Step by step
The question
Question: the same 8 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.
27 minus 25 is 2.
33 minus 30 is 3.
29 minus 28 is 1.
25 minus 22 is 3.
28 minus 27 is 1.
29 minus 26 is 3.
26 minus 24 is 2.
32 minus 29 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 18.
Their mean: 18/8 is 2.25.
For the SD: each difference minus 2.25, squared.
2 gives 0.0625.
3 gives 0.5625.
1 gives 1.5625.
3 gives 0.5625.
1 gives 1.5625.
3 gives 0.5625.
2 gives 0.0625.
3 gives 0.5625.
They add up to 5.5.
A sample SD divides by n minus 1: 5.5/7 is 0.7857.
The square root: about 0.8864.
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: 0.8864 over the square root of 8, about 0.3134.
t is the mean difference minus 0, over 0.3134: about 7.179.
5 The cut-off
The t-table needs df: n minus 1, 7.
One tail: 0.05 in one end.
Row 7, column 0.05.
Row 7, column .05: t* = 1.895.
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, 7.179, 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 1.895 times 0.3134: 0.5939.
The mean difference, 2.25, 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.