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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T04:11:46+00:00 2026-05-16T04:11:46+00:00

I have the following problem. I need to calculate the Shrout & Fleiss ICC’s

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I have the following problem.
I need to calculate the Shrout & Fleiss ICC’s for the situation in which items are judged by a varying number of judges. For example, the competitive nature of an industry is judged for a set of industries, but with a different number of judges per industry. One industry is only judged by 2 judges, whereas the competitive nature of another industry can be judged by up to 12 judges. I have used the ICC (package psych) and icc (package irr) functions for the determination of icc’s, but it is unclear whether they can deal with a varying number of judges. Can anyone help me with this? How to calculate the S&F icc’s when the number of judges is variable? I would prefer to continue using R for this, but if this is better accomplished with another software, I would like to get those suggestions as well.

Thanks, Peter Verbeet

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  1. Editorial Team
    Editorial Team
    2026-05-16T04:11:47+00:00Added an answer on May 16, 2026 at 4:11 am

    I’ve used hierarchical linear modeling for this. I referred to a Prof. Seltman’s HML notes for R to see how. The trick is getting the variance terms for ICC with variable number of judges.

    library(lme4)
    
    # calculate the hierarchical model
    m1 = lmer(score ~ (1|item_id) + (1|judge_id), data=d)
    print(m1)
    
    # helper function to pull out the variances
    xVars <- function(model) {
        exvars = lme4::VarCorr(model)
        vars = c(exvars$item_id[1,1], exvars$judge_id[1,1], attr(exvars,"sc")^2) 
        names(vars) <- c('item var', 'judge var', 'residual var')
        vars
    }
    # helper function for ICC(k) variations
    icck <- function(variances, k=1) {
        icc = variances[1] / (variances[1] + (variances[2] + variances[3]) / k)
        names(icc) = c(paste('ICC', k, sep=''))
        icc
    }
    
    # the output you want
    icc1 = icck(xVars(m1))
    print(icc1)
    icc8 = icck(xVars(m1), 8)
    print(icc8)
    
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