Worked examples
The question
Question: the same 5 people are measured before and after: at the 5% level, did the scores go up?
before: 70, 65, 80, 75, 60; after: 74, 66, 85, 76, 64
The same people are measured twice.
So we look at each person's change: one list of differences.
the same people twice: work with each one's difference
1 The differences
Each row is one person: before and after.
74 minus 70 is 4.
66 minus 65 is 1.
85 minus 80 is 5.
76 minus 75 is 1.
64 minus 60 is 4.
Careful!
The two lists are the same people.
So we test the differences, not two separate groups.
Careful: one list of differences, not two groups
2 Mean and SD of d
The differences add up to 15.
Their mean: 15/5 is 3.
d̄ = 155 = 3
For the SD: each difference minus 3, squared.
4 gives 1.
1 gives 4.
5 gives 4.
1 gives 4.
4 gives 1.
They add up to 14.
A sample SD divides by n minus 1: 14/4 is 3.5.
The square root: about 1.8708.
sd = √145 − 1 = √3.5 ≈ 1.8708
3 Write the hypotheses
H₀ is the "nothing new" idea: it has the equal sign.
Here: the mean change is 0.
H0 : μd = 0
Hₐ is what we want to show.
Here: the mean change is more than 0.
Ha : μd > 0
More than means only the right end counts.
One tail.
one tail: the right end
4 The test number
The SD of the mean difference: 1.8708 over the square root of 5, about 0.8366.
SD of the mean = sd√n = 1.8708√5 ≈ 0.8366
t is the mean difference minus 0, over 0.8366: about 3.586.
t = 3 − 00.8366 ≈ 3.586
5 The cut-off
The t-table needs df: n minus 1, 4.
df = n − 1 = 5 − 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, 3.586, is past the cut-off: in the reject part.
So we reject H zero.
= t = 3.586 > 2.132 → reject H0
7 Say it in words
At the 5% level, there is enough evidence that
In words: at the 5% level, there is enough evidence that the scores went up.
the scores went up.
8 Check
Let's check with the mean difference itself.
The cut-off for it is 0 plus 2.132 times 0.8366: 1.7836.
cut-off for the mean difference: 0 + 2.132 × 0.8366 = 1.7836
The mean difference, 3, is past it.
The same decision.
So it is right!
3: past it, reject
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.
Answer = t = 3.586 > 2.132 → reject H0
▶ Watch it solved
The question
Question: the same 6 people are measured before and after: at the 5% level, did the scores go up?
before: 20, 22, 18, 25, 21, 24; after: 22, 25, 18, 28, 22, 27
The same people are measured twice.
So we look at each person's change: one list of differences.
the same people twice: work with each one's difference
1 The differences
Each row is one person: before and after.
22 minus 20 is 2.
25 minus 22 is 3.
18 minus 18 is 0.
28 minus 25 is 3.
22 minus 21 is 1.
27 minus 24 is 3.
Careful!
The two lists are the same people.
So we test the differences, not two separate groups.
Careful: one list of differences, not two groups
2 Mean and SD of d
The differences add up to 12.
Their mean: 12/6 is 2.
d̄ = 126 = 2
For the SD: each difference minus 2, squared.
2 gives 0.
3 gives 1.
0 gives 4.
3 gives 1.
1 gives 1.
3 gives 1.
They add up to 8.
A sample SD divides by n minus 1: 8/5 is 1.6.
The square root: about 1.2649.
sd = √86 − 1 = √1.6 ≈ 1.2649
3 Write the hypotheses
H₀ is the "nothing new" idea: it has the equal sign.
Here: the mean change is 0.
H0 : μd = 0
Hₐ is what we want to show.
Here: the mean change is more than 0.
Ha : μd > 0
More than means only the right end counts.
One tail.
one tail: the right end
4 The test number
The SD of the mean difference: 1.2649 over the square root of 6, about 0.5164.
SD of the mean = sd√n = 1.2649√6 ≈ 0.5164
t is the mean difference minus 0, over 0.5164: about 3.873.
t = 2 − 00.5164 ≈ 3.873
5 The cut-off
The t-table needs df: n minus 1, 5.
df = n − 1 = 6 − 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, 3.873, is past the cut-off: in the reject part.
So we reject H zero.
= t = 3.873 > 2.015 → reject H0
7 Say it in words
At the 5% level, there is enough evidence that
In words: at the 5% level, there is enough evidence that the scores went up.
the scores went up.
8 Check
Let's check with the mean difference itself.
The cut-off for it is 0 plus 2.015 times 0.5164: 1.0405.
cut-off for the mean difference: 0 + 2.015 × 0.5164 = 1.0405
The mean difference, 2, is past it.
The same decision.
So it is right!
2: past it, reject
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.
Answer = t = 3.873 > 2.015 → reject H0
▶ Watch it solved
The question
Question: the same 7 people are measured before and after: at the 5% level, did the scores go up?
before: 72, 68, 75, 80, 66, 70, 78; after: 75, 70, 76, 84, 70, 73, 82
The same people are measured twice.
So we look at each person's change: one list of differences.
the same people twice: work with each one's difference
1 The differences
Each row is one person: before and after.
75 minus 72 is 3.
70 minus 68 is 2.
76 minus 75 is 1.
84 minus 80 is 4.
70 minus 66 is 4.
73 minus 70 is 3.
82 minus 78 is 4.
Careful!
The two lists are the same people.
So we test the differences, not two separate groups.
Careful: one list of differences, not two groups
2 Mean and SD of d
The differences add up to 21.
Their mean: 21/7 is 3.
d̄ = 217 = 3
For the SD: each difference minus 3, squared.
3 gives 0.
2 gives 1.
1 gives 4.
4 gives 1.
4 gives 1.
3 gives 0.
4 gives 1.
They add up to 8.
A sample SD divides by n minus 1: 8/6 is 1.3333.
The square root: about 1.1547.
sd = √87 − 1 = √1.3333 ≈ 1.1547
3 Write the hypotheses
H₀ is the "nothing new" idea: it has the equal sign.
Here: the mean change is 0.
H0 : μd = 0
Hₐ is what we want to show.
Here: the mean change is more than 0.
Ha : μd > 0
More than means only the right end counts.
One tail.
one tail: the right end
4 The test number
The SD of the mean difference: 1.1547 over the square root of 7, about 0.4364.
SD of the mean = sd√n = 1.1547√7 ≈ 0.4364
t is the mean difference minus 0, over 0.4364: about 6.874.
t = 3 − 00.4364 ≈ 6.874
5 The cut-off
The t-table needs df: n minus 1, 6.
df = n − 1 = 7 − 1 = 6
One tail: 0.05 in one end.
Row 6, column 0.05.
Row 6, column .05: t* = 1.943.
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, 6.874, is past the cut-off: in the reject part.
So we reject H zero.
= t = 6.874 > 1.943 → reject H0
7 Say it in words
At the 5% level, there is enough evidence that
In words: at the 5% level, there is enough evidence that the scores went up.
the scores went up.
8 Check
Let's check with the mean difference itself.
The cut-off for it is 0 plus 1.943 times 0.4364: 0.8479.
cut-off for the mean difference: 0 + 1.943 × 0.4364 = 0.8479
The mean difference, 3, is past it.
The same decision.
So it is right!
3: past it, reject
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.
Answer = t = 6.874 > 1.943 → reject H0
▶ Watch it solved
