of post-test in the control class was in the percentage of 16.0% is categorized low, the the percentage of 76.0 % is categorized medium, while in the percentage of 8.0 % is categorized good.
1. Normality Test of Experimental Class ( X MIPA 1 ) Table 4.13
Normality Test of Experimental Class One-Sample Kolmogorov-Smirnov Test
Unstandardiz ed Residual
N 26
Normal Parametersa,b Mean ,0000000 Std.
Deviation
5,59464029
Most Extreme Differences
Absolute ,095 Positive ,095 Negative -,069
Test Statistic ,095
Asymp. Sig. (2-tailed) ,200c,d
a. Test distribution is Normal.
b. Calculated from data.
c. Lilliefors Significance Correction.
d. This is a lower bound of the true significance.
Based on the calculation table above, showed that sig. 2 tailed of the class was 0.200.
It can be concluding that the data from experimental class (X MIPA 1) was normally
distributed because the asymptotic significance score (2.tailed) was higher than 0.05 ( 0.200
> 0.05 )
2. Normality Test of Control Class ( X MIPA 2) Table 4.14
Normality Test of Control Class One-Sample Kolmogorov-Smirnov Test
Unstandardiz ed Residual
N 25
Normal Parametersa,b Mean ,0000000 Std.
Deviation
6,20735451
Most Extreme Differences
Absolute ,074 Positive ,071 Negative -,074
Test Statistic ,074
Asymp. Sig. (2-tailed) ,200c,d
a. Test distribution is Normal.
b. Calculated from data.
c. Lilliefors Significance Correction.
d. This is a lower bound of the true significance.
Based on the calculation table above, showed that sig. 2 tailed of the class was 0.200.
It can be concluding that the data from experimental class (X MIPA 1) was normally distributed because the asymptotic significance score (2.tailed) was higher than 0.05 ( 0.200
> 0.05 ).
b. Homogeneity Test
Homogeneity test was conducted to determine whether the variant of data distribution on students’ test scores from the two groups for each experimental class and control class were homogenous or not. To calculate the data, the researcher uses the SPSS program version 26. The result of the calculation is as below :
Table 4.15 The Homogeneity Test Test of Homogeneity of Variance
Levene
Statistic df1 df2 Sig.
Student's score Based on Mean ,079 1 48 ,779
Based on Median ,025 1 48 ,875
Based on Median and with adjusted df
,025 1 46,546 ,875
Based on trimmed mean
,060 1 48 ,808
Based on the table of homogeneity test above, the result of data was 0.808. It means
that the calculation of students’ test scores is greater than 0.05 (0.808 > 0.05), it can be concluding that the data in this study has a homogeneous variance.
2. Hypothesis Testing
After the researcher carrying out the normality and homogeneity test, the researcher tested the hypothesis. The researcher used T-test to analyze the data. They are the experimental class and control class. The experimental class was taught using scanning technique while the control class was not given treatment .The researcher has used the T-test to analyzed the data by using SPSS program version 26. The result of the data calculation as table below :
Table 4.16
Mean Score of Experimental and Control Class Group Statistics
Kelompok N Mean
Std.
Deviation
Std. Error Mean Hasil Belajar Experimental Class 26 82.89 6.352 1.246
Control Class 25 58.40 7.320 1.464
Based on the table above, the result of data analysis showed that the means score for students of experimental class is 82.89 while the means score of control class is 58,40. Based on the mean results, there were differences values between the experimental and control class. The score of experimental class was higher than control class. However, the acquisition of students’
test scores in the experimental class needs to be analyzed by using Independent Sample Test to see whether the scores from the experimental class are significant or not.
Table 4.17
Result of T-test Calculation
Independent Samples Test Levene's Test for Equality
of Variances t-test for Equality of Means
F Sig. T df
Sig. (2-tailed)
Mean Differenc e
Std. Error Difference
95% Confidence Interval of the Difference Lower Upper Hasil
Belaja r
Equal variances assumed
.168 .684 12.773 49 .000 24.485 1.917 20.633 28.337
Equal variances not assumed
12.737
47.45 3
.000 24.485 1.922 20.619 28.351
Based on the result of independent sample test above, it is obtained that the value of significant value (2-tailed) p = 0.000<0.05. The significant level (ɑ ) = 0.05, so H0 ia rejected. It means that there was a significant different between the value before the treatment and mean value after the treatment. The value of T-test was 12.737 with the degree of freedom (df) was 49, and the value of ttable on significance 5% was 0.2010. To interpret the data above, the researcher formulated the test of hypothesis below:
- Ha : There is a significant difference between students who taught using scanning technique and students who are not taught without it on their reading comprehension
- H0 : There is no significant difference between students who taught using scanning technique and students who are not taught without it on their reading comprehension.
The criteria of testing hypothesis:
a. Ha: if t test> t table in significant degree 5%
b. Ho : if t test< t table in significant degree 5%.
The result of the researcher showed that the value of t test was higher (12.773> 0,2010). It indicated that Ha was accepted and Ho was rejected. So, There was a significant difference between students who taught using scanning technique and students who are not taught without it on their reading comprehension.
D. Discussion and Interpretation
In this interpretation, researcher compared the result of the data t test with t table. If t test is higher than t table, it means Ho is rejected and Ha is accepted.
There are two hypotheses of this research:
- Ha : There is a significant difference between students who taught using scanning technique and students who are not taught without it on their reading comprehension - H0 : There is no significant difference between students who taught using scanning
technique and students who are not taught without it on their reading comprehension.
Teaching is guiding and facilitating learning, enabling the learner to learn, setting conditions for learning.39 Teaching of reading has an importanrole for academic success. In other words teaching reading is a process for teacher to direct and help students how to build the creating meaning from the students’ reading comprehension. It means, the teacher is someone who transferring their knowledge
39 H. Douglas Brown, Principle Language Learning and Teaching 5th Edition, (Britain: Person Longman,
and information for students in learning process. Teacher also must give facilitating for learning in teaching reading process, because according to Neil J. Anderson reading is an essentiaI skill for Iearners of EngIish as second Ianguage, for most of these Iearners it is the most important skill to master in order to ensure success not only in EngIish is required. With strengthened reading skill, learners will make greater progress and development in all other areas of learning.33
Teaching reading is not easy. many students still lazy when the teacher asked them to read the text. Students are also not responding if the teacher asked about the main idea of the text and then students find difficulties in comprehending English text. They often stop reading because they find that the English text is difficult to understand. It is because they do not understand the meaning of the English text. It makes students feel lazy when reading the process. So, the teacher must make creative learning in teaching reading to make students more enjoyable when reading the process, and the students can more easy to understand the text. There are many technique in teaching reading. One of the teghnique is scanning technique.
Based on the result of the findings, the Scanning technique effectively teach reading comprehension. It can be seen from the students’ mean score in the experimental class X MIPA 1 increases from 49.23 to 82.89, whereas the mean score in the controlled class X MIPA 2 is 45 for the pre-test and 58.40 for the post-test.
Furthermore, it is supported by the result of the test, the test is 12.773, and the t-table is 0.2010; thus the value is higher than t-t-table (12.773 > 0.2010). Related to the result of the t-test, the Ha is accepted, and Ho is rejected. It means there is significant difference between students’ reading comprehension achievement who are taught using the Scanning technique and authentic material and students who are not taught using the strategies.
Through the implementation of the Scanning technique and Authentic material, the students’ reading comprehension achievement is increased. This achievement can be seen from the different post-test mean figures between both classes. In the pre-test, the difference in the mean figure is not significant. However, after the X MIPA 1 class as the experimental class being treated, the mean figure was significantly different.
The post-test mean figure of the X MIPA 2 is 82.89 while the X MIPA 2 is 58.40.
Therefore, in the other word the Scanning Technique and Authentic material is an effective strategies for teaching reading comprehension in SMAN 1 Sambit Ponorogo.
CHAPTER V CLOSING
This chapter presents the conclusion and recommendation. The conclusion is drawn
based on the finding. Then, the recommendation is given by the researcher to everyone who reads this research.