ANOVA & Chi-Square Tests Cheat Sheet

This cheat sheet covers one-way ANOVA and chi-square tests, two common methods for comparing data across groups or categories. Students use ANOVA when they need to compare several population means, not just two means. They use chi-square tests when data are counts in categories and the question is about fit or association.

A compact reference helps students choose the correct test, organize the formulas, and avoid mixing up conditions.

The main ANOVA formula is $F = \frac{MS_{\text{between}}}{MS_{\text{within}}}$ , which compares variation between groups to variation within groups. The main chi-square formula is $\chi^2 = \sum \frac{(O-E)^2}{E}$ , which measures how far observed counts are from expected counts. Both test types use degrees of freedom and a $p$ -value to decide whether results are statistically significant.

The key skill is matching the research question, data type, assumptions, and formula to the correct test.

Key Facts

One-way ANOVA tests whether several population means are equal using hypotheses $H_0: \mu_1 = \mu_2 = \cdots = \mu_k$ and $H_a:$ at least one mean is different.
The ANOVA test statistic is $F = \frac{MS_{\text{between}}}{MS_{\text{within}}}$ , where large values of $F$ give stronger evidence against $H_0$ .
The ANOVA sums of squares are $SS_{\text{between}} = \sum n_i(\bar{x}_i-\bar{x})^2$ and $SS_{\text{within}} = \sum\sum (x_{ij}-\bar{x}_i)^2$ .
For one-way ANOVA with $k$ groups and $N$ total observations, $df_{\text{between}} = k-1$ , $df_{\text{within}} = N-k$ , and $df_{\text{total}} = N-1$ .
Mean squares are found by $MS_{\text{between}} = \frac{SS_{\text{between}}}{df_{\text{between}}}$ and $MS_{\text{within}} = \frac{SS_{\text{within}}}{df_{\text{within}}}$ .
A chi-square goodness-of-fit test uses $\chi^2 = \sum \frac{(O-E)^2}{E}$ to compare observed category counts to expected category counts.
A chi-square test of independence uses $E_{ij} = \frac{(\text{row total})(\text{column total})}{\text{grand total}}$ for each table cell.
For a chi-square independence test with $r$ rows and $c$ columns, the degrees of freedom are $df = (r-1)(c-1)$ .

Vocabulary

One-way ANOVA: A statistical test that compares the means of $k$ independent groups to see whether at least one population mean differs.
F statistic: The ANOVA test statistic $F = \frac{MS_{\text{between}}}{MS_{\text{within}}}$ that compares between-group variation to within-group variation.
Chi-square statistic: The statistic $\chi^2 = \sum \frac{(O-E)^2}{E}$ that measures how far observed counts are from expected counts.
Expected count: The count predicted for a category or table cell if the null hypothesis is true.
Degrees of freedom: The number of independent pieces of information used to find the reference distribution for a test statistic.
p-value: The probability, assuming $H_0$ is true, of getting a test statistic as extreme as or more extreme than the observed result.

Common Mistakes to Avoid

Using several two-sample tests instead of ANOVA: this increases the chance of a Type I error because each extra test adds another opportunity for a false positive.
Treating a significant ANOVA as proof that every group mean is different: ANOVA only shows that at least one mean differs, so follow-up comparisons are needed to identify which ones.
Using a chi-square test with percentages instead of counts: chi-square formulas require observed counts $O$ and expected counts $E$ , not proportions alone.
Forgetting to check expected counts: chi-square results can be unreliable when expected counts are too small, especially when many cells have $E<5$ .
Interpreting a large $p$ -value as proof that $H_0$ is true: a large $p$ -value means there is not enough evidence to reject $H_0$ , not that the null hypothesis has been proven.

Practice Questions

1 For a one-way ANOVA with $k=4$ , $N=36$ , $SS_{\text{between}}=90$ , and $SS_{\text{within}}=210$ , find $df_{\text{between}}$ , $df_{\text{within}}$ , $MS_{\text{between}}$ , $MS_{\text{within}}$ , and $F$ .
2 A fair die is rolled $60$ times with observed counts $(8,12,10,9,11,10)$ . Using $E=10$ for each face, compute $\chi^2 = \sum \frac{(O-E)^2}{E}$ .
3 In a $2 \times 3$ table, the row totals are $40$ and $60$ , the column totals are $30$ , $50$ , and $20$ , and the grand total is $100$ . Find the expected count for row $1$ , column $2$ , and find the degrees of freedom.
4 A student compares test scores from four teaching methods and gets a significant ANOVA result. Explain why this does not automatically show which teaching method is best.

Understanding ANOVA & Chi-Square Tests

ANOVA works by separating two kinds of spread in numerical data. One kind is the ordinary variation among people or measurements within the same group. The other is the variation among the group averages.

If several teaching methods produce average test scores that are far apart, while scores inside each class are fairly similar, the evidence for a real difference is stronger. If every class has a wide mix of scores, different averages may just reflect normal sampling variation. This is why ANOVA compares a measure of between group variation with a measure of within group variation.

A value near one means the two sources of variation are similar. A much larger value suggests that group membership may matter.

A significant ANOVA result does not identify the groups that differ. It only says that the full set of averages is unlikely to be equal. Students often stop too early at this point.

Follow up comparisons are needed to find the specific pair or pairs with meaningful differences. Methods such as Tukey comparisons control the increased chance of a false positive that comes from checking many pairs. It also helps to examine group means, sample sizes, and graphs.

Statistical significance can occur for a very small difference when the sample is large. The size of the difference and its practical effect deserve attention, not just the p-value.

ANOVA relies on useful conditions. Observations should come from independent individuals or independent experimental units. Each group should have a distribution that is not strongly skewed or dominated by extreme outliers.

The groups should have roughly similar spreads, especially when their sample sizes differ greatly. A boxplot can reveal outliers, uneven spread, and unusual shapes before calculations begin. In a fair experiment, subjects are randomly assigned to treatments.

In an observational study, ANOVA can show an association, but it cannot prove that one group caused a change. Hidden factors such as age, prior experience, or income may explain part of a difference.

Chi-square tests begin with counts, so every person or item belongs in one category for the variable being studied. For goodness of fit, expected counts come from a claimed distribution, such as a store expecting equal sales across four colors. For independence, expected counts describe what the table would look like if two categorical variables had no relationship.

Compare observed and expected counts cell by cell. Large differences contribute more when the expected count is small, which makes rare categories especially important to check. Expected counts should usually be at least five in every cell for the standard approximation to work well.

A significant result shows that the pattern differs from the expectation or that variables are associated. It does not show the direction or strength by itself. Inspecting the observed and expected counts reveals which categories drive the result.

Sign in to save

Sign in to save

ANOVA & Chi-Square Tests Cheat Sheet

Related Tools

Related Labs

Related Worksheets

Related Infographics

Study as Flashcards

Key Facts

Vocabulary

Common Mistakes to Avoid

Practice Questions

Understanding ANOVA & Chi-Square Tests