```
#' Reliability of the weighted mean of assessments
#'
#' This function calculates the reliability of a weighted mean of assessments given the
#' reliability coefficient of each assessment, the correlations between them, and the
#' weights.
#'
#' @param rely A vector of reliability coefficients.
#' @param r Either a correlation matrix or a vector of unique correlations. It is
#' recommended to specify the correlations as a matrix to avoid
#' erronous pairings of assessment correlations with reliability coefficients and weights,
#' since this can be confusing if the correlations are supplied as a vector. Presumes that the
#' correlation matrix has the same order of assessments as the reliability and weight vectors.
#' @param w A vector of weights. Will be internally normalized to sum to 1 and presumes the
#' same order of assessments as the correlation matrix and vector of reliabilities.
#' If omitted, it is assumed that all assessments have the same weight.
#'
#' @examples
#' r <- matrix(c( 1, .4, .7,
#' .4, 1, .5,
#' .7, .5, 1), 3,3, byrow=TRUE)
#' reliability_mean(rely=c(.9, .85, .88), r=r, w=c(.5, .3, .2))
#'
#' @export
reliability_mean <- function(rely, r, w=NA) {
# if no weights were provided, create a vector of equal weights
if (is.na(min(w))) {w <- rep(1/length(rely), times=length(rely))}
# check weights
if (min(w)<0) {stop("Weights must be positive")}
# normalize the weights
w <- w / sum(w)
# make sure r is either a vector or matrix
if (!is.vector(r) & !is.matrix(r)) {stop("r must be a correlation matrix or a vector of unique correlations")}
# make sure reliability coefficients are between zero and one
if (min(rely) < 0 | max(rely) >= 1) {stop("rely contains an out-of-range value. reliability coefficients must be between zero and one.")}
if (is.vector(r)) {
# check that r contains valid correlation values
if (min(r) < -1 | max(r) >= 1) {stop("r contains an out-of-range correlation value")}
# if r is supplied as a vector, 1s should not be included
if (max(r) == 1) {warning("r contains one or more values of 1. The vector of unique correlations provided to this function should not include the 1s from the diagonal. Ensure that the values in r are intended")}
# make sure that the length of r is compatible with choose(n,2)
if (!(length(r) %in% choose(seq(2,100),2))) {stop("length of vector r is incorrect")}
# find the number of assessments from the set of correlations
p=1
while (p^2 < 2*length(r)) {
p <- p + 1
}
# check that the lengths of r, weights, and rely are compatible
if (min(c(p, length(rely), length(w)) == rep(length(w), 3))==0) {stop("The number of assessments implied by the length of r, the number of weights, and the number of reliability coefficients must be the same")}
#now build the correlation matrix
cov <- matrix(1, p, p)
cov[lower.tri(cov)] <- r
t.cov <- t(cov)
cov[upper.tri(cov)] <- t.cov[upper.tri(t.cov)]
unique.r <- r
}
if (is.matrix(r)) {
# check that r is square and has 1s on the diagonal
if ((dim(r)[1] != dim(r)[2]) | (max(diag(r) != rep(1, dim(r)[1])))) {stop("r must be a square correlation matrix with ones on the diagonal or a vector of unique correlations")}
# check that r is symmetric
if (!isSymmetric(r)) {stop("the correlation matrix r must be symmetric")}
cov <- r
p <- nrow(cov)
unique.r <- r[lower.tri(r)]
}
# check that correlation matrix is positive definite
if (matrixcalc::is.positive.definite(cov)==FALSE) stop("correlation matrix is not positive definite")
# check that no correlation exceeds the sqrt of the prod of the reliabilities
checkmat <- sqrt(as.matrix(rely) %*% t(as.matrix(rely)))
diag(checkmat) <- 1
if (min(checkmat-cov) < 0) {stop("a correlation is larger than the square root of the product of the involved reliability coefficients")}
#corrs[i] <- sum(r[i,]*w) / sqrt(sum(w^2)+2*sum(combn(w, 2, prod)*unique.r))
rely.m <- (sum(rely*w^2)+2*sum(combn(w,2,prod)*unique.r)) / (sum(w^2)+2*sum(combn(w,2,prod)*unique.r))
return(rely.m)
}
```

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