In the gained scores, the lowest score of the experimental class is 5 and the controlled class is -30. Meanwhile, the medium score of the experimental class is
20 and the controlled class is 15. And then, the highest gained’s score in the
experimental class is 60 and the controlled class is 50. Besides, the mode gained’s
score in the experimental class is 25 and the controlled class is 10 and 20. Meanwhile, the mean of gained score in the experimental class is 25.14 and the
controlled class is 17.29.
B. The Data Analysis
This section is intended to answer the research question Is there any influence of using short story towards students’ mastery of simple past tense?
The statistical analysis done to answer the research question is t-test. Prior to this analysis, t-test was also conducted both the experimental class and the controlled
class by using SPSS Statistic Product and Statistic Solution: a. t-test Analysis of Pre-Test
Table 4.4 The
t-test of pre-test in the experimental and the controlled class Mean Score
t-test Score
The Experimental Class The Controlled Class
M
x
= 62.57 M
y
= 50.14 t
o
= 4.026
The first, table 4.4 reports the t-test score of pre-test in the experimental class and the controlled class. Using t-test it is known there was a significant
difference from measurement score in the experimental class and the controlled class M
x
= 62.57, M
y
= 50.14, SD
x
= 14.265 SD
y
= 11.407, t68= 4.026, p= 0.00. It means that the mean of the experimental class is higher than the controlled class
M
x
M
y
. Therefore, there is a possibility that the mean score of post-test in experimental class will be higher too than the controlled class.
b. t-test Analysis of Post-Test
After the writer got the t-test score of pre-test in the experimental class and the controlled class, the t-test also done for the score of post-test in the
experimental class and the controlled class, as the following table:
Table 4.5 The
t-test of post-test in the experimental and the controlled class Mean Score
t-test Score
The Experimental Class The Controlled Class
M
x
= 87.71 M
y
= 67.43 t
o
= 9.363
Based on the above table, it is known there was a significant difference from measurement post-test scores in the experimental class and the controlled
class M
x
= 87.71, M
y
= 67.43, SD
x
= 7.509, SD
y
= 10.387, t 68 = 9.363, p= 0.00. It shown that the mean post-test in the experimental class is higher than the
controlled class M
x
M
y
. This result is possible because the mean score of pre- test in the experimental class had been higher than the controlled class. However,
it does not indicate that the treatment has the influence for the experimental class because the data is bias. Therefore, the writer also calculated the gained score both
of the experimental and the controlled classes to answer the alternative hypothesis.
c. t-test Analysis of Gained Score There is the t-test measurement of gained score in both of the experimental
and the controlled class, after the writer got the pre-test and post-test score. It is really necessary to know whether there has significant difference between the
experimental class and the controlled class. Moreover, it also answers whether the alternative hypothesis H
a
is accepted or rejected. The t-test calculation by using SPSS is as following table:
Table 4.6 t-test Analysis of Gained Score
Group N
Mean Std.
Deviation Std. Error Mean
Gain Experimental 35
25.14 13.90
2.35 Controlled
35 17.29
17.12 2.89
The above table shows that it was a significant difference from measurement score in the experimental class and the controlled class M
x
= 25.14, M
y
= 17.29, SD
x
= 13.90, SD
y
= 17.12, t 68 = 2.10, p= 0.03. Meanwhile, the writer also made the calculation table to gain Mean and
Deviation Standard from two variables by using manual calculation. The following table is the result of the comparison of the experimental class and the
controlled class:
Table 4.6 t-test Analysis of Gained Score
Levenes Test for Equality
of Variances t-test for Equality of Means
F Sig.
T Df
Sig. 2-
tailed Mean
Differ ence
Std. Error
Differ ence
95 Confidence Interval of the
Difference Lower Upper
Gain Equal varianc
es assume
d 1.20
.27 2.10
68 .03
7.85 3.72
.417 15.29
Equal varianc
es not assume
d 2.10
65.2 55
.03 7.85
3.72 .412
15.30
Table 4.7 The Comparison Score of Each Student in Experimental Class and
Controlled Class
Students X
Y X-MX
Y-MX X-MX
2
Y-MX
2
1 20
20 -5
3 25
9 2
25 35
18 324
3 10
50 -15
33 225
1089 4
20 5
-5 -12
25 144
5 30
5 5
-12 25
144 6
10 10
-15 -7
225 49
7 35
15 10
-2 100
4 8
5 45
-20 28
400 784
9 5
15 -20
-2 400
4 10
45 10
20 -7
400 49
11 15
30 -10
13 100
169 12
25 -10
-27 729
13 15
-30 -10
-47 100
2209 14
20 10
-5 -7
25 49
15 25
30 13
169 16
35 -5
10 -22
100 484
17 45
35 20
18 400
324 18
30 5
5 -12
25 144
19 10
20 -15
3 225
9 20
25 35
18 324
21 25
40 23
529
Students X
Y X-MX
Y-MX X-MX
2
Y-MX
2
22 5
20 -20
3 400
9 23
20 -20
-5 -37
25 1369
24 25
30 13
169 25
30 25
5 8
25 64
26 10
5 -15
-12 225
144 27
25 10
-7 49
28 40
10 15
-7 225
49 29
60 25
35 8
1225 64
30 5
15 -20
-2 400
4 31
55 25
30 8
900 64
32 35
30 10
13 100
169 33
30 15
5 -2
25 4
34 40
25 15
8 225
64 35
25 20
3 9
880 605
6768 10261
Mean 25.14
17.29 13.905
17.122
The procedures of calculation were as follow: 1. The mean of variable X:
M
x
=
M
x
= M
x
= 25.14
2. The mean of variable Y: M
y
=
M
y
= M
y
= 17.29 3. Determining standard of deviation score of variable X:
SD
x
=
√
SD
x
=
√
SD
x
= √
SD
x
= 13.905 4. Determining standard of deviation score of variable Y:
SD
y
=
√
SD
y
=
√
SD
y
= √
SD
y
= 17.122
5. Determining standard error of mean of variable X: SE
x
=
√
SE
x
=
√
SE
x
=
√
SE
x
=
SE
x
= 2.350 6. Determining standard error of mean of variable Y:
SE
y
=
√
SE
y
=
√
SE
y
=
√
SE
y
=
SE
y
= 2.894 7. Determining standard error of different mean of variable X and variable Y:
SE
MX-My
=
√
SE
MX-My
=
√
SE
MX-My
=
√
SE
MX-My
=
√ SE
MX-My
=
3.728 8. Determining t
o
with the formula:
t
o
=
t
o
=
t
o
=
t
o
=
2.10 9. Determining t-table in significance level 5 with degree of freedom df:
df = N
1
+ N
2
– 2 df = 68
df = 35 + 35 – 2
df = 70-2 Therefore, the degree of freedom df is 35+35= 70-2= 68 and the critical
value with the df= 68 at 0.05 alpha level of significance α is 1.65.
C. Testing of Hypothesis
