7 Between-Subjects Student's t-test

between-subjects t-test: Compare two groups or conditions where the participants are different in each group and have not been matched or are only matched on broad demographics, e.g. only age.

7.1 The Worked Example

Here is your data:

Group	N	Mean	SD
A	18	25.58	2.25
B	18	29.07	2.53

Let's look at the main t-test formula:

\[t = \frac{\bar{X}_1 - \bar{X}_2}{s_p \times \sqrt{\frac{1}{N_1} + \frac{1}{N_2}}}\]

Now, from the table above we know:

the mean of Group A is \(\bar{X_1} = 25.58\),
the mean of Group B is \(\bar{X_2} = 29.07\),
the N of Group A is \(N_1 = 18\),
and the N of Group B is \(N_2 = 18\),

which we can put into the equation right now:

\[t = \frac{25.58 - 29.07}{s_p \times \sqrt{\frac{1}{18} + \frac{1}{18}}}\]

And now we can see that the only thing we don't yet know is the pooled standard deviation (\(s_p\)). Let's look at that formula:

Calculating the pooled standard deviation

\[s_p = \sqrt{\frac{(n_1 -1) \times s^2_{X_1} + (n_2 -1)\times s^2_{X_2}}{n_1 + n_2 - 2}}\]

And if we start to fill in some details:

\[s_p = \sqrt{\frac{(18 -1) \times s^2_{X_1} + (18 -1)\times s^2_{X_2}}{18 + 18 - 2}}\]

Now looking at the formula, it is clear we are missing:

\(s^2_{X_1}\) - the variance of Group A (could be written as \(s^2_{A}\))
\(s^2_{X_2}\) - the variance of Group B (could be written as \(s^2_{B}\))

What we do know though, from the table, is the standard deviations of both groups (\(SD_A\) = 2.25; \(SD_B\) = 2.53), and we know that variance of a group is equal to the standard deviation squared. So:

\(s^2_{X_1}\) = \(s^2_A\) = \(SD_A \times SD_A\) = \(2.25 \times 2.25\) = \(5.0625\)
\(s^2_{X_2}\) = \(s^2_B\) = \(SD_B \times SD_B\) = \(2.53 \times 2.53\) = \(6.4009\)

And if we now add those values to our formula we get:

\[s_p = \sqrt{\frac{(18 -1) \times 5.0625 + (18 -1)\times 6.4009}{18 + 18 - 2}}\]

And we can then start working through the formula, taking each stage in turn to make sure we don't make mistakes. Let's get rid of the brackets first:

\[s_p = \sqrt{\frac{(17 \times 5.0625) + (17 \times 6.4009)}{34}}\]

Now we deal with the multiplications:

\[s_p = \sqrt{\frac{86.0625 + 108.8153}{34}}\]

Let's tidy up that top half of the equation (the numerator):

\[s_p = \sqrt{\frac{194.8778}{34}}\]

Which if we then divide the numerator by the denominator (the bottom half), and then take the square root of that we get:

\[s_p = \sqrt{5.7317}\]

Giving a pooled standard deviation of:

\[s_p = 2.3940969\]

Meaning that our pooled standard deviation, rounded to three decimal places, is \(s_p = 2.394\) and we can now add that to the t-test formula to give us:

Calculating the t-value

\[t = \frac{25.58 - 29.07}{2.394 \times \sqrt{\frac{1}{18} + \frac{1}{18}}}\]

And again we just start working through the formula. Let's deal with the fractions relating to sample size first:

\[t = \frac{25.58 - 29.07}{2.394 \times \sqrt{0.0555556 + 0.0555556}}\]

Which we can tidy up a little to give:

\[t = \frac{-3.49}{2.394 \times \sqrt{0.1111111}}\]

And if we sort out the square root on the denominator we are left with:

\[t = \frac{-3.49}{2.394 \times 0.3333333}\]

We can then tidy up the denominator to give us:

\[t = \frac{-3.49}{0.798}\]

Which we can finally solve to give us a t-value, rounded to two decimal places, of \(t = -4.37\)

Degrees of Freedom

Great! Now we just need the degrees of freedom where the formula is:

\[df = (n_1 - 1) + (n_2 - 1)\]

And we already know that:

the N of Group A is \(N_1 = 18\),
and the N of Group B is \(N_2 = 18\),

So putting them into the equation we get:

\[df = (18 - 1) + (18 - 1)\] \[df = 17 + 17\] \[df = 34\]

Effect Size: Cohen's d

And finally Cohen's d, the effect size:

\[d = \frac{2t}{\sqrt{df}}\]

Which, based on the info above, we know:

\(t = -4.37\)
\(df = 34\)

Putting them into the formula we get:

\[d = \frac{2 \times -4.37}{\sqrt{34}}\]

And if we tidy the nominator and the denominator we get:

\[d = \frac{-8.74}{5.8309519}\]

Which we can then solve to learn that \(d = -1.5\)

Determining Significance

If we were to look at a critical values look-up table for \(df = 34\) and \(\alpha = .05\) (two-tailed), we would see that the critical value is \(t_{crit} = 2.032\). Given that our t-value, ignoring polarity and just looking at the absolute value, so \(t = 4.37\), is equal to or larger than our \(t_{crit}\) then we can say our result is significant, and as such would be written up as t(34) = 4.37, p < .05, d = 1.5

Remember: If you were writing this up as a report, and analysis the data in R, then you would see the p-value was actually p = 1.1081944^{-4}, and would be written up as p < .001

7.2 Test Yourself

7.2.1 DataSet 1

Here is your data:

Group	N	Mean	SD
A	5	68.01	0.34
B	5	67.80	0.67

Let's look at the main t-test formula:

\[t = \frac{\bar{X}_1 - \bar{X}_2}{s_p \times \sqrt{\frac{1}{N_1} + \frac{1}{N_2}}}\]

Now, from the table above we know:

the mean of Group A is \(\bar{X_1} = 68.01\),
the mean of Group B is \(\bar{X_2} = 67.8\),
the N of Group A is \(N_1 = 5\),
and the N of Group B is \(N_2 = 5\),

which we can put into the equation right now:

\[t = \frac{68.01 - 67.8}{s_p \times \sqrt{\frac{1}{5} + \frac{1}{5}}}\]

And now we can see that the only thing we don't yet know is the pooled standard deviation (\(s_p\)). Let's look at that formula:

\[s_p = \sqrt{\frac{(n_1 -1) \times s^2_{X_1} + (n_2 -1)\times s^2_{X_2}}{n_1 + n_2 - 2}}\]

And if we start to fill in some details:

\[s_p = \sqrt{\frac{(5 -1) \times s^2_{X_1} + (5 -1)\times s^2_{X_2}}{5 + 5 - 2}}\]

Now looking at the formula, it is clear we are missing:

\(s^2_{X_1}\) - the variance of Group A (could be written as \(s^2_{A}\))
\(s^2_{X_2}\) - the variance of Group B (could be written as \(s^2_{B}\))

What we do know though, from the table, is the standard deviations of both groups (\(SD_A\) = 0.34; \(SD_B\) = 0.67), and we know that variance of a group is equal to the standard deviation squared. So:

\(s^2_{X_1}\) = \(s^2_A\) = \(SD_A \times SD_A\) = \(0.34 \times 0.34\) = \(0.1156\)
\(s^2_{X_2}\) = \(s^2_B\) = \(SD_B \times SD_B\) = \(0.67 \times 0.67\) = \(0.4489\)

And if we now add those values to our formula we get:

\[s_p = \sqrt{\frac{(5 -1) \times 0.1156 + (5 -1)\times 0.4489}{5 + 5 - 2}}\]

And we can then start working through the formula, taking each stage in turn to make sure we don't make mistakes. Let's get rid of the brackets first:

\[s_p = \sqrt{\frac{(4 \times 0.1156) + (4 \times 0.4489)}{8}}\]

Now we deal with the multiplications:

\[s_p = \sqrt{\frac{0.4624 + 1.7956}{8}}\]

Let's tidy up that top half of the equation (the numerator):

\[s_p = \sqrt{\frac{2.258}{8}}\]

Which if we then divide the numerator by the denominator (the bottom half), and then take the square root of that we get:

\[s_p = \sqrt{0.28225}\]

Giving a pooled standard deviation of:

\[s_p = 0.5312721\]

Meaning that our pooled standard deviation, rounded to three decimal places, is \(s_p = 0.531\) and we can now add that to the t-test formula to give us:

\[t = \frac{68.01 - 67.8}{0.531 \times \sqrt{\frac{1}{5} + \frac{1}{5}}}\]

And again we just start working through the formula. Let's deal with the fractions relating to sample size first:

\[t = \frac{68.01 - 67.8}{0.531 \times \sqrt{0.2 + 0.2}}\]

Which we can tidy up a little to give:

\[t = \frac{0.21}{0.531 \times \sqrt{0.4}}\]

And if we sort out the square root on the denominator we are left with:

\[t = \frac{0.21}{0.531 \times 0.6324555}\]

We can then tidy up the denominator to give us:

\[t = \frac{0.21}{0.3358339}\]

Which we can finally solve to give us a t-value, rounded to two decimal places, of \(t = 0.63\)

Great! Now we just need the degrees of freedom where the formula is:

\[df = (n_1 - 1) + (n_2 - 1)\]

And we already know that:

the N of Group A is \(N_1 = 5\),
and the N of Group B is \(N_2 = 5\),

So putting them into the equation we get:

\[df = (5 - 1) + (5 - 1)\] \[df = 4 + 4\] \[df = 8\]

And finally Cohen's d, the effect size:

\[d = \frac{2t}{\sqrt{df}}\]

Which, based on the info above, we know:

\(t = 0.63\)
\(df = 8\)

Putting them into the formula we get:

\[d = \frac{2 \times 0.63}{\sqrt{8}}\]

And if we tidy the nominator and the denominator we get:

\[d = \frac{1.26}{2.8284271}\]

Which we can then solve to learn that \(d = 0.45\)

Determining Significance

If we were to look at a critical values look-up table for \(df = 8\) and \(\alpha = .05\) (two-tailed), we would see that the critical value is \(t_{crit} = 2.306\). Given that our t-value, ignoring polarity and just looking at the absolute value, so \(t = 0.63\), is smaller than our \(t_{crit}\) then we can say our result is non-significant, and as such would be written up as t(8) = 0.63, p > .05, d = 0.45

Remember: If you were writing this up as a report, and analysis the data in R, then you would see the p-value was actually p = 0.5462633, and would be written up as p = 0.548

7.2.2 DataSet 2

Here is your data:

Group	N	Mean	SD
A	47	66.18	1.79
B	47	66.42	1.78

Let's look at the main t-test formula:

\[t = \frac{\bar{X}_1 - \bar{X}_2}{s_p \times \sqrt{\frac{1}{N_1} + \frac{1}{N_2}}}\]

Now, from the table above we know:

the mean of Group A is \(\bar{X_1} = 66.18\),
the mean of Group B is \(\bar{X_2} = 66.42\),
the N of Group A is \(N_1 = 47\),
and the N of Group B is \(N_2 = 47\),

which we can put into the equation right now:

\[t = \frac{66.18 - 66.42}{s_p \times \sqrt{\frac{1}{47} + \frac{1}{47}}}\]

And now we can see that the only thing we don't yet know is the pooled standard deviation (\(s_p\)). Let's look at that formula:

\[s_p = \sqrt{\frac{(n_1 -1) \times s^2_{X_1} + (n_2 -1)\times s^2_{X_2}}{n_1 + n_2 - 2}}\]

And if we start to fill in some details:

\[s_p = \sqrt{\frac{(47 -1) \times s^2_{X_1} + (47 -1)\times s^2_{X_2}}{47 + 47 - 2}}\]

Now looking at the formula, it is clear we are missing:

\(s^2_{X_1}\) - the variance of Group A (could be written as \(s^2_{A}\))
\(s^2_{X_2}\) - the variance of Group B (could be written as \(s^2_{B}\))

What we do know though, from the table, is the standard deviations of both groups (\(SD_A\) = 1.79; \(SD_B\) = 1.78), and we know that variance of a group is equal to the standard deviation squared. So:

\(s^2_{X_1}\) = \(s^2_A\) = \(SD_A \times SD_A\) = \(1.79 \times 1.79\) = \(3.2041\)
\(s^2_{X_2}\) = \(s^2_B\) = \(SD_B \times SD_B\) = \(1.78 \times 1.78\) = \(3.1684\)

And if we now add those values to our formula we get:

\[s_p = \sqrt{\frac{(47 -1) \times 3.2041 + (47 -1)\times 3.1684}{47 + 47 - 2}}\]

And we can then start working through the formula, taking each stage in turn to make sure we don't make mistakes. Let's get rid of the brackets first:

\[s_p = \sqrt{\frac{(46 \times 3.2041) + (46 \times 3.1684)}{92}}\]

Now we deal with the multiplications:

\[s_p = \sqrt{\frac{147.3886 + 145.7464}{92}}\]

Let's tidy up that top half of the equation (the numerator):

\[s_p = \sqrt{\frac{293.135}{92}}\]

Which if we then divide the numerator by the denominator (the bottom half), and then take the square root of that we get:

\[s_p = \sqrt{3.18625}\]

Giving a pooled standard deviation of:

\[s_p = 1.785007\]

Meaning that our pooled standard deviation, rounded to three decimal places, is \(s_p = 1.785\) and we can now add that to the t-test formula to give us:

\[t = \frac{66.18 - 66.42}{1.785 \times \sqrt{\frac{1}{47} + \frac{1}{47}}}\]

And again we just start working through the formula. Let's deal with the fractions relating to sample size first:

\[t = \frac{66.18 - 66.42}{1.785 \times \sqrt{0.0212766 + 0.0212766}}\]

Which we can tidy up a little to give:

\[t = \frac{-0.24}{1.785 \times \sqrt{0.0425532}}\]

And if we sort out the square root on the denominator we are left with:

\[t = \frac{-0.24}{1.785 \times 0.2062842}\]

We can then tidy up the denominator to give us:

\[t = \frac{-0.24}{0.3682174}\]

Which we can finally solve to give us a t-value, rounded to two decimal places, of \(t = -0.65\)

Great! Now we just need the degrees of freedom where the formula is:

\[df = (n_1 - 1) + (n_2 - 1)\]

And we already know that:

the N of Group A is \(N_1 = 47\),
and the N of Group B is \(N_2 = 47\),

So putting them into the equation we get:

\[df = (47 - 1) + (47 - 1)\] \[df = 46 + 46\] \[df = 92\]

And finally Cohen's d, the effect size:

\[d = \frac{2t}{\sqrt{df}}\]

Which, based on the info above, we know:

\(t = -0.65\)
\(df = 92\)

Putting them into the formula we get:

\[d = \frac{2 \times -0.65}{\sqrt{92}}\]

And if we tidy the nominator and the denominator we get:

\[d = \frac{-1.3}{9.591663}\]

Which we can then solve to learn that \(d = -0.14\)

Determining Significance

If we were to look at a critical values look-up table for \(df = 92\) and \(\alpha = .05\) (two-tailed), we would see that the critical value is \(t_{crit} = 1.986\). Given that our t-value, ignoring polarity and just looking at the absolute value, so \(t = 0.65\), is smaller than our \(t_{crit}\) then we can say our result is non-significant, and as such would be written up as t(92) = 0.65, p > .05, d = 0.14

Remember: If you were writing this up as a report, and analysis the data in R, then you would see the p-value was actually p = 0.517312, and would be written up as p = 0.513

7.2.3 DataSet 3

Here is your data:

Group	N	Mean	SD
A	8	31.80	0.42
B	8	32.18	0.57

Let's look at the main t-test formula:

\[t = \frac{\bar{X}_1 - \bar{X}_2}{s_p \times \sqrt{\frac{1}{N_1} + \frac{1}{N_2}}}\]

Now, from the table above we know:

the mean of Group A is \(\bar{X_1} = 31.8\),
the mean of Group B is \(\bar{X_2} = 32.18\),
the N of Group A is \(N_1 = 8\),
and the N of Group B is \(N_2 = 8\),

which we can put into the equation right now:

\[t = \frac{31.8 - 32.18}{s_p \times \sqrt{\frac{1}{8} + \frac{1}{8}}}\]

And now we can see that the only thing we don't yet know is the pooled standard deviation (\(s_p\)). Let's look at that formula:

\[s_p = \sqrt{\frac{(n_1 -1) \times s^2_{X_1} + (n_2 -1)\times s^2_{X_2}}{n_1 + n_2 - 2}}\]

And if we start to fill in some details:

\[s_p = \sqrt{\frac{(8 -1) \times s^2_{X_1} + (8 -1)\times s^2_{X_2}}{8 + 8 - 2}}\]

Now looking at the formula, it is clear we are missing:

\(s^2_{X_1}\) - the variance of Group A (could be written as \(s^2_{A}\))
\(s^2_{X_2}\) - the variance of Group B (could be written as \(s^2_{B}\))

What we do know though, from the table, is the standard deviations of both groups (\(SD_A\) = 0.42; \(SD_B\) = 0.57), and we know that variance of a group is equal to the standard deviation squared. So:

\(s^2_{X_1}\) = \(s^2_A\) = \(SD_A \times SD_A\) = \(0.42 \times 0.42\) = \(0.1764\)
\(s^2_{X_2}\) = \(s^2_B\) = \(SD_B \times SD_B\) = \(0.57 \times 0.57\) = \(0.3249\)

And if we now add those values to our formula we get:

\[s_p = \sqrt{\frac{(8 -1) \times 0.1764 + (8 -1)\times 0.3249}{8 + 8 - 2}}\]

And we can then start working through the formula, taking each stage in turn to make sure we don't make mistakes. Let's get rid of the brackets first:

\[s_p = \sqrt{\frac{(7 \times 0.1764) + (7 \times 0.3249)}{14}}\]

Now we deal with the multiplications:

\[s_p = \sqrt{\frac{1.2348 + 2.2743}{14}}\]

Let's tidy up that top half of the equation (the numerator):

\[s_p = \sqrt{\frac{3.5091}{14}}\]

Which if we then divide the numerator by the denominator (the bottom half), and then take the square root of that we get:

\[s_p = \sqrt{0.25065}\]

Giving a pooled standard deviation of:

\[s_p = 0.5006496\]

Meaning that our pooled standard deviation, rounded to three decimal places, is \(s_p = 0.501\) and we can now add that to the t-test formula to give us:

\[t = \frac{31.8 - 32.18}{0.501 \times \sqrt{\frac{1}{8} + \frac{1}{8}}}\]

And again we just start working through the formula. Let's deal with the fractions relating to sample size first:

\[t = \frac{31.8 - 32.18}{0.501 \times \sqrt{0.125 + 0.125}}\]

Which we can tidy up a little to give:

\[t = \frac{-0.38}{0.501 \times \sqrt{0.25}}\]

And if we sort out the square root on the denominator we are left with:

\[t = \frac{-0.38}{0.501 \times 0.5}\]

We can then tidy up the denominator to give us:

\[t = \frac{-0.38}{0.2505}\]

Which we can finally solve to give us a t-value, rounded to two decimal places, of \(t = -1.52\)

Great! Now we just need the degrees of freedom where the formula is:

\[df = (n_1 - 1) + (n_2 - 1)\]

And we already know that:

the N of Group A is \(N_1 = 8\),
and the N of Group B is \(N_2 = 8\),

So putting them into the equation we get:

\[df = (8 - 1) + (8 - 1)\] \[df = 7 + 7\] \[df = 14\]

And finally Cohen's d, the effect size:

\[d = \frac{2t}{\sqrt{df}}\]

Which, based on the info above, we know:

\(t = -1.52\)
\(df = 14\)

Putting them into the formula we get:

\[d = \frac{2 \times -1.52}{\sqrt{14}}\]

And if we tidy the nominator and the denominator we get:

\[d = \frac{-3.04}{3.7416574}\]

Which we can then solve to learn that \(d = -0.81\)

Determining Significance

If we were to look at a critical values look-up table for \(df = 14\) and \(\alpha = .05\) (two-tailed), we would see that the critical value is \(t_{crit} = 2.145\). Given that our t-value, ignoring polarity and just looking at the absolute value, so \(t = 1.52\), is smaller than our \(t_{crit}\) then we can say our result is non-significant, and as such would be written up as t(14) = 1.52, p > .05, d = 0.81

Remember: If you were writing this up as a report, and analysis the data in R, then you would see the p-value was actually p = 0.1507698, and would be written up as p = 0.144

7.3 Look-Up table

Remembering that the \(t_{crit}\) value is the smallest t-value you need to find a significant effect, find the \(t_{crit}\) for your df, assuming \(\alpha = .05\). If the \(t\) value you calculated is equal to or larger than \(t_{crit}\) then your test is significant.

df	\(\alpha = .05\)
1	12.706
2	4.303
3	3.182
4	2.776
5	2.571
6	2.447
7	2.365
8	2.306
9	2.262
10	2.228
15	2.131
20	2.086
30	2.042
40	2.021
50	2.009
60	2
70	1.994
80	1.99
90	1.987
100	1.984

6 One-Sample t-test

8 Between-Subjects Welch's t-test