From the Table 4.5 above, it can be seen that the student who got 30 was 1 student, the student who got 35 was 1 student, the student who got
40 was 1 student, and so on until the total number of frequency are 41 data.
Chart 4.1 The Pre-test, Post-test and the Gained Scores of the
Experimental Class and Controlled Class
The chart 4.1 above describes the score of both classes in average level. It can be seen that the students’ reading comprehension achievement
in experimental class which was taught by using SQ3R method is higher than controlled class.
In analyzing the data, the researcher used comparative technique by comparing the experiment class and control class. This technique is useful
to statistically prove whether there is any significant difference between the two classes. There are two steps needed to be analyzed before
calculating statistical hypothesis; normality test and homogeneity test. After getting the data which are the results of students reading
comprehension achievement both of two classes, the writer analyzed them by using statistic calculation of the test formula.
B. Analysis of the Data
1. Normality Test
Calculating the normality is important to know whether or not the data has been normally distributed. SPSS software version 22 was used by the
writer to calculate the normality of the data. There are two kinds of testing the normality in SPSS; Kolmogrov Smirnov and Shapiro Wilk. The test
was established by looking at certain criterion as follows: “If respondents ≥ 50, the Kolmogrov Smirnov normality test is used.”
“If respondents ≤ 50, the Shapiro Wilk normality test is used.”
The respondents of this study were 41 students, therefore the Shapiro Wilk test was used to calculate the normality of the data. If the normality
is more t han the level of significance α 0.05, scores will be normally
distributed.
Table 4.11 Normality Pre-test Result between Experimental and Controlled Class
Based on the table above, it can be seen that both experimental and controlled class in the pre-test had normal distribution data because the
sig. score which showed in the Shapiro-Wilk normality test are more than 0.05; 0.064 for experimental class and 0.225 for controlled class.
Table 4.12 Normality Post-test Result Between Experimental and Controlled Class
Based on the table above, it can be seen that both experimental and controlled class in the post-test had normal distribution data because the
sig. score which showed in the Shapiro-Wilk normality test are more than 0.05; 0.014 for experimental class and 0.053 for controlled class.
2. Homogeneity Test
The homogeneity test was used to see whether the data in both classes were homogeneous or heterogeneous. Based on Levene Statistical
in SPSS, the significant score that can be categorized as homogeneous should be higher than 0.05. SPSS version 22 was used in this calculation.
Table 4.13 Homogeneity Pre-test Result between Experimental and Controlled Class
Table 4.14 Homogeneity Post-test Result between Experimental and Controlled Class
Based on the table 4.13 and 4.14 above, it can be seen that the sig. score in the pre-test between experiment and control class is 0.288 and the
sig. score in the post-test between experiment and control class is 0.938. This means that the data had homogeny distribution because the sig. score
are bigger than 0.05. Thus, it could be concluded that pre-test and post-test between experimental class and controlled class had homogenous
distribution data calculated by Levene.
C. Test of Hypothesis
The last calculation was testing the hypothesis. This was the crucial calculation to answer the problem formulation of this research whether there is
significant different between students’ reading achievement in experimental class which was given SQ3R Method
and students’ reading achievement in controlled class which was given conventional method. The writer used the
comparative technique by comparing the experimental and controlled class. After getting the result data of students reading comprehension
achievement in both experiment and control class, the writer analyzed it by using statistic calculation of the t-test formula as follows.
Tabel 4.15 The Comparison Score between Students in Experimental Class and
Students in Controlled Class
No. X
Y x
y 1
20 10
400 100
2.56 -5.13
6.5536 26.3169
2 10
5 100
25 12.56
-9.87 157.7536
97.4169
3 20
400 2.56
4.87 6.5536
23.7169
4 20
10 400
100 2.56
-5.13 6.5536
26.3169
5 35
5 1225
25 -12.44
-0.13 154.7536
0.0169
6 15
10 225
100 7.56
-5.13 57.1536
26.3169
7 20
10 400
100 2.56
-5.13 6.5536
26.3169
8 20
5 400
25 2.56
-0.13 6.5536
0.0169
9 25
10 625
100 -2.44
-5.13 5.9536
26.3169
10 15
5 225
25 7.56
-0.13 57.1536
0.0169
11 35
5 1225
25 -12.44
-0.13 154.7536
0.0169
No. X
Y x
y 12
15 5
225 25
7.56 -0.13
57.1536 0.0169
13 15
15 225
225 7.56
-10.13 57.1536
102.6169
14 20
5 400
25 2.56
-0.13 6.5536
0.0169
15 15
5 225
25 7.56
-0.13 57.1536
0.0169
16 35
10 1225
100 -12.44
-5.13 154.7536
26.3169
17 15
20 225
400 7.56
-15.13 57.1536
228.9169
18 15
225 7.56
4.87 57.1536
23.7169
19 20
5 400
25 2.56
-0.13 6.5536
0.0169
20 30
5 900
25 -7.44
-0.13 55.3536
0.0169
21
25 -15
625 -225
-2.44 19.87
5.9536 394.8169
22 25
10 625
100 -2.44
-5.13 5.9536
26.3169
23 20
5 400
25 2.56
-0.13 6.5536
0.0169
24 15
5 225
25 7.56
-0.13 57.1536
0.0169
25 25
5 625
25 -2.44
-0.13 5.9536
0.0169
26 30
-15 900
-225 -7.44
19.87 55.3536
394.8169
27
25 5
625 25
-2.44 9.87
5.9536 97.4169
28 30
5 900
25 -7.44
-0.13 55.3536
0.0169
29 25
-5 625
-25 -2.44
9.87 5.9536
97.4169
30 20
5 400
25 2.56
-0.13 6.5536
0.0169
31 25
5 625
25 -2.44
-0.13 5.9536
0.0169
32 20
400 2.56
4.87 6.5536
23.7169
33
15 5
225 25
7.56 -0.13
57.1536 0.0169
34 15
5 225
25 7.56
-0.13 57.1536
0.0169
35 20
5 400
25 2.56
-0.13 6.5536
0.0169
36 40
5 1600
25 -17.44
-0.13 304.1536
0.0169
37 30
15 900
225 -7.44
-10.13 55.3536
102.6169
38 40
5 1600
25 -17.44
-0.13 304.1536
0.0169
39
30 5
900 25
-7.44 -0.13
55.3536 0.0169
40 20
-5 400
-25 -17.44
9.87 304.1536
97.4169
41
15 5
225 25
7.56 -0.13
57.1536 0.0169
n=41 925
200 23125
1650 -20.04
-0.07 2553.697
1869.192
The researcher calculated the data based on the data based on the step of the test. The formulation as followed:
1. Determining mean of variable X Experimental Class, with formula: =
=
= 22.56
