n = 7.5 = 7
The result of median is the number 7 + 1 = 8
= 88 From the data of controlled class above, the writer concludes that
the lowest score of the pre-test is 53 and the high score of pre-test is 97. So, average score of the pre-test is 78.43. The lowest score of the post-test
is 64 and the high of post-tests core is 97. And then, the average score for post-test 85.5 and for gained score 7.06. It means that the post-test score is
higher than pre-test
2. The Analysis of Data
The analysis of data from the result of pre-test and post test experiment and controlled class, the writer found that the score of experiment class
was higher than controlled class. See table 2.1 and table 2.4. It happened because the students of experiment class were fully involved in learning
process and they need each other because the material related to each other. In the experiment class, the students were more taking pains in
answer the questions and seriously than controlled class because they often un-careful for answering the questions. It proved that teaching the Present
Continuous Tense through Audio-Lingual Method has significance to the student scores.
The writer also used statistic calculation of the t-test to make the data clearly. The t-test formula with degree of significance 5 looks like as
follows.
Table 2.7 The Result Calculation of the test both Experiment Class and
Control Class
Students Identification
Number X
Y
1. 28
8 -18.11
-0.93 327.97
0.86 2.
3 10
6.88 -2.93
47.33 8.58
3. 7
9.88 1.31
97.61 1.71
4. 11
- 1 -1.11
8.06 1.23
64.96 5.
- 3 5
12.88 2.06
165.89 4.24
6. 6
17 3.88
-9.93 15.05
98.60 7.
1 9.88
6.06 97.61
36.72 8.
16 4
-6.11 3.06
37.33 9.36
9. 25
11 -15.11
-3.93 228.31
15.44 10.
27 10
-17.11 -2.93
292.75 8.58
11. 2
5 7.88
2.06 62.09
4.24 12.
27 -17.11
7.06 292.75
49.84 13.
8 9
1.88 -1.93
3.53 3.72
14. 12
9 -2.11
-1.93 4.45
3.72 15.
8 18
1.88 -10.93
3.53 119.46
16. - 4
9 13.88
-1.93 192.65
3.72 17.
2 -
7.88 -
62.09 -
Total 168
113 0.03
-77 1932.17
433.75 = 168
MX=168:17 =9.88
=113 MY=113:16
= 7.06
Note: X = the students’ gained scores of the experiment class
Y = the students’ gained scores of the controlled class
x = X-MX y = Y-MY
From the total above the writer gets the calculation using t-test formula according to Anas Sudjono p. 315;
a. Determining mean 1 with the formula is:
M
1
=
∑X N
1
M
1
=
168 17
=
9.9
b. Determining Mean 2 the formula is:
M
2
=
∑ Y N
2
M
2
=
113 16
= 7.06
c. Determining Standard of Deviation Score of Variable X the
formula is:
M
2
=
∑ X
2
N
1
M
2
=
. 17
= 10.66
d. Determining Standard of Deviation Score of Variable Y , the
formulation is:
SD
2
=
∑ Y
2
N
2
=
. 15
= 5.38
e. Determining Standard Error Mean of Variable X SE
M1
the formula is:
SE
M 1
=
SD
1
N
1
−1
=
10.66 17
−1
=
10.66 16
= 0.81
f. Determining Standard Error of Mean of Variable Y, the formula is:
SE
M 2
=
SD
2
N
2
− 1
=
5.38 16
−1
=
5.38 15
= 0.59
g. Determining Standard Error of Difference of Mean of Variable X and Y, the formula is:
SE
M 1 −M2
=
= 0.81
2
+ 0.59
2
= 0.6561 + 0.3481
= 1.0042
= 1.00 h. Determining with formula:
=
. − .
1.00
=
2.84 1.00
= 2.84
i. Determining t-table in significance level 5 with Degree of Freedom: df = N1 + N2
– 2 = 17 + 16
– 2 = 33
– 2 = 31
The writer gained Degree of significance 5 = 1.07
t-score t
o
= 2.84
j. The comparison between t- score t and t-table
From calculation above, it is clear that the score of experiment class is higher than the score of controlled class. The writer also got
the result of the comparison between and
2.84 1.07 From the result of statistic calculation, it was found the value of the
was 8.28 and the degree of freedom df was 31. In this research, the writer used the degree of significance 5. It can be seen the df 31
and degree of significance 5 = 1.07 By comparing the value of
= 2.84 and t-table on the degree of significance 5 = 1.07. The writer summarized that
was higher than t-table.
3. The Testing of Hypotheses
