For two-sample lognormal data, let's take a look at the results in R
of three tests (a) Welch t test, (b) Wilcoxon rank sum test,
and (c) Welch t test on logged (transformed) data.
We begin by sampling fictitious data in R and showing brief data summaries. [As usual the parameters of the lognormal distribution are those of the logged distribution.]
set.seed(1029)
x1 = rlnorm(20, 100, 13)
x2 = rlnorm(25, 113, 13)
summary(x1); length(x1); sd(x1)
Min. 1st Qu. Median Mean 3rd Qu. Max.
1.619e+34 5.139e+38 1.494e+43 5.636e+49 3.696e+46 1.124e+51
[1] 20 # sample size
[1] 2.512055e+50 # sample SD
summary(x2); length(x2); sd(x2)
Min. 1st Qu. Median Mean 3rd Qu. Max.
1.963e+39 3.081e+44 1.910e+49 4.193e+57 1.475e+53 9.602e+58
[1] 25
[1] 1.919156e+58
summary(log(x1)); sd(log(x1))
Min. 1st Qu. Median Mean 3rd Qu. Max.
78.77 87.91 97.35 96.77 106.84 117.55
[1] 11.82917
summary(log(x2)); sd(log(x2))
Min. 1st Qu. Median Mean 3rd Qu. Max.
90.48 102.44 113.47 112.75 122.43 135.81
[1] 13.02913
The Welch test on lognormal data finds no significant difference.
t.test(x1, x2)
Welch Two Sample t-test
data: x1 and x2
t = -1.0923, df = 24, p-value = 0.2855
alternative hypothesis:
true difference in means is not equal to 0
95 percent confidence interval:
-1.211453e+58 3.729247e+57
sample estimates:
mean of x mean of y
5.636360e+49 4.192642e+57
The 2-sample Wilcoxon test finds a highly significant difference
in locations. If we judge the two sample to be of 'similar shape'
we might say it's a difference in medians. If not, then x2 dominates x1.
wilcox.test(x1, x2)
Wilcoxon rank sum test
data: x1 and x2
W = 91, p-value = 0.0001614
alternative hypothesis:
true location shift is not equal to 0
The welch test on log-transformed data (which are normal) shows a significant
difference. In terms of the original data we might interpret this
as a ratio of geometric means that is not unity.
t.test(log(x1), log(x2))
Welch Two Sample t-test
data: log(x1) and log(x2)
t = -4.3032, df = 42.262, p-value = 9.764e-05
alternative hypothesis:
true difference in means is not equal to 0
95 percent confidence interval:
-23.470083 -8.486333
sample estimates:
mean of x mean of y
96.77082 112.74903
Simulation of 100,000 sets of such data show that the Welch t test
on transformed data has slightly better power to distinguish between
the two samples than does the Wilcoxon test.
Of course, results may differ for highly right-skewed data other than
lognormal.
mu1 = 100; mu2 = 113; sg = 13
# T TEST ON LOGNORMAL DATA
set.seed(1234)
pv.l = replicate(10^5, t.test(rlnorm(20,mu1,sg),
rlnorm(25,mu2,sg))$p.val)
mean(pv.l <= .05)
[1] 0.0012 # approximate power (nearly 0)
# WICOXON TEST ON LOGNORMAL DATA
set.seed(1234)
pv.w = replicate(10^5, wilcox.test(rlnorm(20,mu1,sg),
rlnorm(25,mu2,sg))$p.val)
mean(pv.w <= .05)
[1] 0.88525 # approximate power
# T TEST ON LOG-TRANSFORMED DATA (NORMAL)
set.seed(1234)
pv.n = replicate(10^5, t.test(log(rlnorm(20,mu1,sg)),
log(rlnorm(25,mu2,sg)))$p.val)
mean(pv.n <= .05)
[1] 0.90322 # approx power (slightly larger than Wilcoxon)
Both 'cures' for highly skewed non-normal data require interpretation
as to what is being compared. The choice between a Wilcoxon test on lognormal data and a Welch t test on transformed data may depend
on which interpretation seems more natural for the situation at hand.
