Student Mathematics Applications & Interpretation…

How Do You Perform a Chi-Squared Test for Independence Using a Contingency Table?

Use the observed frequencies in the contingency table to calculate expected frequencies, then run a chi-squared test for independence on your GDC. Compare the resulting p-value with the significance level to decide whether there is evidence of an association between the two categorical variables.

This is SL 4.11: Hypothesis testing and may be examined on either GDC-active paper.

Step	Required method
1. State hypotheses	$H_0$ : the two variables are independent. $H_1$ : the two variables are not independent, so there is an association.
2. Find expected frequencies	For each cell, use $E=\frac{(\text{row total})(\text{column total})}{\text{grand total}}$ .
3. Check conditions	The data must be frequencies in mutually exclusive categories. In IB SL examination questions, expected frequencies will be greater than $5$ .
4. Find degrees of freedom	For $r$ rows and $c$ columns, $df=(r-1)(c-1)$ .
5. Run the test	Enter the observed matrix and expected matrix into the GDC's $\chi^2$ -test routine. Record $\chi^2$ , (df), and the p-value.
6. Conclude	If $p<\alpha$ , reject $H_0$ . If $p\geq\alpha$ , there is insufficient evidence to reject $H_{0}$ .

For example, if a row total is $40$ , a column total is $30$ , and the grand total is $100$ , then

E=\frac{40\times30}{100}=12.

The test statistic measures the overall difference between observed and expected frequencies:

\chi^2=\sum\frac{(O-E)^2}{E}.

Technology calculates this value, but you must understand what the expected frequencies represent.

Expected frequencies represent the counts predicted by $H_0$ , based on the marginal totals. Large differences between $O$ and $E$ increase $\chi^2$ and usually reduce the p-value, making evidence against independence stronger. Keep full GDC precision until the final reported values.

A common misconception is to say “accept $H_0$ ” when $p\geq\alpha$ . The correct conclusion is that there is insufficient evidence to reject $H_0$ ; this does not prove independence.

Exam technique: State the p-value and significance level explicitly, compare them, and finish with a contextual conclusion about the variables. A numerical result without this comparison and interpretation will lose reasoning marks.

How Do You Perform a Chi-Squared Test for Independence Using a Contingency Table?

This is SL 4.11: Hypothesis testing and may be examined on either GDC-active paper.

Step	Required method
1. State hypotheses	$H_0$ : the two variables are independent. $H_1$ : the two variables are not independent, so there is an association.
2. Find expected frequencies	For each cell, use $E=\frac{(\text{row total})(\text{column total})}{\text{grand total}}$ .
3. Check conditions	The data must be frequencies in mutually exclusive categories. In IB SL examination questions, expected frequencies will be greater than $5$ .
4. Find degrees of freedom	For $r$ rows and $c$ columns, $df=(r-1)(c-1)$ .
5. Run the test	Enter the observed matrix and expected matrix into the GDC's $\chi^2$ -test routine. Record $\chi^2$ , (df), and the p-value.
6. Conclude	If $p<\alpha$ , reject $H_0$ . If $p\geq\alpha$ , there is insufficient evidence to reject $H_{0}$ .

For example, if a row total is $40$ , a column total is $30$ , and the grand total is $100$ , then

E=\frac{40\times30}{100}=12.

The test statistic measures the overall difference between observed and expected frequencies:

\chi^2=\sum\frac{(O-E)^2}{E}.

Technology calculates this value, but you must understand what the expected frequencies represent.

A common misconception is to say “accept $H_0$ ” when $p\geq\alpha$ . The correct conclusion is that there is insufficient evidence to reject $H_0$ ; this does not prove independence.

How Do You Perform a Chi-Squared Test for Independence Using a Contingency Table?

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How Do You Perform a Chi-Squared Test for Independence Using a Contingency Table?