Chi-Square Test for Independence: Test Statistic and p-value

10. Categorical Association: Chi-Square Test for Independence

Chi-Square Test for Independence: Test Statistic and p-value

Data for the Chi-Square Test for Independence

The observed frequency is the number of individuals in the sample that are classified as a particular category and is denoted by $f_o$ .

The expected frequency is the number of individuals that one would expect to be classified as a particular category based on the predictions made by the null hypothesis and is denoted by $f_e$ .

The expected frequency of a cell is calculated with the following formula:
$f_e = \cfrac{f_r \cdot f_c}{n}$

where $f_r$ is frequency total for the row and $f_c$ is the frequency total for the column.

Calculating Expected Frequencies

Consider the following frequency distribution table:

Observed Frequencies
	Apple	Banana	$\blue{\text{Total}}$
Extrovert	$\purple{\text{13}}$	$\purple{\text{37}}$	$\blue{\text{50}}$
Introvert	$\purple{\text{81}}$	$\purple{\text{97}}$	$\blue{\text{178}}$
$\orange{\text{Total}}$	$\orange{\text{94}}$	$\orange{\text{134}}$	228

To calculate the expected frequencies, apply the following formula to each $\purple{\text{cell}}$ in the table:
$f_e = \cfrac{\blue{f_r} \cdot \orange{f_c}}{n}$

where $\blue{f_r}$ is frequency total for the row and $\orange{f_c}$ is the frequency total for the column.

$\begin{array}{llcl} \,\,\,\,\scriptsize{\bullet}&\,\,\normalsize{\text{Extrovert - Apple}}&:&\cfrac{\blue{50}\cdot \orange{94}}{228}=20.61\\ \,\,\,\,\scriptsize{\bullet}&\,\,\normalsize{\text{Extrovert - Banana}}&:&\cfrac{\blue{50}\cdot \orange{134}}{228}=29.39\\ \,\,\,\,\scriptsize{\bullet}&\,\,\normalsize{\text{Introvert - Apple}}&:&\cfrac{\blue{178}\cdot \orange{94}}{228}=73.39\\ \,\,\,\,\scriptsize{\bullet}&\,\,\normalsize{\text{Introvert - Banana}}&:&\cfrac{\blue{178}\cdot \orange{134}}{228}=104.61\\ \end{array}$

Expected Frequencies
	Apple	Banana	Total
Extrovert	20.61	29.39	50
Introvert	73.39	104.61	178
Total	94	134	228

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After the expected frequencies have been calculated, the next step is to calculate the Chi-Square Test for Independence test statistic in order to determine how much the observed frequencies differ from the frequencies expected under the null hypothesis.
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Chi-Square Test Statistic and Distribution

The test statistic for the Chi-Square Test for Independence is denoted by $\chi^2$ and is calculated with the following formula:

$\chi^2=\sum_{\text{all cells}}{\dfrac{(\text{Observed}-\text{Expected})^2}{\text{Expected}}}=\sum_{\text{all cells}}{\dfrac{(f_o-f_e)^2}{f_e}}$
Since the calculation of the test statistic involves adding squared values, a $\chi^2$ -statistic will always have a value of zero or larger.

Assuming the null hypothesis of the Chi-Square Test for Independence is true, the $\chi^2$ -statistic will (approximately) follow a $\chi^2$ -distribution with $df = (r -1)(c-1)$ degrees of freedom, where $r$ is the number of rows and $c$ the number of columns.

Chi-square distributions are positively skewed and the critical region will always entirely be located in the right tail of the distribution.

Calculating the p-value of a Chi-Square Test for Independence

A Chi-Square test is by definition a right-tailed test.

To calculate the $p$ -value of a Chi-Square Test for Independence in Excel, use the following command:
$=1\text{ - }\text{CHISQ.DIST}(\chi^2, df, 1)$

To calculate the $p$ -value of a Chi-Square Test for Independence in R, use the following command:
$\text{pchisq}(\chi^2, df, lower.tail=\text{FALSE})$

Where $df = (r \text{ - }1)(c\text{ - }1)$ .

If $p \lt \alpha$ , reject $H_0$ and conclude $H_a$ . Otherwise, do not reject $H_0$ .

In an effort to assess the impact of funding cuts on pre-school programs, school administrators in a US school district selected a simple random sample of $152$ students in the seventh grade and determined whether or not each student had attended pre-school and whether each student was performing below, at, or above grade level in mathematics.

The distribution was organized in the following two-way frequency table:

	Below grade level	At grade level	Above grade level	Total
Attended pre-school	13	45	26	84
No pre-school	20	31	17	68
Total	33	76	43	152

The researcher plans on using a Chi-Square Test for Independence to determine whether pre-school attendance and mathematical ability are related to one another.

Calculate the $p$ -value of the test and make a decision regarding $H_0$ . Round your answer to $3$ decimal places. Use the $\alpha = 0.08$ significance level.

$p=0.116$

On the basis of this $p$ -value, $H_0$ should not be rejected, because $\,p$ $\gt$ $\alpha$ .

There are a number of different ways we can calculate the $p$ -value of the test. Click on one of the panels to toggle a specific solution.

Excel Calculation

Calculate the expected frequency of all cells in the table with the following formula:

$f_e = \cfrac{f_r \cdot f_c}{n}$

where $f_r$ is frequency total for the row, $f_c$ is the frequency total for the column, and $n$ is the total sample size.

	Below grade level	At grade level	Above grade level	Total
Attended pre-school	18.237	42.0	23.763	84
No pre-school	14.763	34.0	19.237	68
Total	33	76	43	152

Calculate the $\chi^2$ -statistic:
$\begin{array}{rcl} \chi^2&=&\sum\limits_{\text{all cells}}{\dfrac{(f_o-f_e)^2}{f_e}}\\ &=& \cfrac{(13-18.237)^2}{18.237} +\cfrac{(45-42.0)^2}{42.0} +\cfrac{(26-23.763)^2}{23.763} +\cfrac{(20-14.763)^2}{14.763} +\\&&\cfrac{(31-34.0)^2}{34.0} +\cfrac{(17-19.237)^2}{19.237}\\ &=& 4.311 \end{array}$
Determine the degrees of freedom:
$df = (r -1)(c-1) = (2 -1 )(3 - 1)=2$
To calculate the $p$ -value of a $\chi^2$ -test, make use of the following Excel function:

CHISQ.DIST(x, deg_freedom, cumulative)

x: The value at which you wish to evaluate the distribution function.

deg_freedom: An integer indicating the number of degrees of freedom.

cumulative: A logical value that determines the form of the function.

TRUE - uses the cumulative distribution function, $\mathbb{P}(X \leq x)$

FALSE - uses the probability density function

A Chi-Square test is by definition a right-tailed test. Thus, to calculate the $p$ -value of the test, run the following command:
$=1\text{ - }\text{CHISQ.DIST}(\chi^2,(r \text{ - }1)(c\text{ - }1), 1)\\ \downarrow\\ =1\text{ - }\text{CHISQ.DIST}(4.311, 2, 1)$
This gives:
$p = 0.116$
Since $\,p$ $\gt$ $\alpha$ , the null hypothesis of independence should not be rejected.

R Calculation

Calculate the expected frequency of all cells in the table with the following formula:

$f_e = \cfrac{f_r \cdot f_c}{n}$

where $f_r$ is frequency total for the row, $f_c$ is the frequency total for the column, and $n$ is the total sample size.

	Below grade level	At grade level	Above grade level	Total
Attended pre-school	18.237	42.0	23.763	84
No pre-school	14.763	34.0	19.237	68
Total	33	76	43	152

pchisq(q, df, lower.tail)

q: The value at which you wish to evaluate the distribution function.

df: An integer indicating the number of degrees of freedom.

lower.tail: If TRUE (default), probabilities are $\mathbb{P}(X \leq x)$ , otherwise, $\mathbb{P}(X \gt x)$ .

A Chi-Square test is by definition a right-tailed test. Thus, to calculate the $p$ -value of the test, run the following command:
$\text{pchisq}(q = \chi^2, df = (r \text{ - }1)(c\text{ - }1), lower.tail=\text{FALSE})\\ \downarrow\\ \text{pchisq}(q = 4.311, df = 2, lower.tail=\text{FALSE})$
This gives:
$p = 0.116$
Since $\,p$ $\gt$ $\alpha$ , the null hypothesis of independence should not be rejected.

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