Chapter 10.CR, Problem 15CR

To determine

To find:

To calculate the test statistic.

Answer to Problem 15CR

Solution:

a. $H_0$: Education level and region of the U.S are independent.

$H_a$: Education level and region of the U.S are dependent.

b. Chi-Square distribution; $\alpha =0.05$

c. ${\chi }^2=4.9971$.

d. ${\chi }_{0.05}^2=16.9190$; reject $H_0$ if ${\chi }^2\ge 16.919$, Therefore, we fail to reject the null hypothesis.

So, there is sufficient evidence not support the claim that education level and region of the U.S are not independent.

Explanation of Solution

 Consider the following scenario,

“A sociologist wants to study the eduaction levels of people living in various regios of the United States. He surveys a random sample of 100 people in each of the following regios: Northeast, Southeast, Midwest, and West.”

The results obtained are found in the following table is,

		Education Levels
	Less Than a High School Diploma	High School Graduate, No College	Some College or Associate Degree	College Graduate	Total
Northeast	11	29	45	15	100
Southeast	19	28	44	9	100
Midwest	15	25	47	13	100
West	13	31	42	14	100
Total	58	113	178	51	400

 Step 1: To state the null and alternative hypotheses.

Let the null hypothesis be that education level and region of the U.S are independent of one another.

Null hypothesis: $H_0$: Education level and region of the U.S are independent.

Alternative hypothesis: $H_a$: Education level and region of the U.S 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 education level and region of the U.S. 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 $\alpha =0.05$.

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 400.

The expected values of for northeast and education level is given below,

$\begin{array}{rll}{E}_{NortheastandLessThanaHighSchoolDiploma}& =& \frac{100\times 58}{400}\\ & =& \frac{5800}{400}\\ & =& 14.5\\ E_{NortheastandHighSchoolGraduate,NoCollege}& =& \frac{100\times 113}{400}\\ & =& \frac{11300}{400}\\ & =& 28.25\\ E_{NortheastandSomeCollegeorAssociateDegree}& =& \frac{100\times 178}{400}\\ & =& \frac{17800}{400}\\ & =& 44.5\\ E_{NortheastandCollegeGraduate}& =& \frac{100\times 51}{400}\\ & =& \frac{5100}{400}\\ & =& 12.75\end{array}$

The expected values of for southeast and education level is given below,

$\begin{array}{rll}{E}_{SoutheastandLessThanaHighSchoolDiploma}& =& \frac{100\times 58}{400}\\ & =& \frac{5800}{400}\\ & =& 14.5\\ E_{SoutheastandHighSchoolGraduate,NoCollege}& =& \frac{100\times 113}{400}\\ & =& \frac{11300}{400}\\ & =& 28.25\\ E_{SoutheastandSomeCollegeorAssociateDegree}& =& \frac{100\times 178}{400}\\ & =& \frac{17800}{400}\\ & =& 44.5\\ E_{SoutheastandCollegeGraduate}& =& \frac{100\times 51}{400}\\ & =& \frac{5100}{400}\\ & =& 12.75\end{array}$

The expected values of for mid-west and education level is given below,

$\begin{array}{rll}{E}_{MidwestandLessThanaHighSchoolDiploma}& =& \frac{100\times 58}{400}\\ & =& \frac{5800}{400}\\ & =& 14.5\\ E_{MidwestandHighSchoolGraduate,NoCollege}& =& \frac{100\times 113}{400}\\ & =& \frac{11300}{400}\\ & =& 28.25\\ E_{MidwestandSomeCollegeorAssociateDegree}& =& \frac{100\times 178}{400}\\ & =& \frac{17800}{400}\\ & =& 44.5\\ E_{MidwestandCollegeGraduate}& =& \frac{100\times 51}{400}\\ & =& \frac{5100}{400}\\ & =& 12.75\end{array}$

The expected values of for west and education level is given below,

$\begin{array}{rll}{E}_{westandLessThanaHighSchoolDiploma}& =& \frac{100\times 58}{400}\\ & =& \frac{5800}{400}\\ & =& 14.5\\ E_{westandHighSchoolGraduate,NoCollege}& =& \frac{100\times 113}{400}\\ & =& \frac{11300}{400}\\ & =& 28.25\\ E_{westandSomeCollegeorAssociateDegree}& =& \frac{100\times 178}{400}\\ & =& \frac{17800}{400}\\ & =& 44.5\\ E_{westandCollegeGraduate}& =& \frac{100\times 51}{400}\\ & =& \frac{5100}{400}\\ & =& 12.75\end{array}$

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

		Education Levels
	Less Than a High School Diploma	High School Graduate, No College	Some College or Associate Degree	College Graduate	Total
Northeast	14.5	28.25	44.5	12.75	100
Southeast	14.5	28.25	44.5	12.75	100
Midwest	14.5	28.25	44.5	12.75	100
West	14.5	28.25	44.5	12.75	100
Total	58	113	178	51	400

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 400.

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

$\begin{array}{ccc}{\chi }^2& =& {\displaystyle \sum _{i=1}^n\frac{{\left(O_i-E_i\right)}^2}{E_i}}\\ & =& \{\begin{array}{c}\frac{{\left(O_{Northeast}-E_{LessThanaHighSchoolDiploma}\right)}^2}{E_{LessThanaHighSchoolDiploma}}+\frac{{\left(O_{Southeast}-E_{LessThanaHighSchoolDiploma}\right)}^2}{E_{LessThanaHighSchoolDiploma}}\\ +\frac{{\left(O_{Midwest}-E_{LessThanaHighSchoolDiploma}\right)}^2}{E_{LessThanaHighSchoolDiploma}}+\frac{{\left(O_{West}-E_{LessThanaHighSchoolDiploma}\right)}^2}{E_{LessThanaHighSchoolDiploma}}\\ +\frac{{\left(O_{Northeast}-E_{HighSchoolGraduate,NoCollege}\right)}^2}{E_{HighSchoolGraduate,NoCollege}}+\frac{{\left(O_{Southeast}-E_{HighSchoolGraduate,NoCollege}\right)}^2}{E_{HighSchoolGraduate,NoCollege}}\\ +\frac{{\left(O_{Midwest}-E_{HighSchoolGraduate,NoCollege}\right)}^2}{E_{HighSchoolGraduate,NoCollege}}+\frac{{\left(O_{West}-E_{HighSchoolGraduate,NoCollege}\right)}^2}{E_{HighSchoolGraduate,NoCollege}}\\ +\frac{{\left(O_{Northeast}-E_{SomeCollegeorAssociateDegree}\right)}^2}{E_{SomeCollegeorAssociateDegree}}+\frac{{\left(O_{Southeast}-E_{SomeCollegeorAssociateDegree}\right)}^2}{E_{SomeCollegeorAssociateDegree}}\\ +\frac{{\left(O_{Midwest}-E_{SomeCollegeorAssociateDegree}\right)}^2}{E_{SomeCollegeorAssociateDegree}}+\frac{{\left(O_{West}-E_{SomeCollegeorAssociateDegree}\right)}^2}{E_{SomeCollegeorAssociateDegree}}\\ +\frac{{\left(O_{Northeast}-E_{CollegeGraduate}\right)}^2}{E_{CollegeGraduate}}+\frac{{\left(O_{Southeast}-E_{CollegeGraduate}\right)}^2}{E_{CollegeGraduate}}\\ +\frac{{\left(O_{Midwest}-E_{CollegeGraduate}\right)}^2}{E_{CollegeGraduate}}+\frac{{\left(O_{West}-E_{CollegeGraduate}\right)}^2}{E_{CollegeGraduate}}\end{array}\end{array}$

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

Observed values	Expected Values	$\left(O_i-E_i\right)$	${\left(O_i-E_i\right)}^2$	$\frac{{\left(O_i-E_i\right)}^2}{E_i}$
11	14.5	-3.5	12.25	0.84483
19	14.5	4.5	20.25	1.39655
15	14.5	0.5	0.25	0.01724
13	14.5	-1.5	2.25	0.15517
29	28.25	0.75	0.5625	0.01991
28	28.25	-0.25	0.0625	0.00221
25	28.25	-3.25	10.5625	0.37389
31	28.25	2.75	7.5625	0.2677
45	44.5	0.5	0.25	0.00562
44	44.5	-0.5	0.25	0.00562
47	44.5	2.5	6.25	0.14045
42	44.5	-2.5	6.25	0.14045
15	12.75	2.25	5.0625	0.39706
9	12.75	-3.75	14.0625	1.10294
13	12.75	0.25	0.0625	0.0049
14	12.75	1.25	1.5625	0.12255
				${\chi }^2={\displaystyle \sum _{i=1}^n\frac{{\left(O_i-E_i\right)}^2}{E_i}}=4.9971$

 So, the value of test statistic, ${\chi }^2$, for a chi- square test for association is 4.9971.

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 = 4.

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

$\begin{array}{rll}df& =& \left(R-1\right)\cdot \left(C-1\right)\\ & =& \left(4-1\right)\cdot \left(4-1\right)\\ & =& \left(3\right)\cdot \left(3\right)\\ & =& 9\end{array}$

The number of degrees of freedom for this test is 9.

Use the level of significance of $\alpha =0.05$ to find the critical value ${\chi }_{0.10}^2$ of the chi-square distribution with df = 9 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,

${\chi }_{0.05}^2=16.9190$

The test statistic value is ${\chi }^2=4.9971$.

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

$4.9971<16.9190\iff {\chi }^2\le {\chi }_{0.05}^2$. Therefore, we fail to reject the null hypothesis.

So, there is sufficient evidence not support the claim that education level and region of the U.S are not independent.

Final statement:

Thus, there is sufficient evidence not support the claim that education level and region of the U.S are not independent.