--- title: "Package MKpower" author: "Matthias Kohl" date: "`r Sys.Date()`" output: rmarkdown::html_vignette: toc: true bibliography: MKpower.bib vignette: > %\VignetteIndexEntry{Package MKpower} %\VignetteEngine{knitr::rmarkdown} %\VignetteEncoding{utf8} --- ## Introduction The package includes functions for power analysis and sample size calculation for Welch and Hsu [@Hedderich2018] t tests including Monte-Carlo simulations of empirical power and type-I-error. In addition, power and sample size calculation for Wilcoxon rank sum and signed rank tests via Monte-Carlo simulations can be performed. Moreover, power and sample size required for the evaluation of a diagnostic test(-system) [@Flahault2005; @Dobbin2007] as well as for a single proportion [@Fleiss2003], comparing two negative binomial rates [@Zhu2014], ANCOVA [@Shieh2020], reference ranges [@Jennen2005], and multiple primary endpoints [@Sozu2015]. We first load the package. ```{r} library(MKpower) ``` ## Single Proportion The computation of the required sample size for investigating a single proportion can be determined via the respective test or confidence interval [@Fleiss2003]. First, we consider the asymptotic test. ```{r} ## with continuity correction power.prop1.test(p1 = 0.4, power = 0.8) ## without continuity correction power.prop1.test(p1 = 0.4, power = 0.8, cont.corr = FALSE) ``` Next, we compute the sample size via the respective asymptotic confidence interval. ```{r} ## with continuity correction ssize.propCI(prop = 0.4, width = 0.14) ## without continuity correction ssize.propCI(prop = 0.4, width = 0.14, method = "wald") ``` ## Welch Two-Sample t-Test For computing the sample size of the Welch t-test, we only consider the situation of equal group size (balanced design). ```{r} ## identical results as power.t.test, since sd = sd1 = sd2 = 1 power.welch.t.test(n = 20, delta = 1) power.welch.t.test(power = .90, delta = 1) power.welch.t.test(power = .90, delta = 1, alternative = "one.sided") ## sd1 = 0.5, sd2 = 1 power.welch.t.test(delta = 1, sd1 = 0.5, sd2 = 1, power = 0.9) ``` ## Hsu Two-Sample t-Test For computing the sample size of the Hsu t-test [@Hedderich2018], we only consider the situation of equal group size (balanced design). ```{r} ## slightly more conservative than Welch t-test power.hsu.t.test(n = 20, delta = 1) power.hsu.t.test(power = .90, delta = 1) power.hsu.t.test(power = .90, delta = 1, alternative = "one.sided") ## sd1 = 0.5, sd2 = 1 power.welch.t.test(delta = 0.5, sd1 = 0.5, sd2 = 1, power = 0.9) power.hsu.t.test(delta = 0.5, sd1 = 0.5, sd2 = 1, power = 0.9) ``` ## ANCOVA With function power.ancova one can compute power and sample size in ANCOVA designs [@Shieh2020]. The default matrix of contrasts used in the function is ```{r} ## 3 groups cbind(rep(1,2), -diag(2)) ## 4 groups cbind(rep(1,3), -diag(3)) ``` We use the example provided in Table 9.7 of @Maxwell2004. ```{r} ## Example from Maxwell and Delaney (2004) according to Shieh (2020) power.ancova(mu=c(7.5366, 11.9849, 13.9785), var = 29.0898, power = 0.8) power.ancova(n = rep(45/3, 3), mu=c(7.5366, 11.9849, 13.9785), var = 29.0898) power.ancova(mu=c(7.5366, 11.9849, 13.9785), var = 29.0898, power = 0.9) power.ancova(n = rep(57/3, 3), mu=c(7.5366, 11.9849, 13.9785), var = 29.0898) ``` Based on the reported adjusted group means and error variance, a (total) sample size of 45 is required to achieve a power of at least 80%. The calculated power is 82.2%. With a sample size of 57 the power will be at least 90%, where the calculated power is 91.2%. Introducing additional covariates (random covariate model) will increase the required sample size. ```{r} power.ancova(mu=c(7.5366, 11.9849, 13.9785), var = 29.0898, power = 0.8, nr.covs = 2) power.ancova(mu=c(7.5366, 11.9849, 13.9785), var = 29.0898, power = 0.8, nr.covs = 3) power.ancova(mu=c(7.5366, 11.9849, 13.9785), var = 29.0898, power = 0.8, nr.covs = 5) power.ancova(mu=c(7.5366, 11.9849, 13.9785), var = 29.0898, power = 0.8, nr.covs = 10) ``` We can also calculate the per group sample sizes for an imbalanced design. ```{r} power.ancova(mu=c(7.5366, 11.9849, 13.9785), var = 29.0898, power = 0.8, group.ratio = c(1, 1.25, 1.5)) power.ancova(n = c(13, 16, 19), mu=c(7.5366, 11.9849, 13.9785), var = 29.0898, group.ratio = c(1, 1.25, 1.5)) power.ancova(mu=c(7.5366, 11.9849, 13.9785), var = 29.0898, power = 0.8, group.ratio = c(1, 0.8, 2/3)) power.ancova(n = c(17, 14, 12), mu=c(7.5366, 11.9849, 13.9785), var = 29.0898, group.ratio = c(1, 0.8, 2/3)) ``` We get a smaller total sample size with an imbalanced design (43 instead of 45). ## Wilcoxon Rank Sum and Signed Rank Tests For computing the sample size of these tests, we offer a function that performs Monte-Carlo simulations to determine their (empirical) power. ```{r} rx <- function(n) rnorm(n, mean = 0, sd = 1) ry <- function(n) rnorm(n, mean = 0.5, sd = 1) ## two-sample sim.ssize.wilcox.test(rx = rx, ry = ry, n.max = 100, iter = 1000) sim.ssize.wilcox.test(rx = rx, ry = ry, n.min = 65, n.max = 70, step.size = 1, iter = 1000, BREAK = FALSE) power.t.test(delta = 0.5, power = 0.8) ## one-sample sim.ssize.wilcox.test(rx = ry, n.max = 100, iter = 1000, type = "one.sample") sim.ssize.wilcox.test(rx = ry, type = "one.sample", n.min = 33, n.max = 38, step.size = 1, iter = 1000, BREAK = FALSE) power.t.test(delta = 0.5, power = 0.8, type = "one.sample") ## two-sample rx <- function(n) rgamma(n, scale = 2, shape = 1.5) ry <- function(n) rgamma(n, scale = 4, shape = 1.5) # equal shape ## two-sample sim.ssize.wilcox.test(rx = rx, ry = ry, n.max = 100, iter = 1000) sim.ssize.wilcox.test(rx = rx, ry = ry, n.min = 25, n.max = 30, step.size = 1, iter = 1000, BREAK = FALSE) ``` For practical applications we recommend to use a higher number of iterations. For more detailed examples we refer to the help page of the function. ## Two Negative Binomial Rates When we consider two negative binomial rates [@Zhu2014], we can compute sample size or power applying function power.nb.test. ```{r} ## examples from Table III in Zhu and Lakkis (2014) power.nb.test(mu0 = 5.0, RR = 2.0, theta = 1/0.5, duration = 1, power = 0.8, approach = 1) power.nb.test(mu0 = 5.0, RR = 2.0, theta = 1/0.5, duration = 1, power = 0.8, approach = 2) power.nb.test(mu0 = 5.0, RR = 2.0, theta = 1/0.5, duration = 1, power = 0.8, approach = 3) ``` ## Diagnostic Test Given an expected sensitivity and specificity we can compute sample size, power, delta or significance level of diagnostic test [@Flahault2005]. ```{r} ## see n2 on page 1202 of Chu and Cole (2007) power.diagnostic.test(sens = 0.99, delta = 0.14, power = 0.95) # 40 power.diagnostic.test(sens = 0.99, delta = 0.13, power = 0.95) # 43 power.diagnostic.test(sens = 0.99, delta = 0.12, power = 0.95) # 47 ``` The sample size planning for developing binary classifiers in case of high dimensional data, we can apply function ssize.pcc, which is based on the probability of correct classification (PCC) [@Dobbin2007]. ```{r} ## see Table 2 of Dobbin et al. (2008) g <- 0.1 fc <- 1.6 ssize.pcc(gamma = g, stdFC = fc, nrFeatures = 22000) ``` ## Reference Range We can apply function ssize.reference.range to determine the sample size required for a study planned to establish a reference range. The parametric approach assumes a normal distribution whereas the non-parametric approach only assumes a continuous distribution. ```{r} ssize.reference.range(delta = 0.01, ref.prob = 0.95, conf.prob = 0.9, method = "parametric", exact = TRUE) ssize.reference.range(delta = 0.01, ref.prob = 0.95, conf.prob = 0.9, method = "parametric", exact = FALSE) ssize.reference.range(delta = 0.01, ref.prob = 0.95, conf.prob = 0.9, method = "nonparametric", exact = TRUE) ssize.reference.range(delta = 0.01, ref.prob = 0.95, conf.prob = 0.9, method = "nonparametric", exact = FALSE) ``` We can also calculate the sample size for one-sided reference ranges. ```{r} ssize.reference.range(delta = 0.015, ref.prob = 0.95, conf.prob = 0.9, method = "parametric", exact = TRUE, alternative = "one.sided") ``` ## Multiple Primary Endpoints (MPE) We demonstrate how to calculate the sample size for a trial with two co-primary endpoints with known covariance. ```{r} Sigma <- matrix(c(1, 0.8, 0.8, 1), nrow = 2) power.mpe.known.var(K = 2, delta = c(0.25, 0.4), Sigma = Sigma, sig.level = 0.025, power = 0.8) ## equivalent: specify SDs and correlation rho power.mpe.known.var(K = 2, delta = c(0.25, 0.4), SD = c(1,1), rho = 0.8, sig.level = 0.025, power = 0.8) ``` Next, we show how to calculate the sample size for a trial with two co-primary endpoints with unknown covariance. Here, we follow three steps to determine the sample size. + Step 1: As we need starting values for our algorithm that computes the sample size in this case, we first act as if the covariance would be known and compute the sample size by applying our function `power.mpe.known.var`. + Step 2: The resulting value of `n` is considered as lower bound for the sample size in case of unknown covariance and is used as `n.min` in function `power.mpe.unkown.var`. Moreover, we specify a reasonable `n.max`, which must be larger than `n.min`. + Step 3: Finally, by using the arguments from the step 2, we can compute the sample size for the situation with unknown covariance. ```{r} ## Step 1: power.mpe.known.var(K = 2, delta = c(0.5, 0.4), Sigma = Sigma, sig.level = 0.025, power = 0.8) ## Step 2 + 3: Sigma <- matrix(c(1, 0.5, 0.5, 1), nrow = 2) power.mpe.unknown.var(K = 2, delta = c(0.5, 0.4), Sigma = Sigma, sig.level = 0.025, power = 0.8, n.min = 105, n.max = 120) ## equivalent: specify SDs and correlation rho power.mpe.unknown.var(K = 2, delta = c(0.5, 0.4), SD = c(1,1), rho = 0.5, sig.level = 0.025, power = 0.8, n.min = 105, n.max = 120) ``` We finally demonstrate how to calculate the sample size for a trial with two primary endpoints with known covariance, where the trial is designed to find a significant difference for at least one endpoint. ```{r} Sigma <- matrix(c(1, 0.3, 0.3, 1), nrow = 2) power.mpe.atleast.one(K = 2, delta = c(0.2, 0.3), Sigma = Sigma, power = 0.8, sig.level = 0.025) ## equivalent: specify SDs and correlation rho power.mpe.atleast.one(K = 2, delta = c(0.2, 0.3), SD = c(1,1), rho = 0.3, power = 0.8, sig.level = 0.025) ``` ## Power and Type-I-Error Simulations There are quite some discussions and various proposals, which test is the best under which circumstances when we want to compare the location (mean, median) of two groups [@Fagerland2009; @Fagerland2012; @Sezer2017]. In addition, it is questioned whether pre-testing of assumptions is appropriate/useful at least from a practical point of view [@Zimmerman2004; @Rasch2011; @Rochon2012; @Hoekstra2012]. We provide functions to simulate power and type-I-error of classical [@student1908], Welch [@Welch1947] and Hsu [@Hsu1938] t-tests as well as of Wilcoxon-Mann-Whitney tests [@Wilcoxon1945; @Mann1947]. ```{r} ## functions to simulate the data ## group 1 rx <- function(n) rnorm(n, mean = 0, sd = 1) rx.H0 <- rx ## group 2 ry <- function(n) rnorm(n, mean = 3, sd = 3) ry.H0 <- function(n) rnorm(n, mean = 0, sd = 3) ## theoretical results power.welch.t.test(n = 10, delta = 3, sd1 = 1, sd2 = 3) power.hsu.t.test(n = 10, delta = 3, sd1 = 1, sd2 = 3) ## simulation res.t <- sim.power.t.test(nx = 10, rx = rx, rx.H0 = rx.H0, ny = 10, ry = ry, ry.H0 = ry.H0) res.t res.w <- sim.power.wilcox.test(nx = 10, rx = rx, rx.H0 = rx.H0, ny = 10, ry = ry, ry.H0 = ry.H0) res.w ``` For further investigation of the results, we provide some diagnostic plots. ```{r, fig.width=7, fig.height=7} ## t-tests hist(res.t) qqunif(res.t, alpha = 0.05) volcano(res.t, hex = TRUE) ## Wilcoxon-Mann-Whitney tests hist(res.w) qqunif(res.w, alpha = 0.05) ``` We can also generate a volcano plot for the Wilcoxon-Mann-Whitney test, but this would require setting argument conf.int to TRUE, which would lead to a much higher computation time, hence we omitted it here. Furthermore, it is also possible to compute an approximate version of the test by setting argument approximate to TRUE [@Hothorn2008] again by the cost of an increased computation time. ## sessionInfo ```{r} sessionInfo() ``` ## References