6 One-Sample t-test
one-sample t-test: Compare your sample mean to a known test value. For example, to compare your sample IQ to the population norm of 100.
6.1 The Worked Example
The formula is as follows:
\[t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\] And you can translate the symbols as:
- \(\bar{x}\) is the mean value of your sample of interest
- \(\mu\) is the test value to compare against (also called the criterion value)
- \(s\) is the standard deviation of your sample of interest
- \(n\) is the number of observations (e.g. participants) in your sample of interest
So for example, if we had a sample of 25 people, and we knew that the mean and standard deviation of the IQ of was M = 110, SD = 7.5, and that we wanted to compare that sample to a criterion value of \(\mu\) = 100, then we would do the following:
\[t = \frac{110 - 100}{\frac{7.5}{\sqrt{25}}}\]
And if we do the top half of the formula first, that becomes:
\[t = \frac{10}{\frac{7.5}{\sqrt{25}}}\]
And then start to complete the bottom half by taking the sqaure root of n first, (i.e. \(\sqrt{n}\)), we get:
\[t = \frac{10}{\frac{7.5}{5}}\] And so if we then complete the bottom half by dividing the standard deviation of the sample (\(s\)) by the square root of n (\(\sqrt{n}\)), we get:
\[t = \frac{10}{1.5}\]
And finally, if we divide the top half of the formula by the bottom half we get:
\[t = \frac{10}{1.5} = 6.6666667\]
Giving us a t-value of t = 6.6666667 which we can round to two decimal places, giving us t = 6.67
6.2 Degrees of Freedom
The degrees of freedom in a one-sample t-test is calculated as:
\[df = N - 1\]
And if we know that we have 25 participants in our sample then we get:
\[df = N - 1 = 25 - 1 = 24\]
Giving us \(df\) = 24
6.3 Effect Size
A common effect size used for t-test is Cohen's d. The formula for the effect size, in of a one-sample t-test, if you only know the t-value and the number of participants is:
\[d = \frac{t}{\sqrt{N}}\]
And here we know that:
- \(t\) = 6.67
- \(N\) = 25
Which if we put those into the formula we get:
\[d = \frac{6.67}{\sqrt{25}}\] And if we resolve the bottom half first:
\[d = \frac{6.67}{5}\]
And then divide the top by the bottom we get:
\[d = \frac{6.67}{5} = 1.334\]
Giving us an effect size of \(d\) = 1.33, rounded to two decimal places.
6.4 Write-up
If we were to look at a critical values look-up table for \(df = 24\) and \(\alpha = .05\) (two-tailed), we would see that the critical value is \(t_{crit} = 2.064\).
Given that our t-value, ignoring polarity and just looking at the absolute value, so \(t = 6.67\), 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(24) = 6.67, p < .05, d = 1.33
6.5 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 |