What the exam asks
On the AP Statistics exam, you must be able to:
- Identify whether a scenario calls for a chi-square test for homogeneity or a chi-square test for independence based on how the data were collected.
- State the null () and alternative () hypotheses in words and in context.
- Calculate expected counts and degrees of freedom for a two-way table.
- Verify the conditions for inference: randomization, the condition (when sampling without replacement), and the expected counts condition.
- Describe the properties of the chi-square () statistic and the family of distributions.
Core ideas
The chi-square statistic and distributions
A chi-square test compares the observed counts in the cells of a two-way table to the expected counts we would anticipate if the null hypothesis were true. The test statistic is on your formula sheet:
- Meaning of : It measures the overall distance between the observed and expected counts, relative to the expected counts.
- Properties of distributions:
- Because the numerator is squared and the expected counts are positive, values are ().
Homogeneity vs. independence
Although the math for both tests is identical, the procedure name and hypotheses depend entirely on the data collection design.
| Feature | Chi-square test for homogeneity | Chi-square test for independence |
|---|---|---|
| Data collection | Two or more independent random samples from different populations, OR random assignment to two or more treatments in an experiment. | One random sample taken from a single population. |
| Variables | One categorical variable measured across multiple populations or treatment groups. | Two categorical variables measured on each individual in the single sample. |
| Null hypothesis () | There is no difference in the distribution of [categorical variable] across the [populations/treatments]. | There is no association between [variable 1] and [variable 2] in the [population] (or: the variables are independent in the population). |