• The assumption of normality is often difficult to assess in small sample sizes. In a t-test with n = 3 for example, this is virtually impossible

Alternatives:

• One possibility is to transform the data
• A Non-parametric alternative (in the case of the t-test) is the Wilcoxon test

Non-parametric means you are not assuming anything about the parameters of population (‘distribution-free’ tests)

Consider the size of fish (x). A log transformation makes the distribution look much more normal:

hist(x)
hist(log(x))

Consider a variable x that shows a ratio (e.g. mark in %). A square root transformation makes the distribution look much more normal:

hist(x)
hist(sqrt(x))

• Robust against an increased probability for type I and type II errors due to the violation of the assumption of normally distributed samples
• Based on ranks, the absolute numbers are irrelevant
• Therefore we do not require the data to be normally distributed
• Testing the two samples a = c(2, 3, 5) against b = c(3, 6, 8) will yield the very same results than testing x = c(0, 3, 4) against y = c(3, 23, 700). This is why:

a = c(2, 3, 5); b = c(3, 6, 8)
x = c(0, 3, 4); y = c(3, 23, 700)
rank(c(a, b))
[1] 1.0 2.5 4.0 2.5 5.0 6.0
rank(c(x, y))
[1] 1.0 2.5 4.0 2.5 5.0 6.0

If you understand the above code, you will understand how the Wilcoxon test works!

Are x and y from the same population?

• Check for normality of the samples visually

x <- c(1.83,  0.50,  3.64,  2.48, 2.68, 3.88, 1.55, 3.06, 2.30)
y <- c(0.878, 0.647, 0.598, 2.05, 1.06, 1.29, 1.06, 3.14, 1.29)
par(mfrow = c(1, 2))
qqnorm(x)
qqline(x)
qqnorm(y)
qqline(y)

• Check for normality of the samples using the Shapiro-Wilk test

shapiro.test(x)

    Shapiro-Wilk normality test

data:  x
W = 0.97555, p-value = 0.9376
shapiro.test(y)

    Shapiro-Wilk normality test

data:  y
W = 0.81992, p-value = 0.03439

What would you conclude?

If you are unsure, using both a parametric and a non-parametric test is a good idea:

t.test(x, y)

    Welch Two Sample t-test

data:  x and y
t = 2.4915, df = 14.934, p-value = 0.02498
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 0.1587134 2.0428422
sample estimates:
mean of x mean of y 
 2.435556  1.334778 

wilcox.test(x, y)

    Wilcoxon rank sum test with continuity correction

data:  x and y
W = 64, p-value = 0.04205
alternative hypothesis: true location shift is not equal to 0

You are trying to find out whether it is possible to detect a 20% change in transpiration if you apply a certain treatment to forest trees, and your funding restricts you to a sample size of 8 trees:

In a pilot study you find a mean of 4 \(Ld^{-1}\) (liters per day) with a standard deviation of 2 \(Ld^{-1}\). The expected difference is 1 \(Ld^{-1}\)

We simulate a t-test based on these assumptions, what do we conclude?

set.seed(0)
t.test(rnorm(8, mean = 4, sd = 2), rnorm(8, mean = 5, sd = 2))

    Welch Two Sample t-test

data:  rnorm(8, mean = 4, sd = 2) and rnorm(8, mean = 5, sd = 2)
t = -0.6851, df = 13.997, p-value = 0.5045
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -3.124121  1.611485
sample estimates:
mean of x mean of y 
 4.297588  5.053906 

Given the above, a t-test is unlikely to detect a potential difference, so how big would our sample size need to be?

power.t.test(delta = 1, sd = 2, power = 0.8)

     Two-sample t test power calculation 

              n = 63.76576
          delta = 1
             sd = 2
      sig.level = 0.05
          power = 0.8
    alternative = two.sided

NOTE: n is number in *each* group

This tells you that you need a sample size of c. 65 to achieve a power of 80% (i.e. 80% change to detect the difference if there is one)

Conclusion: Don’t even start the experiment if your sample size is restricted to 8, you only have about a 15% chance to detect a potential difference (see next slide)!