Cohen d

Excel file used: ES - Cohen d (one-sample).xlsm.

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without stikpetE add-in

A flowgorithm for Cohen's d (one-sample) in Figure 1.

It takes as input paramaters the scores and the hypothesized mean.

It uses a function for the mean (CEmean), standard deviation (VAsd) which in turn use the MAsumReal function (sums array or real values).

Flowgorithm file: FL-EScohenDos.fprg.

Jupyter Notebook used in video: ES - Cohen d (one-sample) (P).ipynb.

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without stikpetP library

Jupyter Notebook used in video: ES - Cohen d (one-sample) (R).ipynb.

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without stikpetR library

This has become available in SPSS 27. As mentioned on this site, there is no option in earlier versions of SPSS to determine Cohen's D, but it can easily be calculated using by using the output from the t-test and entering the results below in the 'Online calculator' option, or using the 'Compute variable' option as shown in the video.

Enter the sample mean, the hypothesized population mean, and the sample standard deviation:

Formula's

The formula for Cohen's d for a one-sample t-test is:

\(d'=\frac{\bar{x}-\mu_{H_{0}}}{s}\)

Where x̄ is the sample mean, μ_H0 the expected mean in the population (the mean according to the null hypothesis), and s the sample standard deviation.

The formula for the sample standard deviation is:

\(s=\sqrt{\frac{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}}{n-1}}\)

In this formula x_i is the i-th score, x̄ is the sample mean, and n is the sample size.

The sample mean can be calculated using:

\(\bar{x}=\frac{\sum_{i=1}^{n}x_{i}}{n}\)

Example (different example)

We are given the ages of five students, and have an hypothesized population mean of 24. The ages of the students are:

\(X=\left\{18,21,22,19,25\right\}\)

Since there are five students, we can also set n = 5, and the hypothesized population of 24 gives \(\mu_{H_{0}}=24\)

For the standard deviation, we first need to determine the sample mean:

\(\bar{x}=\frac{\sum_{i=1}^{n}x_{i}}{n}=\frac{\sum_{i=1}^{5}x_{i}}{5}=\frac{18+21+22+19+25}{5}\)

\(=\frac{105}{5}=21\)

Then we can determine the standard deviation:

\(s=\sqrt{\frac{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}}{n-1}}=\sqrt{\frac{\sum_{i=1}^{5}\left(x_{i}-21\right)^{2}}{5-1}}=\sqrt{\frac{\sum_{i=1}^{5}\left(x_{i}-21\right)^{2}}{4}}\)

\(=\sqrt{\frac{\left(18-21\right)^{2}}{4}+\frac{\left(21-21\right)^{2}}{4}+\frac{\left(22-21\right)^{2}}{4}+\frac{\left(19-21\right)^{2}}{4}+\frac{\left(25-21\right)^{2}}{4}}\)

\(=\sqrt{\frac{\left(-3\right)^{2}}{4}+\frac{\left(0\right)^{2}}{4}+\frac{\left(1\right)^{2}}{4}+\frac{\left(-2\right)^{2}}{4}+\frac{\left(4\right)^{2}}{4}}\)

\(=\sqrt{\frac{9}{4}+\frac{0}{4}+\frac{1}{4}+\frac{4}{4}+\frac{16}{4}}=\sqrt{\frac{9+0+1+4+16}{4}}=\sqrt{\frac{30}{4}}\)

\(=\sqrt{\frac{15}{2}}=\frac{1}{2}\sqrt{15\times2}=\frac{1}{2}\sqrt{30}\approx2.74\)

Now for Cohen's d:

\(d=\frac{\bar{x}-\mu_{H_{0}}}{s}=\frac{21-24}{\frac{1}{2}\sqrt{30}}=\frac{-3}{\frac{\sqrt{30}}{2}}\)

\(=\frac{-3\times2}{\sqrt{30}}=\frac{-6}{\sqrt{30}}=\frac{-6}{\sqrt{30}}\times\frac{\sqrt{30}}{\sqrt{30}}=\frac{-6\times\sqrt{30}}{\sqrt{30}\times\sqrt{30}}\)

\(=\frac{-6\times\sqrt{30}}{30}=\frac{-6}{30}\times\sqrt{30}=-\frac{1}{5}\sqrt{30}\approx1.0954\)

Excel file: ES - Cohen d and Hedges g (ind samples) (E).xlsm

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To Be Made

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To Be Made

Jupyter Notebook: ES - Cohen d and Hedges g (ind samples) (P).ipynb

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To Be Made

without stikpetP

Jupyter Notebook: ES - Cohen d and Hedges g (ind samples) (R).ipynb

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To Be Made

without stikpetR

Enter the requested information below:

Cohen's d_s formula is:

\(d_{s}=\frac{\bar{x}_{1}-\bar{x}_{1}}{s_{pooled}}\)

In this formula \(\bar{x}_{i}\) is the mean for category i, which can be calculated by:

\(\bar{x}_{i}=\frac{\sum_{j=1}^{n_{i}}x_{i,j}}{n_{i}}\)

In this formula x_i,j is the j-th score in category i, and n_i is the number of cases in category i

s_pooled is the pooled standard deviation. In formula notation this is:

\(s_{pooled}=\sqrt\frac{SS_{1}+SS_{2}}{N-2}\)

In this formula N is the total number of cases (combination of both categories), and SS_i the sum of squared differences with the mean, which in formula notation is:

\(SS_{i}=\sum_{j=1}^{n_{i}}\left(x_{i,j}-\bar{x}_i\right)^{2}\)

An example.
Note: a different example than the one used in the rest of this section, to keep calculations a bit shorter.

Given are the scores of males (category 1) and females (category 2):

\(X_{1}=\left(8,3,2,1,1\right)\)

\(X_{2}=\left(7,7,5,3,9,8\right)\)

The first category, has 5 scores, so n₁ = 5, and for the second category we have n₂ = 6. Let's begin with determining the mean per category:

\(\bar{x}_{1}=\frac{\sum_{j=1}^{n_{1}}x_{1,j}}{n_{1}}=\frac{\sum_{j=1}^{5}x_{1,j}}{5}=\frac{8+3+2+1+1}{5}=\frac{15}{5}=3\)

\(\bar{x}_{2}=\frac{\sum_{j=1}^{n_{2}}x_{2,j}}{n_{2}}=\frac{\sum_{j=1}^{6}x_{2,j}}{6}=\frac{7+7+5+3+9+8}{6}=\frac{39}{6}=\frac{13}{2}=6.5\)

Then we can determine the sum of squares per category. First the male category:

\(SS_{1}=\sum_{j=1}^{n_{1}}\left(x_{1,j}-\bar{x}_1\right)^{2}=\sum_{j=1}^{5}\left(x_{1,j}-3\right)^{2}\)

\(=\left(8-3\right)^{2}+\left(3-3\right)^{2}+\left(2-3\right)^{2}+\left(1-3\right)^{2}+\left(1-3\right)^{2}\)

\(=\left(5\right)^{2}+\left(0\right)^{2}+\left(-1\right)^{2}+\left(-2\right)^{2}+\left(-2\right)^{2}=25+0+1+4+4=34\)

And for the female category:

\(SS_2=\sum_{j=1}^{n_2}\left(x_{2,j}-\bar{x}_2\right)^{2}=\sum_{j=1}^{6}\left(x_{2,j}-\frac{13}{2}\right)^{2}\)

\(=\left(7-\frac{13}{2}\right)^{2}+\left(7-\frac{13}{2}\right)^{2}+\left(5-\frac{13}{2}\right)^{2}+\left(3-\frac{13}{2}\right)^{2}+\left(9-\frac{13}{2}\right)^{2}+\left(8-\frac{13}{2}\right)^{2}\)

\(=\left(\frac{14}{2}-\frac{13}{2}\right)^{2}+\left(\frac{14}{2}-\frac{13}{2}\right)^{2}+\left(\frac{10}{2}-\frac{13}{2}\right)^{2}+\left(\frac{6}{2}-\frac{13}{2}\right)^{2}+\left(\frac{18}{2}-\frac{13}{2}\right)^{2}+\left(\frac{16}{2}-\frac{13}{2}\right)^{2}\)

\(=\left(\frac{14-13}{2}\right)^{2}+\left(\frac{14-13}{2}\right)^{2}+\left(\frac{10-13}{2}\right)^{2}+\left(\frac{6-13}{2}\right)^{2}+\left(\frac{18-13}{2}\right)^{2}+\left(\frac{16-13}{2}\right)^{2}"\)

\(=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{-3}{2}\right)^{2}+\left(\frac{-7}{2}\right)^{2}+\left(\frac{5}{2}\right)^{2}+\left(\frac{3}{2}\right)^{2}\)

\(=\frac{1}{4}+\frac{1}{4}+\frac{9}{4}+\frac{49}{4}+\frac{25}{4}+\frac{9}{4}=\frac{1+1+9+49+25+9}{4}\)

\(=\frac{94}{4}=\frac{47}{2}=23.5\)

The pooled standard deviation therefor is:

\(s_{pooled}=\sqrt\frac{SS_{1}+SS_{2}}{N-2}=\sqrt\frac{34+\frac{47}{2}}{11-2}\)

\(=\sqrt\frac{\frac{68}{2}+\frac{47}{2}}{9}=\sqrt\frac{\frac{68+47}{2}}{9}=\sqrt\frac{\frac{115}{2}}{9}=\sqrt\frac{115}{2\times9}\)

\(=\sqrt\frac{115}{18}=\frac{1}{18}\sqrt{115\times18}=\frac{1}{18}\sqrt{115\times2\times9}=\frac{3}{18}\sqrt{115\times2}\)

And finally Cohen's d_s:\(=\frac{3}{18}\sqrt{115\times2}=\frac{1}{6}\sqrt{230}\approx2.53\)

\(d_s=\frac{\bar{x}_{1}-\bar{x}_{1}}{s_{pooled}}=\frac{3-\frac{13}{2}}{\frac{1}{6}\sqrt{230}}=\frac{\frac{6}{2}-\frac{13}{2}}{\frac{1}{6}\sqrt{230}}=\frac{\frac{6-13}{2}}{\frac{\sqrt{230}}{6}}\)

\(=\frac{\frac{6-13}{2}}{\frac{\sqrt{230}}{6}}=\frac{\frac{-7}{2}}{\frac{\sqrt{230}}{6}}=\frac{-7\times6}{2\times\sqrt{230}}=\frac{-7\times3}{\sqrt{230}}=\frac{-21}{\sqrt{230}}\)

\(=\frac{-21\times\sqrt{230}}{230}=-\frac{21}{230}\times\sqrt{230}\approx-1.38\)

Formulas

\(d_z=\frac{\bar{d}}{s_d}=\frac{t}{\sqrt{n}}\)

\(g=d_z\times{J(df)}\)

\(J(df)=\frac{\Gamma{\left(\frac{df}{2}\right)}}{\sqrt{\frac{df}{2}}\times\Gamma\left(\frac{df-1}{2}\right)}\approx1-\frac{3}{4\times{df}-1}\)

See the manually part in the test section for the formulas for \(\bar{d},s_d,t,df\)

Example

Note: continuation of the example in the manually part of the test section.

We are given the following pairs:

\(P=\left\{(9,8),(2,),(5,3),(4,6),(8,4),(8,3)\right\}\)

Since the second pair is missing a score, it will not be used and removed, so:

\(X=\left\{(9,8),(5,3),(4,6),(8,4),(8,3)\right\}\)

There are five pairs remaining, so n = 5.

In the test section we already calculated the following:

\(\bar{d}=2,s_d=\frac{\sqrt{30}}{2},t=\frac{2\sqrt{6}}{3},df=4\)

Cohen's d is therefor:

\(d_z=\frac{\bar{d}}{s_d}=\frac{2}{\frac{\sqrt{30}}{2}}=\frac{2\times2}{\sqrt{30}}=\frac{4\times\sqrt{30}}{\sqrt{30}\times\sqrt{30}}\)

\(=\frac{4\times\sqrt{30}}{30}=\frac{2\sqrt{30}}{15}\approx0.7303\)

If we had used the alternative formula, we would have gotten the same result:

\(d_z=\frac{t}{\sqrt{n}}=\frac{\frac{2\sqrt{6}}{3}}{\sqrt{5}}=\frac{2\sqrt{6}}{3\times\sqrt{5}}=\frac{2\sqrt{6}\times\sqrt{5}}{3\times\sqrt{5}\times\sqrt{5}}\)

\(=\frac{2\sqrt{6\times5}}{3\times5}=\frac{2\sqrt{30}}{15}\approx0.7303\)

For the Hedges correction we can use the approximation. That would yield:

\(J(df)\approx1-\frac{3}{4\times{df}-1}=1-\frac{3}{4\times{4}-1}=1-\frac{3}{15}=1-\frac{1}{5}\)

\(=\frac{5}{5}-\frac{1}{5}=\frac{5-1}{5}=\frac{4}{5}=0.8\)

Which gives an approximation for Hedges g of:

\(g=d_z\times{J(df)}\approx\frac{2\sqrt{30}}{15}\times\frac{4}{5}=\frac{2\sqrt{30}\times4}{15\times5}=\frac{8\sqrt{30}}{75}\approx0.5842\)