Chapter 11: Chi-Square Tests – Study Notes for Business Statistics

Chi-Square Tests

Introduction

The chi-square test is a fundamental statistical tool used to analyze categorical data. It helps determine whether observed frequencies differ significantly from expected frequencies under a specific hypothesis. In business statistics, chi-square tests are commonly applied to contingency tables, tests of independence, and comparisons of proportions across groups.

Contingency Tables

Definition and Purpose

• Contingency tables (also called cross-classification tables) are used to classify sample observations according to two or more categorical variables.

• They are especially useful for comparing multiple population proportions and examining relationships between categorical variables.

Example: Hand Preference vs. Gender

Suppose we examine a sample of 300 children, classified by hand preference (Left/Right) and gender (Male/Female). This forms a 2 x 2 contingency table:

Gender	Hand Preference: Left	Hand Preference: Right	Total
Female	12	108	120
Male	24	156	180
Total	36	264	300

Chi-Square Test for the Difference Between Two Proportions

Hypotheses

• Null hypothesis (H0): $\pi_1 = \pi_2$ (Proportion of females who are left-handed equals proportion of males who are left-handed)

• Alternative hypothesis (H1): $\pi_1 \neq \pi_2$ (The two proportions are not the same)

Test Statistic

• The chi-square test statistic is:

$\chi^2_{STAT} = \sum_{\text{all cells}} \frac{(f_o - f_e)^2}{f_e}$

• $f_o$ = observed frequency in a cell

• $f_e$ = expected frequency in a cell (if H0 is true)

• For a 2 x 2 table, degrees of freedom = 1

Decision Rule

• If $\chi^2_{STAT} > \chi^2_{\alpha}$, reject H0; otherwise, do not reject H0.

• $\chi^2_{\alpha}$ is the critical value from the chi-square distribution for the chosen significance level $\alpha$.

Computing the Overall Proportion

• The overall proportion of left-handed children is:

$\bar{p} = \frac{X_1 + X_2}{n_1 + n_2} = \frac{36}{300} = 0.12$

• Where $X_1$ and $X_2$ are the counts of left-handed children in each group, $n_1$ and $n_2$ are the group sizes.

Finding Expected Frequencies

• Expected frequency for left-handed females: $0.12 \times 120 = 14.4$

• Expected frequency for left-handed males: $0.12 \times 180 = 21.6$

Observed vs. Expected Frequencies Table

Gender	Left (Observed)	Left (Expected)	Right (Observed)	Right (Expected)	Total
Female	12	14.4	108	105.6	120
Male	24	21.6	156	158.4	180
Total	36		264		300

Example Calculation

• Test statistic:

$\chi^2_{STAT} = \frac{(12-14.4)^2}{14.4} + \frac{(108-105.6)^2}{105.6} + \frac{(24-21.6)^2}{21.6} + \frac{(156-158.4)^2}{158.4} = 0.7576$

• Critical value at $\alpha = 0.05$ and 1 d.f.: $\chi^2_{0.05} = 3.841$

• Since $0.7576 < 3.841$, do not reject H0. There is not sufficient evidence that the two proportions are different at $\alpha = 0.05$.

Using P-value

• P-value = 0.3481 > 0.05, so the null hypothesis is not rejected.

Chi-Square Test for Differences Among More Than Two Proportions

Hypotheses

• Null hypothesis (H0): $\pi_1 = \pi_2 = \cdots = \pi_c$ (All population proportions are equal)

• Alternative hypothesis (H1): Not all $\pi_j$ are equal (for $j = 1, 2, \ldots, c$)

Test Statistic

• The chi-square test statistic is:

$\chi^2_{STAT} = \sum_{\text{all cells}} \frac{(f_o - f_e)^2}{f_e}$

• Degrees of freedom: $(2-1)(c-1) = c-1$

• Each cell should have expected frequency of at least 1.

Example: University Calendar Opinion

Opinion	Administrators	Students	Faculty	Total
Favor	63	20	37	120
Oppose	37	30	13	80
Total	100	50	50	200

• Test statistic: $\chi^2_{STAT} = 12.792$

• Critical value at $\alpha = 0.01$, 2 d.f.: $\chi^2_{0.01} = 9.2103$

• Since $12.792 > 9.2103$, reject H0. There is evidence that at least one group has a different opinion.

• P-value = 0.0017 < 0.05, so the null hypothesis is rejected.

The Marascuilo Procedure

Purpose and Steps

• Used when the null hypothesis of equal proportions is rejected in a chi-square test.

• Allows pairwise comparisons to determine which proportions differ significantly.

• Calculate the observed differences $|p_j - p_{j'}|$ for all pairs, and compare to a critical range.

Critical Range Formula

$\text{Critical range} = \sqrt{\chi^2_{\alpha} \left( \frac{p_j(1-p_j)}{n_j} + \frac{p_{j'}(1-p_{j'})}{n_{j'}} \right)}$

• A pair of proportions is significantly different if $|p_j - p_{j'}| > \text{critical range for } j \text{ and } j'$

Example: Marascuilo Procedure

Opinion	Administrators	Students	Faculty	Total
Favor	63	20	37	120
Oppose	37	30	13	80
Total	100	50	50	200

• At 1% significance, there is evidence of a difference in opinion between administrators & students and students & faculty.

Chi-Square Test of Independence

Definition and Hypotheses

• Tests whether two categorical variables are independent (no relationship) or dependent (there is a relationship).

• Null hypothesis (H0): The two categorical variables are independent.

• Alternative hypothesis (H1): The two categorical variables are dependent.

Test Statistic

$\chi^2_{STAT} = \sum_{\text{all cells}} \frac{(f_o - f_e)^2}{f_e}$

• Degrees of freedom: $(r-1)(c-1)$, where r = number of rows, c = number of columns.

Expected Cell Frequencies

$f_e = \frac{\text{row total} \times \text{column total}}{n}$

• Where n = overall sample size.

Decision Rule

• If $\chi^2_{STAT} > \chi^2_{\alpha}$, reject H0; otherwise, do not reject H0.

• $\chi^2_{\alpha}$ is the critical value from the chi-square distribution with $(r-1)(c-1)$ degrees of freedom.

Example: Meal Plan and Class Standing

Class Standing	20/week	10/week	none	Total
Fresh.	24	32	14	70
Soph.	22	26	12	60
Junior	10	14	6	30
Senior	14	16	10	40
Total	70	88	42	200

• Test statistic: $\chi^2_{STAT} = 12.592$

• Critical value at $\alpha = 0.05$, 6 d.f.: $\chi^2_{0.05} = 12.592$

• P-value = 0.9943 > 0.05, so the null hypothesis is not rejected. There is not sufficient evidence of a relationship between meal plan and class standing.

Summary of Chapter 11

• Application of the chi-square test for the difference between two proportions.

• Application of the chi-square test for differences among more than two proportions.

• Use of the Marascuilo procedure for pairwise comparisons after rejecting the null hypothesis of equal proportions.

• Application of the chi-square test for independence between categorical variables.

Additional info: These notes expand on the original slides and text, providing full definitions, formulas, and context for each statistical procedure. All tables have been reconstructed and formulas provided in LaTeX format for clarity.