Step by step
The question
Question: the same 6 people are measured before and after: at the 5% level, did the times go down?
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 7 is −1.
9 minus 9 is 0.
4 minus 6 is −2.
7 minus 8 is −1.
8 minus 10 is −2.
7 minus 7 is 0.
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: −6/6 is −1.
For the SD: each difference minus −1, squared.
−1 gives 0.
0 gives 1.
−2 gives 1.
−1 gives 0.
−2 gives 1.
0 gives 1.
They add up to 4.
A sample SD divides by n minus 1: 4/5 is 0.8.
The square root: about 0.8944.
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 less than 0.
Less than means only the left end counts.
One tail.
4 The test number
The SD of the mean difference: 0.8944 over the square root of 6, about 0.3651.
t is the mean difference minus 0, over 0.3651: about −2.739.
5 The cut-off
The t-table needs df: n minus 1, 5.
One tail: 0.05 in one end.
Row 5, column 0.05.
Row 5, column .05: t* = 2.015.
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, −2.739, 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 times went down.
8 Check
Let's check with the mean difference itself.
The cut-off for it is 0 minus 2.015 times 0.3651: −0.7357.
The mean difference, −1, 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.