(SPSS does not contain the innate functionality necessary to perform this calculation)
Cohen’s d - (What it is):
Cohen’s d is utilized as a method to assess the magnitude of impact as it relates to two sample groups which are subject to differing conditions. For example, if a two sample t-test was being implemented to test a single group which received a drug, against another group which did not receive the drug, then the p-value of this test would determine whether or not the findings were significant.
Cohen’s d would measure the magnitude of the potential impact.
Cohen’s d - (When to use it):
In your statistics class.
You could also utilize this test to perform post-hoc analysis as it relates to the ANOVA model and the Student’s T-Test. However, I have never witnessed the utilization of this test outside of an academic setting.
Cohen’s d – (How to interpret it):
General Interpretation Guidelines:
Greater than or equal to 0.2 = small
Greater than or equal to 0.5 = medium
Greater than or equal to 0.8 = large
Cohen’s d – (How to state your findings):
The effect size for this analysis (d = x.xx) was found to exceed Cohen’s convention for a [small, medium, large] effect (d = .xx).
Cohen’s d – (How to derive it):
# Within the R-Programming Code Space #
##################################
# length of sample 1 (x) #
lenx <-
# length of sample 2 (y) #
leny <-
# mean of sample 1 (x) #
meanx <-
# mean of sample 2 (y)#
meany <-
# SD of sample 1 (x) #
sdx <-
# SD of sample 2 (y) #
sdy <-
varx <- sdx^2
vary <- sdy^2
lx <- lenx - 1
ly <- leny - 1
md <- abs(meanx - meany) ## mean difference (numerator)
csd <- lx * varx + ly * vary
csd <- csd/(lx + ly)
csd <- sqrt(csd) ## common sd computation
cd <- md/csd ## cohen's d
cd
##################################
# The above code is a modified version of the code found at: #
# https://stackoverflow.com/questions/15436702/estimate-cohens-d-for-effect-size #
Cohen’s d – (Example):
FIRST WE MUST RUN A TEST IN WHICH COHEN’S d CAN BE APPLIED AS AN APPROPRIATE POST-HOC TEST METHODOLOGY.
Two Sample T-Test
This test is utilized if you randomly sample different sets of items from two separate control groups.
Example:
A scientist creates a chemical which he believes changes the temperature of water. He applies this chemical to water and takes the following measurements:
70, 74, 76, 72, 75, 74, 71, 71
He then measures temperature in samples which the chemical was not applied.
74, 75, 73, 76, 74, 77, 78, 75
Can the scientist conclude, with a 95% confidence interval, that his chemical is in some way altering the temperature of the water?
For this, we will use the code:
N1 <- c(70, 74, 76, 72, 75, 74, 71, 71)
N2 <- c(74, 75, 73, 76, 74, 77, 78, 75)
t.test(N2, N1, alternative = "two.sided", var.equal = TRUE, conf.level = 0.95)
Which produces the output:
Two Sample t-test
data: N2 and N1
t = 2.4558, df = 14, p-value = 0.02773
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.3007929 4.4492071
sample estimates:
mean of x mean of y
75.250 72.875
# Note: In this case, the 95 percent confidence interval is measuring the difference of the mean values of the samples. #
# An additional option is available when running a two sample t-test, The Welch Two Sample T-Test. To utilize this option while performing a t-test, the "var.equal = TRUE" must be changed to "var.equal = FALSE". The output produced from a Welch Two Sample t-test is slightly more robust and accounts for differing sample sizes. #
From this output we can conclude:
With a p-value of 0.02773 (.0.02773 < .05), and a corresponding t-value of 2.4558, we can state that, at a 95% confidence interval, that the scientist's chemical is altering the temperature of the water.
Application of Cohen’s d
length(N1) # 8 #
length(N2) # 8 #
mean(N1) # 72.875 #
mean(N2) # 75.25 #
sd(N1) # 2.167124 #
sd(N2) # 1.669046 #
# length of sample 1 (x) #
lenx <- 8
# length of sample 2 (y) #
leny <- 8
# mean of sample 1 (x) #
meanx <- 72.875
# mean of sample 2 (y)#
meany <- 75.25
# SD of sample 1 (x) #
sdx <- 2.167124
# SD of sample 2 (y) #
sdy <- 1.669046
varx <- sdx^2
vary <- sdy^2
lx <- lenx - 1
ly <- leny - 1
md <- abs(meanx - meany) ## mean difference (numerator)
csd <- lx * varx + ly * vary
csd <- csd/(lx + ly)
csd <- sqrt(csd) ## common sd computation
cd <- md/csd ## cohen's d
cd
Which produces the output:
[1] 1.227908
##################################
From this output we can conclude:
The effect size for this analysis (d = 1.23) was found to exceed Cohen’s convention for a large effect (d = .80).
Combining both conclusions, our final written product would resemble:
With a p-value of 0.02773 (.0.02773 < .05), and a corresponding t-value of 2.4558, we can state that, at a 95% confidence interval, that the scientist's chemical is altering the temperature of the water.
The effect size for this analysis (d = 1.23) was found to exceed Cohen’s convention for a large effect (d = .80).
From this output we can conclude:
The effect size for this analysis (d = 1.23) was found to exceed Cohen’s convention for a large effect (d = .80).
Combining both conclusions, our final written product would resemble:
With a p-value of 0.02773 (.0.02773 < .05), and a corresponding t-value of 2.4558, we can state that, at a 95% confidence interval, that the scientist's chemical is altering the temperature of the water.
The effect size for this analysis (d = 1.23) was found to exceed Cohen’s convention for a large effect (d = .80).
And that is it for this article.
Until next time,
-RD
Until next time,
-RD