StatisticsLesson 5 of 5 in the course The Chi-Square Distribution

Chi-Square Tests of Independence for Categorical Data

5 flashcards · 5 quiz questions · 5 match pairs · 4 fill-in sentences · English

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Summary

This lesson explains how to use a chi-square test of independence to investigate whether two categorical variables are related. In the lab, the variables are gender and favorite snack category. You will learn how to organize class survey data, calculate expected counts and a test statistic, and use a p-value to make and explain a decision.

Ask each class member to identify one favorite snack category, then record the responses in a two-way table. The rows represent gender categories and the columns represent snack categories. The entries are observed counts, or the numbers actually recorded in each combination.

All flashcards

What is a chi-square test of independence?
A hypothesis test that uses categorical data in a two-way table to assess whether two variables are associated or independent in the population.
How do you calculate an expected cell count?
Use E=(row total)(column total)grand totalE=\frac{(\text{row total})(\text{column total})}{\text{grand total}}. It is the count anticipated in that cell if the variables are independent.
What does the chi-square statistic measure?
It combines the discrepancies between observed and expected counts across all cells: χ2=∑(O−E)2/E\chi^2=\sum (O-E)^2/E. Larger values indicate greater overall disagreement with independence.
How do you find degrees of freedom for an independence test?
For a table with rr rows and cc columns, use df=(r−1)(c−1)df=(r-1)(c-1). For a two-row, four-column table, the degrees of freedom are 3.
What does failing to reject the null hypothesis mean?
The sample does not provide sufficient evidence of an association at the chosen significance level. It does not prove that the variables are independent.

Quiz questions

  1. A contingency table has two rows and four columns. What are the degrees of freedom for a chi-square test of independence?Answer: 3, because (2−1)(4−1)=3(2-1)(4-1)=3.
  2. Which expression calculates the contribution of one cell to the chi-square statistic?Answer: (O−E)2/E(O-E)^2/E
  3. Why can a chi-square test statistic not be negative?Answer: Each squared difference between an observed and expected count is nonnegative.
  4. According to the lab guidance, what minimum expected count should each cell have?Answer: At least five.
  5. A chi-square test gives p=0.03p=0.03 with significance level α=0.05\alpha=0.05. What is the appropriate decision?Answer: Reject H0H_0; the data provide statistically significant evidence of an association.

Based on Introductory Statistics 2e by Barbara Illowsky, Susan Dean (OpenStax), licensed CC BY 4.0. Adapted into a summary, flashcards and quizzes.

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