T Test for Paired Data
When to use: in situations where the second of each set of scores is
related to the first. The classic case is that of a study where there is
no control group and we are having to gather pre-test and post-test
scores from the same individuals.
Example: A sample of 10 high school seniors has tried two methods of
learning words in a second (used to be called foreign) language. In the
first part of the study they studied in pairs, after which they were
tested for the number of words known. In the second part of the study
they tried it with a second method: alone. They were again tested for
the number of words known. Did the different procedures result in
different achievement? The null hypothesis says there is is no
difference in the number of words known between the two conditions.
Number of words learned
__________________________
2
Studied in pairs
Studied alone
14
14
15
16
14
15
13
10
12
16
__
9
10
9
8
14
7
8
10
13
9
__
139=sum of pairs
97=sum of alones
D
5
4
6
8
0
8
5
0
-1
7
__
42
D
25
16
36
64
0
64
25
0
1
49
__
280
The T test for paired (correlated, dependent) data uses the formula of
_
D
T=
_
where D is the mean of the differences in scores
between pairs of matched subjects
S
diff
_
D is the sum of D divided by N, or
S
S
=
diff
d
_____
N
2
42/10 =4.2.
S =
d
S
280 - (42)
10
____________
9
=
3.39
=
diff
=
3.39
3.16
280- 176.4
__________
9
=
=
103.6
_____
9
=
11.51
= 3.39
1.07
10
T=
4.2
1.07
=
3.93
The T that is required for significance with N-1 or 9 pairs of data at
the .05 level (two-tailed test) is 2.262. Our T of 3.93 is greater than
the table value of T, so we reject the null hypothesis; studying in pairs
is more effective in learning than studying alone. If we repeated the
experiment 100 times, we could expect this trend in the data 95 times.
Five times out of 100 studying alone would result in the retention of
more words in a second language.