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ANOVA

Welch's ANOVA — Unequal Variances Alternative

Compare three course formats where one group has 200x the variance of the others. Standard ANOVA misses the effect (p=.080); Welch's ANOVA correctly detects it (F=182.99, p<.001).

📊 F_Welch(2,7.12) = 182.99, p < .001 (standard F=3.14, p=.080)
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📊 28 · Welch's ANOVA — When Group Variances Are Unequal

Comparing Three Course Formats With Very Different Score Spread
ANOVA
Research Question: 15 students were split into three exam-score groups (n=5 each) across course formats: Self-Paced Online, Cohort-Based Online, and In-Person. The cohort-based group shows wild score spread while the other two are tight. Does format affect scores once we account for this unequal spread?
Self-Paced Online (A) Cohort-Based Online (B) In-Person (C)
585078
6210082
595579
619581
607580
Standard (pooled) F, for comparison
F = M S B M S W = S S B / ( k − 1 ) S S W / ( N − k )
Welch's F — weights each group by the inverse of its own variance
F W = ∑ w i ( x ¯ i − x ¯ w ) 2 / ( k − 1 ) 1 + 2 ( k − 2 ) k 2 − 1 ∑ ( 1 − w i / ∑ w ) 2 n i − 1 , w i = n i s i 2
  1. 1
    Group means and variances:
    x ¯ A = 60.00 ,   s A 2 = 2.50 x ¯ B = 75.00 ,   s B 2 = 512.50 x ¯ C = 80.00 ,   s C 2 = 2.50
    Group B's variance (512.50) is over 200× larger than A's or C's — a red flag for the equal-variance assumption.
  2. 2
    Check Levene's test first:
    F L e v e n e ( 2 , 12 ) = 12.96 , p = .001
    p = .001 < .05 → variances are significantly unequal → the standard (pooled) ANOVA's equal-variance assumption is violated. This is the trigger to prefer Welch's ANOVA.
  3. 3
    Standard (pooled) ANOVA anyway, for contrast:
    S S B = 1083.33 ,   S S W = 2070.00 ,   F ( 2 , 12 ) = 541.67 172.50 = 3.14 ,   p = .080
    Taken at face value this says 'no significant difference' — but that conclusion is untrustworthy, because it relies on an assumption we just showed is violated. Group B's huge within-group variance is inflating the pooled error term and masking the real separation between A and C.
  4. 4
    Welch's ANOVA — weight by 1/variance:
    w A = 2.00 ,   w B = 0.0098 ,   w C = 2.00 , x ¯ w = 70.01 F W ( 2 ,   7.12 ) = 200.12 1.0936 = 182.99 , p < .001
    Because B's variance is so large, Welch's weighting effectively down-weights it to almost nothing (w_B ≈ 0.01 vs. w_A=w_C=2.00) — the test becomes, in effect, a clean, low-variance comparison of A against C, which are 20 points apart. That is why F jumps from 3.14 to 182.99.
  5. 5
    Decision rule: Because Levene's test flagged unequal variances, report Welch's F, not the standard F. Reporting the standard result here would have been a Type II error — concluding 'no effect' when a large, real effect exists.
182.99
F_Welch(2,7.12)
<.001
p (Welch)
3.14
F standard (misleading)
.080
p standard (n.s.)
🔴 Significant by Welch's ANOVA — F(2, 7.12) = 182.99, p < .001 — despite the misleading standard ANOVA (p = .080).
APA-7
Because Levene's test indicated unequal variances, F(2, 12) = 12.96, p = .001, a Welch's ANOVA was conducted. There was a significant effect of course format on exam scores, F(2, 7.12) = 182.99, p < .001. (The standard ANOVA alone would have misleadingly suggested no effect, F(2, 12) = 3.14, p = .080.)

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