Step by step
The question
Question: the same 5 people are measured before and after: at the 5% level, did the times 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.
4 minus 3 is 1.
5 minus 5 is 0.
6 minus 4 is 2.
7 minus 6 is 1.
4 minus 2 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: 6/5 is 1.2.
For the SD: each difference minus 1.2, squared.
1 gives 0.04.
0 gives 1.44.
2 gives 0.64.
1 gives 0.04.
2 gives 0.64.
They add up to 2.8.
A sample SD divides by n minus 1: 2.8/4 is 0.7.
The square root: about 0.8367.
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: 0.8367 over the square root of 5, about 0.3742.
t is the mean difference minus 0, over 0.3742: about 3.207.
5 The cut-off
The t-table needs df: n minus 1, 4.
Two tails: 0.05 is split, 0.025 in each end.
Row 4, column 0.025.
Row 4, column .025: t* = 2.776.
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, 3.207, 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 changed.
8 Check
Let's check with the mean difference itself.
The cut-off for it is 0 plus or minus 2.776 times 0.3742: −1.0388 and 1.0388.
The mean difference, 1.2, 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.