--- title: "One-Way Analysis of Variance" description: "Compare population means across groups defined by one factor. Split total squared variation into between-group and within-group parts, then compare their mean squares using an F test." canonical: https://lightmysky.com/learn/mathematics/one-way-analysis-of-variance-mt_WYGK3Ktnqz source: https://lightmysky.com/learn/mathematics/one-way-analysis-of-variance-mt_WYGK3Ktnqz.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # One-Way Analysis of Variance Compare population means across groups defined by one factor. Split total squared variation into between-group and within-group parts, then compare their mean squares using an F test. Subject: Mathematics · Area: Data & Statistics · Ages 21 to 22 Page: https://lightmysky.com/learn/mathematics/one-way-analysis-of-variance-mt_WYGK3Ktnqz ## Ready when they can - Identify a one-factor numerical-response design and partition the total sum of squares - Compute and interpret an F ratio and its two degrees of freedom, checking the classical model assumptions - Explain pair counting, family-wise error and adjusted follow-up comparisons ## Lesson: Many means, one test One-way analysis of variance (ANOVA) compares population means across groups defined by one categorical factor, using a numerical response. The null says all population means are equal; the alternative says at least one pair differs. The classical F test assumes independent observations, normal populations within groups and equal population variances. Check the study design and these assumptions before interpreting the test. A sum of squares (SS) adds squared deviations. Total SS uses observations' distances from the grand mean, the mean of all observations. Within-group SS uses distances from each observation's own group mean. Between-group SS measures variation among group means, weighted by group size. Total SS = between-group SS + within-group SS. Find a missing component by subtracting the known component from the total. *(drawing: The two components add to the total: 18 + 42 = 60.)* To calculate between-group SS, subtract the grand mean from each group mean, square each difference, multiply by that group's size, and add. For equal group sizes, the grand mean is also the average of the group means. Equal sample means do not force within-group SS to be zero: observations can still vary around those means. Let k be the number of groups and N the total number of observations. Between-group degrees of freedom (df) = k - 1; within-group df = N - k. A mean square (MS) is SS divided by its df. Calculate F = MS between / MS within when MS within is positive. Dividing the raw sums of squares skips the required scaling. F measures MS between relative to MS within. Here F = 9 / 3 = 3, so MS between is three times MS within. Large F values give evidence against equal population means, but chance can also produce large values. Under the null and the model assumptions, the reference curve is an F distribution with numerator df k - 1 and denominator df N - k. The F distribution is nonnegative and right-skewed. This ANOVA test rejects only in the right tail. At a significance level chosen in advance, reject when F exceeds the matching critical value, or when its right-tail p-value is at most that level. Otherwise, do not reject. A small F does not support rejection, and nonrejection does not prove equal population means. There are k(k - 1)/2 pairs among k groups. Five groups give 5 x 4 / 2 = 10 pairs. Unadjusted pairwise tests at 5% can raise the chance of at least one false rejection above 5%. That chance is the family-wise error rate. Pairwise tests share data, so their errors need not be independent. One ANOVA tests the joint null at the chosen level; it does not protect later unadjusted tests. An ANOVA rejection gives evidence that not all population means are equal. It does not identify the differing pairs or show that every pair differs. To compare all pairs, use family-wise error control, such as Tukey's procedure or adjusted pairwise tests. Planned comparisons can also be valid without first rejecting ANOVA when their error control is specified in advance. **Recap.** Split total SS into between and within parts. Divide each by its df to form F. Check assumptions, use the right-tail decision rule, and control error across follow-up comparisons. *This lesson, its practice questions, worked solutions and flashcards were written for this dataset.* ## Practice 17 questions on this page, each with its working shown. ## Needs first - [Comparing Two Populations](https://lightmysky.com/learn/mathematics/comparing-two-populations-mt_v34rZkMfqr) - [Chi-Square Tests for Categorical Data](https://lightmysky.com/learn/mathematics/chi-square-tests-for-categorical-data-mt_yxc87pzaXX) ## Opens up - [Simple Linear Regression and Inference on the Slope](https://lightmysky.com/learn/mathematics/simple-linear-regression-and-inference-on-the-slope-mt_W2huNTKsMj)