11.4 – Two sample effect size

Introduction

An effect size is a measure of the strength of the difference between two samples. The effect size statistic is calculated by subtracting one sample mean from the other and dividing by the pooled standard deviation.

Measures of effect size, Cohen’s d

    \begin{align*} d = \frac{\bar{X}_{1} + \bar{X}_{2}}{s_{pooled}} \end{align*}

where spooled is the pooled standard deviation for the two sample means. An equation for pooled standard deviation was provided in Chapter 3.3, but we’ll give it again here.

    \begin{align*} s_{pooled} = \sqrt{\frac{s_{1}^2 + s_{2}^2}{2}} \end{align*}

An alternative version of Cohen’s d is available for the t-test test statistic value

    \begin{align*} d = \frac{2t}{\sqrt{df}} \end{align*}

A d of one (1) indicates the effect size is equal to one standard deviation; a d of two (2) indicates the effect size between two sample means is equal to two standard deviations, and so on. Note that effect sizes complement inferential statistics such as p-values.

What makes a large effect size?

Cohen cautiously suggested that values of d

0.2 – small effect size

0.5 – medium effect size

0.8 – large effect size

That is, if the two group means don’t differ by much more than 0.2 standard deviations, than the magnitude of the treatment effect is small and unlikely to be biologically important, whereas a d = 0.8 or more would indicate a difference of 0.8 standard deviations between the sample means and, thus, likely to be an important treatment effect. Cohen (1992) provided these guidelines based on the following argument. The small effect 0.2 comes from the idea that it is much worse to conclude there is an effect when in fact there is no effect of the treatment rather than the converse (conclude no effect when there is an effect). The ratio of the Type II error (0.2) divided by the Type I error (0.05) gives us the penalty of 4. Similarly, for a moderate effect, 0.5/0.05} equals 10. Clearly, these are only guidelines (see Lakens 2013).

Examples

The difference in average body size between 6 week old females of two strains of lab mice is 0.4 g (Table 1), and increases to 1.38 g by 16 weeks (Table 2).

Table 1. Average body weights of 6 week old female mice of two different inbred strains.†

Strain \bar{X} s
C57BL/6J 18.5 0.9
CBA/J 18.1 1.27

†Source: Jackson Laboratories: C57BL/6J; CBA/J

Table 2. Average body weights of 16 week old female mice of two different inbred strains.†

Strain \bar{X} s
C57BL/6J 23.9 2.3
CBA/J 25.38 3.76

†Source: Jackson Laboratories: C57BL/6J; CBA/J

The descriptive statistics are based on weights of 360 individuals in each strain (Jackson Labs).

The differences are both statistically significant from a independent t-test, i.e., p-value less than 0.05. I’ll show you how to calculate independent t-test given summary statistics (means, standard deviations), for Table 1 data, then will ask you to do this on your own in Questions.

Write an R script, example data from Table 1

sdd1 = 0.9
var1 = sdd1^2
sdd2 = 1.27
var2 = sdd2^2
mean1 = 18.5
mean2 = 18.1
n1 = 360
n2 = 360
dff = n1+n2-2
pooledSD <-sqrt((var1+var2)/2)
pooledSEM <-sqrt(var1/n1 + var2/n2); pooledSEM
tdiff<-(mean1-mean2)/pooledSEM; tdiff
pt(tdiff, df=dff, lower.tail=FALSE)
#get two-tailed p-value
2*0.0000006675956 
#get cohen's d
2*tdiff/sqrt(dff)

Results from the calculations we report (value of the test statistic, degrees of freedom, p-value), and the effect size, then are

t = 4.875773, df = 718, p-value = 0.0000006675956
cohen's d = 0.364

Now, I’m from the school of “don’t reinvent the wheel” or “someone has already solved your problems” (Freeman et al 2008), when it comes to coding problems. And, as you would expect, of course someone has written a function to calculate the t-test given summary statistics. In addition to base R and the pwr package (see Chapter 11.5), the package BSDA contains several nice functions for power calculations.

To follow this example, install BSDA, then run the following code

require(BSDA)
tsum.test(mean1, sdd1, n1, mean2, sdd2, n2, alternative = "two.sided", mu = 0, var.equal = TRUE, conf.level = 0.95)

R output

Standard Two-Sample t-Test

data: Summarized x and y
t = 4.8758, df = 718, p-value = 0.000001335
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.2389364 0.5610636
sample estimates:
mean of x mean of y 
18.5 18.1

Similarly, Cohen’s d is available from a package called effsize.

One reason to “re-invent the wheel,” I only needed the one function; the BSDA package contains more 330 different objects/functions. A simple way to check how many objects in a package, e.g., BSDA, run

ls("package:BSDA")

BSDA stands for “Basic Statistics and Data Analysis,” and was intended to accompany the 2002 book of the same title by Larry Kitchens.

And of course, if using some else’s code, give proper citation!

Questions

  1. We needed an equation to calculate pooled standard error of the mean (pooledSEM in the R code). Read the code and write the equation used to calculate the pooled SEM.
  2. Calculate the t-test and the effect size for the Table 1 data, but at three smaller sample sizes. Change from 360 for n1 = n2 = 20, repeat for n1 = n2 = 50, and finally, repeat for n1 = n2 = 100. Use your own code, or use the tsum.test function from the BSDA package.
  3. Calculate Cohen’s effect size d for each new calculation based on different sample size.
  4. Create a table to report the p-values from the t-tests, the effect size, for each of the four n1 = n2 = (20, 50, 100, 360).
  5. True or false. The mean difference between sample means remains unaffected by sample size.
  6. True or false. The effect size between sample means remains unaffected by sample size.
  7. Based on comparisons in your table, what can you conclude about p-value and “statistical significance?” About effect size?
  8. Repeat questions 2 – 7 for Table 2.

Chapter 11 contents