Overview

Common Statistical Tests in R - Part I

T-statistic

Let [math]\displaystyle{ \scriptstyle\hat\beta }[/math] be an estimator of parameter β in some statistical model. Then a t-statistic for this parameter is any quantity of the form

[math]\displaystyle{ t_{\hat{\beta}} = \frac{\hat\beta - \beta_0}{\mathrm{s.e.}(\hat\beta)}, }[/math]

where β₀ is a non-random, known constant, and [math]\displaystyle{ \scriptstyle s.e.(\hat\beta) }[/math] is the standard error of the estimator [math]\displaystyle{ \scriptstyle\hat\beta }[/math].

Two sample test assuming equal variance

The t statistic (df = [math]\displaystyle{ n_1 + n_2 - 2 }[/math]) to test whether the means are different can be calculated as follows:

[math]\displaystyle{ t = \frac{\bar {X}_1 - \bar{X}_2}{s_{X_1X_2} \cdot \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} }[/math]

where

[math]\displaystyle{ s_{X_1X_2} = \sqrt{\frac{(n_1-1)s_{X_1}^2+(n_2-1)s_{X_2}^2}{n_1+n_2-2}}. }[/math]

[math]\displaystyle{ s_{X_1X_2} }[/math] is an estimator of the common/pooled standard deviation of the two samples. The square-root of a pooled variance estimator is known as a pooled standard deviation.

Pooled variance from Wikipedia
The pooled sample variance is an unbiased estimator of the common variance if Xi and Yi are following the normal distribution.
(From minitab) The pooled standard deviation is the average spread of all data points about their group mean (not the overall mean). It is a weighted average of each group's standard deviation. The weighting gives larger groups a proportionally greater effect on the overall estimate.
Type I error rates in two-sample t-test by simulation

Pooled variance vs overall variance

Overall variance is typically larger than the pooled variance unless the group means are identical.
This is because overall variance captures both within-group variability and between-group variability, while pooled variance only accounts for within-group variability.
Pooled variance does not directly capture between-group variability, only within-group variability under the assumption of homogeneous variance.
Overall variance reflects the influence of different group means and different variances.

Examples

foo <- function(n1, n2, seed=1, mu1=0, mu2=0, s1=1, s2=1) {
  set.seed(seed)
  x1 <- rnorm(n1, mu1, s1); x2 <- rnorm(n2, mu2, s2)
  cat("var(all)", round(var(c(x1, x2)), 3), ", ")
  pv <- ((n1-1)*var(x1) + (n2-1)*var(x2)) / (n1+n2-2)
  cat("var(pooled)", round(pv, 3) , "\n")
}
foo(10, 10)               # var(all) 0.834 , var(pooled) 0.877 
foo(100, 100, 1)          # var(all) 0.863 , var(pooled) 0.862 
foo(100, 100, 1234)       # var(all) 1.042 , var(pooled) 1.037 
foo(100, 100, 1234, 0, 1) # var(all) 1.393 , var(pooled) 1.037
foo(100, 100, 1234, s1=1, s2=3) # var(all) 5.292 , var(pooled) 5.299

Assumptions

The Four Assumptions Made in a T-Test
How to Check ANOVA Assumptions. ANOVA also assumes variances of the populations that the samples come from are equal.

T-tests that assume equal variances of the two populations aren't valid when the two populations have different variances, & it's worse for unequal sample sizes. If the smallest sample size is the one with highest variance the test will have inflated Type I error). Small and unbalanced sample sizes for two groups - what to do?

type1err <- function(n1=10, n2=100, sigma1=1, sigma2=1, n_simulations=10000) {
  set.seed(1)
  
  # Set the population means and standard deviations for the two groups
  mu1 <- 0
  mu2 <- 0

  # Initialize a counter for the number of significant results
  n_significant <- 0
  
  # Run the simulations
  for (i in 1:n_simulations) {
    # Generate random samples from two normal distributions
    sample1 <- rnorm(n1, mean = mu1, sd = sigma1)
    sample2 <- rnorm(n2, mean = mu2, sd = sigma2)
    
    # Perform a two-sample t-test
    t_test <- t.test(sample1, sample2, var.equal = FALSE)
    
    # Check if the result is significant at the 0.05 level
    if (t_test$p.value < 0.05) {
      n_significant <- n_significant + 1
    }
  }
  
  # Calculate the proportion of significant results
  prop_significant <- n_significant / n_simulations
  
  cat(sprintf('Proportion of significant results: %.4f', prop_significant))
}

type1err(10, 100) # 0.0521
type1err(28, 1371, 1, 1) # 0.0511
type1err(28, 1371, 1, 2) # 0.0000

type1err(10, 100, var.equal = FALSE) # 0.0519
type1err(28, 1371, 1, 1, var.equal = FALSE) # 0.0503
type1err(28, 1371, 1, 2, var.equal = FALSE) # 0.0509

type2err <- function(n1=10, n2=100, delta=1, sigma1=1, sigma2=1, n_simulations=10000) {
  set.seed(1)
  
  # Set the population means and standard deviations for the two groups
  mu1 <- 0
  mu2 <- delta
  
  # Initialize a counter for the number of significant results
  n_significant <- 0
  
  # Run the simulations
  for (i in 1:n_simulations) {
    # Generate random samples from two normal distributions
    sample1 <- rnorm(n1, mean = mu1, sd = sigma1)
    sample2 <- rnorm(n2, mean = mu2, sd = sigma2)
    
    # Perform a two-sample t-test
    t_test <- t.test(sample1, sample2, var.equal = FALSE)
    
    # Check if the result is significant at the 0.05 level
    if (t_test$p.value >= 0.05) {
      n_significant <- n_significant + 1
    }
  }
  
  # Calculate the proportion of p-values greater than or equal to alpha (i.e., the estimated type II error rate)
  type2_error_rate <- n_significant / n_simulations
  
  cat("The estimated type II error rate is", type2_error_rate)
}

type2err() # 0.2209
type2err(delta=3) # 0
                  # A larger value of delta will increase the power 
                  # of the test and result in a lower type II error rate.

A deflated Type I error rate could occur in situations where the significance level is set too low, making it more difficult to reject the null hypothesis even when it is false. This could result in an increased rate of Type II errors, or false negatives, where the null hypothesis is not rejected even though it is false.

Compare to unequal variance

	denominator	df
Same variance Student t-test	[math]\displaystyle{ \begin{align} & S_p\sqrt{1/n_1+1/n_2}\\ &S_p^2=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}\end{align} }[/math] (complicated)	[math]\displaystyle{ n_1+n_2-2 }[/math] (simple)
Unequal variance Welch t-test	[math]\displaystyle{ \sqrt{{s_1^2 \over n_1} + {s_2^2 \over n_2}} }[/math] (simple)	Satterthwaite's approximation (complicated)

Two sample test assuming unequal variance

The t statistic (Behrens-Welch test statistic) to test whether the population means are different is calculated as:

[math]\displaystyle{ t = {\overline{X}_1 - \overline{X}_2 \over s_{\overline{X}_1 - \overline{X}_2}} }[/math]

where

[math]\displaystyle{ s_{\overline{X}_1 - \overline{X}_2} = \sqrt{{s_1^2 \over n_1} + {s_2^2 \over n_2}}. }[/math]

Here s² is the unbiased estimator of the variance of the two samples.

The degrees of freedom is evaluated using the Satterthwaite's approximation

[math]\displaystyle{ df = { ({s_1^2 \over n_1} + {s_2^2 \over n_2})^2 \over {({s_1^2 \over n_1})^2 \over n_1-1} + {({s_2^2 \over n_2})^2 \over n_2-1} }. }[/math]

See an example The Satterthwaite Approximation: Definition & Example.

Different versions of t-test

Student's t-test in R and by hand: how to compare two groups under different scenarios?

Paired test, Wilcoxon signed-rank test

t.test(, paired = T), wilcox.test(, paired = T). ?t.test, ?wilcox.test
Have you ever asked yourself, "how should I approach the classic pre-post analysis?"
How to best visualize one-sample test?
Can R visualize the t.test or other hypothesis test results?, gginference package
Wilcoxon Test in R
How Do We Perform a Paired t-Test When We Don’t Know How to Pair? 2022

Relation to ANOVA

Paired Samples T-test in R, Can I use DESeq2 to analyze paired samples? from DESeq2 vignette

# Weight of the mice before treatment
before <-c(200.1, 190.9, 192.7, 213, 241.4, 196.9, 172.2, 185.5, 205.2, 193.7)
# Weight of the mice after treatment
after <-c(392.9, 393.2, 345.1, 393, 434, 427.9, 422, 383.9, 392.3, 352.2)

my_data <- data.frame( 
     group = rep(c("before", "after"), each = 10),
     weight = c(before,  after),
     subject = factor(rep(1:10, 2)))

aov(weight ~ subject + group, data  = my_data) |> summary()
#             Df Sum Sq Mean Sq F value  Pr(>F)    
# subject      9   6946     772    1.78   0.202    
# group        1 189132  189132  436.11 6.2e-09 ***
# Residuals    9   3903     434                    
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

t.test(weight ~ group, data = my_data, paired = TRUE)
# 	Paired t-test
#
# data:  weight by group
# t = 20.883, df = 9, p-value = 6.2e-09

t.test(after-before)
# 	One Sample t-test
# 
# data:  after - before
# t = 20.883, df = 9, p-value = 6.2e-09

Or using repeated measure ANOVA with Error() function to incorporate a random factor. See How to Perform a Nested ANOVA in R (Step-by-Step) adn How to Perform a Repeated Measures ANOVA in R.

aov(weight ~ group + Error(subject), data  = my_data) |> summary()
# Assume same 10 random subjects for both group1 and group2
# 
# Error: subject
#           Df Sum Sq Mean Sq F value Pr(>F)
# Residuals  9   6946   771.8               
# 
# Error: Within
#           Df Sum Sq Mean Sq F value  Pr(>F)    
# group      1 189132  189132   436.1 6.2e-09 ***
# Residuals  9   3903     434                    

# Below is a nested design, not repeated measure.
aov(weight ~ group + Error(subject/group), data  = my_data) |> summary() 
# Assume subjects 1-10 and 11-20 in group1 and group2 are different. 
# So this model is not right in this case.
# 
# Error: subject
#           Df Sum Sq Mean Sq F value Pr(>F)
# Residuals  9   6946   771.8               
#
# Error: subject:group
#           Df Sum Sq Mean Sq F value  Pr(>F)    
# group      1 189132  189132   436.1 6.2e-09 ***
# Residuals  9   3903     434

Paired t-test as a special case of linear mixed-effect modeling

na.action = "na.pass"

error-prone t.test(<formula>, paired = TRUE)

Z-value/Z-score

If the population parameters are known, then rather than computing the t-statistic, one can compute the z-score.

Check assumptions

To test or not to test: Preliminary assessment of normality when comparing two independent samples 2012
Parametric tests are used when the data being analyzed meet certain assumptions, such as normality and homogeneity of variance.
- To check for normality, you can use graphical methods such as a histogram or a Q-Q plot. You can also use statistical tests such as the Shapiro-Wilk test or the Anderson-Darling test. The shapiro.test() function can be used to perform a Shapiro-Wilk test in R.
- To check for homogeneity of variance, you can use graphical methods such as a box plot or a scatter plot. You can also use statistical tests such as Levene’s test or Bartlett’s test. The leveneTest() function from the car package can be used to perform Levene’s test in R
- Here is why I don’t care about the Levene’s test

mydata <- PlantGrowth

# Check for normality
shapiro.test(mydata$weight)

# Check for homogeneity of variance
library(car)
leveneTest(weight ~ group, data = mydata)

Nonparametric test: Wilcoxon rank-sum test/Mann–Whitney U test for 2-sample data

Mann–Whitney U test/Wilcoxon rank-sum test or Wilcoxon–Mann–Whitney test
Sensitive to differences in location
wilcox.test()
Wilcoxon test in R: how to compare 2 groups under the non-normality assumption
Wilcoxon-Mann-Whitney test and a small sample size. if performing a WMW test comparing S1=(1,2) and S2=(100,300) it wouldn’t differ of comparing S1=(1,2) and S2=(4,5). Therefore when having a small sample size this is a great loss of information.

Nonparametric test: Kolmogorov-Smirnov test

Sensitive to difference in shape and location of the distribution functions of two groups

Kolmogorov–Smirnov test. The Kolmogorov–Smirnov statistic for a given cumulative distribution function F(x) is

[math]\displaystyle{ D_n= \sup_x |F_n(x)-F(x)| }[/math]

where sup_x is the supremum of the set of distances.

Gene set enrichment analysis The Enrichment score ES is is the maximum deviation from zero of P(hit)-P(miss). For a randomly distributed S, ES will be relatively small, but if it is concentrated at the top or bottom of the list, or otherwise nonrandomly distributed, then ES(S) will be correspondingly high. When p=0, ES reduces to the standard Kolmogorov–Smirnov statistic. c; when p=1, we are weighting the genes in S by their correlation with C normalized by the sum of the correlations over all of the genes in S.

[math]\displaystyle{ P_{hit}(S, i) = \sum \frac{|r_j|^p}{N_R}, P_{miss}(S, i)= \sum \frac{1}{(N-N_H)} }[/math] where [math]\displaystyle{ N_R = \sum_{g_j \in S} |r_j|^p }[/math].

Limma: Empirical Bayes method

Some Bioconductor packages: limma, RnBeads, IMA, minfi packages.
The moderated T-statistics used in Limma is defined on Limma's user guide.
- Limma assumes that the variances of the genes from a certain distribution (Gamma distribution) and estimates this distribution from the data. The variance of each gene is then "shrunk" towards the mean of this distribution. This shrinkage is more pronounced for genes with low counts or high variability, thereby stabilizing their variance estimates.
- The log fold changes are then calculated based on these shrunken variances. This has the effect of making the log fold changes more reliable and the resulting p-values more accurate. This is particularly important when you are dealing with multiple testing correction, as is often the case in differential expression analysis.

From PennStat
Interpretation of the moderated t-statistics and B-statistics
不同方法的结果比较转录组差异分析（DESeq2+limma+edgeR+t-test/wilcox-test）总结.

Diagram of usage ?makeContrasts, ?contrasts.fit, ?eBayes

           lmFit        contrasts.fit           eBayes       topTable
        x ------> fit ------------------> fit2  -----> fit2  --------->
                   ^                       ^
                   |                       |
    model.matrix   |    makeContrasts      |
class ---------> design ----------> contrasts

Moderated t-test mod.t.test() from MKmisc package

Examples of contrasts (search contrasts.fit and/or model.matrix from the user guide)

# Ex 1 (Single channel design):
design <- model.matrix(~ 0+factor(c(1,1,1,2,2,3,3,3))) # number of samples x number of groups
colnames(design) <- c("group1", "group2", "group3")
fit <- lmFit(eset, design)
contrast.matrix <- makeContrasts(group2-group1, group3-group2, group3-group1, 
                                 levels=design)        # number of groups x number of contrasts
fit2 <- contrasts.fit(fit, contrast.matrix)
fit2 <- eBayes(fit2)
topTable(fit2, coef=1, adjust="BH")
topTable(fit2, coef=1, sort = "none", n = Inf, adjust="BH")$adj.P.Val

# Ex 2 (Common reference design):
targets <- readTargets("runxtargets.txt")
design <- modelMatrix(targets, ref="EGFP")
contrast.matrix <- makeContrasts(AML1,CBFb,AML1.CBFb,AML1.CBFb-AML1,AML1.CBFb-CBFb,
                                 levels=design)
fit <- lmFit(MA, design)
fit2 <- contrasts.fit(fit, contrasts.matrix)
fit2 <- eBayes(fit2)

# Ex 3 (Direct two-color design):
design <- modelMatrix(targets, ref="CD4")
contrast.matrix <- cbind("CD8-CD4"=c(1,0),"DN-CD4"=c(0,1),"CD8-DN"=c(1,-1))
rownames(contrast.matrix) <- colnames(design)
fit <- lmFit(eset, design)
fit2 <- contrasts.fit(fit, contrast.matrix)

# Ex 4 (Single channel + Two groups):
fit <- lmFit(eset, design)
cont.matrix <- makeContrasts(MUvsWT=MU-WT, levels=design)
fit2 <- contrasts.fit(fit, cont.matrix)
fit2 <- eBayes(fit2)

# Ex 5 (Single channel + Several groups):
f <- factor(targets$Target, levels=c("RNA1","RNA2","RNA3"))
design <- model.matrix(~0+f)
colnames(design) <- c("RNA1","RNA2","RNA3")
fit <- lmFit(eset, design)
contrast.matrix <- makeContrasts(RNA2-RNA1, RNA3-RNA2, RNA3-RNA1,
                                 levels=design)
fit2 <- contrasts.fit(fit, contrast.matrix)
fit2 <- eBayes(fit2)

# Ex 6 (Single channel + Interaction models 2x2 Factorial Designs) :
cont.matrix <- makeContrasts(
  SvsUinWT=WT.S-WT.U,
  SvsUinMu=Mu.S-Mu.U,
  Diff=(Mu.S-Mu.U)-(WT.S-WT.U),
  levels=design)
fit2 <- contrasts.fit(fit, cont.matrix)
fit2 <- eBayes(fit2)

Example from user guide 17.3 (Mammary progenitor cell populations)

setwd("~/Downloads/IlluminaCaseStudy")
url <- c("http://bioinf.wehi.edu.au/marray/IlluminaCaseStudy/probe%20profile.txt.gz",
  "http://bioinf.wehi.edu.au/marray/IlluminaCaseStudy/control%20probe%20profile.txt.gz",
  "http://bioinf.wehi.edu.au/marray/IlluminaCaseStudy/Targets.txt")
for(i in url)  system(paste("wget ", i))
system("gunzip probe%20profile.txt.gz")
system("gunzip control%20probe%20profile.txt.gz")

source("http://www.bioconductor.org/biocLite.R")
biocLite("limma")
biocLite("statmod")
library(limma)
targets <- readTargets()
targets

x <- read.ilmn(files="probe profile.txt",ctrlfiles="control probe profile.txt",
  other.columns="Detection")
options(digits=3)
head(x$E)
boxplot(log2(x$E),range=0,ylab="log2 intensity")
y <- neqc(x)
dim(y)
expressed <- rowSums(y$other$Detection < 0.05) >= 3
y <- y[expressed,]
dim(y) # 24691 12
plotMDS(y,labels=targets$CellType)

ct <- factor(targets$CellType)
design <- model.matrix(~0+ct)
colnames(design) <- levels(ct)
dupcor <- duplicateCorrelation(y,design,block=targets$Donor) # need statmod
dupcor$consensus.correlation

fit <- lmFit(y, design, block=targets$Donor, correlation=dupcor$consensus.correlation)
contrasts <- makeContrasts(ML-MS, LP-MS, ML-LP, levels=design)
fit2 <- contrasts.fit(fit, contrasts)
fit2 <- eBayes(fit2, trend=TRUE)
summary(decideTests(fit2, method="global"))
topTable(fit2, coef=1) # Top ten differentially expressed probes between ML and MS
#                 SYMBOL TargetID logFC AveExpr     t  P.Value adj.P.Val    B
# ILMN_1766707     IL17B     <NA> -4.19    5.94 -29.0 2.51e-12  5.19e-08 18.1
# ILMN_1706051      PLD5     <NA> -4.00    5.67 -27.8 4.20e-12  5.19e-08 17.7
# ...
tT <- topTable(fit2, coef=1, number = Inf)
dim(tT)
# [1] 24691     8

Three groups comparison (What is the difference of A vs Other AND A vs (B+C)/2?). Contrasts comparing one factor to multiple others

library(limma)
set.seed(1234)
n <- 100
testexpr <- matrix(rnorm(n * 10, 5, 1), nc= 10)
testexpr[, 6:7] <- testexpr[, 6:7] + 7  # mean is 12

design1 <- model.matrix(~ 0 + as.factor(c(rep(1,5),2,2,3,3,3)))
design2 <- matrix(c(rep(1,5),rep(0,5),rep(0,5),rep(1,5)),ncol=2)
colnames(design1) <- LETTERS[1:3]
colnames(design2) <- c("A", "Other")

fit1 <- lmFit(testexpr,design1)
contrasts.matrix1 <- makeContrasts("AvsOther"=A-(B+C)/2, levels = design1)
fit1 <- eBayes(contrasts.fit(fit1,contrasts=contrasts.matrix1))

fit2 <- lmFit(testexpr,design2)
contrasts.matrix2 <- makeContrasts("AvsOther"=A-Other, levels = design2)
fit2 <- eBayes(contrasts.fit(fit2,contrasts=contrasts.matrix2))

t1 <- topTable(fit1,coef=1, number = Inf)
t2 <- topTable(fit2,coef=1, number = Inf)

rbind(head(t1, 3), tail(t1, 3))
#        logFC  AveExpr         t      P.Value    adj.P.Val         B
# 92 -5.293932 5.810926 -8.200138 1.147084e-15 1.147084e-13 26.335702
# 81 -5.045682 5.949507 -7.815607 2.009706e-14 1.004853e-12 23.334600
# 37 -4.720906 6.182821 -7.312539 7.186627e-13 2.395542e-11 19.625964
# 27 -2.127055 6.854324 -3.294744 1.034742e-03 1.055859e-03 -1.141991
# 86 -1.938148 7.153142 -3.002133 2.776390e-03 2.804434e-03 -2.039869
# 75 -1.876490 6.516004 -2.906626 3.768951e-03 3.768951e-03 -2.314869
rbind(head(t2, 3), tail(t2, 3))
#        logFC  AveExpr          t    P.Value adj.P.Val         B
# 92 -4.518551 5.810926 -2.5022436 0.01253944 0.2367295 -4.587080
# 81 -4.500503 5.949507 -2.4922492 0.01289503 0.2367295 -4.587156
# 37 -4.111158 6.182821 -2.2766414 0.02307100 0.2367295 -4.588728
# 27 -1.496546 6.854324 -0.8287440 0.40749644 0.4158127 -4.595601
# 86 -1.341607 7.153142 -0.7429435 0.45773401 0.4623576 -4.595807
# 75 -1.171366 6.516004 -0.6486690 0.51673851 0.5167385 -4.596008

var(as.numeric(testexpr[, 6:10]))
# [1] 12.38074
var(as.numeric(testexpr[, 6:7]))
# [1] 0.8501378
var(as.numeric(testexpr[, 8:10]))
# [1] 0.9640699

As we can see the p-values returned from the first contrast are very small (large mean but small variance) but the p-values returned from the 2nd contrast are large (still large mean but very large variance). The variance from the "Other" group can be calculated from a mixture distribution ( pdf = .4 N(12, 1) + .6 N(5, 1), VarY = E(Y^2) - (EY)^2 where E(Y^2) = .4 (VarX1 + (EX1)^2) + .6 (VarX2 + (EX2)^2) = 73.6 and EY = .4 * 12 + .6 * 5 = 7.8; so VarY = 73.6 - 7.8^2 = 12.76).

Correct assumptions of using limma moderated t-test and the paper Should We Abandon the t-Test in the Analysis of Gene Expression Microarray Data: A Comparison of Variance Modeling Strategies.
- Evaluation: statistical power (figure 3, 4, 5), false-positive rate (table 2), execution time and ease of use (table 3)
- Limma presents several advantages
- RVM inflates the expected number of false-positives when sample size is small. On the other hand the, RVM is very close to Limma from either their formulas (p3 of the supporting info) or the Hierarchical clustering (figure 2) of two examples.
- Slides

Use Limma to run ordinary T tests, Limma Moderated and Ordinary t-statistics

# where 'fit' is the output from lmFit() or contrasts.fit().
unmod.t <- fit$coefficients/fit$stdev.unscaled/fit$sigma
pval <- 2*pt(-abs(unmod.t), fit$df.residual)

# Following the above example
t.test(testexpr[1, 1:5], testexpr[1, 6:10], var.equal = T)
# Two Sample t-test
# 
# data:  testexpr[1, 1:5] and testexpr[1, 6:10]
# t = -1.2404, df = 8, p-value = 0.25
# alternative hypothesis: true difference in means is not equal to 0
# 95 percent confidence interval:
#   -7.987791  2.400082
# sample estimates:
#   mean of x mean of y 
# 4.577183  7.371037 
fit2$coefficients[1] / (fit2$stdev.unscaled[1] * fit2$sigma[1]) # Ordinary t-statistic
# [1] -1.240416
fit2$coefficients[1] / (fit2$stdev.unscaled[1] * sqrt(fit2$s2.post[1])) # moderated t-statistic
# [1] -1.547156
topTable(fit2,coef=1, sort.by = "none")[1,]
#       logFC  AveExpr         t   P.Value adj.P.Val         B
# 1 -2.793855 5.974110 -1.547156 0.1222210 0.2367295 -4.592992

# Square root of the pooled variance
fit2$sigma[1]
# [1] 3.561284
(((5-1)*var(testexpr[1, 1:5]) + (5-1)*var(testexpr[1, 6:10]))/(5+5-2)) %>% sqrt()
# [1] 3.561284

Comparison of ordinary T-statistic, RVM T-statistic and Limma/eBayes moderated T-statistic.

	Test statistic for gene g
Ordinary T-test	[math]\displaystyle{ \frac{\overline{y}_{g1} - \overline{y}_{g2}}{S_g^{Pooled}/\sqrt{1/n_1 + 1/n_2}} }[/math]	[math]\displaystyle{ (S_g^{Pooled})^2 = \frac{(n_1-1)S_{g1}^2 + (n_2-1)S_{g2}^2}{n1+n2-2} }[/math]
RVM	[math]\displaystyle{ \frac{\overline{y}_{g1} - \overline{y}_{g2}}{S_g^{RVM}/\sqrt{1/n_1 + 1/n_2}} }[/math]	[math]\displaystyle{ (S_g^{RVM})^2 = \frac{(n_1+n_2-2)S_{g}^2 + 2a(ab)^{-1}}{n1+n2-2+2a} }[/math]
Limma	[math]\displaystyle{ \frac{\overline{y}_{g1} - \overline{y}_{g2}}{S_g^{Limma}/\sqrt{1/n_1 + 1/n_2}} }[/math]	[math]\displaystyle{ (S_g^{Limma})^2 = \frac{d_0 S_0^2 + d_g S_g^2}{d_0 + d_g} }[/math]

In Limma,
- [math]\displaystyle{ \sigma_g^2 }[/math] assumes an inverse Chi-square distribution with mean [math]\displaystyle{ S_0^2 }[/math] and [math]\displaystyle{ d_0 }[/math] degrees of freedom
- [math]\displaystyle{ d_0 }[/math] (fit$df.prior) and [math]\displaystyle{ d_g }[/math] are, respectively, prior and residual/empirical degrees of freedom.
- [math]\displaystyle{ S_0^2 }[/math] (fit$s2.prior) is the prior distribution and [math]\displaystyle{ S_g^2 }[/math] is the pooled variance.
- [math]\displaystyle{ (S_g^{Limma})^2 }[/math] can be obtained from fit$s2.post.
Empirical Bayes estimation of normal means, accounting for uncertainty in estimated standard errors Lu 2019

t.test() vs aov() vs sva::f.pvalue() vs genefilter::rowttests()

t.test() & aov() & sva::f.pvalue() & AT function will be equivalent if we assume equal variances in groups (not the default in t.test)
See examples in gist.

# Method 1:
tmp <- data.frame(groups=groups.gem, 
                  x=combat_edata3["TDRD7", 29:40])
anova(aov(x ~ groups, data = tmp))
# Analysis of Variance Table
# 
# Response: x
#           Df Sum Sq Mean Sq F value Pr(>F)
# groups     1 0.0659 0.06591  0.1522 0.7047

# Method 2:
t.test(combat_edata3["TDRD7", 29:40][groups.gem == "CR"], 
       combat_edata3["TDRD7", 29:40][groups.gem == "PD"])
# 0.7134
t.test(combat_edata3["TDRD7", 29:40][groups.gem == "CR"], 
       combat_edata3["TDRD7", 29:40][groups.gem == "PD"], var.equal = TRUE)
# 0.7047

# Method 3:
require(sva)
pheno <- data.frame(groups.gem = groups.gem)
mod = model.matrix(~as.factor(groups.gem), data = pheno)
mod0 = model.matrix(~1, data = pheno)

f.pvalue(combat_edata3[c("TDRD7", "COX7A2"), 29:40], 
         mod, mod0)
#     TDRD7    COX7A2 
# 0.7046624 0.2516682

# Method 4: 
# load some functions from AT randomForest plugin
tmp2 <- Vat2(combat_edata3[c("TDRD7", "COX7A2"), 29:40], 
             ifelse(groups.gem == "CR", 0, 1), 
             grp=c(0,1), rvm = FALSE)
tmp2$tp
#     TDRD7    COX7A2 
# 0.7046624 0.2516682 

# Method 5: https://stats.stackexchange.com/a/474335
# library(genefilter)

# Method 6:
# library(limma)

sapply + lm()

using lm() in R for a series of independent fits. Gene expression is on response part (Y). We have nG columns on Y. We'll fit a regression for each column of Y and the same X.

Treatment effect estimation

Model-assisted inference for treatment effects using regularized calibrated estimation with high-dimensional data

Permutation test

https://en.wikipedia.org/wiki/Permutation_test
Permutation Statistical Methods with R by Berry, et al.
7 Modern classical statistics from "Modern Statistics with R" by Thulin

run a permutation test in R to compare the means of two groups. See also Permutation tests notes by Rice & Lumley

# example data
group1 <- c(1, 2, 3, 4, 5)
group2 <- c(6, 7, 8, 9, 10)

# calculate observed test statistic
observed_statistic <- abs(mean(group1) - mean(group2))

# combine data into one vector
combined_data <- c(group1, group2)

# number of permutations
n_permutations <- 1000

# initialize vector to store permuted test statistics
permuted_statistics <- numeric(n_permutations)
set.seed(1)
# run permutation test
for (i in seq_len(n_permutations)) {
    # permute combined data
    permuted_data <- sample(combined_data)
    
    # split permuted data into two groups
    permuted_group1 <- permuted_data[seq_along(group1)]
    permuted_group2 <- permuted_data[(length(group1) + 1):length(combined_data)]
    
    # calculate permuted test statistic
    permuted_statistics[i] <- abs(mean(permuted_group1) - mean(permuted_group2))
}

# calculate p-value
p_value <- mean(permuted_statistics >= observed_statistic)
p_value
# [1] 0.013

t.test(group1, group2)$p.value * 0.5
# [1] 0.0005264129

Permutation tests may not always be as powerful as parametric t-tests when the assumptions of the t-test are met. This means that they may require larger sample sizes to achieve the same level of statistical power.

The smallest possible p-value is equal to 1 / (number of permutations + 1) if the observed test statistic is included in permutation. For example, if 1000 permutations are performed, the smallest possible p-value is 1 / (1000 + 1) = 0.000999. Since the observed test statistic is included in one of permutations, the smallest possible p-value is 1 / (number of permutations + 1), where +1 accounts for the observed test statistic. Fewer permutations, more accurate P-values (pseudo count).
Could permutation test p-value be zero? Permutation P-values Should Never Be Zero: Calculating Exact P-values When Permutations Are Randomly Drawn by Phipson and Smyth
When Your Permutation Test is Doomed to Fail
Permutation P-values should never be zero: calculating exact P-values when permutations are randomly drawn Phipson & Smyth 2010

getPvals() from singscore package. The null hypothesis is that the gene-set is not enriched in the sample. Question: it seems the permutation p-value is wrong. It should be calculated by using the tail values not by one-sided because the scores can be positive or negative.

# observed: 1 x nsamples
# permuteResults: B x nsamples
# pvals: 1 x nsamples
pvals <- colSums(permuteResult > 
                matrix(1, nrow = nrow(permuteResult),
                          ncol = 1) %*% observed) / B
# Consider B=1000, nsamples=5
# e.g. pvals = (0.996, 0.000, 0.999, 0.000, 0.530) 
pvals <- sapply(pvals, max, 1/B)
# e.g. pvals = (0.996, 0.001, 0.999, 0.001, 0.530)

range(permuteResult[,1])
# [1] -0.09813482  0.10853650
round(scoredf[,"TotalScore"], 3)
# [1] -0.088  0.287 -0.099  0.271 -0.002  0.176  0.017  0.188 -0.062 -0.065

In ArrayTools, LS score (= -sum (log p_j)) is always > 0. So the permutation p-value is calculated using one-sided.
GSA package. Correct p-value in GSA (Gene set enrichment) permutation tests? Why it uses one-sided?
Two-sided permutation test vs. two one-sided
- FAQ: What are the differences between one-tailed and two-tailed tests?
- Why do we use a one-tailed test F-test in analysis of variance (ANOVA)?
https://faculty.washington.edu/kenrice/sisg/SISG-08-06.pdf
How about the fgsea package?
Roughly estimate the number of permutations based on the desired FDR GSEA for RNA-seq analysis

A/B testing

Degrees of freedom

https://en.wikipedia.org/wiki/Degrees_of_freedom, Degrees of freedom (statistics)
t distribution. When the sample size is large, the df becomes very large and t distribution approximates the normal distribution.
Chi-squared distribution. When the sample size is large, standardized chi-squared distribution approximates the normal distribution.
The degrees of freedom for a chi-squared test depend on the type of test being performed. Here are some common examples:
- Chi-squared goodness-of-fit test: The degrees of freedom are equal to the number of categories minus 1. For example, if you are testing whether a six-sided die is fair, there are six categories (one for each side of the die), so the degrees of freedom would be 6 - 1 = 5.
- Chi-squared test of independence: The degrees of freedom are equal to (number of rows - 1) * (number of columns - 1). For example, if you have a contingency table with 3 rows and 4 columns, the degrees of freedom would be (3 - 1) * (4 - 1) = 6.
- Chi-squared test for homogeneity: The degrees of freedom are equal to (number of groups - 1) * (number of categories - 1). For example, if you have three groups and four categories, the degrees of freedom would be (3 - 1) * (4 - 1) = 6.
By its definition, chi-squared distribution is the same of squared normal distribution.
The square of t distribution follows F(1, df). The F distribution is the ratio of two chi-squared random variables, each divided by its degrees of freedom.

ANOVA

Practical Regression and Anova using R by Julian J. Faraway, 2002
A simple ANOVA
Repeated measures ANOVA in R Exercises
Mixed models for ANOVA designs with one observation per unit of observation and cell of the design
afex package, afex_plot(): Publication-Ready Plots for Factorial Designs
Experiment designs for Agriculture
ANOVA in R from statsandr.com
Design and Analysis of Experiments by Montgomery
- R Companion to Montgomery’s Design and Analysis of Experiments (2005)
https://cran.r-project.org/web/views/ExperimentalDesign.html

Partition of sum of squares

https://en.wikipedia.org/wiki/Partition_of_sums_of_squares

Extract p-values

my_anova <- aov(response_variable ~ factor_variable, data = my_data)

# Extract the p-value
p_value <- summary(my_anova)[[1]]$'Pr(>F)'[1]

aov() allows multiple response variables. In a single response variable case, we still need to use [[1]] to subset the result.

F-test in anova

How is the F-statistic computed in anova() when there are multiple models?

F-test and likelihood ratio test

Since F is a monotone function of the likelihood ratio statistic, the F-test is a likelihood ratio test; See Wikipedia.

Common tests are linear models

https://lindeloev.github.io/tests-as-linear/

ANOVA Vs Multiple Comparisons

Descriptive Analysis by Groups

compareGroups: Descriptive Analysis by Groups. Cf tableone package.

Post-hoc test

Determine which levels have significantly different means.

http://jamesmarquezportfolio.com/one_way_anova_with_post_hocs_in_r.html
pairwise.t.test() for one-way ANOVA
Post-hoc Pairwise Comparisons of Two-way ANOVA using TukeyHSD().

The "TukeyHSD()" function is then used to perform a post-hoc multiple comparisons test to compare the treatment means, taking into account the effect of the block variable.

summary(fm1 <- aov(breaks ~ wool + tension, data = warpbreaks))
#             Df Sum Sq Mean Sq F value  Pr(>F)   
# wool         1    451   450.7   3.339 0.07361 . 
# tension      2   2034  1017.1   7.537 0.00138 **
# Residuals   50   6748   135.0                   

TukeyHSD(fm1, "tension", ordered = TRUE)
#   Tukey multiple comparisons of means
#    95% family-wise confidence level
#    factor levels have been ordered
#
# Fit: aov(formula = breaks ~ wool + tension, data = warpbreaks)
#
# $tension
#          diff        lwr      upr     p adj
# M-H  4.722222 -4.6311985 14.07564 0.4474210
# L-H 14.722222  5.3688015 24.07564 0.0011218
# L-M 10.000000  0.6465793 19.35342 0.0336262

tapply(warpbreaks$breaks, warpbreaks$tension, mean)
#        L        M        H 
# 36.38889 26.38889 21.66667 
26.38889 - 21.66667 # [1] 4.72222   #  M-H
36.38889 - 21.66667 # [1] 14.72222  #  L-H
36.38889 - 26.38889 # [1] 10        #  L-M

plot(TukeyHSD(fm1, "tension"))

post-hoc tests: pairwise.t.test versus TukeyHSD test
How to Perform Dunnett’s Test in R
How to do Pairwise Comparisons in R?
[https://www.statology.org/tukey-vs-bonferroni-vs-scheffe/

TukeyHSD (Honestly Significant Difference), diagnostic checking

https://datascienceplus.com/one-way-anova-in-r/, Tukey HSD for Post-Hoc Analysis (detailed explanation including the type 1 error problem in multiple testings)

TukeyHSD for the pairwise tests

You can’t just perform a series of t tests, because that would greatly increase your likelihood of a Type I error.
compute something analogous to a t score for each pair of means, but you don’t compare it to the Student’s t distribution. Instead, you use a new distribution called the studentized range (from Wikipedia) or q distribution.
Suppose that we take a sample of size n from each of k populations with the same normal distribution N(μ, σ) and suppose that [math]\displaystyle{ \bar{y} }[/math]_min is the smallest of these sample means and [math]\displaystyle{ \bar{y} }[/math]_max is the largest of these sample means, and suppose S² is the pooled sample variance from these samples. Then the following random variable has a Studentized range distribution: [math]\displaystyle{ q = \frac{\overline{y}_{\max} - \overline{y}_{\min}}{S/\sqrt{n}} }[/math]

One-Way ANOVA Test in R from sthda.com.

res.aov <- aov(weight ~ group, data = PlantGrowth)
summary(res.aov)
#              Df Sum Sq Mean Sq F value Pr(>F)  
#  group        2  3.766  1.8832   4.846 0.0159 *
#  Residuals   27 10.492  0.3886                 
TukeyHSD(res.aov)
# Tukey multiple comparisons of means
# 95% family-wise confidence level
# 
# Fit: aov(formula = weight ~ group, data = PlantGrowth)
# 
# $group
#             diff        lwr       upr     p adj
# trt1-ctrl -0.371 -1.0622161 0.3202161 0.3908711
# trt2-ctrl  0.494 -0.1972161 1.1852161 0.1979960
# trt2-trt1  0.865  0.1737839 1.5562161 0.0120064

# Extra:
# Check your data
my_data <- PlantGrowth
levels(my_data$group)
set.seed(1234)
dplyr::sample_n(my_data, 10)

# compute the summary statistics by group
library(dplyr)
group_by(my_data, group) %>%
  summarise(
    count = n(),
    mean = mean(weight, na.rm = TRUE),
    sd = sd(weight, na.rm = TRUE)
  )

Or we can use Benjamini-Hochberg method for p-value adjustment in pairwise comparisons

library(multcomp)
pairwise.t.test(my_data$weight, my_data$group,
                 p.adjust.method = "BH")
#      ctrl  trt1 
# trt1 0.194 -    
# trt2 0.132 0.013
#
# P value adjustment method: BH

Shapiro-Wilk test for normality

# Extract the residuals
aov_residuals <- residuals(object = res.aov )
# Run Shapiro-Wilk test
shapiro.test(x = aov_residuals )

Bartlett test and Levene test for the homogeneity of variances across the groups

Repeated measure

How to do Repeated Measures ANOVAs in R
Cross-over Repeated Measure Designs
Cross-over study design with a major constraint
Repeated Measures of ANOVA in R Complete Tutorial
How to write the error term in repeated measures ANOVA in R: Error(subject) vs Error(subject/time)
Specifying the Error() term in repeated measures ANOVA in R
Simple example of using Error().
```
aov(y ~ x + Error(random_factor), data=mydata)

# y=yield of a crop (measured in bushels per acre)
# x=fertilizer: fertilizer treatment (1 = control, 2 = treatment A, 3 = treatment B)
# random_factor=field (1, 2, 3, ...)
# H1: there is a difference in yield due to fertilizer treatment, 
# while accounting for the fact that the fields may have different yields.
```
- The Error() function is used to specify the random factor, which is assumed to be nested within the other factors in the model. This means that the levels of the random factor are not assumed to be independent of one another, and that the error term should be adjusted accordingly.
- The Error() function is used when you want to specify the random factor in your model. This function is used when you want to specify the random factor in your model. A random factor is a factor that is assumed to be nested within another factor in the model. The error term is adjusted accordingly.
- This aov() model assumes that the yield from different fields might be different, and the error term should be adjusted accordingly. By including the random factor "field" in the Error() function, we are accounting for the fact that the fields may have different yields. The analysis of variance will test whether there is a significant difference in yield due to fertilizer treatment, while adjusting for the difference in yields between fields.
More complicate example
```
aov(y ~ x + Error(subject/x) )

# y: test scores (measured in percentage)
# x: teaching method (1 = traditional, 2 = online)
# subject: student ID
# H1: We want to test whether there is a difference in test scores due to teaching method, 
# while accounting for the fact that different students may have different baseline test scores.
```
- The formula "y ~ x + Error(subject/x)" means that there is a fixed factor "x" and a random factor "subject", which is nested within the levels of "x". The term "subject/x" in the Error() function is specifying that the random factor "subject" is nested within the levels of the fixed factor "x".
- By including the random factor "subject" in the Error() function, we are accounting for the fact that the subjects may have different baseline levels of "y" and that this might affect the results.
- This aov() model assumes that the test scores from different students might be different, and the error term should be adjusted accordingly. By including the random factor "subject" in the Error() function, we are accounting for the fact that different students may have different baseline test scores and that this might affect the results.
- The term "subject/x" in the Error() function is specifying that the random factor "subject" is nested within the levels of the fixed factor "x". In other words, each subject is tested under different teaching methods. By including the random factor "subject" in the Error() function, we are accounting for the fact that the subjects may have different baseline levels of "y" and different teaching methods might affect them differently.
The error term is adjusted accordingly.
How to Perform a Nested ANOVA in R (Step-by-Step). Note the number of subjects (technicians, random variable, they are nested within fertilizer/fixed random variable) are different for each level of the fixed random variable.

Nested

See the repeated measure ANOVA section.

Combining insignificant factor levels

COMBINING AUTOMATICALLY FACTOR LEVELS IN R

Omnibus tests

https://en.wikipedia.org/wiki/Omnibus_test
Understanding the definition of omnibus tests Tests are refereed to as omnibus if after rejecting the null hypothesis you do not know where the differences assessed by the statistical test are. In the case of F tests they are omnibus when there is more than one df in the numerator (3 or more groups) it is omnibus.

One-way ANOVA

https://www.mathstat.dal.ca/~stat2080/Fall14/Lecturenotes/anova1.pdf

Randomized block design

What is a randomized block design?
- In a randomized block design, the subjects are first divided into blocks based on their similarities, and then they are randomly assigned to the treatment groups within each block.
How to interpret the result from a randomized block design?
- If the results are statistically significant, it means that there is a significant difference between the treatment groups in terms of the response variable. This indicates that the treatment had an effect on the response variable, and that this effect is not likely due to chance alone.
How to incorporate the block variable in the interpretation?
- In a randomized block design, the block variable is a characteristic that is used to group the subjects or experimental units into blocks. The goal of using a block variable is to control for the effects of this characteristic, so that the effects of the experimental variables can be more accurately measured.
- To incorporate the block variable into the interpretation of the results, you will need to consider whether the block variable had an effect on the response variable.
- If the block variable had a significant effect on the response variable, it means that the results may be confounded by the block variable. In this case, it may be necessary to take the block variable into account when interpreting the results. For example, you might need to consider whether the treatment effects are different for different blocks, or whether the block variable is interacting with the treatment in some way.
What does that mean the result is confounded by the variable?
- If the results of an experiment are confounded by a variable, it means that the variable is influencing the results in some way and making it difficult to interpret the effects of the experimental treatment. This can occur when the variable is correlated with both the treatment and the response variable, and it can lead to incorrect conclusions about the relationship between the treatment and the response.
- For example, consider an experiment in which the treatment is a drug that is being tested for its ability to lower blood pressure. If the subjects are divided into blocks based on their age, and the results show that the drug is more effective in younger subjects than in older subjects, this could be because the drug is more effective in younger subjects, or it could be because blood pressure tends to be higher in older subjects and the effect of the drug is being masked. In this case, the results would be confounded by age, and it would be difficult to draw conclusions about the effectiveness of the drug without taking age into account.
- To avoid confounding, it is important to carefully control for any variables that could potentially influence the results of the experiment. This may involve stratifying the sample, using a matching or blocking design, or controlling for the variable in the statistical analysis. By controlling for confounding variables, you can more accurately interpret the results of the experiment and draw valid conclusions about the relationship between the treatment and the response.
What are the advantages of a randomized block design?
- Control for extraneous variables: By grouping subjects into blocks based on a similar characteristic, a randomized block design can help to control for the effects of this characteristic on the response variable. This makes it easier to accurately measure the effects of the experimental variables.
- Increased precision: Randomizing the subjects within each block helps to reduce the variance among the treatment groups. This increase the power of the statistical analysis and makes it more likely to detect true differences between the treatment groups.
- Reduced sample size: Randomized block designs typically require a smaller sample size than completely randomized designs to achieve the same level of precision. This makes it more cost-effective and logistically easier to conduct the research
- Identify the characteristics that affect the response: The block design can help to identify the characteristics of subjects or experimental units that affect the response variable. This can help to identify important factors that should be controlled for in future experiments or can be used to improve the understanding of the phenomena being studied.
- Multiple comparison: with multiple blocks allows to perform multiple comparison of the treatment groups, which can help to identify specific variables or groups that are contributing to the overall results and to understand the mechanism of the effect.
What are the disadvantages of randomized block design
- Complexity
- Difficulty in identifying blocks
- Increases experimental control
- Difficulty in interpretation: When multiple blocks are used it can be difficult to interpret the results of the experiment and it's crucial to have sufficient sample size within each block to detect the effects of the experimental variable.
Does the power of the test change with the number of levels in the block variable in randomized block design analysis?
- The analysis for a randomized block design does not change with the number of levels in the block variable, but the number of levels in the block variable does affect the power of the analysis. As the number of levels in the block variable increases, the power of the test to detect differences among treatments also increases. However, increasing the number of levels in the block variable also increases the number of blocks required, which can make the study more complex and costly to conduct.
How does the block variable affect the interpretation of the significant of the main effect in a randomized block design?
- In a randomized block design, the block variable is used to control for sources of variation that are not of interest in the study, but that may affect the response variable. By blocking, we are trying to make sure that any differences in the response variable among the treatments are due to the treatment and not due to other sources of variation.
- The main effect of the block variable represents the difference in the response variable among the blocks, regardless of the treatment. If the main effect of the block variable is significant, it means that there are systematic differences in the response variable among the blocks, and that the blocks are not exchangeable. This can affect the interpretation of the main effect of the treatment variable.
- When the main effect of the block variable is significant, it means that the blocks are not homogeneous and that the treatment effect may be different among the blocks. In this case, we cannot conclude that the treatment effect is the same across all blocks. Therefore, it's important to examine the interaction between the block and treatment variables, which tells us whether the treatment effect is the same across all blocks. If the interaction is not significant, it means that the treatment effect is the same across all blocks, and we can conclude that the treatment has a consistent effect on the response variable. However, if the interaction is significant, it means that the treatment effect varies among the blocks, and we cannot conclude that the treatment has a consistent effect on the response variable.
- In summary, when the main effect of the block variable is significant, it means that the blocks are not homogeneous, and it may affect the interpretation of the main effect of the treatment variable. The effect of treatment may vary across the blocks and it's important to examine the interaction between the block and treatment variables to understand whether the treatment effect is consistent across all blocks.
Visualize the data. Both treatment and block variables are significant.