Chart 4.2.1 The Score Pre-Test and Post-Test of Controlled Class
The highest score for the controlled class is 85; there is one student who got the highest score. The lowest score for controlled class is 45; there is one
student who has the lowest score. The total of controlled class is 1160. From the differences between table 4.1 and 4.2 above, it showed that the
experimental class is higher than the controlled class in the posttest result, which the total experimental class score 1460 is higher than the total of controlled class,
1160. It can be concluded that there was a positive influence of using pre-reading activities toward students‟ reading skill.
B. Data Analysis
Before calculating the t value of the observation, the writer would like to compare the result between two classes.
To know the result of the test, the writer makes table of students score for each class, both experiment class and controlled class.
10 20
30 40
50 60
70 80
90
Pre-Test Post-Test
Table 4.3 The Calculation of Comparison of the Experimental Class and Controlled
Class Test Student
Experimental Class X
Controlled Class Y
x X-MX
y Y-MY
X
2
Y
2
Student 1 20
25 8
33.75 64 1139.0625
Student 2 25
-5 13
3.75 169
14.0625 Student 3
15 -5
3 3.75
9 14.0625
Student 4 15
-25 3
-16.25 9
264.0625 Student 5
25 -20
13 -11.25
169 126.5625
Student 6 5
-12 13.75
144 189.0625
Student 7 20
-10 8
-1.25 64
1.5625 Student 8
10 15
-2 23.75
4 564.0625
Student 9 20
-15 8
-6.25 64
39.0625 Student 10
15 -10
3 -1.25
9 1.5625
Student 11 20
-25 8
-16.25 64
264.0625 Student 12
-10 -12
-1.25 144
1.5625 Student 13
5 10
-7 18.75
49 351.5625
Student 14 20
-25 8
-16.25 64
264.0625 Student 15
-20 -12
-11.25 144
126.5625 Student 16
-15 -12
-6.25 144
39.0625 Student 17
10 -10
-2 -1.25
4 1.5625
Student 18 10
-25 -2
-16.25 4
264.0625 Student 19
10 20
-2 28.75
4 826.5625
Student 20 -30
-12 -21.25
144 451.5625
∑X= 240 ∑Y= -175 ∑x=0
∑y= 0 ∑x
2
= 1470
∑y
2
= 4943.75
After making the table, the writer wants to calculate the t observation to be compared with t table. Below are the steps:
Based on data in the table 3, it has bee n calculated the result of ∑X= 240
and ∑Y= -175. Then the writer tries to find out the mean variable X variable Y withformula:
∑
= 240 20
=12
∑ = -175
20 = -8.75
Those mean result showed that there is differentiation between the mean of experiment class score and mean of controlled class. The mean gain score of
experiment class is 12 and the mean gain score of controlled class is -8.75. It means that the mean of experiment class score is higher than the mean of
controlled class. This score indicates that pre-reading activity is effective in teaching reading recount text.
Based on the table, it h as been known the result of ∑
x 2
= 14 70 and ∑
y 2
= 4943.75
which is the sum of each x
2
and the sum of y
2
. After we know that, the
values of deviation standard are calculated by using this formula: √∑
= √
= 8.57 √∑
= √
= 15.72
Standard deviation describes how variable the data itself, it represents the deviation of data from mean. If the standard deviation has higher value, it means
that the research data has higher variability and lower homogeneity. In contrary, if the standard deviation has lower value, it means that the research data tend to be
very close to homogeneity.
1
It can be interpreted that the sample of research is good for representing the population, because the research data is homogenous.
Therefore, the homogeneity test is connected with the value of Standard Deviation of two classes. Then, the writer conducted the homogeneity
test See the appendix 7. The result of homogeneity test showed, the F observation is 1.06. Based on df in significance degree of 5 is 2.12, it concludes
that F
o
F
table
1.06 2.12. From this explanation, it can be concluded that the variance between experiment class and controlled class is homogeneous.
1
Prof. Dr. Anas Sudijono, Pengantar Statistik Pendidikan, Jakarta: PT Raja Grafindo Persada, 2005, p. 170
After the value of standard deviation from two classes has been taken, the next procedure of calculation is determining Standard Error Mean variable X and
variable Y, with this formula: √
=
√
=
√
=1.97 √
=
√
=
√
= 3.61 The variability of mean sample is caused by sampling error.
2
It means that there is variability of mean data when the sample was taken. The value of
sampling error can be known by calculating Standard Error of the Mean SE
M
. The standard error mean of experiment class is 1.97 and standard error mean of
controlled class is 3.61. It means that the value of sampling error from experimental class is lower than the value of sampling error from controlled class.
After the values of Standard Error Mean of variable X and variable Y have been taken, the writer will use those values to calculate the Standard Error of the
Mean Difference between experiment class X and controlled class Y with the formula:
= √
= √
= 4.11 The calculation of Standard Error of the Mean Difference of variable X
and variable Y showed 4.11. This value will be used to calculate t
o
. After conducting Standard Error Mean of variable X and variable Y, the
last calculation of this research in determining the value of t
o
t observation with formula:
2
Sudijono, op. cit,. p. 281
= 12 – -8.75
4.11 = 20.75
4.11 = 5.05
At the last calculation, finally t
o
t observation was obtained that is 5.05. This result of t
o
will be compared with t-
table.
The interpretation of t
o
will be explained in the test of hypothesis.
C. Test of Hypothesis
