Author
David Gerard
There are no base R functions that do power and sample size calculations. But I created some for you:
#' Power/sample size calculation of 1-sample proportion test#' #' Uses central limit theorem, so make sure `p0 * (1 - p0) * n >= 5`#' #' Exactly one of `n`, `power`, `p0`, `p1`, or `alpha` needs to be `NULL`.#' #' @param n The sample size#' @param power The power#' @param p0 The null proportion#' @param p1 The alternative proportion#' @param alpha The significance level#' @param TOL Tolerance level#' #' @author David Gerard#' #' @examples#' b1power(n = 500, power = NULL, p0 = 0.02, p1 = 0.05, alpha = 0.05)#' b1power(n = NULL, power = 0.9, p0 = 0.02, p1 = 0.05, alpha = 0.05)#' #' ## two p1's#' b1power(n = 500, power = 0.9, p0 = 0.02, p1 = NULL, alpha = 0.05)#' b1power(n = 500, power = NULL, p0 = 0.02, p1 = 0.00406, alpha = 0.05)$power#' b1power(n = 500, power = NULL, p0 = 0.02, p1 = 0.044, alpha = 0.05)$powerb1power <- function( n = NULL, power = NULL, p0 = NULL, p1 = NULL, alpha = 0.05, TOL = 1e-6) { if (is.null(n) + is.null(power) + is.null(p0) + is.null(p1) + is.null(alpha) != 1) { stop("exactly one of n, power, p0, p1, and alpha need to be NULL") } oout <- list(n = n, power = power, p0 = p0, p1 = p1, alpha = alpha) pfun <- function(n, p0, p1, alpha) { za2 <- stats::qnorm(alpha / 2) stats::pnorm(sqrt(p0 * (1 - p0) / (p1 * (1 - p1))) * (za2 + abs(p0 - p1) * sqrt(n) / sqrt(p0 * (1 - p0)))) } if (is.null(power)) { oout$power <- pfun(n = n, p0 = p0, p1 = p1, alpha = alpha) } else if (is.null(n)) { z1a2 <- stats::qnorm(1 - alpha / 2) zp <- stats::qnorm(power) oout$n <- p0 * (1 - p0) * (z1a2 + zp * sqrt(p1 * (1 - p1) / (p0 * (1 - p0))))^2 / (p1 - p0)^2 oout$n <- ceiling(oout$n) } else if (is.null(p0)) { rp0 <- function(p0) {power - pfun(n = n, p0 = p0, p1 = p1, alpha = alpha)} if (sign(rp0(p0 = TOL)) * sign(rp0(p0 = p1)) < 0) { r1 <- stats::uniroot(f = rp0, interval = c(TOL, p1)) } else { r1 <- list(root = NA) } if (sign(rp0(p0 = 1 - TOL)) * sign(rp0(p0 = p1)) < 0) { r2 <- stats::uniroot(f = rp0, interval = c(p1, 1 - TOL)) } else { r2 <- list(root = NA) } oout$p0 <- c(r1$root, r2$root) } else if (is.null(p1)) { rp1 <- function(p1) {power - pfun(n = n, p0 = p0, p1 = p1, alpha = alpha)} if (sign(rp1(p1 = TOL)) * sign(rp1(p1 = p0)) < 0) { r1 <- stats::uniroot(f = rp1, interval = c(TOL, p0)) } else { r1 <- list(root = NA) } if (sign(rp1(p1 = 1 - TOL)) * sign(rp1(p1 = p0)) < 0) { r2 <- stats::uniroot(f = rp1, interval = c(p0, 1 - TOL)) } else { r2 <- list(root = NA) } oout$p1 <- c(r1$root, r2$root) } else if (is.null(alpha)) { ralpha <- function(alpha) {power - pfun(n = n, p0 = p0, p1 = p1, alpha = alpha)} rout <- stats::uniroot(f = ralpha, interval = c(TOL, 1-TOL)) oout$alpha <- rout$root } if (any(oout$p0[!is.na(oout$p0)] * (1 - oout$p0[!is.na(oout$p0)]) * oout$n < 5)) { warning("too small sample size") } return(oout)}Assumes the sample size is large enough to use the central limit theorem (\(np_0(1-p_0) \geq 5\)).
Suppose we wish to test the hypothesis that women with a sister history of breast cancer are at higher risk of developing breast cancer themselves. Suppose the prevalence rate of breast cancer is 2% among 50 to 54 year-old US women, whereas it is 5% among women with a sister history. We wish to interval 500 women 50 to 54 years old with a sistory history of the disease. What is the power of such a study assuming that we conduct a two-sided test with \(\alpha = 0.05\)?
# 0.9655b1power(n = 500, power = NULL, p0 = 0.02, p1 = 0.05, alpha = 0.05)$n[1] 500$power[1] 0.9655$p0[1] 0.02$p1[1] 0.05$alpha[1] 0.05How many women should we interview in the study proposed to achieve 90% power?
# 341b1power(n = NULL, power = 0.9, p0 = 0.02, p1 = 0.05, alpha = 0.05)$n[1] 341$power[1] 0.9$p0[1] 0.02$p1[1] 0.05$alpha[1] 0.05
