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Assumption Check

Levene's Test for Equality of Variances

Test the homogeneity-of-variance assumption before an independent t-test using novice vs. expert typist reaction times. Levene's test as an ANOVA run on absolute deviations.

📊 F(1,12) = 12.67, p = .004 → variances unequal
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🧪 29 · Levene's Test for Equality of Variances

Checking Equal Variances Before an Independent-Samples t-test
Assumption Check
Research Question: Before comparing reaction times between novice (n=7) and expert (n=7) typists with an independent-samples t-test, is it reasonable to assume the two groups have equal variances?
Novice (ms) Expert (ms)
812430
940465
705410
1050455
880398
690448
965421
Levene's test (using deviations from each group's mean)
z i j = | x i j − x ¯ i | , F = ∑ i n i ( z ¯ i − z ¯ ) 2 / ( k − 1 ) ∑ i ∑ j ( z i j − z ¯ i ) 2 / ( N − k )
  1. 1
    Group means and spread:
    x ¯ N o v i c e = 863.1 ,   S D N o v i c e = 134.8 x ¯ E x p e r t = 432.4 ,   S D E x p e r t = 24.6
    Experts are not only faster on average but far more consistent — their SD is roughly 1/5 of the novices'.
  2. 2
    Transform to absolute deviations from each group's own mean (z), then run a one-way ANOVA on z: This is exactly what Levene's test is: a one-way ANOVA performed on the absolute deviations instead of the raw scores.
  3. 3
    F:
    F ( 1 , 12 ) = 12.67 , p = .0039
  4. 4
    Decision rule: p = .004 < .05 → reject the null of equal variances → the equal-variance assumption for a standard (pooled) t-test is violated. Report the Welch t-test instead — which MindStat runs unconditionally for every independent-samples comparison, precisely so this decision never has to gate the test manually.
12.67
F(1,12)
.0039
p
134.8 / 24.6
SD ratio
🔴 Variances significantly unequal — F(1,12) = 12.67, p = .004. Use Welch's t-test, not the pooled t-test.
APA-7
A Levene's test indicated that the assumption of homogeneity of variance was violated, F(1, 12) = 12.67, p = .004; a Welch's t-test was therefore used for the group comparison.

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