Step by step
The question
Question: the same 4 people are measured before and after: at the 5% level, did the scores change?
The same people are measured twice.
So we look at each person's change: one list of differences.
1 The differences
Each row is one person: before and after.
103 minus 100 is 3.
102 minus 104 is −2.
101 minus 98 is 3.
106 minus 102 is 4.
Careful!
The two lists are the same people.
So we test the differences, not two separate groups.
2 Mean and SD of d
The differences add up to 8.
Their mean: 8/4 is 2.
For the SD: each difference minus 2, squared.
3 gives 1.
−2 gives 16.
3 gives 1.
4 gives 4.
They add up to 22.
A sample SD divides by n minus 1: 22/3 is 7.3333.
The square root: about 2.708.
3 Write the hypotheses
H₀ is the "nothing new" idea: it has the equal sign.
Here: the mean change is 0.
Hₐ is what we want to show.
Here: the mean change is different from 0.
Different from means too big or too small: both ends of the bell count.
Two tails.
4 The test number
The SD of the mean difference: 2.708 over the square root of 4, about 1.354.
t is the mean difference minus 0, over 1.354: about 1.477.
5 The cut-off
The t-table needs df: n minus 1, 3.
Two tails: 0.05 is split, 0.025 in each end.
Row 3, column 0.025.
Row 3, column .025: t* = 3.182.
The t curve, with the cut-offs and our t.
Past the cut-offs is the reject part: the rare samples.
6 Decide
Our t, 1.477, is not past the cut-off.
So we do not reject H zero.
7 Say it in words
In words: at the 5% level, there is not enough evidence that the scores changed.
This does not prove H zero is true.
We just cannot rule it out.
8 Check
Let's check with the mean difference itself.
The cut-off for it is 0 plus or minus 3.182 times 1.354: −4.3084 and 4.3084.
The mean difference, 2, is not past it.
The same decision.
So it is right!
Let's look back.
First: the differences.
Then: mean and SD of d.
Then: write the hypotheses.
Then: the test number.
Then: the cut-off.
Then: decide.
Then: say it in words.
Last: check.