One Mean Sample

Equality

Hypothesis

The Null and Alternative hypotheses are \[ \begin{array}{l} H_{0}: \mu=\mu_{0} \\ H_{1}: \mu \neq \mu_{0} \end{array} \] Formulas compute sample size and power, respectively: \[n=\left(\sigma \frac{z_{1-\alpha / 2}+z_{1-\beta}}{\mu-\mu_{0}}\right)^{2}\] \[ 1-\beta=\Phi\left(z-z_{1-\alpha / 2}\right)+\Phi\left(-z-z_{1-\alpha / 2}\right) \quad, \quad z=\frac{\mu-\mu_{0}}{\sigma / \sqrt{n}} \]

Manual Calculation

mu=2
mu0=1.5
sd=1
alpha=0.05
beta=0.20
(n=(sd*(qnorm(1-alpha/2)+qnorm(1-beta))/(mu-mu0))^2)
ceiling(n) 

z=(mu-mu0)/sd*sqrt(n)
(Power=pnorm(z-qnorm(1-alpha/2))+pnorm(-z-qnorm(1-alpha/2)))

Non-Inferiority or Superiority

Hypothesis

\[ \begin{array}{l} H_{0}: \mu-\mu_{0} \leq \delta \\ H_{1}: \mu-\mu_{0}>\delta \end{array} \] Sample Size \[n=\left(\sigma \frac{z_{1-\alpha}+z_{1-\beta}}{\mu-\mu_{0}-\delta}\right)^{2}\] Power \[1-\beta=\Phi\left(z-z_{1-\alpha}\right)+\Phi\left(-z-z_{1-\alpha}\right) \quad, \quad z=\frac{\mu-\mu_{0}-\delta}{\sigma / \sqrt{n}}\]

Equivalence

Hypothesis

\[ \begin{array}{l} H_{0}:\left|\mu-\mu_{0}\right| \geq \delta \\ H_{1}:\left|\mu-\mu_{0}\right|<\delta \end{array} \] Sample Size \[n=\left(\sigma \frac{z_{1-\alpha}+z_{1-\beta / 2}}{\delta-\left|\mu-\mu_{0}\right|}\right)^{2}\] Power \[1-\beta=2\left[\Phi\left(z-z_{1-\alpha}\right)+\Phi\left(-z-z_{1-\alpha}\right)\right]-1 \quad, \quad z=\frac{\left|\mu-\mu_{0}\right|-\delta}{\sigma / \sqrt{n}}\]

Manual Calculation

mu=2
mu0=2
delta=0.05
sd=0.10
alpha=0.05
beta=0.20
(n=(sd*(qnorm(1-alpha)+qnorm(1-beta/2))/(delta-abs(mu-mu0)))^2)
ceiling(n) 


z=(abs(mu-mu0)-delta)/sd*sqrt(n)
(Power=2*(pnorm(z-qnorm(1-alpha))+pnorm(-z-qnorm(1-alpha)))-1)

Two Mean Samples

Z-Test

Hypothesis

Equality

Two Sided

\[ \begin{array}{l} H_{0}: \mu_{A}-\mu_{B}=0 \\ H_{1}: \mu_{A}-\mu_{B} \neq 0 \end{array} \] where the ratio between the sample sizes of the two groups is \[ \kappa=\frac{n_{A}}{n_{B}} \]

Sample size \[ n_{A}=\kappa n_{B} \] \[ n_{B}=\left(1+\frac{1}{\kappa}\right)\left(\sigma \frac{z_{1-\alpha / 2}+z_{1-\beta}}{\mu_{A}-\mu_{B}}\right)^{2} \] Power \[ 1-\beta=\Phi\left(z-z_{1-\alpha / 2}\right)+\Phi\left(-z-z_{1-\alpha / 2}\right) \quad, \quad z=\frac{\mu_{A}-\mu_{B}}{\sigma \sqrt{\frac{1}{n_{A}}+\frac{1}{n_{B}}}} \]

One Sided

\[ \begin{array}{l} H_{0}: \mu_{A}=\mu_{B} \\ H_{1}: \mu_{A}>\mu_{B} \end{array} \] where the ratio between the sample sizes of the two groups is \[ \kappa=\frac{n_{B}}{n_{A}} \] Sample size and power, respectively: \[ \begin{array}{c} n_{A}=\left(\sigma_{A}^{2}+\sigma_{B}^{2} / \kappa\right)\left(\frac{z_{1-\alpha}+z_{1-\beta}}{\mu_{A}-\mu_{B}}\right)^{2} \\ n_{B}=\kappa n_{A} \\ 1-\beta=\Phi\left(\frac{\left|\mu_{A}-\mu_{B}\right| \sqrt{n_{A}}}{\sqrt{\sigma_{A}^{2}+\sigma_{B}^{2} / \kappa}}-z_{1-\alpha}\right) \end{array} \]

Non-Inferiority or Superiority

\[ \begin{array}{l} H_{0}: \mu_{A}-\mu_{B} \leq \delta \\ H_{1}: \mu_{A}-\mu_{B}>\delta \end{array} \] where \(\delta\) is the superiority or non-inferiority margin and the ratio between the sample sizes of the two groups is \[ \kappa=\frac{n_{A}}{n_{B}} \] Formulas of sample size and power, respectively: \[ \begin{array}{c} n_{A}=\kappa n_{B} \\ n_{B}=\left(1+\frac{1}{\kappa}\right)\left(\sigma \frac{z_{1-\alpha}+z_{1-\beta}}{\mu_{A}-\mu_{B}-\delta}\right)^{2} \\ 1-\beta=\Phi\left(z-z_{1-\alpha}\right)+\Phi\left(-z-z_{1-\alpha}\right) \quad, \quad z=\frac{\mu_{A}-\mu_{B}-\delta}{\sigma \sqrt{\frac{1}{n_{A}}+\frac{1}{n_{B}}}} \end{array} \]

Equivalence (TOST)

\[ \begin{array}{l} H_{0}:\left|\mu_{A}-\mu_{B}\right| \geq \delta \\ H_{1}:\left|\mu_{A}-\mu_{B}\right|<\delta \end{array} \] where \(\delta\) is the superiority or non-inferiority margin and the ratio between the sample sizes of the two groups is \[ \kappa=\frac{n_{1}}{n_{2}} \] Formulas of sample size and power, respectively: \[ \begin{array}{c} n_{A}=\kappa n_{B} \\ n_{B}=\left(1+\frac{1}{\kappa}\right)\left(\sigma \frac{z_{1-\alpha}+z_{1-\beta / 2}}{\left|\mu_{A}-\mu_{B}\right|-\delta}\right)^{2} \\ 1-\beta=2\left[\Phi\left(z-z_{1-\alpha}\right)+\Phi\left(-z-z_{1-\alpha}\right)\right]-1 \quad, \quad z=\frac{\left|\mu_{A}-\mu_{B}\right|-\delta}{\sigma \sqrt{\frac{1}{n_{A}}+\frac{1}{n_{B}}}} \end{array} \]

muA=5
muB=4
delta=5
kappa=1
sd=10
alpha=0.05
beta=0.20
(nB=(1+1/kappa)*(sd*(qnorm(1-alpha)+qnorm(1-beta/2))/(abs(muA-muB)-delta))^2)
ceiling(nB)

z=(abs(muA-muB)-delta)/(sd*sqrt((1+1/kappa)/nB))
(Power=2*(pnorm(z-qnorm(1-alpha))+pnorm(-z-qnorm(1-alpha)))-1)

Calculation in SAS

***********************************************************************
* This is a program that illustrates the use of PROC POWER to         *
* calculate sample size when comparing two normal means in an         *
* equivalence trial.                                                  *
***********************************************************************;

proc power;
twosamplemeans dist=normal groupweights=(1 1) alpha=0.05 power=0.9 stddev=0.75 
   lower=-0.10 upper=0.10 meandiff=0.05 test=equiv_diff ntotal=.;
plot min=0.1 max=0.9;
title "Sample Size Calculation for Comparing Two Normal Means (1:1 Allocation)in an Equivalence Trial"; 
run;

Calculation in R

Equility (two-sided)

ssc.twosamplemean.ztest(design = 1L, ratio = 1, 
                        alpha = 0.05, power = 0.8, 
                        sd = 0.1, theta = 0.2)
## 
## Sample Size Calculation for Clinical Trials: Test for equilit 
## Alpha 0.05 
## Power 80 %
## Ratio between subjects in the Treatment Group and in the Control Group: 1 
## Standard deviation: 0.1 
## Mean difference between two arms: 0.2

Equility (one-sided)

ssc.twosamplemean.ztest(design = 1L, ratio = 1,
                        alpha = 0.025, power = 0.8, 
                        sd = 0.1, theta = 0.2)
## 
## Sample Size Calculation for Clinical Trials: Test for equilit 
## Alpha 0.025 
## Power 80 %
## Ratio between subjects in the Treatment Group and in the Control Group: 1 
## Standard deviation: 0.1 
## Mean difference between two arms: 0.2

Superioriry

ssc.twosamplemean.ztest(design = 2L, ratio = 1, 
                        alpha = 0.05, power = 0.8, 
                        sd = 0.1, theta = 0, delta =  0.05)
## 
## Sample Size Calculation for Clinical Trials: Test for superiority 
## Alpha 0.05 
## Power 80 %
## Ratio between subjects in the Treatment Group and in the Control Group: 1 
## Standard deviation: 0.1 
## Mean difference between two arms: 0 
## Superiority Margin 0.05

Non-Inferiority

ssc.twosamplemean.ztest(design = 3L, ratio = 1,
                        alpha = 0.05, power = 0.8, 
                        sd = 0.1, theta = 0, delta = -0.05)
## 
## Sample Size Calculation for Clinical Trials: Test for non-inferiority 
## Alpha 0.05 
## Power 80 %
## Ratio between subjects in the Treatment Group and in the Control Group: 1 
## Standard deviation: 0.1 
## Mean difference between two arms: 0 
## Non-inferiority margin -0.05

Equivalence

ssc.twosamplemean.ztest(design = 4L, ratio = 1, 
                        alpha = 0.05, power = 0.8, 
                        sd = 0.1, theta = 0, delta =  0.05)
## 
## Sample Size Calculation for Clinical Trials: Test for equivalence 
## Alpha 0.05 
## Power 80 %
## Ratio between subjects in the Treatment Group and in the Control Group: 1 
## Standard deviation: 0.1 
## Mean difference between two arms: 0 
## Equivalence margin 0.05

T-Test

Calculation in R

  • pwr.t.test: Power calculations for t-tests of means (one sample, two samples and paired samples)
  • pwr.t2n.test: Power calculations for two samples (different sizes) t-tests of means
  • pwrss.t.2means: Difference between Two Means (t or z Test for Independent or Paired Samples)
pwr.t.test(n = NULL, d = NULL, sig.level = 0.05, power = NULL, 
    type = c("two.sample"),
    alternative = c("two.sided", "less", "greater"))
    
pwr.t2n.test(n1 = NULL, n2= NULL, d = NULL, 
             sig.level = 0.05, power = NULL,
             alternative = c("two.sided", "less","greater"))

Equility

TTest <- function(alpha,mean,std,power,side){
  d <- mean/std
  # d: Effect size (Cohen's d) - difference between the means divided by the pooled standard deviation
  samplesize <- pwr.t.test(d=d, 
                           power=power, 
                           sig.level=alpha, 
                           type="two.sample",
                           alternative=side)
  CI.Left  <- mean-qnorm(1-alpha/2)*std/sqrt(samplesize$n)
  CI.Right <- mean+qnorm(1-alpha/2)*std/sqrt(samplesize$n)
  CI <- paste("[",round(CI.Left,3), ",", round(CI.Right,3), "]", 
              sep = "",collapse = NULL)
  results <- data.frame(alpha = alpha,
                      mean = mean,
                      sd = std,
                      power = samplesize$power,
                      side = samplesize$alternative,
                      n = samplesize$n,
                      CI = CI)
  return(results)
}

TTest(alpha = 0.05, mean = 0.42, std = 0.7, power = 0.8, side = "two.sided")
pwr.t.test(d=0.42/0.7, 
           power=0.8, 
           sig.level=0.05, 
           type="two.sample", 
           alternative="two.sided")
## 
##      Two-sample t test power calculation 
## 
##               n = 44.58577
##               d = 0.6
##       sig.level = 0.05
##           power = 0.8
##     alternative = two.sided
## 
## NOTE: n is number in *each* group
## 0.7 is pooled variance
## pwrss.t.2means
ssc.twosamplemean.ttest(mu1 = 29.42, mu2 = 29, sd1 = 0.7, kappa = 1,
                        power = .80, alpha = 0.05,
                        alternative = "not equal")
##  Difference between Two means 
##  (Independent Samples t Test) 
##  H0: mu1 = mu2 
##  HA: mu1 != mu2 
##  ------------------------------ 
##   Statistical power = 0.8 
##   n1 = 45 
##   n2 = 45 
##  ------------------------------ 
##  Alternative = "not equal" 
##  Degrees of freedom = 88 
##  Non-centrality parameter = 2.846 
##  Type I error rate = 0.05 
##  Type II error rate = 0.2
## 0.42/0.7 as Cohen'd
## pwrss.t.2means
ssc.twosamplemean.ttest(mu1 = 0.42/0.7, kappa = 1,
                        power = .80, alpha = 0.05,
                        alternative = "not equal")
##  Difference between Two means 
##  (Independent Samples t Test) 
##  H0: mu1 = mu2 
##  HA: mu1 != mu2 
##  ------------------------------ 
##   Statistical power = 0.8 
##   n1 = 45 
##   n2 = 45 
##  ------------------------------ 
##  Alternative = "not equal" 
##  Degrees of freedom = 88 
##  Non-centrality parameter = 2.846 
##  Type I error rate = 0.05 
##  Type II error rate = 0.2
## pwrss.t.2means
ssc.twosamplemean.ttest(mu1 = 29.42, mu2 = 29, sd1 = 0.7, kappa = 1,
                        power = .80, alpha = 0.025,
                        alternative = "greater")
##  Difference between Two means 
##  (Independent Samples t Test) 
##  H0: mu1 = mu2 
##  HA: mu1 > mu2 
##  ------------------------------ 
##   Statistical power = 0.8 
##   n1 = 45 
##   n2 = 45 
##  ------------------------------ 
##  Alternative = "greater" 
##  Degrees of freedom = 88 
##  Non-centrality parameter = 2.846 
##  Type I error rate = 0.025 
##  Type II error rate = 0.2

Superioriry

## Cohen'd = 0.10
pwrss.t.2means(mu1 = 0.25, mu2 = 0.15,
                margin = 0.05, power = 0.80,
                alternative = "superior")
##  Difference between Two means 
##  (Independent Samples t Test) 
##  H0: mu1 - mu2 <= margin 
##  HA: mu1 - mu2 > margin 
##  ------------------------------ 
##   Statistical power = 0.8 
##   n1 = 4947 
##   n2 = 4947 
##  ------------------------------ 
##  Alternative = "superior" 
##  Degrees of freedom = 9892 
##  Non-centrality parameter = 2.487 
##  Type I error rate = 0.05 
##  Type II error rate = 0.2

Non-Inferiority

## Cohen'd = 0.10
pwrss.t.2means(mu1 = 0.25, mu2 = 0.15,
                margin = -0.05, power = 0.80,
                alternative = "non-inferior")
##  Difference between Two means 
##  (Independent Samples t Test) 
##  H0: mu1 - mu2 <= margin 
##  HA: mu1 - mu2 > margin 
##  ------------------------------ 
##   Statistical power = 0.8 
##   n1 = 551 
##   n2 = 551 
##  ------------------------------ 
##  Alternative = "non-inferior" 
##  Degrees of freedom = 1100 
##  Non-centrality parameter = 2.49 
##  Type I error rate = 0.05 
##  Type II error rate = 0.2

Equivalence

## Cohen'd = 0
pwrss.t.2means(mu1 = 0.25, mu2 = 0.25,
                margin = 0.05, power = 0.80,
                alternative = "equivalent")
##  Difference between Two means 
##  (Independent Samples t Test) 
##  H0: |mu1 - mu2| >= margin 
##  HA: |mu1 - mu2| < margin 
##  ------------------------------ 
##   Statistical power = 0.8 
##   n1 = 6852 
##   n2 = 6852 
##  ------------------------------ 
##  Alternative = "equivalent" 
##  Degrees of freedom = 13702 
##  Non-centrality parameter = -2.927 
##  Type I error rate = 0.05 
##  Type II error rate = 0.2

T-Test (paired)

pwr.t.test(n = NULL, d = NULL, sig.level = 0.05, power = NULL, 
    type = c("paired"),
    alternative = c("two.sided", "less", "greater"))

Non-Inferiority for Ratio of Two Means

Hypothesis

Suppose you are designing a study hoping to show that a new (less expensive) manufacturing process does not produce appreciably more pollution than the current process. Quantifying “appreciably worse” as \(10 \%\), you seek to show that the mean pollutant level from the new process is less than \(110 \%\) of that from the current process. In standard hypothesis testing notation, you seek to reject \[ H_0: \frac{\mu_{\text {new }}}{\mu_{\text {current }}} \geq 1.10 \] in favor of \[ H_A: \frac{\mu_{\text {new }}}{\mu_{\text {current }}}<1.10 \]

An appropriate test for this situation is the common two-group \(t\) test on log-transformed data. The hypotheses become \[ \begin{aligned} H_0: \log \left(\mu_{\text {new }}\right)-\log \left(\mu_{\text {current }}\right) & \geq \log (1.10) \\ H_A: \log \left(\mu_{\text {new }}\right)-\log \left(\mu_{\text {current }}\right) & <\log (1.10) \end{aligned} \] Measurements of the pollutant level will be taken by using laboratory models of the two processes and will be treated as independent lognormal observations with a coefficient of variation \((\sigma / \mu)\) between 0.5 and 0.6 for both processes. You will end up with 300 measurements for the current process and 180 for the new one. It is important to avoid a Type I error here, so you set the Type I error rate to 0.01 . Your theoretical work suggests that the new process will actually reduce the pollutant by about \(10 \%\) (to \(90 \%\) of current), but you need to compute and graph the power of the study if the new levels are actually between \(70 \%\) and \(120 \%\) of current levels.

Calculation in SAS

reference document

ods graphics on;

proc power;
   twosamplemeans test=ratio
      meanratio = 0.7 to 1.2 by 0.1
      nullratio = 1.10
      sides     = L
      alpha     = 0.01
      cv        = 0.5 0.6
      groupns   = (300 180)
      power     = .;
   plot x=effect step=0.05;
run;

ods graphics off;

Cross Over Design

Supported Designs of Package PowerTOST

Designs: Codes of designs follow this pattern: treatments x sequences x periods
design name df
parallel 2 parallel groups n-2
2x2 2x2 crossover n-2
2x2x2 2x2x2 crossover n-2
3x3 3x3 crossover 2*n-4
3x6x3 3x6x3 crossover 2*n-4
4x4 4x4 crossover 3*n-6
2x2x3 2x2x3 replicate crossover 2*n-3
2x2x4 2x2x4 replicate crossover 3*n-4
2x4x4 2x4x4 replicate crossover 3*n-4
2x3x3 partial replicate (2x3x3) 2*n-3
2x4x2 Balaam’s (2x4x2) n-2
2x2x2r Liu’s 2x2x2 repeated x-over 3*n-2
paired paired means n-1

For various methods power can be calculated based on

  • nominal α, coefficient of variation (CV), deviation of test from reference (θ0), acceptance limits {θ1, θ2}, sample size (n), and design.

For all methods the sample size can be estimated based on

  • nominal α, coefficient of variation (CV), deviation of test from reference (θ0), acceptance limits {θ1, θ2}, target (i.e., desired) power, and design.

By Defaults

Parameter Argument Purpose Default
\(\small{\alpha}\) alpha Nominal level of the test 0.05
\(\small{\pi}\) targetpower Minimum desired power 0.80
logscale logscale Analysis on log-transformed or original scale? TRUE
\(\small{\theta_0}\) theta0 ‘True’ or assumed deviation of T from R see below
\(\small{\theta_1}\) theta1 Lower BE limit see below
\(\small{\theta_2}\) theta2 Upper BE limit see below
CV CV CV none
design design Planned design "2x2"
method method Algorithm "exact"
robust robust ‘Robust’ evaluation (Senn’s basic estimator) FALSE
print print Show information in the console? TRUE
details details Show details of the sample size search? FALSE
imax imax Maximum number of iterations 100

Defaults depending on the argument logscale:

Parameter Argument logscale = TRUE logscale = FALSE
\(\small{\theta_0}\) theta0 0.95 +0.05
\(\small{\theta_1}\) theta1 0.80 −0.20
\(\small{\theta_2}\) theta2 1.25 +0.20 
power.TOST(CV = 0.35, n = c(52, 49), design = "parallel")
## [1] 0.8011186
sampleN.TOST(CV = 0.30, details = FALSE, print = FALSE)[["Sample size"]]
## [1] 40

Note that sampleN.TOST() is not vectorized. If we are interested in combinations of assumed values:

sampleN.TOST.vectorized <- function(CVs, theta0s, ...) {
  n <- power <- matrix(ncol = length(CVs), nrow = length(theta0s))
  for (i in seq_along(theta0s)) {
    for (j in seq_along(CVs)) {
      tmp         <- sampleN.TOST(CV = CVs[j], theta0 = theta0s[i], ...)
      n[i, j]     <- tmp[["Sample size"]]
      power[i, j] <- tmp[["Achieved power"]]
    }
  }
  DecPlaces <- function(x) match(TRUE, round(x, 1:15) == x)
  fmt.col <- paste0("CV %.",    max(sapply(CVs, FUN = DecPlaces),
                                    na.rm = TRUE), "f")
  fmt.row <- paste0("theta %.", max(sapply(theta0s, FUN = DecPlaces),
                                    na.rm = TRUE), "f")
  colnames(power) <- colnames(n) <- sprintf(fmt.col, CVs)
  rownames(power) <- rownames(n) <- sprintf(fmt.row, theta0s)
  res <- list(n = n, power = power)
  return(res)
}
CVs     <- seq(0.20, 0.40, 0.05)
theta0s <- seq(0.90, 0.95, 0.01)
x       <- sampleN.TOST.vectorized(CV = CVs, theta0 = theta0s,
                                   details = FALSE, print = FALSE)
cat("Sample size\n"); print(x$n); cat("Achieved power\n"); print(signif(x$power, digits = 5))
## Sample size
##            CV 0.20 CV 0.25 CV 0.30 CV 0.35 CV 0.40
## theta 0.90      38      56      80     106     134
## theta 0.91      32      48      66      88     112
## theta 0.92      28      40      56      76      96
## theta 0.93      24      36      50      66      84
## theta 0.94      22      32      44      58      74
## theta 0.95      20      28      40      52      66
## Achieved power
##            CV 0.20 CV 0.25 CV 0.30 CV 0.35 CV 0.40
## theta 0.90 0.81549 0.80358 0.80801 0.80541 0.80088
## theta 0.91 0.81537 0.81070 0.80217 0.80212 0.80016
## theta 0.92 0.82274 0.80173 0.80021 0.80678 0.80238
## theta 0.93 0.81729 0.81486 0.81102 0.80807 0.80655
## theta 0.94 0.83063 0.81796 0.81096 0.80781 0.80740
## theta 0.95 0.83468 0.80744 0.81585 0.80747 0.80525

k Mean Samples

One-Way ANOVA

Hypothesis

One-Way ANOVA Pairwise, 2-Sided Equality using Bonferroni Adjustment

In more general terms, we may have \(k\) groups, meaning there are a total of \(K \equiv\left(\begin{array}{c}k \\ 2\end{array}\right)=k(k-1) / 2\) possible pairwise comparisons. When we test \(\tau \leq K\) of these pairwise comparisons, we have \(\tau\) hypotheses of the form \[ \begin{array}{l} H_{0}: \mu_{A}=\mu_{B} \\ H_{1}: \mu_{A} \neq \mu_{B} \end{array} \] where \(\mu_{A}\) and \(\mu_{B}\) represent the means of two of the \(k\) groups, groups ‘A’ and ‘B’. We’ll compute the required sample size for each of the \(\tau\) comparisons, and total sample size needed is the largest of these. In the formula below, \(n\) represents the sample size in any one of these \(\tau\) comparisons; that is, there are \(n / 2\) people in the ‘A’ group, and \(n / 2\) people in the ‘B’ group.

Formulas to compute sample size and power, respectively: \[ \begin{array}{c} n=2\left(\sigma \frac{z_{1-\alpha /(2 \tau)}+z_{1-\beta}}{\mu_{A}-\mu_{B}}\right)^{2} \\ 1-\beta=\Phi\left(z-z_{1-\alpha /(2 \tau)}\right)+\Phi\left(-z-z_{1-\alpha /(2 \tau)}\right) \quad, \quad z=\frac{\mu_{A}-\mu_{B}}{\sigma \sqrt{\frac{2}{n}}} \end{array} \]

See more under multiple tests.

Calculation in R

Perform power analysis on balanced one-way analysis of variance, where k is the number of groups, and n is the sample size in each group.

Now do a one-way analysis of variance for the five groups to achieve a power of 0.8, an effect value of 0.25, and a significance level of 0.05 to calculate the sample size required for each group

Where effect value is calculated using \[f=\sqrt{\frac{\sum_{i=1}^k p_i \times (\mu_i - \mu)^2}{\sigma^2}}\]

pwr.anova.test(k = 5, f = 0.25, sig.level = 0.05, power = 0.8)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 5
##               n = 39.1534
##               f = 0.25
##       sig.level = 0.05
##           power = 0.8
## 
## NOTE: n is number in each group
# Sample sizes for detecting significant effects in a One-Way ANOVA

es <- seq(0.1, 0.5, 0.01)
nes <- length(es)
samsize <- NULL
for (i in 1:nes) {
    result <- pwr.anova.test(k = 5, f = es[i], 
        sig.level = 0.05, 
        power = 0.9)
    samsize[i] <- ceiling(result$n)
}
plot(samsize, es, type = "l", lwd = 2, col = "red", 
    ylab = "Effect Size", xlab = "Sample Size (per cell)", 
    main = "One Way ANOVA with Power=.90 and Alpha=.05")