This vignette replicates several of the examples found in the G*Power manual (version 3.1). It is not meant to be exhaustive but instead demonstrates how power analyses could be computed and extended using simulation methodology by either editing the default functions found within the package or by creating a new user-defined function for yet-to-be-defined statistical analysis contexts.

Correlation

Power associated with the hypotheses \[H_0:\, \rho-\rho_0=0\] \[H_1:\, \rho-\rho_0\ne 0\]

where \(\rho\) is the population correlation and \(\rho_0\) the null hypothesis constant.

Example 3.3; Difference from constant (one sample case)

Sample size estimate to reject \(H_0:\, \rho_0=.60\) in correlation analysis with \(1-\beta=.95\) probability when \(\rho=.65\).

p_r(n = NA, r = .65, rho = .60) |> 
    Spower(power = .95, interval=c(500,3000))

## 
## Design conditions: 
## 
## # A tibble: 1 × 5
##       n     r   rho sig.level power
##   <dbl> <dbl> <dbl>     <dbl> <dbl>
## 1    NA  0.65   0.6      0.05  0.95
## 
## Estimate of n: 1931.4
## 95% Prediction Interval: [1901.1, 1958.2]

G*power estimates \(n\) to be 1929 using the same Fisher z-transformation approximation.

Test against constant \(\rho_0=0\)

Unlike the previous section, this is the more canonical version of the hypotheses involving correlation coefficients. Power associated with \(\rho = .3\) with 100 pairs of observations, tested against \(\rho_0=0\).

p_r(n = 100, r = .3) |> Spower()

## 
## Design conditions: 
## 
## # A tibble: 1 × 4
##       n     r sig.level power
##   <dbl> <dbl>     <dbl> <lgl>
## 1   100   0.3      0.05 NA   
## 
## Estimate of power: 0.861
## 95% Confidence Interval: [0.854, 0.867]

Sample size estimate to reject \(H_0:\, \rho_0=0\) in correlation analysis with \(1-\beta=.95\) probability when \(\rho=.3\).

p_r(n = NA, r = .3) |> 
    Spower(power = .95, interval=c(50,1000))

## 
## Design conditions: 
## 
## # A tibble: 1 × 4
##       n     r sig.level power
##   <dbl> <dbl>     <dbl> <dbl>
## 1    NA   0.3      0.05  0.95
## 
## Estimate of n: 138.3
## 95% Prediction Interval: [136.1, 140.4]

Compare to approximate result from pwr package.

pwr::pwr.r.test(r=.3, power=.95, n=NULL)

## 
##      approximate correlation power calculation (arctangh transformation) 
## 
##               n = 137.8
##               r = 0.3
##       sig.level = 0.05
##           power = 0.95
##     alternative = two.sided

Example 27.3; Correlation - inequality of two independent Pearson r’s

n <- 206
n2_n1 <- 51/n
p_2r(n=n, r.ab1=.75, r.ab2=.88, n2_n1=n2_n1) |> Spower()

## 
## Design conditions: 
## 
## # A tibble: 1 × 4
##   r.ab1 r.ab2 sig.level power
##   <dbl> <dbl>     <dbl> <lgl>
## 1  0.75  0.88      0.05 NA   
## 
## Estimate of power: 0.727
## 95% Confidence Interval: [0.718, 0.735]

G*power gives power of .726.

Example 16.3; Point-biserial correlation

# solution per group
out <- p_t.test(r = .25, n = NA, two.tailed=FALSE) |> 
    Spower(power = .95, interval=c(50, 200))
out

## 
## Design conditions: 
## 
## # A tibble: 1 × 5
##       n     r two.tailed sig.level power
##   <dbl> <dbl> <lgl>          <dbl> <dbl>
## 1    NA  0.25 FALSE           0.05  0.95
## 
## Estimate of n: 81.1
## 95% Prediction Interval: [79.4, 82.7]

# total sample size required
ceiling(out$n) * 2

## [1] 164

G*power gives the result \(N=164\).

Relatedly, one can specify \(d\), Cohen’s standardized mean-difference effect size, instead of \(r\) since \(d\) is easily converted to \(r\).

Example 31.3; tetrachoric correlation

F <- matrix(c(203, 186, 167, 374), 2, 2)
N <- sum(F)
(marginal.x <- colSums(F)/N)

## [1] 0.4183 0.5817

(marginal.y <- rowSums(F)/N)

## [1] 0.3978 0.6022

# converted to intercepts
tauX <- qnorm(marginal.x)[2]
tauY <- qnorm(marginal.y)[2]
c(tauX, tauY)

## [1] 0.2063 0.2589

G*power gives \(n=463\), though uses the SE value at the null (Score test). p_r.cat(), on the other hand, defaults to the Wald approach where the SE is at the MLE. To switch, use score=FALSE, though note that this requires twice as many computations.

p_r.cat(n=NA, r=0.2399846, tauX=tauX, tauY=tauY, 
        score=FALSE, two.tailed=FALSE) |> 
    Spower(power = .95, interval=c(100, 500), 
           parallel=TRUE, check.interval=FALSE)

## 
## Design conditions: 
## 
## # A tibble: 1 × 8
##       n     r  tauX  tauY score two.tailed sig.level power
##   <dbl> <dbl> <dbl> <dbl> <lgl> <lgl>          <dbl> <dbl>
## 1    NA 0.240 0.206 0.259 FALSE FALSE           0.05  0.95
## 
## Estimate of n: 462.9
## 95% Prediction Interval: [458.5, 466.6]

Proportions

Example 4.3; One sample proportion tests

One sample, one-tailed proportions test given data generated from a population tested against \(\pi_0 = .65\) with \(g=.15\) (hence, \(\pi = .65+g=.80\)) with \(n=20\).

pi <- .65
g <- .15
p <- pi + g

p_prop.test(n=20, prop=p, pi=pi, two.tailed=FALSE) |>
    Spower()

## 
## Design conditions: 
## 
## # A tibble: 1 × 4
##       n two.tailed sig.level power
##   <dbl> <lgl>          <dbl> <lgl>
## 1    20 FALSE           0.05 NA   
## 
## Estimate of power: 0.423
## 95% Confidence Interval: [0.413, 0.432]

G*power gives the estimate \(1-\beta=.4112\).

Fisher’s exact test is also supported by using the argument exact = TRUE.

# Fisher exact test
p_prop.test(n=20, prop=p, pi=pi, exact=TRUE, 
            two.tailed=FALSE) |> Spower()

## 
## Design conditions: 
## 
## # A tibble: 1 × 5
##       n two.tailed exact sig.level power
##   <dbl> <lgl>      <lgl>     <dbl> <lgl>
## 1    20 FALSE      TRUE       0.05 NA   
## 
## Estimate of power: 0.415
## 95% Confidence Interval: [0.405, 0.425]

Example 22.1; Sign test

Standard Normal/Gaussian as the parent distribution (default).

p_wilcox.test(n=649, d=.1, type='one.sample', two.tailed=FALSE) |> 
    Spower()

## 
## Design conditions: 
## 
## # A tibble: 1 × 6
##       n     d type       two.tailed sig.level power
##   <dbl> <dbl> <chr>      <lgl>          <dbl> <lgl>
## 1   649   0.1 one.sample FALSE           0.05 NA   
## 
## Estimate of power: 0.794
## 95% Confidence Interval: [0.786, 0.802]

G*power gives the power estimate of .800.

The following partially recreates the simulation results in Figure 29 (which itself was partially extracted from Shieh, Jan, and Randles, 2007) for the Gaussian(0,1) distribution with varying sample sizes and effect sizes. The target was to obtain “approximate power of \(1-\beta = .80\)”, though how these sample sizes were decided upon was not specified. Spower()’s stochastic root-solving approach would likely get closer to more optimal \(N\) estimates were these the target of the analyses.

combos <- data.frame(n=c(649, 164, 42, 20, 12, 9),
                     d=c(.1, .2, .4, .6, .8, 1.0))

out <- lapply(1:nrow(combos), \(i){
    p_wilcox.test(n=combos$n[i], d=combos$d[i], 
                  type='one.sample', two.tailed=FALSE) |> Spower(replications = 50000)
})

# For Gaussian(0,1)
combos$power <- sapply(out, \(x) x$power)
combos

##     n   d  power
## 1 649 0.1 0.7983
## 2 164 0.2 0.8014
## 3  42 0.4 0.7966
## 4  20 0.6 0.8096
## 5  12 0.8 0.8026
## 6   9 1.0 0.8461

Laplace(0,1) version

A one-sample sign test with Laplace distribution as the parent:

library(VGAM)

# generate data with scale 0-1 for d effect size to be same as mean
# VAR = 2*b^2, so scale should be 1 = 2*b^2 -> sqrt(1/2)
parent <- function(n, d, scale=sqrt(1/2)) 
    VGAM::rlaplace(n, d, scale=scale)

p_wilcox.test(n=11, d=.8, parent1=parent, type='one.sample',
              two.tailed=FALSE, correct = FALSE) |> Spower()

## 
## Design conditions: 
## 
## # A tibble: 1 × 7
##       n     d type       correct two.tailed sig.level power
##   <dbl> <dbl> <chr>      <lgl>   <lgl>          <dbl> <lgl>
## 1    11   0.8 one.sample FALSE   FALSE           0.05 NA   
## 
## Estimate of power: 0.797
## 95% Confidence Interval: [0.789, 0.805]

G*power gives the estimate .830, which seems somewhat high (see below).

The following partially recreates the simulation results in Figure 29 for the Laplace(0,1) distribution with varying sample sizes and effect sizes. The target was to obtain “approximate power of \(1-\beta = .80\)”, though how these sample sizes were decided upon was not specified. Spower()’s stochastic root-solving approach would likely get closer to more optimal \(N\) estimates were these the target of the analyses.

combos <- data.frame(n=c(419, 109, 31, 16, 11, 8),
                          d=c(.1, .2, .4, .6, .8, 1.0))

out <- lapply(1:nrow(combos), \(i){
    p_wilcox.test(n=combos$n[i], d=combos$d[i], parent1=parent, type='one.sample',
              two.tailed=FALSE, correct = FALSE) |> Spower(replications = 50000)
})

# For Laplace(0,1)
combos$power <- sapply(out, \(x) x$power)
combos

##     n   d  power
## 1 419 0.1 0.8017
## 2 109 0.2 0.8012
## 3  31 0.4 0.8027
## 4  16 0.6 0.7979
## 5  11 0.8 0.8043
## 6   8 1.0 0.7738

Example 5.3; Two dependent proportions test (McNemar’s test)

obrien2002 <- matrix(c(.54, .32, .08, .06), 2, 2,
                     dimnames = list('Treatment' = c('Yes', 'No'),
                                    'Standard' = c('Yes', 'No')))
obrien2002

##          Standard
## Treatment  Yes   No
##       Yes 0.54 0.08
##       No  0.32 0.06

p_mcnemar.test(n=50, prop=obrien2002, two.tailed=FALSE) |>
        Spower(replications=30000)

## 
## Design conditions: 
## 
## # A tibble: 1 × 4
##       n two.tailed sig.level power
##   <dbl> <lgl>          <dbl> <lgl>
## 1    50 FALSE           0.05 NA   
## 
## Estimate of power: 0.837
## 95% Confidence Interval: [0.833, 0.841]

G*Power gives .839 (\(\alpha = .032\)). Slightly more power can be achieved when not using the continuity correction, though in general this is not recommended in practice.

p_mcnemar.test(n=50, prop=obrien2002, two.tailed=FALSE, correct=FALSE) |> 
    Spower()

## 
## Design conditions: 
## 
## # A tibble: 1 × 5
##       n two.tailed correct sig.level power
##   <dbl> <lgl>      <lgl>       <dbl> <lgl>
## 1    50 FALSE      FALSE        0.05 NA   
## 
## Estimate of power: 0.888
## 95% Confidence Interval: [0.882, 0.894]

Multiple Linear Regression (Random IVs)

Example 7.3

p_lm.R2(n=95, R2=.1, k=5, fixed=FALSE) |> Spower()

## 
## Design conditions: 
## 
## # A tibble: 1 × 6
##       n    R2     k fixed sig.level power
##   <dbl> <dbl> <dbl> <lgl>     <dbl> <lgl>
## 1    95   0.1     5 FALSE      0.05 NA   
## 
## Estimate of power: 0.660
## 95% Confidence Interval: [0.651, 0.669]

G*power gives 0.662 using a one-tailed test criterion.

Multiple Linear Regression (Fixed IVs)

Example 13.1

p_lm.R2(n=95, R2=.1, k=5) |> Spower()

## 
## Design conditions: 
## 
## # A tibble: 1 × 5
##       n    R2     k sig.level power
##   <dbl> <dbl> <dbl>     <dbl> <lgl>
## 1    95   0.1     5      0.05 NA   
## 
## Estimate of power: 0.670
## 95% Confidence Interval: [0.661, 0.680]

G*power gives \(1-\beta = .673\).

Example 14.3

Note that k is total IVs, k.R2_0 is number of IVs for baseline model.

p_lm.R2(n=90, R2=.3, k=9, R2_0=.25, k.R2_0=5) |> Spower(sig.level=.01)

## 
## Design conditions: 
## 
## # A tibble: 1 × 7
##       n    R2     k  R2_0 k.R2_0 sig.level power
##   <dbl> <dbl> <dbl> <dbl>  <dbl>     <dbl> <lgl>
## 1    90   0.3     9  0.25      5      0.01 NA   
## 
## Estimate of power: 0.235
## 95% Confidence Interval: [0.227, 0.243]

G*power gives \(1-\beta = .241\). Solving the sample size to achieve 80% power

p_lm.R2(n=NA, R2=.3, R2_0 = .25, k=9, k.R2_0=5) |> 
        Spower(sig.level=.01, power=.8, interval=c(100, 400))

## 
## Design conditions: 
## 
## # A tibble: 1 × 7
##       n    R2     k  R2_0 k.R2_0 sig.level power
##   <dbl> <dbl> <dbl> <dbl>  <dbl>     <dbl> <dbl>
## 1    NA   0.3     9  0.25      5      0.01   0.8
## 
## Estimate of n: 242.4
## 95% Prediction Interval: [240.9, 244.0]

G*power gives \(n = 242\).

Example 14.3b

Compare model with 12 IVs to model with 9 IVs.

p_lm.R2(n=200, R2=.16, R2_0 = .1, k=12, k.R2_0=9, R2.resid=.8) |> 
    Spower(sig.level=.01)

## 
## Design conditions: 
## 
## # A tibble: 1 × 8
##       n    R2     k  R2_0 k.R2_0 R2.resid sig.level power
##   <dbl> <dbl> <dbl> <dbl>  <dbl>    <dbl>     <dbl> <lgl>
## 1   200  0.16    12   0.1      9      0.8      0.01 NA   
## 
## Estimate of power: 0.765
## 95% Confidence Interval: [0.756, 0.773]

G*power gives \(1-\beta = .767\).

Fixed effects ANOVA - One way (F-test)

Example 10.3

p_anova.test(n=NA, k=10, f=.25) |>  Spower(power=.95, interval=c(20, 300))

## 
## Design conditions: 
## 
## # A tibble: 1 × 5
##       n     k     f sig.level power
##   <dbl> <dbl> <dbl>     <dbl> <dbl>
## 1    NA    10  0.25      0.05  0.95
## 
## Estimate of n: 38.5
## 95% Prediction Interval: [37.7, 39.5]

G*power gives the estimate \(n=39\).

Fixing \(n=200\) in total (hence, \(n=200/k=20\)) and performing a compromise analysis assuming \(q=\frac{\beta}{\alpha}=1\),

p_anova.test(n=20, k=10, f=.25) |> Spower(beta_alpha=1, replications=30000)

## 
## Design conditions: 
## 
## # A tibble: 1 × 6
##       n     k     f sig.level power beta_alpha
##   <dbl> <dbl> <dbl>     <dbl> <lgl>      <dbl>
## 1    20    10  0.25        NA NA             1
## 
## Estimate of Type I error rate (alpha/sig.level): 0.159
## 95% Confidence Interval: [0.155, 0.163]
## 
## Estimate of power (1-beta): 0.841
## 95% Confidence Interval: [0.837, 0.845]

G*Power gives \(\alpha=\beta=0.159\).

\(t\)-test: Linear regression (two groups)

Test coefficients across distinct datasets with similar form. In this case

\[Y_1 = \beta_0 + \beta_1 X_1 + \epsilon\] \[Y_2 = \beta_0^* + \beta_1^* X_2 + \epsilon\]

where the null of interest is

\[H_0:\, \beta_1 - \beta_1^* = 0\]

To do this a multiple linear regression model is setup with three variables

\[Y = \beta_0 + \beta_1 X + \beta_2 S + \beta_3 (S\cdot X) + \epsilon\] where \(Y=[Y_1, Y_2]\), \(X = [X_1, X_2]\), and \(S\) is a binary indicator variable indicating whether the observations were in the second sample.

When \(S = 0\) the first group’s parameterization will be recovered, while when \(S=1\) the second group’s parameterization will be recovered as the potentially non-zero \(\beta_2\) reflects a change in the intercept (\(\beta_0^* = \beta_0 + \beta_2\)) while the change in the slope for the second group will be reflected by the \(\beta_3\) (\(\beta_1^*=\beta_1 + \beta_3\)). Hence, the null hypothesis that the two groups have the same slope can be evaluated using this augmented model by testing

\[H_0:\, \beta_3 = 0\]

Example 17.3

We start by defining the population generating model to replace the gen_glm() function that is the default in p_glm(). This generating function is organized such that a data.frame is returned with the columns y, X, and S, where the interaction effect reflects the magnitude of the difference between the \(\beta\) coefficients across the independent samples.

gen_twogroup <- function(n, dbeta, sdx1, sdx2, sigma, n2_n1 = 1, ...){
    X1 <- rnorm(n, sd=sdx1)
    X2 <- rnorm(n*n2_n1, sd=sdx2)
    X <- c(X1, X2)
    N <- length(X)
    S <- c(rep(0, n), rep(1, N-n))
    y <- dbeta * X*S + rnorm(N, sd=sigma)
    dat <- data.frame(y, X, S)
    dat
}

To demonstrate, the post-hoc power for the described example in G*Power is the following.

p_glm(formula=y~X*S, test="X:S = 0",
      n=28, n2_n1=44/28, sdx1=9.02914, sdx2=11.86779, dbeta=0.01592,
      sigma=0.5578413, gen_fun=gen_twogroup) |> Spower()

## 
## Design conditions: 
## 
## # A tibble: 1 × 8
##   test      sigma     n   sdx1   sdx2   dbeta sig.level power
##   <chr>     <dbl> <dbl>  <dbl>  <dbl>   <dbl>     <dbl> <lgl>
## 1 X:S = 0 0.55784    28 9.0291 11.868 0.01592      0.05 NA   
## 
## Estimate of power: 0.195
## 95% Confidence Interval: [0.187, 0.202]

For the a priori power analysis to achieve a power of .80

p_glm(formula=y~X*S, test="X:S = 0",
      n=NA, n2_n1=44/28, sdx1=9.02914, sdx2=11.86779, dbeta=0.01592,
      sigma=0.5578413, gen_fun=gen_twogroup) |> 
    Spower(power=.8, interval=c(100, 1000))

## 
## Design conditions: 
## 
## # A tibble: 1 × 8
##   test      sigma     n   sdx1   sdx2   dbeta sig.level power
##   <chr>     <dbl> <dbl>  <dbl>  <dbl>   <dbl>     <dbl> <dbl>
## 1 X:S = 0 0.55784    NA 9.0291 11.868 0.01592      0.05   0.8
## 
## Estimate of n: 165.2
## 95% Prediction Interval: [163.0, 167.6]

G*Power gives the estimate for \(n\) to be 163 (and therefore 256 in the second group given \(n2_n1\)).

Variance tests

Example 26.3; Difference from constant (one sample case)

# solve n for variance ratio of 1/1.5 = 2/3, one.tailed, 80% power
p_var.test(n=NA, vars=1, sigma2=1.5, two.tailed=FALSE) |> 
    Spower(power=.80, interval=c(10, 200))

## 
## Design conditions: 
## 
## # A tibble: 1 × 6
##       n  vars sigma2 two.tailed sig.level power
##   <dbl> <dbl>  <dbl> <lgl>          <dbl> <dbl>
## 1    NA     1    1.5 FALSE           0.05   0.8
## 
## Estimate of n: 79.7
## 95% Prediction Interval: [78.9, 80.6]

G*power gives sample size of 81.

Example 15.3; Two-sample variance test

For a two-sample equality of variance test with equal sample sizes,

# solve n for variance ratio of 1/1.5 = 2/3, two.tailed, 80% power
p_var.test(n=NA, vars=c(1, 1.5), two.tailed=TRUE) |> 
    Spower(power=.80, interval=c(50, 300))

## 
## Design conditions: 
## 
## # A tibble: 1 × 4
##       n two.tailed sig.level power
##   <dbl> <lgl>          <dbl> <dbl>
## 1    NA TRUE            0.05   0.8
## 
## Estimate of n: 194.1
## 95% Prediction Interval: [191.5, 196.6]

G*Power gives estimate of 193 per group.

t-tests

Estimate sample size (\(n\)) per group in independent samples \(t\)-test, one-tailed, medium effect size (\(d=0.5\)), \(\alpha=0.05\), 95% power (\(1-\beta = 0.95\)), equal sample sizes (\(\frac{n_2}{n_1}=1\)).

(out <- p_t.test(n = NA, d = .5, two.tailed=FALSE) |>  
               Spower(power = .95, interval=c(10,500)))

## 
## Design conditions: 
## 
## # A tibble: 1 × 5
##       n     d two.tailed sig.level power
##   <dbl> <dbl> <lgl>          <dbl> <dbl>
## 1    NA   0.5 FALSE           0.05  0.95
## 
## Estimate of n: 87.4
## 95% Prediction Interval: [84.8, 89.3]

G*power estimate is 88 per group, Spower estimate is 87.4101 with the 95% CI [84.7763, 89.287].

Example 19.3; Paired samples t-test

p_t.test(n=50, d=0.421637, type = 'paired') |> Spower(replications=50000)

## 
## Design conditions: 
## 
## # A tibble: 1 × 5
##       n       d type   sig.level power
##   <dbl>   <dbl> <chr>      <dbl> <lgl>
## 1    50 0.42164 paired      0.05 NA   
## 
## Estimate of power: 0.841
## 95% Confidence Interval: [0.838, 0.844]

G*power gives power estimate of .832, though Cohen reported a value closer to .840. When \(d=0.2828427\) this leads to

p_t.test(n=50, d=.2828427, type = 'paired') |> Spower(replications=50000)

## 
## Design conditions: 
## 
## # A tibble: 1 × 5
##       n       d type   sig.level power
##   <dbl>   <dbl> <chr>      <dbl> <lgl>
## 1    50 0.28284 paired      0.05 NA   
## 
## Estimate of power: 0.508
## 95% Confidence Interval: [0.504, 0.513]

In this case G*Power gives the estimate .500. To answer the question “How many subjects would we need to arrive at a power of about 0.832114 in a two-group design?” this is specified within Spower() and where n is set to NA.

p_t.test(n=NA, d=0.2828427, type = 'paired') |> 
    Spower(power=0.832114, interval=c(100,300))

## 
## Design conditions: 
## 
## # A tibble: 1 × 5
##       n       d type   sig.level   power
##   <dbl>   <dbl> <chr>      <dbl>   <dbl>
## 1    NA 0.28284 paired      0.05 0.83211
## 
## Estimate of n: 107.2
## 95% Prediction Interval: [106.5, 108.0]

G*power reports that around \(N=110\) is required, though this is estimated visually using interpolation.

Example 20.3; One-sample t-test

p_t.test(n=NA, d=.625, two.tailed=FALSE, type='one.sample') |>  
    Spower(power=.95, interval=c(10, 100))

## 
## Design conditions: 
## 
## # A tibble: 1 × 6
##       n     d type       two.tailed sig.level power
##   <dbl> <dbl> <chr>      <lgl>          <dbl> <dbl>
## 1    NA 0.625 one.sample FALSE           0.05  0.95
## 
## Estimate of n: 28.9
## 95% Prediction Interval: [28.1, 29.7]

G*power gives sample size of \(n=30\).

p_t.test(n=NA, d=.1, type='one.sample') |>  
    Spower(power=.9,sig.level=.01, interval=c(100,2000))

## 
## Design conditions: 
## 
## # A tibble: 1 × 5
##       n     d type       sig.level power
##   <dbl> <dbl> <chr>          <dbl> <dbl>
## 1    NA   0.1 one.sample      0.01   0.9
## 
## Estimate of n: 1491.8
## 95% Prediction Interval: [1473.0, 1508.2]

G*power gives sample size of \(n=1492\).

Wilcox tests

Example 22.3; One-sample test with normal distribution

p_wilcox.test(n=649, d=.1, type='one.sample', two.tailed=FALSE) |> Spower()

## 
## Design conditions: 
## 
## # A tibble: 1 × 6
##       n     d type       two.tailed sig.level power
##   <dbl> <dbl> <chr>      <lgl>          <dbl> <lgl>
## 1   649   0.1 one.sample FALSE           0.05 NA   
## 
## Estimate of power: 0.801
## 95% Confidence Interval: [0.793, 0.809]

G*power provide power estimate of .800.

Two-sample test with Laplace distributions

library(VGAM)

parent1 <- function(n, d) VGAM::rlaplace(n, d, scale=sqrt(1/2))
parent2 <- function(n, d) VGAM::rlaplace(n, scale=sqrt(1/2))

nr <- 134/67
p_wilcox.test(n=67, n2_n1=nr, d=0.375, parent1=parent1, parent2=parent2) |> 
    Spower()

## 
## Design conditions: 
## 
## # A tibble: 1 × 4
##       n     d sig.level power
##   <dbl> <dbl>     <dbl> <lgl>
## 1    67 0.375      0.05 NA   
## 
## Estimate of power: 0.850
## 95% Confidence Interval: [0.843, 0.857]

G*power gives power of .847.

G*Power examples evaluated with Spower

Phil Chalmers

2025-06-28