Step by step
The question
Question: the same 6 people are measured before and after: at the 5% level, did the weights 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.
200 minus 210 is −10.
190 minus 195 is −5.
228 minus 230 is −2.
199 minus 205 is −6.
210 minus 220 is −10.
212 minus 215 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 −36.
Their mean: −36/6 is −6.
For the SD: each difference minus −6, squared.
−10 gives 16.
−5 gives 1.
−2 gives 16.
−6 gives 0.
−10 gives 16.
−3 gives 9.
They add up to 58.
A sample SD divides by n minus 1: 58/5 is 11.6.
The square root: about 3.4059.
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: 3.4059 over the square root of 6, about 1.3905.
t is the mean difference minus 0, over 1.3905: about −4.315.
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, −4.315, 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 weights went down.
8 Check
Let's check with the mean difference itself.
The cut-off for it is 0 minus 2.015 times 1.3905: −2.8019.
The mean difference, −6, 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.