Chi-Squared Tests: Ace AP Statistics Like a Pro

Next Topic - Setting Up a Chi-Square Test for Homogeneity or Independence

Chi-Squared Tests for Two-Way Tables

Hey there, future AP Stats superstar! 👋 Let's break down chi-squared (χ²) tests for two-way tables. These tests are super common, so nailing them is key. Remember, we're using these to see if there's a relationship between categorical variables. Let's dive in!

What's a Two-Way Table?

Think of a two-way table as a grid that shows how two categorical variables are related. For example, car type (SUV vs. sports car) and gender (male vs. female). Here's a visual:

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Quick Fact

Two-way tables help us organize and analyze categorical data to see if there's a relationship between the variables.

Types of Chi-Squared Tests for Two-Way Tables

Okay, here's where it gets a bit tricky. We have two main types of χ² tests for two-way tables, and it's crucial to know which one to use. Don't worry, I've got your back!

Chi-Squared Test for Homogeneity

What it does: Compares the distribution of a categorical variable across two or more independent groups or populations. 🍞
Goal: To see if the proportions of categories are the same in all groups.
When to use it: When you're comparing different populations to see if they have different distributions for a categorical variable.
Example: Comparing the proportion of people who prefer coffee vs. tea in different age groups.

Memory Aid

Homogeneity = Same-ness. Think of it as testing if different populations are homogeneous (similar) in their distribution of a categorical variable.

Chi-Squared Test for Independence

What it does: Examines the relationship between two categorical variables within a single population. 💥
Goal: To see if the two variables are independent (one doesn't affect the probability of the other).
When to use it: When you're working with one population and want to see if two categorical variables are associated.
Example: Seeing if there's a relationship between a person's political affiliation and their favorite type of music.

Memory Aid

Independence = No Relationship. Think of it as testing if two variables are independent or if they are associated with each other.

Key Concept

Key Difference: Homogeneity compares across populations, while independence looks within a single population. This is a super common point of confusion, so make sure you've got this down! 💡

Calculating Expected Counts

No matter which test you're doing, you'll need to calculate the expected counts. This is what we expect to see if there's no association (independence) or no difference (homogeneity). Here's how: 😀

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Formula: Expected Count = (Row Total * Column Total) / Table Total

Example

Let's use the SUV and sports car example. For the 'Male & SUV' cell:

Row Total (Males): 60
Column Total (SUV): 156
Table Total: 240
Expected Count = (60 * 156) / 240 = 39

Exam Tip

Make sure to calculate the expected count for every cell in the table. This is a crucial step for both homogeneity and independence tests.

Expected Counts Table:

	SUV	Sports Car
Male	39	21
Female	117	63

Next Steps

Now that you know the difference between the two tests and how to calculate expected counts, you're ready to set up and perform the full chi-squared tests! We'll get into the nitty-gritty of hypothesis testing, degrees of freedom, and interpreting results in the next sections. Stay tuned! 😉

Practice Question

Multiple Choice Questions

A researcher is studying the relationship between hair color and eye color in a population. Which test is appropriate? (A) Chi-squared test for homogeneity (B) Chi-squared test for independence (C) Two-sample t-test (D) Paired t-test
A school wants to know if the distribution of favorite subjects is the same for freshmen and seniors. Which test is appropriate? (A) Chi-squared test for homogeneity (B) Chi-squared test for independence (C) One-sample z-test (D) One-sample t-test
In a two-way table, the expected count for a cell is calculated as (row total * column total) / table total. If the row total is 80, the column total is 50, and the table total is 200, what is the expected count? (A) 10 (B) 20 (C) 30 (D) 40

Free Response Question

A survey was conducted to investigate the relationship between political affiliation (Democrat, Republican, Independent) and opinion on a certain policy (Support, Oppose, Neutral). The following data was collected:

	Support	Oppose	Neutral	Total
Democrat	60	20	20	100
Republican	30	50	20	100
Independent	40	30	30	100
Total	130	100	70	300

(a) State the null and alternative hypotheses for this test. (b) Calculate the expected counts for each cell in the table. (c) Calculate the chi-squared test statistic. (d) Determine the degrees of freedom. (e) Based on your calculations, what can you conclude about the relationship between political affiliation and opinion on the policy? (Assume α = 0.05).

Answer Key

Multiple Choice

(B)
(A)
(B)

Free Response

(a) Hypotheses * Null Hypothesis (H₀): There is no association between political affiliation and opinion on the policy. They are independent. * Alternative Hypothesis (Hₐ): There is an association between political affiliation and opinion on the policy. They are not independent.

(b) Expected Counts

	Support	Oppose	Neutral
Democrat	43.33	33.33	23.33
Republican	43.33	33.33	23.33
Independent	43.33	33.33	23.33

Expected count = (Row Total * Column Total) / Table Total
Example: Democrat & Support: (100 * 130) / 300 = 43.33

(c) Chi-Squared Test Statistic

$χ^2 = \sum \frac{(Observed - Expected)^2}{Expected}$

$χ^2 = \frac{(60-43.33)^2}{43.33} + \frac{(20-33.33)^2}{33.33} + \frac{(20-23.33)^2}{23.33} + \frac{(30-43.33)^2}{43.33} + \frac{(50-33.33)^2}{33.33} + \frac{(20-23.33)^2}{23.33} + \frac{(40-43.33)^2}{43.33} + \frac{(30-33.33)^2}{33.33} + \frac{(30-23.33)^2}{23.33} = 32.25$

(d) Degrees of Freedom

$df = (number\ of\ rows - 1) * (number\ of\ columns - 1)$

$df = (3 - 1) * (3 - 1) = 4$

(e) Conclusion

Using a chi-squared distribution table or calculator, with df = 4 and χ² = 32.25, we find a very small p-value (p < 0.001).
Since the p-value is less than α = 0.05, we reject the null hypothesis.
Conclusion: There is sufficient evidence to suggest that there is an association between political affiliation and opinion on the policy.

Previous Topic - Carrying Out a Chi Square Goodness of Fit Test Next Topic - Setting Up a Chi-Square Test for Homogeneity or Independence

Expected Counts in Two Way Tables

Isabella Lopez