Step by step
The question
Question: the same 4 people are measured before and after: at the 5% level, did the scores go up?
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.
61 minus 60 is 1.
60 minus 58 is 2.
64 minus 65 is −1.
65 minus 62 is 3.
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 5.
Their mean: 5/4 is 1.25.
For the SD: each difference minus 1.25, squared.
1 gives 0.0625.
2 gives 0.5625.
−1 gives 5.0625.
3 gives 3.0625.
They add up to 8.75.
A sample SD divides by n minus 1: 8.75/3 is 2.9167.
The square root: about 1.7078.
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 more than 0.
More than means only the right end counts.
One tail.
4 The test number
The SD of the mean difference: 1.7078 over the square root of 4, about 0.8539.
t is the mean difference minus 0, over 0.8539: about 1.464.
5 The cut-off
The t-table needs df: n minus 1, 3.
One tail: 0.05 in one end.
Row 3, column 0.05.
Row 3, column .05: t* = 2.353.
The t curve, with the cut-off and our t.
Past the cut-off is the reject part: the rare samples.
6 Decide
Our t, 1.464, 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 went up.
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 2.353 times 0.8539: 2.0092.
The mean difference, 1.25, 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.