Akaike information criterion. AICs analysis ( sample size =5, No of parameters (not counting the error variance) = 3), in result AICc = infinite.Could you please recommend another formula for AIC correction for small sample size (n/k < 40)
Not possible. Sample size is VERY MUCH TOO SMALL. Sorry. Best, David Booth PS see the link:
https://www.google.com/search?client=firefox-b-1&ei=0XhwW5C8JIvRjwS32qzwDw&q=aic+vs+r2&oq=AIC&gs_l=psy-ab.1.3.0i67k1l2j0j0i67k1l4j0l3.842773.847216.0.857491.40.12.0.0.0.0.356.1228.0j7j0j1.8.0..2..0...1.1.64.psy-ab..36.4.719...0i22i30k1j0i131k1.0.rflT2SxUF08
For small sample sizes (n/K < ≈ 40), use the second-order AIC: AICc = -2(log-likelihood) + 2K + (2K(K+1)/(n-K-1)) Where:
- n = sample size,
- K= number of model parameters,
- Log-likelihood is a measure of model fit.
As AICc approximates AIC for large samples, it is recommended to us AICc in all cases. For simplicity, we use the name “AIC” for all variants of it.
Ujjaval Srivastava could you tell me how to calculate AICs when N=5 and k= 4 because value will be infinite due to divide by zero ?
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