*(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

##################################

##################################

# 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 #

# 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

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)

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

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

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).

And that is it for this article.

Until next time,

-RD

Until next time,

-RD

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