{8800}{440}\\ & =& 20\\ E_{\text{Grade D and English, Male }}& =& \frac{80\times 110}{440}\\ & =& \frac{8800}{440}\\ & =& 20\end{array}$

The expected values of for the Grade A and the combination of subjects and gender is given below,

$\begin{array}{rll}{E}_{\text{Grade F and Math, Female }}& =& \frac{27\times 110}{440}\\ & =& \frac{29.70}{440}\\ & =& 6.75\\ E_{\text{Grade F and Math, Male }}& =& \frac{27\times 110}{440}\\ & =& \frac{29.70}{440}\\ & =& 6.75\end{array}$

$\begin{array}{rll}{E}_{\text{Grade F and English, Female }}& =& \frac{27\times 110}{440}\\ & =& \frac{29.70}{440}\\ & =& 6.75\\ E_{\text{Grade F and English, Male }}& =& \frac{27\times 110}{440}\\ & =& \frac{29.70}{440}\\ & =& 6.75\end{array}$

The table represents the contingency table of expected values by using the above formula is given below,

	Expected values
	A	B	C	D	F	Total
Math, Female	10.75	33.75	38.75	20	6.75	110
Math, Male	10.75	33.75	38.75	20	6.75	110
English, Female	10.75	33.75	38.75	20	6.75	110
English, Male	10.75	33.75	38.75	20	6.75	110
Total	43	135	155	80	27	440

Formula for calculating test statistic for a Chi-Square test for association:

The test statistic for a chi-square test for association is given below,

${\chi }^2={\displaystyle \sum _{i=1}^n\frac{{\left(O_i-E_i\right)}^2}{E_i}}$

Where O_i, is the observed frequency for the i^th possible outcome and E_i is the expected frequency for the i^th possible outcome and n is the sample size.

From the above table represent the value of n is 440.

The test statistic, ${\chi }^2$, for a chi- square test for association using the above contingency tables is given below,

$\begin{array}{rll}{\chi }^2& =& {\displaystyle \sum _{i=1}^n\frac{{\left(O_i-E_i\right)}^2}{E_i}}\\ & =& \{\begin{array}{l}\frac{{\left(O_A-E_{Math,Female}\right)}^2}{E_{Math,Female}}+\frac{{\left(O_B-E_{Math,Female}\right)}^2}{E_{Math,Female}}+\frac{{\left(O_C-E_{Math,Female}\right)}^2}{E_{Math,Female}}\\ +\frac{{\left(O_D-E_{Math,Female}\right)}^2}{E_{Math,Female}}+\frac{{\left(O_F-E_{Math,Female}\right)}^2}{E_{Math,Female}}+\frac{{\left(O_A-E_{Math,Male}\right)}^2}{E_{Math,Male}}\\ +\frac{{\left(O_B-E_{Math,Male}\right)}^2}{E_{Math,Male}}+\frac{{\left(O_C-E_{Math,Male}\right)}^2}{E_{Math,Male}}+\frac{{\left(O_D-E_{Math,Male}\right)}^2}{E_{Math,Male}}\\ +\frac{{\left(O_F-E_{Math,Male}\right)}^2}{E_{Math,Female}}+\frac{{\left(O_A-E_{English,Female}\right)}^2}{E_{English,Female}}+\frac{{\left(O_B-E_{English,Female}\right)}^2}{E_{English,Female}}\\ +\frac{{\left(O_C-E_{English,Female}\right)}^2}{E_{English,Female}}+\frac{{\left(O_D-E_{English,Female}\right)}^2}{E_{English,Female}}+\frac{{\left(O_F-E_{English,Female}\right)}^2}{E_{English,Female}}\\ +\frac{{\left(O_A-E_{English,Male}\right)}^2}{E_{English,Male}}+\frac{{\left(O_B-E_{English,Male}\right)}^2}{E_{English,Male}}+\frac{{\left(O_C-E_{English,Male}\right)}^2}{E_{English,Male}}\\ +\frac{{\left(O_D-E_{English,Male}\right)}^2}{E_{English,Male}}+\frac{{\left(O_F-E_{English,Male}\right)}^2}{E_{English,Male}}\end{array}\end{array}$

The following table represents the chi square test statistic value s given below,

Observed	Expected	$O_i-E_i$	${\left(O_i-E_i\right)}^2$	$\frac{{\left(O_i-E_i\right)}^2}{E_i}$
12	10.75	1.25	1.5625	0.14535
5	10.75	-5.75	33.0625	3.07558
16	10.75	5.25	27.5625	2.56395
10	10.75	-0.75	0.5625	0.05233
30	33.75	-3.75	14.0625	0.41667
24	33.75	-9.75	95.0625	2.81667
41	33.75	7.25	52.5625	1.55741
40	33.75	6.25	39.0625	1.15741
44	38.75	5.25	27.5625	0.71129
37	38.75	-1.75	3.0625	0.07903
39	38.75	0.25	0.0625	0.00161
35	38.75	-3.75	14.0625	0.3629
15	20	-5	25	1.25
36	20	16	256	12.8
10	20	-10	100	5
19	20	-1	1	0.05
9	6.75	2.25	5.0625	0.75
8	6.75	1.25	1.5625	0.23148
4	6.75	-2.75	7.5625	1.12037
6	6.75	-0.75	0.5625	0.08333
				${\chi }^2={\displaystyle \sum _{i=1}^n\frac{{\left(O_i-E_i\right)}^2}{E_i}}=34.2254$

So, the value of test statistic, ${\chi }^2$, for a chi- square test for association is 34.2254.

Step 4:

Draw a conclusion and interpret the decision.

Degrees of freedom in a Chi-square test for association:

In a chi-square test for association the number of degrees of freedom for the chi-square distribution of the test statistic is given by,

$df=\left(R-1\right)\cdot \left(C-1\right)$

Where R is the number of rows of the data in the contingency table (not including the row total) and C is the number of columns of data in the contingency table (not including the column totals).

Rejection Region for Chi-Square Test for Association:

Reject the null hypothesis, $H_0$, if ${\chi }^2\ge {\chi }_{\alpha }^2$.

From the given information R = 4 and C = 5.

Substitute the above values in the formula of degrees of freedom to get the following,

$\begin{array}{rll}df& =& \left(R-1\right)\cdot \left(C-1\right)\\ & =& \left(4-1\right)\cdot \left(5-1\right)\\ & =& \left(3\right)\cdot \left(4\right)\\ & =& 12\end{array}$

The number of degrees of freedom for this test is 12.

Use the level of significance of $\alpha =0.005$ to find the critical value ${\chi }_{0.005}^2$ of the chi-square distribution with df = 12 in order to make a decision about the null hypothesis.

Use the “Area to the right of the critical value ${\chi }^2$” table to get the following,

${\chi }_{0.005}^2=28.2995$

The test statistic value is ${\chi }^2=34.2254$.

Comparing the test statistic and critical value to get the following,

$34.2254\ge 28.2995\iff {\chi }^2\ge {\chi }_{0.005}^2$. Therefore, we reject the null hypothesis and conclude that assumptions are independent.

So, there is a relationship between grades and the combination of particular subject and gender for student in the state.

Final statement:

There is a relationship between grades and the combination of particular subject and gender for student in the state.