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.
6 minus 5 is 1.
8 minus 7 is 1.
8 minus 6 is 2.
10 minus 8 is 2.
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 6.
Their mean: 6/4 is 1.5.
For the SD: each difference minus 1.5, squared.
1 gives 0.25.
1 gives 0.25.
2 gives 0.25.
2 gives 0.25.
They add up to 1.
A sample SD divides by n minus 1: 1/3 is 0.3333.
The square root: about 0.5773.
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: 0.5773 over the square root of 4, about 0.2887.
t is the mean difference minus 0, over 0.2887: about 5.196.
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, 5.196, is past the cut-off: in the reject part.
So we reject H zero.
7 Say it in words
In words: at the 5% level, there is enough evidence that the scores went up.
8 Check
Let's check with the mean difference itself.
The cut-off for it is 0 plus 2.353 times 0.2887: 0.6793.
The mean difference, 1.5, is 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.