#### Documented in var_mean

```#' The variance of the weighted mean of assessments
#'
#' This function calculates the variance of the weighted mean of a set of unit-standardized
#' assessments. When the assessments are imperfectly correlated, this variance will
#' be less than one. The identification cut score must be adjusted accordingly to
#' maintain the desired percentile cutoff.
#'
#' The value returned by this function can be interpreted as the shrinkage factor
#' of the variance of the weighted mean. The square root of this value is the shrinkage
#' factor of the standard deviation of the weighted mean.
#'
#' @param r Either a correlation matrix or a vector of unique correlations. If the weights
#'  are not all equal, it is recommended to specify the correlations as a matrix to avoid
#'  erronous pairings of assessment correlations and weights, since this can be confusing
#'  if the correlations are supplied as a vector.
#' @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. If omitted, it is assumed that all
#'  assessments have the same weight.
#'
#' @examples
#' var_mean(r=c(.4, .7, .9), w=c(1,2,3))
#'
#' var_mean(r=matrix(c( 1,.4,.7,
#'                       .4, 1,.9,
#'                       .7,.9, 1), 3,3, byrow=TRUE))
#'
#' @export

var_mean <- function(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")}

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 and weights are compatible
if (min(c(p, length(w)) == rep(length(w), 2))==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")

var.m <- sum(w^2)+2*sum(combn(w,2,prod)*unique.r)

return(var.m)
}
```
mcbeem/giftedCalcs documentation built on Nov. 20, 2019, 5:32 p.m.