One-Way ANOVA Method Card Template (Statistics, R and Python)

One-Way ANOVA Method Card: [Your study question or course]

A method card for comparing the means of three or more independent groups: the logic of between-group versus within-group variance, the F statistic and the ANOVA table by hand, the assumptions and their checks, Welch's ANOVA for unequal variances, post hoc comparisons (Tukey HSD and Games-Howell), planned contrasts, and the effect sizes η² and ω² with a confidence interval. It is for students preparing for a statistics exam, master's and PhD students comparing conditions in thesis data, and analysts comparing several groups at once. It speeds up learning by putting the decisions in order (classic or Welch, which follow-up, which effect size, which sentence), by showing one example fully solved, one half solved and one left for you, and by ending with self-test questions whose answers sit in a separate key. The worked examples use fictional data; replace them with your own. Source basis: Fisher (1925), Welch (1951), Tukey (1949, 1953), Games and Howell (1976), Hays (1963) for ω², Cohen (1988), the NIST/SEMATECH e-Handbook of Statistical Methods, and the R, SciPy and statsmodels documentation, all written up in original words and examples.

Document control

When to use it

One-sentence rule: use a one-way ANOVA when you want to know whether the mean of a continuous outcome differs across three or more independent groups defined by one factor, and you want one overall test before comparing pairs.

Why not just run t-tests on every pair? With three groups there are three pairs, with five groups ten. Each t-test at α = 0.05 has a 5% false-positive risk, so the chance of at least one false positive across all pairs grows quickly (about 14% for three independent tests). ANOVA tests all means at once with a single α, and post hoc procedures then control the family-wise error rate across the pairs.

Classic or Welch? The classic F test assumes equal population variances. Welch's ANOVA does not, and it costs almost nothing when variances are in fact equal. A sensible default, matching the Welch-by-default rule for t-tests, is: report Welch's ANOVA with Games-Howell comparisons unless equal variances are justified by design; report the classic ANOVA with Tukey HSD when the group SDs are similar and group sizes are equal. Do not choose between them on the basis of a variance test's p-value alone.

Alternatives. Kruskal-Wallis with Dunn's test for ranks (academic/stats-nonparametric-tests); a permutation F test (academic/stats-resampling-methods); linear regression with a categorical predictor, which gives the identical F (academic/stats-linear-regression).

The question and the data

Research question (Example A). At Lakemoor University (fictional), 24 STAT 101 volunteers were randomly assigned to one of three revision methods for one week: rereading notes, practice tests, or spaced flashcards. Does the mean score on a 100-point quiz differ by method?

The group means with their 95% confidence intervals (each from its own group SD) show the pattern before any test. Notice that the Flashcard interval is much wider: its SD is about 2.6 times the Reread SD, a warning about the equal-variance assumption.

Assumptions and how to check each

In Example A the residuals look normal (Shapiro-Wilk W = 0.989, p = .994), but the SD ratio is 7.41 / 2.88 = 2.6 and Levene's median-centred test gives F(2, 21) = 3.41, p = .052. The p-value is just above 0.05, which shows why you should not let a variance test decide for you: the SD ratio alone justifies Welch's ANOVA here. car::leveneTest centres on the median by default; scipy.stats.levene also defaults to center="median", so both report the same Brown-Forsythe version.

How it works

Plain language. Every score differs from the grand mean for two reasons: its group's mean differs from the grand mean (the between-group part, which the factor might explain), and the score differs from its own group's mean (the within-group part, which is noise). ANOVA splits the total variation into these two pieces. If H0 is true, the group means differ only because of noise, so the between-group variation per degree of freedom should be about the same size as the within-group variation per degree of freedom. Their ratio, F, should be near 1. If the group means really differ, the between part is inflated and F is large.

Formulas. With $k$ groups, $n_j$ units in group $j$, group means $\bar y_j$, grand mean $\bar y$ and total sample size $N$:

The ANOVA table:

Welch's ANOVA weights each group by $w_j = n_j/s_j^2$, so precise groups count more, and replaces the denominator df with an estimate based on the group variances (Welch 1951). Its F and df are not the classic ones; read them from the output.

Effect sizes. η² is the share of total variation explained by the factor in this sample; it is biased upwards. ω² corrects that bias and estimates the population share:

Planned contrast. A contrast is a weighted sum of means with weights $c_j$ that add to zero, tested with the pooled error:

Tukey HSD compares every pair using the studentised range distribution, which allows for the fact that the largest and smallest of $k$ means are compared. With unequal group sizes the Tukey-Kramer version (Kramer 1956) is used, which R's TukeyHSD, statsmodels' pairwise_tukeyhsd and SciPy's tukey_hsd apply automatically. Games-Howell does the same with Welch-type standard errors and df for each pair, so it does not assume equal variances.

Worked example A (fully solved): revision methods

Hypotheses. H0: the three population mean quiz scores are equal. H1: at least one differs. α = 0.05. Planned contrast (stated before data collection): the two active methods (Practice, Flashcard) versus Reread, weights (−1, 0.5, 0.5).

Checks. Random assignment, one score per student: independence holds. Residuals are close to normal (Shapiro-Wilk p = .994). Group SDs range from 2.88 to 7.41 (ratio 2.6), so we report the classic ANOVA for learning and Welch's ANOVA with Games-Howell as the primary analysis.

R code and output.

d <- data.frame(
  method = factor(rep(c("Reread", "Practice", "Flashcard"), each = 8),
                  levels = c("Reread", "Practice", "Flashcard")),
  score = c(58, 62, 55, 64, 60, 57, 61, 59,
            70, 66, 74, 63, 78, 69, 72, 68,
            65, 72, 58, 80, 61, 75, 69, 64))
fit <- aov(score ~ method, data = d)
summary(fit)                                   # classic ANOVA
oneway.test(score ~ method, data = d)          # Welch ANOVA: var.equal = FALSE is the default
shapiro.test(residuals(fit))
# install.packages("car")
car::leveneTest(score ~ method, data = d)      # median-centred (Brown-Forsythe)
TukeyHSD(fit)

(output abridged: headers, significance codes and blank lines removed)
            Df Sum Sq Mean Sq F value  Pr(>F)
method       2  497.3  248.67   8.762 0.00171 **
Residuals   21  596.0   28.38

	One-way analysis of means (not assuming equal variances)
F = 15.796, num df = 2.000, denom df = 12.452, p-value = 0.0003837

Shapiro-Wilk: W = 0.98934, p-value = 0.9945
Levene's Test (center = median): F(2, 21) = 3.4123, Pr(>F) = 0.0521

$method                diff       lwr       upr     p adj
Practice-Reread        10.5  3.785985 17.214015 0.0020668
Flashcard-Reread        8.5  1.785985 15.214015 0.0117201
Flashcard-Practice     -2.0 -8.714015  4.714015 0.7364417

Games-Howell is not in base R or in car; packages such as rstatix and PMCMRplus provide it. This short function implements it from the formula with base R's studentised range distribution (ptukey, qtukey). Run it in the same R session as the block above (it uses d):

games_howell <- function(y, g) {
  lv <- levels(g); out <- NULL
  for (i in 1:(length(lv) - 1)) for (j in (i + 1):length(lv)) {
    a <- y[g == lv[i]]; b <- y[g == lv[j]]
    va <- var(a) / length(a); vb <- var(b) / length(b)
    diff <- mean(b) - mean(a); se <- sqrt(va + vb)
    df <- (va + vb)^2 / (va^2 / (length(a) - 1) + vb^2 / (length(b) - 1))
    p <- ptukey(abs(diff) / se * sqrt(2), length(lv), df, lower.tail = FALSE)
    crit <- qtukey(0.95, length(lv), df) / sqrt(2)
    out <- rbind(out, data.frame(pair = paste(lv[j], "-", lv[i]), diff = diff,
      lwr = round(diff - crit * se, 2), upr = round(diff + crit * se, 2),
      df = round(df, 2), p = round(p, 4)))
  }
  out
}
games_howell(d$score, d$method)

                  pair diff    lwr   upr    df      p
1    Practice - Reread 10.5   5.29 15.71 11.62 0.0005
2   Flashcard - Reread  8.5   0.67 16.33  9.07 0.0345
3 Flashcard - Practice -2.0 -10.28  6.28 11.84 0.7985

Effect sizes, the 95% CI for η² (by inverting the noncentral F distribution, Steiger 2004) and the planned contrast. This block continues the R session above (it reuses d and fit):

tab <- summary(fit)[[1]]
ssb <- tab[1, "Sum Sq"]; ssw <- tab[2, "Sum Sq"]; msw <- tab[2, "Mean Sq"]
eta2 <- ssb / (ssb + ssw); omega2 <- (ssb - 2 * msw) / (ssb + ssw + msw)
Fobs <- tab[1, "F value"]; N <- 24
ncp_lo <- uniroot(function(n) pf(Fobs, 2, 21, n) - 0.975, c(0, 200))$root
ncp_hi <- uniroot(function(n) pf(Fobs, 2, 21, n) - 0.025, c(0, 200))$root
c(eta2 = eta2, omega2 = omega2, lo = ncp_lo / (ncp_lo + N), hi = ncp_hi / (ncp_hi + N))
m <- tapply(d$score, d$method, mean); w <- c(-1, 0.5, 0.5)
L <- sum(w * m); se <- sqrt(msw * sum(w^2 / 8))
c(L = L, SE = se, t = L / se, p = 2 * pt(-abs(L / se), 21))

      eta2     omega2         lo         hi
0.45487805 0.39276617 0.09719016 0.62976107
           L           SE            t            p
9.5000000000 2.3068221803 4.1182194628 0.0004897752

Python code and output.

import numpy as np
import pandas as pd
from scipy import stats
from scipy.optimize import brentq
import statsmodels.formula.api as smf
from statsmodels.stats.anova import anova_lm
from statsmodels.stats.oneway import anova_oneway
from statsmodels.stats.multicomp import pairwise_tukeyhsd

d = pd.DataFrame({
    "method": ["Reread"] * 8 + ["Practice"] * 8 + ["Flashcard"] * 8,
    "score": [58, 62, 55, 64, 60, 57, 61, 59,
              70, 66, 74, 63, 78, 69, 72, 68,
              65, 72, 58, 80, 61, 75, 69, 64]})
names = ["Reread", "Practice", "Flashcard"]
groups = [d.score[d.method == m].to_numpy() for m in names]

fit = smf.ols("score ~ C(method)", data=d).fit()
print(anova_lm(fit))                                        # classic ANOVA
welch = anova_oneway(d.score, d.method, use_var="unequal")  # Welch ANOVA
print(welch.statistic, welch.df_denom, welch.pvalue)
print(stats.levene(*groups, center="median"), stats.shapiro(fit.resid))
print(pairwise_tukeyhsd(d.score, d.method))                 # Tukey HSD
gh = stats.tukey_hsd(*groups, equal_var=False)              # Games-Howell (recent SciPy; run on 1.18.1)
ci = gh.confidence_interval()
for i, j in [(1, 0), (2, 0), (2, 1)]:
    print(names[i], "-", names[j], groups[i].mean() - groups[j].mean(),
          round(ci.low[i, j], 2), round(ci.high[i, j], 2), round(gh.pvalue[i, j], 4))

def eta2_ci(F, df1, df2, level=0.95):
    a = (1 - level) / 2
    cdf = lambda ncp: stats.ncf.cdf(F, df1, df2, ncp)
    lo = 0.0 if stats.f.cdf(F, df1, df2) < 1 - a else brentq(lambda n: cdf(n) - (1 - a), 1e-9, 500)
    hi = brentq(lambda n: cdf(n) - a, 1e-9, 500)
    N = df1 + df2 + 1
    return round(lo / (lo + N), 3), round(hi / (hi + N), 3)

tab = anova_lm(fit)
ssb, ssw, msw = tab.sum_sq.iloc[0], tab.sum_sq.iloc[1], tab.mean_sq.iloc[1]
print(round(ssb / (ssb + ssw), 3), round((ssb - 2 * msw) / (ssb + ssw + msw), 3),
      eta2_ci(tab.F.iloc[0], 2, 21))
w = np.array([-1, 0.5, 0.5]); m = np.array([g.mean() for g in groups])
L = w @ m; se = np.sqrt(msw * np.sum(w**2 / 8))
print(L, round(se, 3), round(L / se, 3), round(2 * stats.t.sf(abs(L / se), 21), 4))

             df      sum_sq     mean_sq         F    PR(>F)
C(method)   2.0  497.333333  248.666667  8.761745  0.001711
Residual   21.0  596.000000   28.380952       NaN       NaN
15.796387430609785 12.45234664226742 0.0003837347334856708
LeveneResult(statistic=np.float64(3.4122562674094707), pvalue=np.float64(0.0520970541855913)) ShapiroResult(statistic=np.float64(0.9893375295174673), pvalue=np.float64(0.9944708622838384))
  Multiple Comparison of Means - Tukey HSD, FWER=0.05
========================================================
  group1   group2  meandiff p-adj   lower  upper  reject
--------------------------------------------------------
Flashcard Practice      2.0 0.7364  -4.714  8.714  False
Flashcard   Reread     -8.5 0.0117 -15.214 -1.786   True
 Practice   Reread    -10.5 0.0021 -17.214 -3.786   True
--------------------------------------------------------
Practice - Reread 10.5 5.29 15.71 0.0005
Flashcard - Reread 8.5 0.67 16.33 0.0345
Flashcard - Practice -2.0 -10.28 6.28 0.7985
0.455 0.393 (0.097, 0.63)
9.5 2.307 4.118 0.0005

R and Python agree on every number. Two presentation differences matter: statsmodels' pairwise_tukeyhsd sorts the groups alphabetically and reports group2 minus group1 (so "Flashcard, Reread, −8.5" means Reread is 8.5 lower), while R's TukeyHSD follows the factor level order. And R's oneway.test is Welch unless you set var.equal = TRUE, whereas scipy.stats.f_oneway and statsmodels' anova_lm give the classic F; for Welch in Python, ask for use_var="unequal" explicitly.

By hand. Grand mean = (59.5 + 70 + 68) / 3 = 65.833 (equal n). Between groups: $SS_B = 8[(59.5-65.833)^2 + (70-65.833)^2 + (68-65.833)^2] = 8(40.11 + 17.36 + 4.69) = 497.33$. Within groups: $SS_W = 7(2.878^2 + 4.690^2 + 7.407^2) = 7(8.29 + 22.00 + 54.86) = 596.0$. Then $MS_B = 497.33/2 = 248.67$, $MS_W = 596/21 = 28.38$, and $F = 248.67/28.38 = 8.76$ with df (2, 21).

Effect size with CI. η² = 497.33 / 1093.33 = 0.455, 95% CI [0.097, 0.630]; ω² = (497.33 − 2 x 28.38) divided by (1093.33 + 28.38) = 0.393. The CI is wide because n = 8 per group: the data are consistent with anything from a small to a very large share of variance explained.

95% or 90%? The code above gives a two-sided 95% interval (2.5% in each tail of the noncentral F). Steiger (2004) shows that a 90% interval matches the F test for η², and several effect-size texts recommend it, because η² cannot be negative and the F test is one-sided: a 90% interval excludes 0 exactly when the F test is significant at α = 0.05, so the interval and the test agree. With level = 0.90 the same function gives [0.143, 0.600] here. Either is defensible; state the level you used, and do not compare a 90% interval from one paper with a 95% interval from another. Report ω² (less biased) as the main effect size, or the mean differences with their CIs, which are easier to interpret.

Interpretation. Revision method made a difference to quiz scores. Welch's ANOVA, which is appropriate given the unequal SDs, gives F(2, 12.45) = 15.80, p < .001. The planned contrast shows that the two active methods together beat rereading by 9.5 points (t(21) = 4.12, p < .001). Games-Howell comparisons show that Practice beat Reread by 10.5 points (95% CI [5.29, 15.71]) and Flashcard beat Reread by 8.5 points (95% CI [0.67, 16.33]), while Practice and Flashcard did not differ detectably (−2.0, 95% CI [−10.28, 6.28]). That last interval is wide: it does not show the two active methods are equal. Welch's F is larger than the classic F here because the low-variance Reread group, which differs most from the others, gets more weight.

APA-style report. Quiz scores differed by revision method, Welch's F(2, 12.45) = 15.80, p < .001, ω² = .39 (classic ANOVA: F(2, 21) = 8.76, p = .002, η² = .46, 95% CI [.10, .63]). Students who used practice tests (M = 70.0, SD = 4.7) or flashcards (M = 68.0, SD = 7.4) scored higher than those who reread notes (M = 59.5, SD = 2.9); Games-Howell: Practice − Reread = 10.5, 95% CI [5.3, 15.7], p < .001; Flashcard − Reread = 8.5, 95% CI [0.7, 16.3], p = .035; Practice − Flashcard = 2.0, 95% CI [−6.3, 10.3], p = .799.

Worked example B (half solved): Brightwell Bakery ovens

Question. Brightwell Bakery (fictional) baked six loaves from the same dough batch in each of three ovens and measured loaf volume (cm³). Do the ovens differ in mean volume?

b <- data.frame(oven = factor(rep(c("A", "B", "C"), each = 6)),
  volume = c(1510, 1495, 1520, 1505, 1500, 1515,
             1530, 1525, 1540, 1520, 1535, 1530,
             1500, 1490, 1505, 1495, 1510, 1500))
fb <- aov(volume ~ oven, b); summary(fb); oneway.test(volume ~ oven, b); TukeyHSD(fb)

import pandas as pd
from scipy import stats
from statsmodels.stats.oneway import anova_oneway
from statsmodels.stats.multicomp import pairwise_tukeyhsd

b = pd.DataFrame({"oven": ["A"] * 6 + ["B"] * 6 + ["C"] * 6,
                  "volume": [1510, 1495, 1520, 1505, 1500, 1515,
                             1530, 1525, 1540, 1520, 1535, 1530,
                             1500, 1490, 1505, 1495, 1510, 1500]})
print(stats.f_oneway(*[b.volume[b.oven == o] for o in "ABC"]))
print(anova_oneway(b.volume, b.oven, use_var="unequal"))
print(pairwise_tukeyhsd(b.volume, b.oven))

(output abridged: R printouts condensed)
            Df Sum Sq Mean Sq F value   Pr(>F)
oven         2 2925.0  1462.5    23.4 2.44e-05 ***
Residuals   15  937.5    62.5
Welch: F = 26.696, num df = 2.000, denom df = 9.863, p-value = 0.0001047
       diff       lwr        upr     p adj
B-A    22.5  10.64422  34.355775 0.0005018
C-A    -7.5 -19.35578   4.355775 0.2588712
C-B   -30.0 -41.85578 -18.144225 0.0000250

Python gives the same values (f_oneway: statistic 23.4, p = 2.44e-05; Welch statistic 26.696, df_denom 9.863).

1. Assumption check: SD ratio = 9.35 / 7.07 = [ ]. Is the classic ANOVA reasonable here? [ ]
2. Grand mean = [ ]. $SS_B = 6[(1507.5 - \bar y)^2 + (1530 - \bar y)^2 + (1500 - \bar y)^2]$ = [ ] (check: 2925).
3. $SS_W = 5(9.35^2 + 7.07^2 + 7.07^2)$ = [ ]. $MS_W$ = [ ] with df = [ ].
4. F = [ ] with df ([ ], [ ]); p [ ].
5. η² = [ ]; ω² = [ ].
6. Which pairs differ (Tukey, α = 0.05)? [ ]
7. Report sentence: [ ]

Exercise C (unsolved): Harbour City Library reshelving times

Harbour City Library (fictional) timed how long it took to reshelve a trolley of 50 books (minutes) at three branches, five trolleys per branch. Only the summaries were recorded:

Tasks:

1. State H0 and H1.
2. Compute $SS_B$, $SS_W$, the mean squares and F. Fill a complete ANOVA table.
3. Find the p-value (software: pf(F, 2, 12, lower.tail = FALSE) in R or stats.f.sf(F, 2, 12) in Python).
4. Compute η² and ω².
5. Which follow-up procedure would you use, and why? Write the APA sentence.

Reading software output

Reporting template

Fill-in sentence. [Outcome] [differed / did not differ detectably] across [factor] groups, [Welch's / classic] F([df1], [df2]) = [value], p = [.xxx], ω² = [.xx] [95% CI for η² [lo, hi]]. [Follow-up procedure] showed that [group A] (M = [ ], SD = [ ]) scored [higher / lower] than [group B] (M = [ ], SD = [ ]), difference = [ ], 95% CI [[lo], [hi]], p = [.xxx].

Results checklist:

• Design, group sizes and how units were assigned
• Descriptives per group (n, mean, SD) in a table or figure with CIs
• Assumption checks reported briefly, and the reason for classic or Welch
• Omnibus F with both df and p in APA style
• Effect size: ω² (or η² with CI), and mean differences with CIs
• Planned contrasts or a named post hoc procedure with adjusted p-values
• Non-significant pairs described as "not detectably different", with their CIs
• Causal language only if units were randomly assigned

Common mistakes and misconceptions

1. Running all pairwise t-tests without adjustment. The family-wise false-positive rate grows with the number of pairs; use Tukey, Games-Howell or planned contrasts.
2. Stopping at a significant F. F says "some means differ", not which. Report the follow-up comparisons and effect sizes.
3. Choosing classic or Welch from Levene's p-value. A non-significant Levene test with small groups is not evidence of equal variances (Example A: p = .052 with an SD ratio of 2.6).
4. "A non-significant pair means the methods are equally good." Practice versus Flashcard has a 95% CI from about −6 to 10 points: the data cannot rule out a meaningful difference.
5. Checking normality of the raw outcome pooled across groups. Check the residuals (or each group); a mixture of shifted normal groups is not normal.
6. Reporting η² as if it were unbiased. In small samples η² overstates the population effect; ω² is smaller (0.393 vs 0.455 here).
7. Post hoc tests chosen after looking at which one gives significance. Choose the procedure in your analysis plan.
8. Using one-way ANOVA for repeated measures. The same students in every condition violate independence and waste the pairing.
9. Forgetting that software defaults differ. R's oneway.test is Welch by default; SciPy's f_oneway is classic; state which you used.
10. "F = 8.76 means the active methods are 8.76 times better." F is a ratio of variance estimates, not of means.

Self-test questions

1. A study has 4 groups of 10. What are the df for the classic F test?
2. Explain in two sentences why F is close to 1 when H0 is true.
3. Group SDs are 3, 4 and 9, with n = 6 per group. Which omnibus test and which post hoc procedure do you choose?
4. SS_B = 120, SS_W = 380, k = 3, N = 33. Compute F, η² and ω².
5. Why can the ANOVA be significant while no Tukey pair is significant?
6. What is the difference between a planned contrast and a post hoc comparison, and why does it matter for error control?

My notes and next steps

Sources and further reading

Structure aligned with the method-card shape of this statistics library; methods based on the original papers and the public-domain NIST/SEMATECH e-Handbook, written in original words with fictional examples.

• Fisher, R. A. (1925). Statistical Methods for Research Workers. Oliver and Boyd.
• Welch, B. L. (1951). On the comparison of several mean values: an alternative approach. Biometrika, 38(3–4), 330–336.
• Tukey, J. W. (1949). Comparing individual means in the analysis of variance. Biometrics, 5(2), 99–114.
• Tukey, J. W. (1953). The Problem of Multiple Comparisons. Unpublished manuscript.
• Kramer, C. Y. (1956). Extension of multiple range tests to group means with unequal numbers of replications. Biometrics, 12(3), 307–310.
• Games, P. A., and Howell, J. F. (1976). Pairwise multiple comparison procedures with unequal n's and/or variances: a Monte Carlo study. Journal of Educational Statistics, 1(2), 113–125.
• Brown, M. B., and Forsythe, A. B. (1974). Robust tests for the equality of variances. Journal of the American Statistical Association, 69(346), 364–367.
• Hays, W. L. (1963). Statistics for Psychologists. Holt, Rinehart and Winston (the usual cited source of ω²).
• Cohen, J. (1988). Statistical Power Analysis for the Behavioral Sciences (2nd ed.). Lawrence Erlbaum Associates.
• Steiger, J. H. (2004). Beyond the F test: effect size confidence intervals and tests of close fit in the analysis of variance and contrast analysis. Psychological Methods, 9(2), 164–182.
• NIST/SEMATECH e-Handbook of Statistical Methods, sections 7.4.3 Are the means equal? and 7.4.7 How can we make multiple comparisons?, https://www.itl.nist.gov/div898/handbook/ (accessed 2026-10-25). Public domain.
• R Core Team. Documentation for aov, oneway.test, TukeyHSD, ptukey (R 4.5.1); Fox and Weisberg, car package documentation for leveneTest (3.1-5).
• SciPy documentation for f_oneway, levene, tukey_hsd (1.18.1); statsmodels documentation for anova_lm, anova_oneway, pairwise_tukeyhsd (0.15.0).
• Further reading (by name only): de Smith, M. J. (2024). Statistical Analysis Handbook, 2024 edition. StatsRef, https://www.statsref.com/; OpenIntro Statistics; APA Publication Manual (7th ed.) and APA JARS-Quant.

Answer key

Worked example B.

1. SD ratio = 1.32; with equal n and similar SDs, the classic ANOVA with Tukey HSD is reasonable (Welch agrees: F(2, 9.86) = 26.70, p < .001).
2. Grand mean = 1512.5; $SS_B = 6(25 + 306.25 + 156.25) = 6 \times 487.5 = 2925$.
3. $SS_W = 5(87.5 + 50 + 50) = 937.5$; $MS_W = 937.5 / 15 = 62.5$ with df = 15.
4. $MS_B = 2925 / 2 = 1462.5$; F = 1462.5 / 62.5 = 23.4 with df (2, 15); p < .001 (2.44 x 10⁻⁵).
5. η² = 2925 / 3862.5 = 0.757; ω² = (2925 − 125) / (3862.5 + 62.5) = 0.713.
6. B differs from A (22.5 cm³, 95% CI [10.6, 34.4], p < .001) and from C (30.0 cm³, 95% CI [18.1, 41.9], p < .001); A and C do not differ detectably (7.5 cm³, 95% CI [−4.4, 19.4], p = .259).
7. Loaf volume differed by oven, F(2, 15) = 23.40, p < .001, ω² = .71; Tukey HSD showed that oven B produced larger loaves than ovens A and C, which did not differ detectably.

Exercise C. H0: the three branch means are equal. Grand mean = 15.0. $SS_B = 5(9 + 0 + 9) = 90$, df = 2, $MS_B = 45$. $SS_W = 4(4 + 9 + 6.25) = 77$, df = 12, $MS_W = 6.417$. F = 45 / 6.417 = 7.01, p = .010 (0.0096). η² = 90 / 167 = 0.539; ω² = (90 − 12.83) / (167 + 6.42) = 0.445. With equal n and SDs from 2.0 to 3.0, Tukey HSD is a reasonable follow-up (Games-Howell if you prefer not to assume equal variances; with only the summaries, you can compute either by hand from the means, SDs and n). Sentence: Reshelving time differed by branch, F(2, 12) = 7.01, p = .010, ω² = .45.

Self-test.

1. df = (k − 1, N − k) = (3, 36).
2. Under H0 both the between-group mean square and the within-group mean square estimate the same error variance, so their ratio fluctuates around 1. Real differences in means add to the between-group mean square only.
3. The SD ratio is 3 with small equal groups, so Welch's ANOVA and Games-Howell comparisons.
4. $MS_B = 60$, $MS_W = 380/30 = 12.67$, F(2, 30) = 4.74; η² = 120 / 500 = 0.24; ω² = (120 − 25.33) / (500 + 12.67) = 0.185.
5. The omnibus F can detect a pattern spread over several means (for example a linear trend, or one mean against the average of the others) that no single pair captures after the multiple-comparison adjustment. Planned contrasts test such patterns directly.
6. A planned contrast is specified before seeing the data and tests one focused hypothesis, so it uses its own α (adjust only if there are several). Post hoc comparisons look at all pairs after seeing the data, so they need a family-wise adjustment such as Tukey or Games-Howell.

Revision history