Chi-Square Tests for Independence: Contingency Tables and Statistical Inference

Chi-Square Tests and the F-Distribution

Section 10.2: Independence

This section introduces the use of contingency tables and the chi-square test for independence, which is a fundamental method in statistics for determining whether two categorical variables are related.

Contingency Tables

Definition and Structure

• Contingency Table (r × c table): A table that displays the observed frequencies for two categorical variables, arranged in r rows and c columns.

• The intersection of a row and a column is called a cell.

Example: Contingency Table

Gender	Cup	Cone	Sundae	Sandwich	Other
Male	600	288	204	24	84
Female	410	340	180	20	50

Example: The table above shows the favorite way to eat ice cream by gender for a sample of 2200 adults.

Expected Frequencies in Contingency Tables

Calculating Expected Frequencies

• Assuming the two variables are independent, the expected frequency for each cell can be calculated using the marginal totals.

• Marginal Frequency: The total frequency for a single category of one variable (row or column total).

• Joint Frequency: The observed frequency in the interior cells of the table (intersection of categories).

Formula for Expected Frequency:

$E_{r,c} = \frac{(\text{Sum of row } r) \times (\text{Sum of column } c)}{\text{Sample size}}$

Example: Calculating Expected Frequencies

Gender	Cup	Cone	Sundae	Sandwich	Other	Total
Male	600	288	204	24	84	1200
Female	410	340	180	20	50	1000
Total	1010	628	384	44	134	2200

For the cell (Male, Cup):

$E_{1,1} = \frac{1200 \times 1010}{2200} = 550.91$

Other expected frequencies are calculated similarly for each cell.

Summary Table: Observed vs. Expected Frequencies

Gender	Cup	Cone	Sundae	Sandwich	Other
Male (Observed)	600	288	204	24	84
Male (Expected)	550.91	342.55	209.45	24	73.09
Female (Observed)	410	340	180	20	50
Female (Expected)	459.09	285.45	174.55	20	60.91

Chi-Square Test for Independence

Purpose and Application

• The chi-square independence test is used to determine whether two categorical variables are independent.

• It assesses whether the occurrence of one variable affects the probability of the occurrence of the other variable.

Conditions for Use

• The observed frequencies must be obtained from a random sample.

• Each expected frequency must be at least 5.

Test Statistic and Degrees of Freedom

• The sampling distribution for the test is approximated by a chi-square distribution with degrees of freedom:

$df = (r-1)(c-1)$

• The test statistic is:

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

where $O$ is the observed frequency and $E$ is the expected frequency for each cell.

Steps for Performing a Chi-Square Independence Test

1. Verify that the observed frequencies are from a random sample and each expected frequency is at least 5.

2. State the null hypothesis ($H_0$) and alternative hypothesis ($H_a$):

   • $H_0$: The variables are independent.

   • $H_a$: The variables are dependent.

3. Specify the level of significance ($\alpha$).

4. Determine the degrees of freedom: $df = (r-1)(c-1)$.

5. Find the critical value from the chi-square distribution table.

6. Determine the rejection region (values of $\chi^2$ greater than the critical value).

7. Calculate the test statistic $\chi^2$ using the observed and expected frequencies.

8. Make a decision: If $\chi^2$ is in the rejection region, reject $H_0$; otherwise, fail to reject $H_0$.

9. Interpret the result in the context of the original claim.

Worked Example: Ice Cream Preferences and Gender

Step-by-Step Solution

• Hypotheses:

  • $H_0$: Favorite way to eat ice cream and gender are independent.

  • $H_a$: Favorite way to eat ice cream and gender are dependent.

• Significance Level: $\alpha = 0.01$

• Degrees of Freedom: $df = (2-1)(5-1) = 4$

• Critical Value: $\chi^2_{0.01,4} = 13.277$

• Test Statistic Calculation:

  • Plug observed and expected frequencies into the formula:

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

  For this example, $\chi^2 = 32.630$

• Decision: Since $32.630 > 13.277$, reject $H_0$.

• Conclusion: There is enough evidence at the 1% significance level to conclude that favorite way to eat ice cream and gender are dependent.

Additional Example: Exercise Frequency and Gender

Contingency Table

Gender	0-1	2-3	4-5	6-7	Total
Male	40	53	26	6	125
Female	34	68	37	11	150
Total	74	121	63	17	275

Hypotheses and Test

• $H_0$: Number of days of exercise per week is independent of gender.

• $H_a$: Number of days of exercise per week depends on gender.

• $\alpha = 0.05$

• $df = (2-1)(4-1) = 3$

• Critical value: $\chi^2_{0.05,3} = 7.815$

Calculation and Decision

• Calculate expected frequencies for each cell using the formula above.

• Compute the test statistic $\chi^2 = 1.9042$

• Since $1.9042 < 7.815$, fail to reject $H_0$.

• Conclusion: There is not enough evidence at the 5% significance level to conclude that the number of days a student exercises per week is related to gender.

Summary Table: Steps for Chi-Square Test for Independence

Step	Description	Symbol/Formula
1	Verify random sample and expected frequencies ≥ 5
2	State hypotheses	$H_0$, $H_a$
3	Specify significance level	$\alpha$
4	Determine degrees of freedom	$df = (r-1)(c-1)$
5	Find critical value	Chi-square table
6	Determine rejection region
7	Calculate test statistic	$\chi^2 = \sum \frac{(O - E)^2}{E}$
8	Make decision	Compare $\chi^2$ to critical value
9	Interpret result

Key Terms: contingency table, marginal frequency, joint frequency, expected frequency, chi-square test for independence, degrees of freedom, significance level, rejection region.

Example Applications: Testing whether preferences, behaviors, or outcomes are associated with demographic variables such as gender, age group, or treatment group in survey or experimental data.