Step by step
The question
Question: the same 5 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.
78 minus 80 is −2.
72 minus 75 is −3.
89 minus 90 is −1.
80 minus 85 is −5.
66 minus 70 is −4.
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 −15.
Their mean: −15/5 is −3.
For the SD: each difference minus −3, squared.
−2 gives 1.
−3 gives 0.
−1 gives 4.
−5 gives 4.
−4 gives 1.
They add up to 10.
A sample SD divides by n minus 1: 10/4 is 2.5.
The square root: about 1.5811.
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: 1.5811 over the square root of 5, about 0.7071.
t is the mean difference minus 0, over 0.7071: about −4.243.
5 The cut-off
The t-table needs df: n minus 1, 4.
One tail: 0.05 in one end.
Row 4, column 0.05.
Row 4, column .05: t* = 2.132.
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, −4.243, 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 2.132 times 0.7071: −1.5075.
The mean difference, −3, 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.