Power: Difference between revisions

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* [https://www.rdatagen.net/post/using-simulation-for-power-analysis-an-example/ An example based on a stepped wedge study design]
* [https://www.rdatagen.net/post/using-simulation-for-power-analysis-an-example/ An example based on a stepped wedge study design]
* [https://www.rdatagen.net/post/2021-03-16-framework-for-power-analysis-using-simulation/ Framework for power analysis using simulation]
* [https://www.rdatagen.net/post/2021-03-16-framework-for-power-analysis-using-simulation/ Framework for power analysis using simulation]
== Pearson correlation ==
<pre>
set.seed(0)
simulate_p_value <- function(sample_size) {
  # Simulate data
  x <- rnorm(sample_size)
  y <- x + rnorm(sample_size)
 
  # Calculate correlation and p-value
  cor.test(x, y)$p.value
}
simulate_p_value_ttest <- function(sample_size) {
  # Simulate data
  x <- rnorm(sample_size, mean = 0)
  y <- rnorm(sample_size, mean = 0.5)
 
  # Calculate p-value
  t.test(x, y)$p.value
}
sample_sizes <- c(10, 30, 50, 100, 300, 500, 1000)
p_values <- sapply(sample_sizes, simulate_p_value)
plot(sample_sizes, p_values, log = "y", type = "o",
    xlab = "Sample Size", ylab = "p-value",
    main = "Effect of Sample Size on p-value")
</pre>
= T-test case =
T-test variance could be large if the sample size is small. Below is a case of 30 vs 5 and effect size .5.
<pre>
set.seed(13)
hist(replicate(100, t.test(rnorm(30), rnorm(5)+.5)$p.value),
    main='', xlab='pvalue')
set.seed(13); summary(replicate(100, t.test(rnorm(30), rnorm(5)+.5)$p.value))
#      Min.  1st Qu.    Median      Mean  3rd Qu.      Max.
# 0.0000157 0.1730494 0.3178627 0.3970372 0.6592276 0.9999164
</pre>


= Power analysis and sample size calculation for Agriculture =
= Power analysis and sample size calculation for Agriculture =
Line 149: Line 190:


[https://www.rdatagen.net/post/unbalanced-randomization-can-improve-power-in-some-situations/ Yes, unbalanced randomization can improve power, in some situations]
[https://www.rdatagen.net/post/unbalanced-randomization-can-improve-power-in-some-situations/ Yes, unbalanced randomization can improve power, in some situations]
= Prediction =
[https://onlinelibrary.wiley.com/doi/10.1002/sim.9025 Minimum sample size for external validation of a clinical prediction model with a binary outcome] Riley 2021
= High dimensional data =
* [https://hbiostat.org/bbr/hdata.html#sec-hdata-simor Simulation To Understand Needed Sample Sizes] from the online book '''Biostatistics for Biomedical Research'''
* [https://cran.r-project.org/web/packages/MKmisc/index.html MKmisc] package
** [https://typeset.io/pdf/sample-size-planning-for-developing-classifiers-using-high-1xapsr5vtv.pdf Sample size planning for developing classifiers using high-dimensional DNA microarray data] Biostatistics 2007 & [https://brb.nci.nih.gov/brb/samplesize/samplesize4GE.html online calculator].
*** Goal: Determine the sample size required to ensure that the expected correct classification probability for a predictor developed from training data is within γ of the optimal expected correct classification probability for a linear classification problem (p6).
*** Input: standard fold-change, number of genes, prevalence in the larger group, tolerance (cannot control). The standard fold-change is calculated by <math> 0.8 * \max|{t_g}| * \sqrt{1/n1 + 1/n2} </math> where <math>t_g</math> is the T-statistic for gene <math>g</math>.
** [https://search.r-project.org/CRAN/refmans/MKmisc/html/ssize.pcc.html ssize.pcc()]. Input: standard fold-change, number of genes, prevalence in the larger group, tolerance, number of significant genes.
** Vignette [https://cran.r-project.org/web/packages/MKmisc/vignettes/MKmisc.html#sample-size-calculation  ssize.pcc, which is based on the probability of correct classification (PCC)]
= Survival data =
[https://onlinelibrary.wiley.com/doi/10.1002/sim.9931 Sample size and predictive performance of machine learning methods with survival data: A simulation study] 2023. R code is in the supplementary.

Latest revision as of 08:02, 22 October 2024

Power analysis/Sample Size determination

Binomial distribution

Power analysis for default Bayesian t-tests

http://daniellakens.blogspot.com/2016/01/power-analysis-for-default-bayesian-t.html

Using simulation for power analysis

Pearson correlation

set.seed(0)

simulate_p_value <- function(sample_size) {
  # Simulate data
  x <- rnorm(sample_size)
  y <- x + rnorm(sample_size)
  
  # Calculate correlation and p-value
  cor.test(x, y)$p.value
}

simulate_p_value_ttest <- function(sample_size) {
  # Simulate data
  x <- rnorm(sample_size, mean = 0)
  y <- rnorm(sample_size, mean = 0.5)
  
  # Calculate p-value
  t.test(x, y)$p.value
}

sample_sizes <- c(10, 30, 50, 100, 300, 500, 1000)
p_values <- sapply(sample_sizes, simulate_p_value)

plot(sample_sizes, p_values, log = "y", type = "o",
     xlab = "Sample Size", ylab = "p-value",
     main = "Effect of Sample Size on p-value")

T-test case

T-test variance could be large if the sample size is small. Below is a case of 30 vs 5 and effect size .5.

set.seed(13)
hist(replicate(100, t.test(rnorm(30), rnorm(5)+.5)$p.value), 
     main='', xlab='pvalue')
set.seed(13); summary(replicate(100, t.test(rnorm(30), rnorm(5)+.5)$p.value))
#      Min.   1st Qu.    Median      Mean   3rd Qu.      Max. 
# 0.0000157 0.1730494 0.3178627 0.3970372 0.6592276 0.9999164 

Power analysis and sample size calculation for Agriculture

http://r-video-tutorial.blogspot.com/2017/07/power-analysis-and-sample-size.html

Power calculation for proportions (shiny app)

https://juliasilge.shinyapps.io/power-app/

Derive the formula/manual calculation

[math]\displaystyle{ \begin{align} Power & = P_{\mu_1-\mu_2 = \Delta}(\frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\sigma^2/n + \sigma^2/n}} \gt Z_{\alpha /2}) + P_{\mu_1-\mu_2 = \Delta}(\frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\sigma^2/n + \sigma^2/n}} \lt -Z_{\alpha /2}) \\ & \approx P_{\mu_1-\mu_2 = \Delta}(\frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\sigma^2/n + \sigma^2/n}} \gt Z_{\alpha /2}) \\ & = P_{\mu_1-\mu_2 = \Delta}(\frac{\bar{X}_1 - \bar{X}_2 - \Delta}{\sqrt{2 * \sigma^2/n}} \gt Z_{\alpha /2} - \frac{\Delta}{\sqrt{2 * \sigma^2/n}}) \\ & = \Phi(-(Z_{\alpha /2} - \frac{\Delta}{\sqrt{2 * \sigma^2/n}})) \\ & = 1 - \beta =\Phi(Z_\beta) \end{align} }[/math]

Therefore

[math]\displaystyle{ \begin{align} Z_{\beta} &= - Z_{\alpha/2} + \frac{\Delta}{\sqrt{2 * \sigma^2/n}} \\ Z_{\beta} + Z_{\alpha/2} & = \frac{\Delta}{\sqrt{2 * \sigma^2/n}} \\ 2 * (Z_{\beta} + Z_{\alpha/2})^2 * \sigma^2/\Delta^2 & = n \\ n & = 2 * (Z_{\beta} + Z_{\alpha/2})^2 * \sigma^2/\Delta^2 \end{align} }[/math]
# alpha = .05, delta = 200, n = 79.5, sigma=450
1 - pnorm(1.96 - 200*sqrt(79.5)/(sqrt(2)*450)) + pnorm(-1.96 - 200*sqrt(79.5)/(sqrt(2)*450))
# [1] 0.8
pnorm(-1.96 - 200*sqrt(79.5)/(sqrt(2)*450))
# [1] 9.58e-07
1 - pnorm(1.96 - 200*sqrt(79.5)/(sqrt(2)*450)) 
# [1] 0.8

Calculating required sample size in R and SAS

pwr package is used. For two-sided test, the formula for sample size is

[math]\displaystyle{ n_{\mbox{each group}} = \frac{2 * (Z_{\alpha/2} + Z_\beta)^2 * \sigma^2}{\Delta^2} = \frac{2 * (Z_{\alpha/2} + Z_\beta)^2}{d^2} }[/math]

where [math]\displaystyle{ Z_\alpha }[/math] is value of the Normal distribution which cuts off an upper tail probability of [math]\displaystyle{ \alpha }[/math], [math]\displaystyle{ \Delta }[/math] is the difference sought, [math]\displaystyle{ \sigma }[/math] is the presumed standard deviation of the outcome, [math]\displaystyle{ \alpha }[/math] is the type 1 error, [math]\displaystyle{ \beta }[/math] is the type II error and (Cohen's) d is the effect size - difference between the means divided by the pooled standard deviation.

# An example from http://www.stat.columbia.edu/~gelman/stuff_for_blog/c13.pdf#page=3
# Method 1.
require(pwr)
pwr.t.test(d=200/450, power=.8, sig.level=.05,
           type="two.sample", alternative="two.sided")
#
#     Two-sample t test power calculation 
#
#              n = 80.4
#              d = 0.444
#      sig.level = 0.05
#          power = 0.8
#    alternative = two.sided
#
# NOTE: n is number in *each* group

# Method 2.
2*(qnorm(.975) + qnorm(.8))^2*450^2/(200^2)
# [1] 79.5
2*(1.96 + .84)^2*450^2 / (200^2)
# [1] 79.4

And stats::power.t.test() function.

power.t.test(n = 79.5, delta = 200, sd = 450, sig.level = .05,
             type ="two.sample", alternative = "two.sided")
#
#     Two-sample t test power calculation 
#
#              n = 79.5
#          delta = 200
#             sd = 450
#      sig.level = 0.05
#          power = 0.795
#    alternative = two.sided
#
# NOTE: n is number in *each* group

R package related to power analysis

CRAN Task View: Design of Experiments

RNA-seq

ScRNA-seq

scRNA-seq

Russ Lenth Java applets

https://homepage.divms.uiowa.edu/~rlenth/Power/index.html

Bootstrap method

The upstrap Crainiceanu & Crainiceanu, Biostatistics 2018

Multiple Testing Case

Optimal Sample Size for Multiple Testing The Case of Gene Expression Microarrays

Unbalanced randomization

Can unbalanced randomization improve power?

Yes, unbalanced randomization can improve power, in some situations

Prediction

Minimum sample size for external validation of a clinical prediction model with a binary outcome Riley 2021

High dimensional data

Survival data

Sample size and predictive performance of machine learning methods with survival data: A simulation study 2023. R code is in the supplementary.