When is this used?
When you have two values for the same
Paired Samples t-tests
person
You have a “pair” of data
Every participant must be in both samples
An Experiment
The main idea
It is just like a single sample t test
but instead of using individual
scores we use “difference scores”
All of the calculations are based off
these difference scores
First
Four patients are tested on a treatment for
depression
Scores on a depression scale are given for before
and after for each patient
Is this study conducted between subjects or
within subjects?
Patient
Before
Treatment
After Treatment
A
30
40
B
10
15
C
45
45
D
25
30
Paired samples t test formula
Identify your population and write
your hypotheses
t=
M Difference − µ Difference
s Difference
To conduct a paired samples t test
Just like we do with raw data…
Compute the difference scores
Notice ALL of your calculations will be done using the
difference scores
Your book suggests to cross out the regular scores once
you have difference scores so that you don’t mistakenly
use the raw data for anything
Calculate the squared deviations
Patient
Before
Treatment
After
Treatment
Difference
Score
A
30
40
10
B
10
15
5
C
45
45
0
D
25
30
5
Sum each of these as well
Patient
Difference
Score
Difference –
mean
difference
Squared
Deviation
A
10
5
25
B
5
0
0
C
0
-5
25
D
5
0
0
Calculate the needed statistics
What is the t statistic?
Mdifference = 5
s = sqrt (SS / n-1) = sqrt (50 / 3) = sqrt (16.67) = 4.08
t = (M - µ) / sm = Mdifference – 0 / sm =
5/2.04 = 2.45
What was our critical value?
sm = s / sqrt(n) = 4.08 / sqrt (4) = 4.08 / 2 = 2.04
df = 3
Patient
Difference
Score
Difference –
mean
difference
Squared
Deviation
A
10
5
25
B
5
0
0
C
0
-5
25
D
5
0
0
A study skills experiment
Struggling students are often unsure of how to raise
their grade. One possibility is that educating them in
proper study techniques may help. In your notes and
on the next slide is a table of before and after exam
scores for a group of students that attempted to take a
study skills course to help raise their grades. Evaluate
the benefits of the study skills course.
t = 2.353
Our data do show a significant difference
between the before and after treatment
groups
We would reject the null hypothesis here
Data for the experiment
Student
Score before the Study
Skills Class
Score after the Study
Skills Class
1
60
63
2
72
71
3
55
50
4
60
64
5
57
80
6
71
71
7
62
61
8
56
58
9
55
54
10
70
68
Calculate Difference Scores
Student
Score before the Score after the
Difference
Study Skills Class Study Skills Class Scores
1
60
63
3
2
72
71
3
55
50
4
60
5
Calculate Squared Deviations
Student
Difference
Scores
Difference Score
– Mean
difference
Squared
Deviations
-1
1
3
.8
.64
-5
2
-1
-3.2
10.24
64
4
3
-5
-7.2
51.84
57
80
23
4
4
1.8
3.24
6
71
71
0
5
23
20.8
432.64
7
62
61
-1
6
0
-2.2
4.84
8
56
58
2
7
-1
-3.2
10.24
9
55
54
-1
8
2
-.2
.04
10
70
68
-2
9
-1
-3.2
10.24
10
-2
-4.2
17.64
Next
What is the t statistic?
Mdifference = 2.2
t = (M - µ) / sm = Mdifference – 0 / sm = 2.2/2.45 =
.896
What was our critical value?
df = 9
t = 1.833
Our data do NOT show a significant difference
between the before and after treatment groups
We would fail to reject the null hypothesis here
s = sqrt (SS / n-1) = sqrt (541.6/9) = sqrt
(60.18) = 7.76
sm = s / sqrt(n) = 7.76 / sqrt (10) = 7.76 / 3.16
= 2.45
Your turn
Chronically bad drivers (those with 3 + tickets in two
years) were targeted and allowed to take an enhanced
driver’s training course. Below are their scores on a
driving simulator before the extra training course and
after the extra training course. Does it appear that the
training course helps?
Subject
Before training
After training
1
68
72
2
72
82
3
74
74
4
70
68
5
72
80