Question 1

What is Cohen's d and how do I interpret it?

Accepted Answer

Cohen's d measures the difference between two group means in standard deviation units. Jacob Cohen's benchmarks (1988): d = 0.2 (small), d = 0.5 (medium), d = 0.8 (large). A d of 1.0 means the groups differ by one full standard deviation. These are rough guidelines — in educational research, d = 0.4 is often considered meaningful; in clinical settings, even d = 0.2 may be clinically significant if the outcome is serious.

Question 2

How is r as an effect size different from Pearson r for correlation?

Accepted Answer

Pearson r can be used as an effect size for any analysis, not just correlations. When converting from a t-test: r = √(t²/(t²+df)). Cohen's benchmarks for r: 0.1 (small), 0.3 (medium), 0.5 (large). The value r² (coefficient of determination) tells you the proportion of variance explained — r = 0.3 means 9% of variance is explained. For comparing two groups, r = d/√(d²+4) converts between the two measures.

Question 3

What is eta-squared and when should I use it?

Accepted Answer

Eta-squared (η²) is the proportion of total variance explained by a factor in ANOVA: η² = SS_between / SS_total. Cohen's benchmarks: η² = 0.01 (small), 0.06 (medium), 0.14 (large). Note: η² overestimates the population effect because it is based on sample SS. Partial eta-squared (η²p) is preferred in multi-factor ANOVA as it accounts for other factors. Omega-squared (ω²) is a less biased estimate of the population η².

Question 4

Can I compare effect sizes across different studies?

Accepted Answer

Yes — this is the primary purpose of standardised effect sizes. A meta-analysis combines effect sizes across studies to estimate the overall effect. Cohen's d and Pearson r are on universal scales, allowing direct comparison even when studies use different measurement instruments, sample sizes, or populations. This is why reporting standardised effect sizes is now mandatory in APA format publications.

Question 5

What is the pooled standard deviation in Cohen's d?

Accepted Answer

The pooled SD combines the standard deviations of both groups: s_pooled = √[(s₁²(n₁−1) + s₂²(n₂−1)) / (n₁+n₂−2)]. This weights each group's variance by its degrees of freedom. If sample sizes are equal, s_pooled simplifies to √[(s₁²+s₂²)/2]. For one-sample d (comparing to a reference value), use the single group's SD in the denominator.

Question 6

What effect size should I target when designing a study?

Accepted Answer

Power analysis requires specifying an expected effect size. In applied research, use the smallest effect that would be practically meaningful — not Cohen's arbitrary benchmarks. If a 5-point IQ improvement is meaningful, calculate d based on IQ SD ≈ 15: d = 5/15 ≈ 0.33. To detect d = 0.33 with 80% power at α = 0.05 (two-tailed, two groups), you need approximately 116 participants per group. Use a sample size calculator to translate effect size to required n.

Interpretation	Cohen's d	r	η² (ANOVA)
Negligible	< 0.2	< 0.1	< 0.01
Small	0.2 – 0.5	0.1 – 0.3	0.01 – 0.06
Medium	0.5 – 0.8	0.3 – 0.5	0.06 – 0.14
Large	≥ 0.8	≥ 0.5	≥ 0.14

Analysis type	Effect size	Notes
Two independent group means (t-test)	Cohen's d	Most common; units = SDs
One-sample vs reference mean	Cohen's d (one-sample)	d = (x̄ − μ₀) / s
Paired before/after means	Cohen's d (paired)	d = mean difference / SD of differences
Correlation between two variables	Pearson r	r² = proportion of variance shared
t-test → effect size	r = √(t²/(t²+df))	Convert directly from t output
One-way ANOVA	Eta-squared η²	SS_between / SS_total
Multi-factor ANOVA	Partial η²p	Excludes other factors from denominator

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