9 Within-Subjects t-test
within-subjects t-test: Compare two conditions where the participants are the same in both conditions (or more rarely are different participants that have been highly matched on a number of demographics such as IQ, reading ability, etc - must be matched on a number of demographics).
9.1 The Worked Example
Let's say that this is our starting data:
| Participants | PreTest | PostTest |
|---|---|---|
| 1 | 60 | 68 |
| 2 | 64 | 75 |
| 3 | 56 | 62 |
| 4 | 82 | 85 |
| 5 | 74 | 73 |
| 6 | 79 | 85 |
| 7 | 63 | 64 |
| 8 | 59 | 59 |
| 9 | 72 | 73 |
| 10 | 66 | 70 |
The first thing we need to do is calculate the difference between the PostTest and the PreTest for each participant, based on \(D = PostTest - PreTest\). So for example:
- Participant 1 would be: 68 - 60 = 8
- Participant 2 would be: 75 - 64 = 11
- etc
And if we do that for each Participant and added a column of the differences (\(D\)) then we would see:
| Participants | PreTest | PostTest | D |
|---|---|---|---|
| 1 | 60 | 68 | 8 |
| 2 | 64 | 75 | 11 |
| 3 | 56 | 62 | 6 |
| 4 | 82 | 85 | 3 |
| 5 | 74 | 73 | -1 |
| 6 | 79 | 85 | 6 |
| 7 | 63 | 64 | 1 |
| 8 | 59 | 59 | 0 |
| 9 | 72 | 73 | 1 |
| 10 | 66 | 70 | 4 |
Now, the within-subjects t-test formula is:
\[t = \frac{\bar{D}}{\frac{SD_{D}}{\sqrt{N}}}\]
We can see that \(N = 10\), but we need to calculate \(\bar{D}\) (called D-Bar, the mean of the \(D\) column) and \(SD_{D}\).
Calculating D-bar
So the \(\bar{D}\) formula is the same as the mean formula:
\[\bar{D} = \frac{\sum{D}}{N}\]
Where \(D\) is \(PostTest - PreTest\) for each Participant.
Then:
\[\bar{D} = \frac{(68 - 60) + (75 - 64) + (62 - 56) + (85 - 82) + (73 - 74) + \\ (85- 79) + (64 - 63) + (59 - 59) + (73 - 72) + (70 - 66)}{10}\]
Which if we resolve all the brackets becomes:
\[\bar{D} = \frac{8 + 11 + 6 + 3 + -1 + 6 + 1 + 0 + 1 + 4}{10}\]
And if we sum the top half together
\[\bar{D} = \frac{39}{10}\]
Leaving us with:
\[\bar{D} = 3.9\]
So we find that \(\bar{D}\) = 3.9, which is the mean difference between the Post test and Pre test values.
The Standard Deviation of D
The standard deviation formula is:
\[SD = \sqrt\frac{\sum(X - \bar{X})^2}{N-1}\]
Which if we translate to using D, becomes:
\[SD_{D} = \sqrt\frac{\sum(D - \bar{D})^2}{N-1}\]
Then:
\[SD_{D} =\sqrt\frac{(8 - 3.9)^2 + (11 - 3.9)^2 + (6 - 3.9)^2 + (3 - 3.9)^2 + (-1 - 3.9)^2 + \\ (6 - 3.9)^2 + (1 - 3.9)^2 + (0 - 3.9)^2 + (1 - 3.9)^2 + (4 - 3.9)^2}{10 - 1}\]
And if we start stepping through this analysis by dealing with the brackets:
\[SD_{D} =\sqrt\frac{(4.1)^2 + (7.1)^2 + (2.1)^2 + (-0.9)^2 + (-4.9)^2 + \\ (2.1)^2 + (-2.9)^2 + (-3.9)^2 + (-2.9)^2 + (0.1)^2}{10 - 1}\]
And then we square those brackets
\[SD_{D} =\sqrt\frac{16.81 + 50.41 + 4.41 + 0.81 + 24.01 + 4.41 + 8.41 + 15.21 + 8.41 + 0.01}{10 - 1}\]
And sum up all the values on the top half:
\[SD_{D} =\sqrt\frac{132.9}{10 - 1}\]
And then we need to sort out the bottom half
\[SD_{D} =\sqrt\frac{132.9}{9}\]
Which by dividing the top half by the bottom half reduces down to:
\[SD_{D} =\sqrt{14.7666667}\]
And then we finally take the square root, which leaves us with:
\[SD_{D} =3.8427421\]
And so we find that the \(SD_{D}\) = 3.8427421 which to two decimal places would be, \(SD_{D} = 3.84\)
Calculating the t-value
And finally the t-test formula is:
\[t = \frac{\bar{D}}{\frac{SD_{D}}{\sqrt{N}}}\]
Then if we start filling in the values we know from above we see:
\[t = \frac{3.9}{\frac{3.8427421}{\sqrt{10}}} \]
And if we deal with the square root first:
\[t = \frac{3.9}{\frac{3.8427421}{3.1622777}} \]
And divide \(SD_{D}\) by \(\sqrt{N}\) - tidying up the bottom of the formula:
\[t = \frac{3.9}{1.2151817} \]
And then solve for \(t\) by dividing the top half by the bottom half gives us:
\[t = 3.2093965 \]
And so, rounding to two decimal places, we find that \(t = 3.21\)
Degrees of Freedom
Great! Now we just need the degrees of freedom where the formula is:
\[df = N - 1\]
And we already know that \(N= 10\) and putting them into the equation we get:
\[df = 10 - 1\]
Which reduces to:
\[df = 9\]
Meaning that we find a \(df = 9\)
Effect Size - Cohen's d
And finally Cohen's d, the effect size. One of the common formulas based on knowing the t-value and the N is:
\[d = \frac{t}{\sqrt{N}}\]
Which, based on the info above, we know:
- \(t = 3.21\)
- \(N = 10\)
And putting those into the formula we get:
\[d = \frac{3.21}{\sqrt{10}}\]
Which gives us:
\[d = \frac{3.21}{3.1622777}\]
And so:
\[d = 1.0150911\]
Meaning that the effect size, to two decimal places, is d = 1.01.
Determining Significance
If we were to look at a critical values look-up table for \(df = 9\) and \(\alpha = .05\) (two-tailed), we would see that the critical value is \(t_{crit} = 2.262\). Given that our t-value, ignoring polarity and just looking at the absolute value, so \(t = 3.21\), 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(9) = 3.21, p < .05, d = 1.01.
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 = 0.011, and would be written up as p = 0.011
9.2 Test Yourself
9.2.1 DataSet 1
Let's say that this is our starting data:
| Participants | PreTest | PostTest |
|---|---|---|
| 1 | 71 | 64 |
| 2 | 56 | 54 |
| 3 | 75 | 70 |
| 4 | 50 | 46 |
| 5 | 51 | 52 |
| 6 | 61 | 58 |
| 7 | 62 | 61 |
| 8 | 72 | 74 |
| 9 | 66 | 60 |
| 10 | 78 | 78 |
The first thing we need to do is calculate the difference between the PostTest and the PreTest for each participant, based on \(D = PostTest - PreTest\). So for example:
- Participant 1 would be: 64 - 71 = -7
- Participant 2 would be: 54 - 56 = -2
- etc
And if we do that for each Participant and added a column of the differences (\(D\)) then we would see:
| Participants | PreTest | PostTest | D |
|---|---|---|---|
| 1 | 71 | 64 | -7 |
| 2 | 56 | 54 | -2 |
| 3 | 75 | 70 | -5 |
| 4 | 50 | 46 | -4 |
| 5 | 51 | 52 | 1 |
| 6 | 61 | 58 | -3 |
| 7 | 62 | 61 | -1 |
| 8 | 72 | 74 | 2 |
| 9 | 66 | 60 | -6 |
| 10 | 78 | 78 | 0 |
Now, the within-subjects t-test formula is:
\[t = \frac{\bar{D}}{\frac{SD_{D}}{\sqrt{N}}}\]
We can see that \(N = 10\), but we need to calculate \(\bar{D}\) (called D-Bar, the mean of the \(D\) column) and \(SD_{D}\).
Calculating D-bar
So the \(\bar{D}\) formula is the same as the mean formula:
\[\bar{D} = \frac{\sum{D}}{N}\]
Where \(D\) is \(PostTest - PreTest\) for each Participant.
Then:
\[\bar{D} = \frac{(64 - 71) + (54 - 56) + (70 - 75) + (46 - 50) + (52 - 51) + \\ (58- 61) + (61 - 62) + (74 - 72) + (60 - 66) + (78 - 78)}{10}\]
Which if we resolve all the brackets becomes:
\[\bar{D} = \frac{-7 + -2 + -5 + -4 + 1 + -3 + -1 + 2 + -6 + 0}{10}\]
And if we sum the top half together
\[\bar{D} = \frac{-25}{10}\]
Leaving us with:
\[\bar{D} = -2.5\]
So we find that \(\bar{D}\) = -2.5, which is the mean difference between the Post test and Pre test values.
The Standard Deviation of D
The standard deviation formula is:
\[SD = \sqrt\frac{\sum(X - \bar{X})^2}{N-1}\]
Which if we translate to using D, becomes:
\[SD_{D} = \sqrt\frac{\sum(D - \bar{D})^2}{N-1}\]
Then:
\[SD_{D} =\sqrt\frac{(-7 - -2.5)^2 + (-2 - -2.5)^2 + (-5 - -2.5)^2 + (-4 - -2.5)^2 + \\ (1 - -2.5)^2 + (-3 - -2.5)^2 + (-1 - -2.5)^2 + (2 - -2.5)^2 + \\ (-6 - -2.5)^2 + (0 - -2.5)^2}{10 - 1}\]
And if we start stepping through this analysis by dealing with the brackets:
\[SD_{D} =\sqrt\frac{(-4.5)^2 + (0.5)^2 + (-2.5)^2 + (-1.5)^2 + (3.5)^2 + \\ (-0.5)^2 + (1.5)^2 + (4.5)^2 + (-3.5)^2 + (2.5)^2}{10 - 1}\]
And then we square those brackets
\[SD_{D} =\sqrt\frac{20.25 + 0.25 + 6.25 + 2.25 + 12.25 + \\ 0.25 + 2.25 + 20.25 + 12.25 + 6.25}{10 - 1}\]
And sum up all the values on the top half:
\[SD_{D} =\sqrt\frac{82.5}{10 - 1}\]
And then we need to sort out the bottom half
\[SD_{D} =\sqrt\frac{82.5}{9}\]
Which by dividing the top half by the bottom half reduces down to:
\[SD_{D} =\sqrt{9.1666667}\]
And then we finally take the square root, which leaves us with:
\[SD_{D} =3.0276504\]
And so we find that the \(SD_{D}\) = 3.0276504 which to two decimal places would be, \(SD_{D} = 3.03\)
Calculating the t-value
And finally the t-test formula is:
\[t = \frac{\bar{D}}{\frac{SD_{D}}{\sqrt{N}}}\]
Then if we start filling in the values we know from above we see:
\[t = \frac{-2.5}{\frac{3.0276504}{\sqrt{10}}} \]
And if we deal with the square root first:
\[t = \frac{-2.5}{\frac{3.0276504}{3.1622777}} \]
And divide \(SD_{D}\) by \(\sqrt{N}\) - tidying up the bottom of the formula:
\[t = \frac{-2.5}{0.9574271} \]
And then solve for \(t\) by dividing the top half by the bottom half gives us:
\[t = -2.6111648 \]
And so, rounding to two decimal places, we find that \(t = -2.61\)
Degrees of Freedom
Great! Now we just need the degrees of freedom where the formula is:
\[df = N - 1\]
And we already know that \(N= 10\) and putting them into the equation we get:
\[df = 10 - 1\]
Which reduces to:
\[df = 9\]
Meaning that we find a \(df = 9\)
Effect Size - Cohen's d
And finally Cohen's d, the effect size. One of the common formulas based on knowing the t-value and the N is:
\[d = \frac{t}{\sqrt{N}}\]
Which, based on the info above, we know:
- \(t = -2.61\)
- \(N = 10\)
And putting those into the formula we get:
\[d = \frac{-2.61}{\sqrt{10}}\]
Which gives us:
\[d = \frac{-2.61}{3.1622777}\]
And so:
\[d = -0.8253545\]
Meaning that the effect size, to two decimal places, is d = -0.83.
Determining Significance
If we were to look at a critical values look-up table for \(df = 9\) and \(\alpha = .05\) (two-tailed), we would see that the critical value is \(t_{crit} = 2.262\). Given that our t-value, ignoring polarity and just looking at the absolute value, so \(t = 2.61\), is equal to or larger than our \(t_{crit}\) then we can say our result is not significant, and as such would be written up as t(9) = 2.61, p < .05, d = 0.83.
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 = 0.028, and would be written up as p = 0.028
9.2.2 DataSet 2
Let's say that this is our starting data:
| Participants | PreTest | PostTest |
|---|---|---|
| 1 | 73 | 72 |
| 2 | 60 | 54 |
| 3 | 53 | 52 |
| 4 | 74 | 74 |
| 5 | 58 | 54 |
| 6 | 55 | 52 |
| 7 | 54 | 49 |
| 8 | 57 | 58 |
| 9 | 59 | 60 |
| 10 | 72 | 72 |
The first thing we need to do is calculate the difference between the PostTest and the PreTest for each participant, based on \(D = PostTest - PreTest\). So for example:
- Participant 1 would be: 72 - 73 = -1
- Participant 2 would be: 54 - 60 = -6
- etc
And if we do that for each Participant and added a column of the differences (\(D\)) then we would see:
| Participants | PreTest | PostTest | D |
|---|---|---|---|
| 1 | 73 | 72 | -1 |
| 2 | 60 | 54 | -6 |
| 3 | 53 | 52 | -1 |
| 4 | 74 | 74 | 0 |
| 5 | 58 | 54 | -4 |
| 6 | 55 | 52 | -3 |
| 7 | 54 | 49 | -5 |
| 8 | 57 | 58 | 1 |
| 9 | 59 | 60 | 1 |
| 10 | 72 | 72 | 0 |
Now, the within-subjects t-test formula is:
\[t = \frac{\bar{D}}{\frac{SD_{D}}{\sqrt{N}}}\]
We can see that \(N = 10\), but we need to calculate \(\bar{D}\) (called D-Bar, the mean of the \(D\) column) and \(SD_{D}\).
Calculating D-bar
So the \(\bar{D}\) formula is the same as the mean formula:
\[\bar{D} = \frac{\sum{D}}{N}\]
Where \(D\) is \(PostTest - PreTest\) for each Participant.
Then:
\[\bar{D} = \frac{(72 - 73) + (54 - 60) + (52 - 53) + (74 - 74) + (54 - 58) + \\ (52- 55) + (49 - 54) + (58 - 57) + (60 - 59) + (72 - 72)}{10}\]
Which if we resolve all the brackets becomes:
\[\bar{D} = \frac{-1 + -6 + -1 + 0 + -4 + -3 + -5 + 1 + 1 + 0}{10}\]
And if we sum the top half together
\[\bar{D} = \frac{-18}{10}\]
Leaving us with:
\[\bar{D} = -1.8\]
So we find that \(\bar{D}\) = -1.8, which is the mean difference between the Post test and Pre test values.
The Standard Deviation of D
The standard deviation formula is:
\[SD = \sqrt\frac{\sum(X - \bar{X})^2}{N-1}\]
Which if we translate to using D, becomes:
\[SD_{D} = \sqrt\frac{\sum(D - \bar{D})^2}{N-1}\]
Then:
\[SD_{D} =\sqrt\frac{(-1 - -1.8)^2 + (-6 - -1.8)^2 + (-1 - -1.8)^2 + (0 - -1.8)^2 + \\ (-4 - -1.8)^2 + (-3 - -1.8)^2 + (-5 - -1.8)^2 + (1 - -1.8)^2 + \\ (1 - -1.8)^2 + (0 - -1.8)^2}{10 - 1}\]
And if we start stepping through this analysis by dealing with the brackets:
\[SD_{D} =\sqrt\frac{(0.8)^2 + (-4.2)^2 + (0.8)^2 + (1.8)^2 + (-2.2)^2 + \\ (-1.2)^2 + (-3.2)^2 + (2.8)^2 + (2.8)^2 + (1.8)^2}{10 - 1}\]
And then we square those brackets
\[SD_{D} =\sqrt\frac{0.64 + 17.64 + 0.64 + 3.24 + 4.84 + \\ 1.44 + 10.24 + 7.84 + 7.84 + 3.24}{10 - 1}\]
And sum up all the values on the top half:
\[SD_{D} =\sqrt\frac{57.6}{10 - 1}\]
And then we need to sort out the bottom half
\[SD_{D} =\sqrt\frac{57.6}{9}\]
Which by dividing the top half by the bottom half reduces down to:
\[SD_{D} =\sqrt{6.4}\]
And then we finally take the square root, which leaves us with:
\[SD_{D} =2.5298221\]
And so we find that the \(SD_{D}\) = 2.5298221 which to two decimal places would be, \(SD_{D} = 2.53\)
Calculating the t-value
And finally the t-test formula is:
\[t = \frac{\bar{D}}{\frac{SD_{D}}{\sqrt{N}}}\]
Then if we start filling in the values we know from above we see:
\[t = \frac{-1.8}{\frac{2.5298221}{\sqrt{10}}} \]
And if we deal with the square root first:
\[t = \frac{-1.8}{\frac{2.5298221}{3.1622777}} \]
And divide \(SD_{D}\) by \(\sqrt{N}\) - tidying up the bottom of the formula:
\[t = \frac{-1.8}{0.8} \]
And then solve for \(t\) by dividing the top half by the bottom half gives us:
\[t = -2.25 \]
And so, rounding to two decimal places, we find that \(t = -2.25\)
Degrees of Freedom
Great! Now we just need the degrees of freedom where the formula is:
\[df = N - 1\]
And we already know that \(N= 10\) and putting them into the equation we get:
\[df = 10 - 1\]
Which reduces to:
\[df = 9\]
Meaning that we find a \(df = 9\)
Effect Size - Cohen's d
And finally Cohen's d, the effect size. One of the common formulas based on knowing the t-value and the N is:
\[d = \frac{t}{\sqrt{N}}\]
Which, based on the info above, we know:
- \(t = -2.25\)
- \(N = 10\)
And putting those into the formula we get:
\[d = \frac{-2.25}{\sqrt{10}}\]
Which gives us:
\[d = \frac{-2.25}{3.1622777}\]
And so:
\[d = -0.7115125\]
Meaning that the effect size, to two decimal places, is d = -0.71.
Determining Significance
If we were to look at a critical values look-up table for \(df = 9\) and \(\alpha = .05\) (two-tailed), we would see that the critical value is \(t_{crit} = 2.262\). Given that our t-value, ignoring polarity and just looking at the absolute value, so \(t = 2.25\), is smaller than our \(t_{crit}\) then we can say our result is not significant, and as such would be written up as t(9) = 2.25, p > .05, d = 0.71.
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 = 0.051, and would be written up as p = 0.051
9.2.3 DataSet 3
Let's say that this is our starting data:
| Participants | PreTest | PostTest |
|---|---|---|
| 1 | 62 | 56 |
| 2 | 75 | 72 |
| 3 | 56 | 55 |
| 4 | 80 | 79 |
| 5 | 55 | 55 |
| 6 | 70 | 69 |
| 7 | 79 | 79 |
| 8 | 50 | 52 |
| 9 | 51 | 49 |
| 10 | 78 | 77 |
The first thing we need to do is calculate the difference between the PostTest and the PreTest for each participant, based on \(D = PostTest - PreTest\). So for example:
- Participant 1 would be: 56 - 62 = -6
- Participant 2 would be: 72 - 75 = -3
- etc
And if we do that for each Participant and added a column of the differences (\(D\)) then we would see:
| Participants | PreTest | PostTest | D |
|---|---|---|---|
| 1 | 62 | 56 | -6 |
| 2 | 75 | 72 | -3 |
| 3 | 56 | 55 | -1 |
| 4 | 80 | 79 | -1 |
| 5 | 55 | 55 | 0 |
| 6 | 70 | 69 | -1 |
| 7 | 79 | 79 | 0 |
| 8 | 50 | 52 | 2 |
| 9 | 51 | 49 | -2 |
| 10 | 78 | 77 | -1 |
Now, the within-subjects t-test formula is:
\[t = \frac{\bar{D}}{\frac{SD_{D}}{\sqrt{N}}}\]
We can see that \(N = 10\), but we need to calculate \(\bar{D}\) (called D-Bar, the mean of the \(D\) column) and \(SD_{D}\).
Calculating D-bar
So the \(\bar{D}\) formula is the same as the mean formula:
\[\bar{D} = \frac{\sum{D}}{N}\]
Where \(D\) is \(PostTest - PreTest\) for each Participant.
Then:
\[\bar{D} = \frac{(56 - 62) + (72 - 75) + (55 - 56) + (79 - 80) + (55 - 55) + \\ (69- 70) + (79 - 79) + (52 - 50) + (49 - 51) + (77 - 78)}{10}\]
Which if we resolve all the brackets becomes:
\[\bar{D} = \frac{-6 + -3 + -1 + -1 + 0 + -1 + 0 + 2 + -2 + -1}{10}\]
And if we sum the top half together
\[\bar{D} = \frac{-13}{10}\]
Leaving us with:
\[\bar{D} = -1.3\]
So we find that \(\bar{D}\) = -1.3, which is the mean difference between the Post test and Pre test values.
The Standard Deviation of D
The standard deviation formula is:
\[SD = \sqrt\frac{\sum(X - \bar{X})^2}{N-1}\]
Which if we translate to using D, becomes:
\[SD_{D} = \sqrt\frac{\sum(D - \bar{D})^2}{N-1}\]
Then:
\[SD_{D} =\sqrt\frac{(-6 - -1.3)^2 + (-3 - -1.3)^2 + (-1 - -1.3)^2 + (-1 - -1.3)^2 + \\ (0 - -1.3)^2 + (-1 - -1.3)^2 + (0 - -1.3)^2 + (2 - -1.3)^2 + \\ (-2 - -1.3)^2 + (-1 - -1.3)^2}{10 - 1}\]
And if we start stepping through this analysis by dealing with the brackets:
\[SD_{D} =\sqrt\frac{(-4.7)^2 + (-1.7)^2 + (0.3)^2 + (0.3)^2 + (1.3)^2 + \\ (0.3)^2 + (1.3)^2 + (3.3)^2 + (-0.7)^2 + (0.3)^2}{10 - 1}\]
And then we square those brackets
\[SD_{D} =\sqrt\frac{22.09 + 2.89 + 0.09 + 0.09 + 1.69 + \\ 0.09 + 1.69 + 10.89 + 0.49 + 0.09}{10 - 1}\]
And sum up all the values on the top half:
\[SD_{D} =\sqrt\frac{40.1}{10 - 1}\]
And then we need to sort out the bottom half
\[SD_{D} =\sqrt\frac{40.1}{9}\]
Which by dividing the top half by the bottom half reduces down to:
\[SD_{D} =\sqrt{4.4555556}\]
And then we finally take the square root, which leaves us with:
\[SD_{D} =2.1108187\]
And so we find that the \(SD_{D}\) = 2.1108187 which to two decimal places would be, \(SD_{D} = 2.11\)
Calculating the t-value
And finally the t-test formula is:
\[t = \frac{\bar{D}}{\frac{SD_{D}}{\sqrt{N}}}\]
Then if we start filling in the values we know from above we see:
\[t = \frac{-1.3}{\frac{2.1108187}{\sqrt{10}}} \]
And if we deal with the square root first:
\[t = \frac{-1.3}{\frac{2.1108187}{3.1622777}} \]
And divide \(SD_{D}\) by \(\sqrt{N}\) - tidying up the bottom of the formula:
\[t = \frac{-1.3}{0.6674995} \]
And then solve for \(t\) by dividing the top half by the bottom half gives us:
\[t = -1.9475671 \]
And so, rounding to two decimal places, we find that \(t = -1.95\)
Degrees of Freedom
Great! Now we just need the degrees of freedom where the formula is:
\[df = N - 1\]
And we already know that \(N= 10\) and putting them into the equation we get:
\[df = 10 - 1\]
Which reduces to:
\[df = 9\]
Meaning that we find a \(df = 9\)
Effect Size - Cohen's d
And finally Cohen's d, the effect size. One of the common formulas based on knowing the t-value and the N is:
\[d = \frac{t}{\sqrt{N}}\]
Which, based on the info above, we know:
- \(t = -1.95\)
- \(N = 10\)
And putting those into the formula we get:
\[d = \frac{-1.95}{\sqrt{10}}\]
Which gives us:
\[d = \frac{-1.95}{3.1622777}\]
And so:
\[d = -0.6166441\]
Meaning that the effect size, to two decimal places, is d = -0.62.
Determining Significance
If we were to look at a critical values look-up table for \(df = 9\) and \(\alpha = .05\) (two-tailed), we would see that the critical value is \(t_{crit} = 2.262\). Given that our t-value, ignoring polarity and just looking at the absolute value, so \(t = 1.95\), is smaller than our \(t_{crit}\) then we can say our result is not significant, and as such would be written up as t(9) = 1.95, p > .05, d = 0.62.
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 = 0.083, and would be written up as p = 0.083
9.3 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 |