📏 Composite Reliability (CR) in Research
Composite Reliability (CR) is a measure of internal consistency of a construct in structural equation modeling (SEM) and other latent variable models. It tells you how well a group of items (indicators) consistently reflect the same underlying construct.
It is similar to Cronbach’s alpha, but more accurate in SEM because CR:
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Takes into account actual factor loadings of each item,
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Does not assume equal item reliability, unlike Cronbach’s alpha.
🧮 Formula for Composite Reliability
CR=(∑λi)2(∑λi)2+∑θiCR = \frac{(\sum \lambda_i)^2}{(\sum \lambda_i)^2 + \sum \theta_i}CR=(∑λi)2+∑θi(∑λi)2
Where:
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λi\lambda_iλi = standardized factor loading for each item,
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θi\theta_iθi = error variance for each item (1 – λi2\lambda_i^2λi2).
✅ Interpretation of CR Values
| CR Value | Interpretation |
|---|---|
| ≥ 0.70 | Acceptable (good reliability) |
| 0.60 – 0.70 | Acceptable in exploratory research |
| < 0.60 | Poor reliability (problematic) |
Some scholars accept 0.60–0.70 as adequate for exploratory studies, but for confirmatory studies, a CR ≥ 0.70 is usually expected.
🧠 Why is Composite Reliability Important?
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It ensures that the latent construct is measured reliably by the observed variables.
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CR is part of construct validity testing, especially in Confirmatory Factor Analysis (CFA).
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It is used alongside:
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Average Variance Extracted (AVE) for convergent validity,
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Discriminant validity tests like Fornell-Larcker criterion.
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🔄 CR vs. Cronbach’s Alpha
| Aspect | Composite Reliability | Cronbach’s Alpha |
|---|---|---|
| Based on factor loadings | ✅ Yes | ❌ No (assumes equal) |
| Accuracy in SEM | ✅ Higher | ❌ Lower |
| Use in CFA/SEM | ✅ Preferred | ❌ Less preferred |
📌 Example:
You have a construct “Customer Satisfaction” measured by 4 items with factor loadings: 0.80, 0.85, 0.78, and 0.83.
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Compute ∑λi=0.80+0.85+0.78+0.83=3.26\sum \lambda_i = 0.80 + 0.85 + 0.78 + 0.83 = 3.26∑λi=0.80+0.85+0.78+0.83=3.26
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Compute ∑θi=(1−0.802)+(1−0.852)+…=0.36+0.28+0.39+0.31=1.34\sum \theta_i = (1 – 0.80^2) + (1 – 0.85^2) + … = 0.36 + 0.28 + 0.39 + 0.31 = 1.34∑θi=(1−0.802)+(1−0.852)+…=0.36+0.28+0.39+0.31=1.34
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Apply the CR formula:
CR=(3.26)2(3.26)2+1.34=10.6310.63+1.34=10.6311.97≈0.89CR = \frac{(3.26)^2}{(3.26)^2 + 1.34} = \frac{10.63}{10.63 + 1.34} = \frac{10.63}{11.97} ≈ 0.89CR=(3.26)2+1.34(3.26)2=10.63+1.3410.63=11.9710.63≈0.89
✅ Result: Very good composite reliability