---
title: "G*Power examples evaluated with Spower"
author: Phil Chalmers
date: "2025-11-27"
output: 
  bookdown::html_document2:
    base_format: rmarkdown::html_vignette
    number_sections: false
    toc: true
vignette: >
  %\VignetteIndexEntry{G*Power examples evaluated with Spower}
  %\VignetteEncoding{UTF-8}
  %\VignetteEngine{knitr::rmarkdown}
editor_options: 
  markdown: 
    wrap: 72
  chunk_output_type: console
---







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 the presented power analyses can 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. Unless
otherwise specified, the following analyses assume that the
"significance level" (`sig.level` in `Spower()`) is set to
$\alpha = .05$.

# Correlation

Correlation analyses require evaluating the 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)

The following estimates the sample size required to reject
$H_0:\, \rho_0=.60$ in correlation analysis with $1-\beta=.95$
probability when $\rho=.65$.


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

```
## 
## Execution time (H:M:S): 00:00:21
## 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% Predicted Confidence Interval: [1901.1, 1958.2]
```

G\*power estimates this $n$ to be 1929 using the Fisher
$z$-transformation approximation, which is what is used by the `Spower`
definition as well.

### Test against constant $\rho_0=0$

The more canonical version hypotheses involving correlation coefficients
appear when $rho_0=0$, as these do not require the Fisher approximation.
For instance, the power associated with $\rho = .3$ with 100 pairs of
observations, tested against $\rho_0=0$, results in the following.


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

```
## 
## Execution time (H:M:S): 00:00:11
## 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]
```

Next, the sample sample size estimate required to reject
$H_0:\, \rho_0=0$ in correlation analysis with $1-\beta=.95$ probability
when $\rho=.3$ is expressed as


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

```
## 
## Execution time (H:M:S): 00:00:16
## 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% Predicted Confidence Interval: [136.1, 140.4]
```

G\*power 3.1 provides the same estimate as the `pwr` package in this
case, which for comparison is presented below.


``` r
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

Were the correlation between two independent samples to be compared, the
`p_2r()` simulation can be adopted. Below a sample of $N_1=206$
observations appeared in the first sample ($r=.75$), while the second
sample ($r=.88$) contained only $N_2=51$ observations (hence, the ratio
$N_2/N_1=51/206$). This results in the post-hoc/observed power of


``` r
p_2r(n=206, r.ab=.75, r.ab2=.88, n2_n1=51/206) |> Spower()
```

```
## 
## Execution time (H:M:S): 00:00:26
## Design conditions: 
## 
## # A tibble: 1 × 6
##       n  r.ab r.ab2   n2_n1 sig.level power
##   <dbl> <dbl> <dbl>   <dbl>     <dbl> <lgl>
## 1   206  0.75  0.88 0.24757      0.05 NA   
## 
## Estimate of power: 0.727
## 95% Confidence Interval: [0.718, 0.735]
```

G\*power 3.1 returns the power of .726 in this context.

### Example 28.3.1; Correlation - inequality of two dependent Pearson r's (no common index)

The following two examples assume the correlation matrix 

``` r
# From Gpower 3.1 manual
Cp <- matrix(c(1, .5, .4, .1, 
			   .5, 1, .2, -.4, 
			   .4, .2, 1, .8, 
			   .1, -.4, .8, 1), 4, 4)

# rearrange rows for convenience
Cp <- Cp[c(1,4,2,3), c(1,4,2,3)]
colnames(Cp) <- rownames(Cp) <- c('x1', 'y1', 'x2', 'y2')
Cp
```

```
##     x1   y1   x2  y2
## x1 1.0  0.1  0.5 0.4
## y1 0.1  1.0 -0.4 0.8
## x2 0.5 -0.4  1.0 0.2
## y2 0.4  0.8  0.2 1.0
```
is the population structure. For the no common index tests all of these elements are required, while for the common index form only a $3\times 3$ subset is needed.

Evaluating the null hypothesis that 
$$H_0: \rho_{ab} = \rho_{cd}$$
where in this case $\rho_{ab} = .5$ and $\rho_{cd} = .8$ can be explored using the `p_2r()` function. The following performs an a priori analyses to determine the sample size ($N$) required to achieve 80% power using Steiger's (1980) inferential $z$ approach. 


``` r
p_2r(n=NA, r.ab=.1, r.ac=.5, r.ad=.4, r.bc=-.4, r.bd=.8, r.cd=.2, 
	 two.tailed=FALSE) |> Spower(power = .80, interval=c(500, 2000))
```

```
## 
## Execution time (H:M:S): 00:01:35
## Design conditions: 
## 
## # A tibble: 1 × 10
##       n  r.ab  r.ac  r.bc  r.ad  r.bd  r.cd two.tailed sig.level power
##   <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <lgl>          <dbl> <dbl>
## 1    NA   0.1   0.5  -0.4   0.4   0.8   0.2 FALSE           0.05   0.8
## 
## Estimate of n: 886.2
## 95% Predicted Confidence Interval: [875.3, 897.3]
```

G\*power 3.1 returns the required sample size of $N=886$.


### Example 28.3.2; Correlation - inequality of two dependent Pearson r's (common index)

The information in this example is the same as `Example 28.3.1`, however it is assumed that there is a common index between the correlation measures instead of a complete overlap. As such, the previous `Cp` object may be further subset to see what type of correlation structure is required for the common index setup.


``` r
Cp[c(4,3,1),c(4,3,1)]
```

```
##     y2  x2  x1
## y2 1.0 0.2 0.4
## x2 0.2 1.0 0.5
## x1 0.4 0.5 1.0
```

The null under instigation in this case is 
$$H_0:\rho_{ab} = \rho_{ac}$$
where $\rho_{ab} = .2$ and $\rho_{ac} = .4$. In `Spower`, this equates to the following inputs, which again use Steiger's (1980) inferential $z$ approach by default.


``` r
p_2r(n=NA, r.ab=.4, r.ac=.2, r.bc=.5, two.tailed=FALSE) |> 
	Spower(power = .80, interval=c(50, 500))
```

```
## 
## Execution time (H:M:S): 00:01:18
## Design conditions: 
## 
## # A tibble: 1 × 7
##       n  r.ab  r.ac  r.bc two.tailed sig.level power
##   <dbl> <dbl> <dbl> <dbl> <lgl>          <dbl> <dbl>
## 1    NA   0.4   0.2   0.5 FALSE           0.05   0.8
## 
## Estimate of n: 134.7
## 95% Predicted Confidence Interval: [133.1, 136.3]
```

G\*power 3.1 returns the required sample size of $N=144$, which interestingly is slightly higher than the simulation version from `Spower`. Providing $N=144$ to the above to obtain the power estimate gives the following:


``` r
p_2r(n=144, r.ab=.4, r.ac=.2, r.bc=.5, two.tailed=FALSE) |> Spower()
```

```
## 
## Execution time (H:M:S): 00:00:22
## Design conditions: 
## 
## # A tibble: 1 × 7
##       n  r.ab  r.ac  r.bc two.tailed sig.level power
##   <dbl> <dbl> <dbl> <dbl> <lgl>          <dbl> <lgl>
## 1   144   0.4   0.2   0.5 FALSE           0.05 NA   
## 
## Estimate of power: 0.831
## 95% Confidence Interval: [0.824, 0.839]
```


### Example 28.3.3; sensitivity analysis 

It is also possible to perform a sensitivity analyses rather than the above a priori power analysis. Below fixes $N=144$, while `r.ac` is solved to obtain 80% power. G\*power 3.1 reports that $\rho_{ac} = 0.047702$, which is confirmed using the simulation below.


``` r
# confirm solution obtained by G*power (post hoc power estimate)
p_2r(n=144, r.ab=.4, r.ac=0.047702, r.bc=-0.6, two.tailed=FALSE) |> Spower()
```

```
## 
## Execution time (H:M:S): 00:00:22
## Design conditions: 
## 
## # A tibble: 1 × 7
##       n  r.ab     r.ac  r.bc two.tailed sig.level power
##   <dbl> <dbl>    <dbl> <dbl> <lgl>          <dbl> <lgl>
## 1   144   0.4 0.047702  -0.6 FALSE           0.05 NA   
## 
## Estimate of power: 0.815
## 95% Confidence Interval: [0.808, 0.823]
```

Obtaining a similar estimate using `Spower()` requires the following sensitivity analysis structure:


``` r
# note that interval is specified as c(upper, lower) as higher values
# of r.ac result in lower power in this context
p_2r(n=144, r.ab=.4, r.ac=NA, r.bc=-0.6, two.tailed=FALSE) |> 
	Spower(power = .80, interval=c(.4, .001))
```

```
## 
## Execution time (H:M:S): 00:01:37
## Design conditions: 
## 
## # A tibble: 1 × 7
##       n  r.ab  r.ac  r.bc two.tailed sig.level power
##   <dbl> <dbl> <dbl> <dbl> <lgl>          <dbl> <dbl>
## 1   144   0.4    NA  -0.6 FALSE           0.05   0.8
## 
## Estimate of r.ac: 0.048
## 95% Predicted Confidence Interval: [0.046, 0.050]
```

For this example, `Spower` and G\*power 3.1 seem to agree. 

### Example 16.3; Point-biserial correlation

The following estimates the sample size required to obtain a power of
$1-\beta=.95$ given that $r=.25$ is the true correlation, evaluated
under the null $H_0:\rho\le0$ (hence, is one-tailed) with
$\alpha = .05$.


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

```
## 
## Execution time (H:M:S): 00:00:10
## 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: 81.7
## 95% Predicted Confidence Interval: [79.1, 85.4]
```

``` r
# 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 $d$ will be converted to $r$ inside
the `p_t.test()` function.

### Example 31.3; tetrachoric correlation

For tetrachoric and polychoric correlations, the experiment definition
in `p_r.cat()` can be used. This requires specifying the associated
$\tau$ threshold coefficients for the population normal truncation
processes, as well as the bivariate correlation itself prior to the
truncation.


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

```
## [1] 0.4183 0.5817
```

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

```
## [1] 0.3978 0.6022
```

``` r
# converted to intercepts
tauX <- qnorm(1-marginal.x)[2]
tauY <- qnorm(1-marginal.y)[2]
c(tauX, tauY)
```

```
## [1] -0.2063 -0.2589
```

These $\tau$ values correspond to where along the assumed normal p.d.f.
the truncation took place, which for the $X$ variable can be seen in the
following graphic.



![Tetrachoric](tetrachoric.png)

Finally, assuming that the untruncated $r=0.2399846$, and a Score test
were used to evaluate the null hypothesis of interest (`score = TRUE`),
the sample size required to reject the null hypothesis that the
tetrachoric correlation is less than or equal to 0 in this population
(one-tailed) is expressed as


``` r
p_r.cat(n=NA, r=0.2399846, tauX=tauX, tauY=tauY, 
		score=TRUE, 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]
```

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 maximum-likelihood estimate (MLE); hence,
`score = FALSE` by default. To switch, use `score=TRUE`, though note
that this requires twice as many computations as a second set of data is
generated and analyzed at $r=r_0$ to obtain the required $SE_0$
estimate.

# Proportions

### Example 4.3; One sample proportion tests

A one sample, one-tailed proportion test given data generated from a
population with $\pi = .80$ and tested against the null hypothesis
$H_0:\pi_0\le.65$ with $n=20$ is presented in the following. Note that
G\*power requires a term $g$ to be specified as the proportion
*difference* from the null instead (hence, $g = .80-.65=.15$), though
`p_prop.teset()` accepts the null and alternative probability values
as-is.


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

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

```
## 
## Execution time (H:M:S): 00:00:04
## 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: 0.416
## 95% Confidence Interval: [0.406, 0.425]
```

G\*power gives the estimate $1-\beta=.4112$. Note that with
`p_prop.test()`, the Fisher's exact version of this test is also
supported by passing the argument `exact = TRUE`.


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

```
## 
## Execution time (H:M:S): 00:00:04
## 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: 0.411
## 95% Confidence Interval: [0.402, 0.421]
```

### Example 22.1; Wilcoxon signed-rank test

The following performed a one-sample, one-tailed Wilcoxon signed rank
test given $N=649$, $d=.1$, where the parent distribution is assumed to
follow a Normal/Gaussian shape (default).


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

```
## 
## Execution time (H:M:S): 00:00:20
## 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: 0.812
## 95% Confidence Interval: [0.805, 0.820]
```

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($\mu$,1)* distribution with varying sample sizes and
effect sizes. The target was to obtain the "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.


``` r
# For Gaussian(d,1)
out <- p_wilcox.test(type='one.sample', two.tailed=FALSE) |> 
	SpowerBatch(n=c(649, 164, 42, 20, 12, 9),
				d=c(.1, .2, .4, .6, .8, 1.0), replications = 50000, fully.crossed=FALSE)
as.data.frame(out)
```

```
##     n   d       type two.tailed sig.level  power CI_2.5 CI_97.5
## 1 649 0.1 one.sample      FALSE      0.05 0.8012 0.7977  0.8047
## 2 164 0.2 one.sample      FALSE      0.05 0.8027 0.7992  0.8062
## 3  42 0.4 one.sample      FALSE      0.05 0.8043 0.8009  0.8078
## 4  20 0.6 one.sample      FALSE      0.05 0.8070 0.8035  0.8105
## 5  12 0.8 one.sample      FALSE      0.05 0.8018 0.7983  0.8053
## 6   9 1.0 one.sample      FALSE      0.05 0.8467 0.8435  0.8498
```

#### Laplace($\mu$, 1) version

A one-sample Wilcoxon signed rank test with Laplace distribution as the
parent. Note that this requires defining the parent distribution
manually, accepting arguments such as `n` and `d`.


``` r
library(extraDistr)

# 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, sigma=sqrt(1/2)) 
	extraDistr::rlaplace(n, d, sigma=sigma)

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

```
## 
## Execution time (H:M:S): 00:00:03
## 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: 0.801
## 95% Confidence Interval: [0.793, 0.809]
```

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($\mu$, 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.


``` r
# For Laplace(0,1)
out <- p_wilcox.test(parent1=parent, type='one.sample',
			  two.tailed=FALSE) |> 
	SpowerBatch(n=c(419, 109, 31, 16, 11, 8),
				d=c(.1, .2, .4, .6, .8, 1.0), replications=50000, fully.crossed=FALSE)
as.data.frame(out)
```

```
##     n   d       type two.tailed sig.level  power CI_2.5 CI_97.5
## 1 419 0.1 one.sample      FALSE      0.05 0.8021 0.7986  0.8056
## 2 109 0.2 one.sample      FALSE      0.05 0.7992 0.7957  0.8028
## 3  31 0.4 one.sample      FALSE      0.05 0.8031 0.7996  0.8065
## 4  16 0.6 one.sample      FALSE      0.05 0.8007 0.7972  0.8042
## 5  11 0.8 one.sample      FALSE      0.05 0.8032 0.7998  0.8067
## 6   8 1.0 one.sample      FALSE      0.05 0.7771 0.7735  0.7808
```

### Example 5.3; Two dependent proportions test (McNemar's test)

The following performs a proportions test between two dependent groups
using McNemar's test. The data is from O'Brien (2002, p. 161-163).


``` r
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
```

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

```
## 
## Execution time (H:M:S): 00:00:02
## 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: 0.836
## 95% Confidence Interval: [0.828, 0.843]
```

Alternatively, specifying the inputs not in terms of proportions but
rather as the odds ratio ($OR=\pi_{12}/\pi_{21}=.08/.32=.25$) and
proportions of discordant pairs
($\pi_D=\pi_{12} + \pi_{21}=.08 + .32=.40$) can be supplied


``` r
OR <- obrien2002[1,2] / obrien2002[2,1]
disc <- obrien2002[1,2] + obrien2002[2,1]
p_mcnemar.test(n=50, OR=OR, prop.disc=disc, two.tailed=FALSE) |> Spower()
```

```
## 
## Execution time (H:M:S): 00:00:02
## 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: 0.841
## 95% Confidence Interval: [0.834, 0.848]
```

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.


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

```
## 
## Execution time (H:M:S): 00:00:02
## 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: 0.887
## 95% Confidence Interval: [0.881, 0.893]
```

# Multiple Linear Regression (Fixed IVs)

### Example 13.1

Evaluating $R^2=.1$ generated data for a linear regression model given
the null hypothesis $H_0:R^2_0=0$. When evaluated using $N=95$
observations with $k=5$ predictor variables gives the estimate.


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

```
## 
## Execution time (H:M:S): 00:00:42
## 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: 0.664
## 95% Confidence Interval: [0.655, 0.674]
```

G\*power gives $1-\beta = .673$.

### Example 14.3

Similarly, comparing nested models for changes in $R^2$. For the
following, note that `k` is total IVs (in this case, 9), while `k.R2_0`
is number of IVs for baseline model (in this case, 5). At $\alpha=.01$
and a change of $\Delta R^2=.05$ from the baseline $R^2_0=.25$ gives


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

```
## 
## Execution time (H:M:S): 00:00:54
## 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: 0.238
## 95% Confidence Interval: [0.230, 0.247]
```

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


``` r
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))
```

```
## 
## Execution time (H:M:S): 00:03:06
## 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: 242.6
## 95% Predicted Confidence Interval: [240.5, 244.6]
```

G\*power gives $n = 242$.

### Example 14.3b

Nested model comparison for changes in $R^2$ for models with 12 IVs
versus 9 IVs. Requires the specification of the $R^2_{residual}$.


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

```
## 
## Execution time (H:M:S): 00:01:15
## 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: 0.756
## 95% Confidence Interval: [0.748, 0.765]
```

G\*power gives $1-\beta = .767$.

# Multiple Linear Regression (Random IVs)

### Example 7.3

Same as in Example 13.1 above, however assuming that the IVs are
randomly sampled instead of fixed.


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

```
## 
## Execution time (H:M:S): 00:00:15
## 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: 0.659
## 95% Confidence Interval: [0.650, 0.669]
```

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

# Simple linear regression

## Example 12.3

Evaluate post-hoc power for simple linear regression model null
hypothesis $H_0:\beta_1=0$ given $\sigma_x = 7.5$, $\sigma_y$,
$\beta_1=-0.0667$, and $N=100$.


``` r
p_slr(n=100, beta=-0.0667, sd_x=7.5, sd_y = 4) |> Spower()
```

```
## 
## Execution time (H:M:S): 00:00:29
## Design conditions: 
## 
## # A tibble: 1 × 6
##       n    beta  sd_x  sd_y sig.level power
##   <dbl>   <dbl> <dbl> <dbl>     <dbl> <lgl>
## 1   100 -0.0667   7.5     4      0.05 NA   
## 
## Estimate of power: 0.243
## 95% Confidence Interval: [0.234, 0.251]
```

G\*power returns the power estimate $1-\beta = 0.2389$.

# Fixed effects ANOVA - One way (F-test)

### Example 10.3

One-way ANOVA example to solve $n$ per group (of which there are
$k=10$), using Cohen's $f=.25$, to achieve a power of $1-\beta=.95$.


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

```
## 
## Execution time (H:M:S): 00:00:23
## 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: 38.7
## 95% Predicted Confidence Interval: [38.1, 39.2]
```

G\*power gives the estimate $n=39$.

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


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

```
## 
## Execution time (H:M:S): 00:00:43
## 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): 0.160
## 95% Confidence Interval: [0.156, 0.164]
## 
## Estimate of power (1-beta): 0.840
## 95% Confidence Interval: [0.836, 0.844]
```

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 and 18.3

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.


``` r
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.


``` r
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()
```

```
## 
## Execution time (H:M:S): 00:00:33
## Design conditions: 
## 
## # A tibble: 1 × 9
##   test      sigma     n  n2_n1   sdx1   sdx2   dbeta sig.level power
##   <chr>     <dbl> <dbl>  <dbl>  <dbl>  <dbl>   <dbl>     <dbl> <lgl>
## 1 X:S = 0 0.55784    28 1.5714 9.0291 11.868 0.01592      0.05 NA   
## 
## Estimate of power: 0.199
## 95% Confidence Interval: [0.191, 0.207]
```

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


``` r
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))
```

```
## 
## Execution time (H:M:S): 00:01:58
## Design conditions: 
## 
## # A tibble: 1 × 9
##   test      sigma     n  n2_n1   sdx1   sdx2   dbeta sig.level power
##   <chr>     <dbl> <dbl>  <dbl>  <dbl>  <dbl>   <dbl>     <dbl> <dbl>
## 1 X:S = 0 0.55784    NA 1.5714 9.0291 11.868 0.01592      0.05   0.8
## 
## Estimate of n: 164.9
## 95% Predicted Confidence Interval: [163.3, 166.8]
```

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

# Variance tests

### Example 26.3; Difference from constant (one sample case)

Solve $n$ for variance ratio of $1/1.5 = 2/3$ using a one-tailed
variance ratio test, assuming that the target power is $1-\beta=.80$.


``` r
p_var.test(n=NA, vars=1, sigma2=1.5, two.tailed=FALSE) |> 
	Spower(power=.80, interval=c(10, 200))
```

```
## 
## Execution time (H:M:S): 00:00:31
## 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: 80.6
## 95% Predicted Confidence Interval: [79.3, 82.2]
```

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,


``` r
# 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))
```

```
## 
## Execution time (H:M:S): 00:01:36
## Design conditions: 
## 
## # A tibble: 2 × 5
##       n  vars two.tailed sig.level power
##   <dbl> <dbl> <lgl>          <dbl> <dbl>
## 1    NA   1   TRUE            0.05   0.8
## 2    NA   1.5 TRUE            0.05   0.8
## 
## Estimate of n: 193.4 
## Estimate of n: 193.7
## 95% Predicted Confidence Interval: [191.5, 195.4]
```

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 ($\frac{n_2}{n_1}=1$).


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

```
## 
## Execution time (H:M:S): 00:00:09
## 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: 87.1
## 95% Predicted Confidence Interval: [85.5, 88.6]
```



G\*power estimate is 88 per group, `Spower` estimate is 87.0506 with
the 95% CI [85.4503, 88.6126].

### Example 19.3; Paired samples t-test

Paired-samples $t$-test, assuming the generated difference is the
repeated measures Cohen's $d_r=.421637$ (e.g., were the unadjusted
$d=.4$, while $r_{xy} = .55$, then this results in the repeated
$d_r=\frac{|\mu_x-\mu_y|}{\sqrt{\sigma^2_x + \sigma^2_y - 2\rho_{xy}\sigma_x\sigma_y}}$).


``` r
p_t.test(n=50 * 2, d=0.421637, type = 'paired') |> Spower(replications=50000)
```

```
## 
## Execution time (H:M:S): 00:00:20
## Design conditions: 
## 
## # A tibble: 1 × 5
##       n       d type   sig.level power
##   <dbl>   <dbl> <chr>      <dbl> <lgl>
## 1   100 0.42164 paired      0.05 NA   
## 
## Estimate of power: 0.840
## 95% Confidence Interval: [0.836, 0.843]
```

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


``` r
p_t.test(n=50 * 2, d=.2828427, type = 'paired') |> Spower(replications=50000)
```

```
## 
## Execution time (H:M:S): 00:00:20
## Design conditions: 
## 
## # A tibble: 1 × 5
##       n       d type   sig.level power
##   <dbl>   <dbl> <chr>      <dbl> <lgl>
## 1   100 0.28284 paired      0.05 NA   
## 
## Estimate of power: 0.506
## 95% Confidence Interval: [0.502, 0.510]
```

In this case G\*Power 3.1 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`.


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

```
## 
## Execution time (H:M:S): 00:00:25
## 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: 216.9
## 95% Predicted Confidence Interval: [215.2, 218.7]
```

G\*power reports that around $N=110*2=220$ pairs are required, though
this is estimated visually using interpolation.

### Example 20.3; One-sample t-test

Evaluating the hypotheses for the mean expression $$H_0:\mu\le\mu_0$$
$$H_a:\mu>\mu_0$$

using a one-sample $t$-test. The following estimates $n$ given a
one-tailed $d=.625$ to achieve $1-\beta=.95$.


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

```
## 
## Execution time (H:M:S): 00:00:07
## 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: 29.5
## 95% Predicted Confidence Interval: [28.9, 30.1]
```

G\*power gives sample size of $n=30$. Similarly, though with different
inputs.


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

```
## 
## Execution time (H:M:S): 00:00:15
## 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: 1493.8
## 95% Predicted Confidence Interval: [1468.0, 1519.9]
```

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

## Wilcoxon tests

### Example 22.3; One-sample test with normal distribution

Same as Example 22.1 above.


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

```
## 
## Execution time (H:M:S): 00:00:20
## 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: 0.806
## 95% Confidence Interval: [0.798, 0.813]
```

G\*power 3.1 provides a power estimate of .800, agreeing with `Spower`.

Similarly, assuming that the distribution for the one-sample followed a
Laplace distribution, and that $N=11$ were used instead. This requires
defining an alternative parent distribution, which below uses the
`rlaplace` function from the `extraDistr` package.


``` r
library(extraDistr)
parent1 <- function(n, d) extraDistr::rlaplace(n, mu=d, sigma=sqrt(1/2))

# properties of sampled distribution
psych::describe(parent1(n=100000, d=0.8))
```

```
##    vars     n mean sd median trimmed  mad   min  max range skew kurtosis se
## X1    1 1e+05  0.8  1    0.8     0.8 0.73 -5.99 8.63 14.61 0.05     2.85  0
```

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

```
## 
## Execution time (H:M:S): 00:00:03
## Design conditions: 
## 
## # A tibble: 1 × 6
##       n     d type       two.tailed sig.level power
##   <dbl> <dbl> <chr>      <lgl>          <dbl> <lgl>
## 1    11   0.8 one.sample FALSE           0.05 NA   
## 
## Estimate of power: 0.800
## 95% Confidence Interval: [0.792, 0.808]
```

Interestingly, G\*power 3.1 reports this power to be 0.830.

### Two-sample test with Laplace distributions

Two-sample Wilcoxon test comparing Laplace distributions with different
central tendencies.


``` r
library(extraDistr)

parent1 <- function(n, d) extraDistr::rlaplace(n, mu=d, sigma=sqrt(1/2))
parent2 <- function(n, d) extraDistr::rlaplace(n, sigma=sqrt(1/2))

# properties of sampled distributions
psych::describe(parent1(n=100000, d=0.375))
```

```
##    vars     n mean sd median trimmed  mad   min  max range  skew kurtosis se
## X1    1 1e+05 0.38  1   0.38    0.38 0.73 -8.65 6.85  15.5 -0.04     2.83  0
```

``` r
psych::describe(parent1(n=100000, d=0))
```

```
##    vars     n mean sd median trimmed  mad   min  max range skew kurtosis se
## X1    1 1e+05 0.01  1      0    0.01 0.73 -8.44 7.65 16.09 0.04     3.13  0
```

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

```
## 
## Execution time (H:M:S): 00:00:10
## 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: 0.843
## 95% Confidence Interval: [0.836, 0.850]
```

Unlike before with the Laplace distribution, G\*power 3.1 seems to agree
with `Spower`, where a power of .847 is reported. This seems to raise questions about the consistency of the results.

### Example 23.3: Paired-samples test with Laplace distributions

Finally, paired-samples approach using Wilcoxon test with $N=10$.


``` r
parent1 <- function(n, d) extraDistr::rlaplace(n, mu=d, sigma=sqrt(1/2))
parent2 <- function(n, d) extraDistr::rlaplace(n, sigma=sqrt(1/2))

psych::describe(parent1(n=100000, d=1.13842))
```

```
##    vars     n mean sd median trimmed  mad   min  max range skew kurtosis se
## X1    1 1e+05 1.14  1   1.14    1.14 0.72 -6.68 9.92  16.6 0.02     3.02  0
```

``` r
psych::describe(parent1(n=100000, d=0))
```

```
##    vars     n mean sd median trimmed  mad   min  max range  skew kurtosis se
## X1    1 1e+05    0  1      0       0 0.73 -8.64 6.81 15.45 -0.04        3  0
```

``` r
p_wilcox.test(n=10*2, d=1.13842, type = 'paired',
			  parent1=parent1, parent2=parent2) |> Spower()
```

```
## 
## Execution time (H:M:S): 00:00:04
## Design conditions: 
## 
## # A tibble: 1 × 5
##       n      d type   sig.level power
##   <dbl>  <dbl> <chr>      <dbl> <lgl>
## 1    20 1.1384 paired      0.05 NA   
## 
## Estimate of power: 0.932
## 95% Confidence Interval: [0.927, 0.937]
```

Again, the simulation approach and G\*power 3.1 differ in their outputs,
where in G\*power 3.1 the reported power is 0.853.