Parametic Test

T-test for Dependent Samples

A. Definition / Description

T-test for dependent samples (T-test dependent)

- compares the means between two related groups on the same
continuous dependent variable.

- used to test if a significant difference exists between means

before and after treatment (pre-test and post test) or when we compare
results of match-paired groups.

- used to determine if there is a change from one measurement

(group) to the other.

B. Assumptions for the T-test Dependent

1. Your dependent variable should be measures on a continuous scale

(interval/ratio).
2. Related samples/groups. This means that the subjects in the first
group are also in the second group.
3. No significant outliers( a data point that differs significantly from
other observations) in the two groups.
4. Normality. The distribution of the difference in the dependent
variable between the two related groups be approximately normally
distributed.

C. Sample Problem 1: T-test for Dependent Samples

Pre-test and post test were administered to the ten(10) BEEd students. After
the administration of the new teaching strategy for one(1) month, same test was
administered again. Does the new strategy significantly increase the scores of
the students?
Pre-test Post test
x
( 1) ( x 2)

0 14
0 6
3 4
20 15
0 3
0 3
1 6
1 5
1 6
0 3

D. T-test for Dependent Samples Formula

∑D
t= Where in

√ n ∑ D2 −(∑ D)2
n−1
study
D is the difference between pairs of scores
n is the number of pairs of subjects in the
E. Application of the T-test dependent
* to answer the given problem, steps may vary from book to book

1. State the hypotheses.

H 0 : μ1=μ2 H a : μ1 ≠ μ2

H 0: There is no significant H a : There is significant difference

difference between the pre-test between the pre-test and post test
and post test scores of the BEEd scores of the BEEd students.
2. students. Find the critical values.

α =0.05
Degree of freedom(df )=n − 1=10 −1=9

Critical value= 1.833

3. Compute for the test value(t)

A. Subtract pre-test and post test to determine D

B. Find the value of D 2 by multiplying the value of D by itself

C. Find the value of ∑ D and ∑ D2.

Pre-test Post test D D

2

( x 1) ( x 2) ( x 1 − x2 )

0 14 -14 196
0 6 -6 36
3 4 -1 1
20 15 5 25
0 3 -3 9
0 3 -3 9
1 6 -5 25
1 5 -4 16
1 6 -5 25
0 3 -3 9
∑ D=− 39 2
∑ D =351

D. Substitute the values on the given formula

∑ D=− 39
2
∑ D =351
n = 10

∑D
t=

√
2 2
n ∑ D −(∑ D)
n−1

−39
t=

√
2
(10)(351)−(− 39)
10 −1

E. Solve the value of t.

−39
t=

√ 3510 −1521
9

−39
t=

√ 1989
9

− 39
t=
√221
 − 39
t=
14.866
t=−2.623

The value of t-test dependent is -2.623.

4. Write the decision rule for rejecting the null hypothesis.

t-value= |− 2.623| = 2.623

t-critical value = 1.833

2.623 ≥1.833

Reject H 0 :t − value ≥t − critical value

* We will be rejecting the null hypothesis if the absolute value of t-value
is greater than or equal to t-critical value.

Accept H 0 :t − value<t − critical value

* We will be accepting the null hypothesis if the absolute value of the t-
value is less than the t-critical value.

5. Conclusion

Since, t-value is greater than the t-critical value, we reject the null
hypothesis. Therefore, there is a significant difference between the pre-test aand
post test of the BEEd students. The new strategy did significant increase on the
post test scores of the students.
Sample Problem 2: T-test for Dependent Sample

A composition teacher wishes to see whether a new grammar program will

reduce the number of grammatical errors her students make when writing a two-
page essay. The data are shown here. At α =0.05 . Can it be concluded that the
number of errors have been reduced?

Student Errors before Errors after

1 12 9
2 9 6
3 0 1
4 5 3
5 4 2
6 3 3

T-test for Dependent Sample Formula

∑D
t=

√
2 2
n ∑ D −(∑ D)
n−1

Application of the T-test dependent

1. State the hypotheses.

: There is no significant difference between :There is significant difference between the

the average number of errors before and average number of errors before and after
after the new grammar program. the new grammar program.

2.
Find
the critical values.

α =0.05
Degree of freedom(df )=n − 1=6 −1=5

Critical Value = 2.015

3. Compute for the test value(t)

A. Subtract the number of grammatical errors before the new program

and the errors after to determine D

B. Find the value of D2 by multiplying the value of D by itself

C. Find the value of ∑ D and ∑ D2.

Student Errors Errors after D D

2

before (x 1 − x2 )
1 12 9 3 9
2 9 6 3 9
3 0 1 -1 1
4 5 3 2 4
5 4 2 2 4
6 3 3 0 0
∑ D=9 2
∑ D =27

D. Substitute the values on the given formula.

∑ D=9
2
∑ D =27
n=6

F. Solve the value of t.

9
t=
4.025

t=2.236

The value of t-test dependent is 2.236.

4. Write the decision rule for rejecting the null hypothesis.

t-value= |2.236| = 2.236

t-critical value = 2.015

2.236 ≥ 2.015

Reject H 0 :t − value ≥t − critical value

* We will be rejecting the null hypothesis if the absolute value of t-value
is greater than or equal to t-critical value.

Accept H 0 :t − value<t − critical value

* We will be accepting the null hypothesis if the absolute value of the t-
value is less than the t-critical value.

5. Conclusion

Since, t-value is greater than the t-critical value, we reject the null
hypothesis. Therefore, there is a significant difference between the average
number of errors before and after the new grammar program.
Sample Problem 3: T-test for Dependent Sample

A Mathematics teacher wanted to determine the effectiveness of using a new

method of teaching a lesson in Mathematics. Before teaching the subject matter
he gave a 20-item test to the group and after that, he exposed the group to the
new method. At the end of the unit he gave the same test to the group. The
data collected are reflective in the table that follows

Pretest and Posttest: Scores of 9 Students in Math

Student Pretest Scores Posttest Scores

1 12 15
2 14 17
3 10 13
4 9 9
5 11 12
6 13 16
7 14 18
8 9 12
9 8 13

Is there a significant difference between the pretest meean score and the
posttest mean score of the experimental group? Test at 0.05 level.

T-test for Dependent Sample Formula

∑D
t=

√ n ∑ D2 −(∑ D)2
n−1

Application of the T-test dependent

1. State the hypotheses.

2.
: There is no significant difference between :There is significant difference between the
the pretest and posttest mean scores of pretest and posttest mean scores of the
the experimental group. experimental group.

Determine the Scale of Measurement: Interval

3. Level of Significance: 5 %=0.05(α )

4. Test Statistics: T-test for dependent samples

5. Computation:(Based on the given data)

The table shows the SPSS statistical package software output for this data.

Paired Samples Test

Sig.(2-
Paired Differences t d tailed)
Mean Std. Std. 5% Confidence
Interval of the
f
Deviatio Error Difference

n Mean Lower Upper

Pair 1 -2.7778 1.48137 .49379 -3.9165 -1.6391 -5.625 8 .000

PRETEST-
POSTTEST

6. Statistical Decision:

Reject H 0 if t compounded >2.306 , otherwise, accept H 0.

Since the computed t-value of 5.625 exceeded the table value of t .05 for 8
degrees of freedom, H 0 is rejected. (There is evidence to support the rejection of
H 0) or, since the p-value of .000* is less than .05, the null hypothesis is rejected.

* A p-value of .000 is not equal to 0. It means that it is very, very small.

7. Conclusion:

Since H 0 is rejected, there is a significant difference between the pretest and

posttest mean scores of the experimental group. Since the posttest mean is
higher than the pretest mean score, it implies that the performance of the group
improved. It means further, that the new method of teaching was effective.

Manual Computation

1. State the hypotheses.

: There is no significant difference between :There is significant difference between the

the pretest and posttest mean scores of pretest and posttest mean scores of the
the experimental group. experimental group.

2. Find the critical values.

α =0.05
Degree of freedom(df )=n − 1=9 −1=8

Critical Value= 1.860

3. Compute for the test value(t).

A. Subtract the number of grammatical errors before the new program and the
errors after to determine D.

B. Find the value of D 2 by multiplying the value of D by itself.

C. Find the value of ∑ D and ∑ D2.

Student Pretest Posttest D D

2

Scores Scores (x 1 − x2 )
1 12 15 -3 9
2 14 17 -3 9
3 10 13 -3 9
4 9 9 0 0
5 11 12 -1 1
6 13 16 -3 9
7 14 18 -4 16
8 9 12 -3 9
9 8 13 -5 25
∑ D=− 25 ∑ D2=¿ 87

D. Substitute the values on the given formula

∑ D=− 25
2
∑ D =87
n=9
∑D
t=

√
n ∑ D2 −(∑ D)2
n−1

− 25
t=

√ (9)( 87)−(− 25)2

9− 1

E. Solve for the value of T. −25

t=

√ 783 −625
8
 −25
t=

√ 158
8

− 25
t=
√19.75
t=−5.625

The value of t-test dependent is -5.625.

4. Write the decision rule for rejecting the null hypothesis.

t-value= |− 5.625| = 5.625

T-critical value= 1.860
5.625 ¿ 1.860

Reject H 0 :t − value ≥t − critical value

* We will be rejecting the null hypothesis if the absolute value of t-value
is greater than or equal to t-critical value.

Accept H 0 :t − value<t − critical value

* We will be accepting the null hypothesis if the absolute value of the t-
value is less than the t-critical value.

5. Conclusion

Since, t-value is greater than the t-critical value, we reject the null hypothesis.
Therefore, there is significant difference between the pretest and post test mean
scores of the experimental group.