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CHAPTER IV RESEARCH FINDING AND DISCUSSION

A. Research Findings

1. Data Description

This section shows the data description of the students’ scores in both

of the experimental class and control class. Furthermore, the data were collected from students’ score of pre-test and post-test from both of the class.

The followings are description:

42

a. Pre-test score

Table 4.1

The Student’s Pretest Score of Control and Experiment Class4344 Name Score of Control Class Score of Experiment Class

S-1 36 48

S-2 32 40

S-3 32 48

S-4 40 40

S-5 36 36

S-6 44 40

S-7 36 36

S-8 36 44

S-9 36 48

S-10 36 36

S-11 44 48

S-12 32 36

S-13 44 48

S-14 32 36

S-15 40 36

S-16 36 44

S-17 40 48

S-18 44 40

S-19 40 40

S-20 40 44

S-21 36 40

S-22 32 48

S-23 40 44

S-24 40 44

Total 904 1012

Mean 37.6 42.1

Max 44 48

Min 32 36

The explanation of the data can be seen in the appendix. Related to the table above the writer describe as follow:

43 , Test, Montong-Gamang, 04 November 2019

Table 4.2

The Data Description of Pretest

No Class The Total of

Students

Mean

1 Control Class 24 37.6

2 Experiment Class 24 42.17

Based on data above, explains that related to the pretest of control class the higher score was 44 and the lower score was 32with mean was 37,6and the standard deviation was 4.07182. Furthermore, the score of posttest experiment class, the higher score was 48 and the lower score was 36 with the mean score was 50and the standard deviation was 4.71.

The mean value is obtained by using the formula =AVERAGE (data range) in the Microsoft Excel box then press ENTER.

Related to the table 4.2 inform the scores of pre-test can beclarified that the value between control and experimental classeswith the total number of the students was 48. It can be seen from the mean score of experiment class has higher score than the control class.

b. Treatment

In this section, the researcher was conducted 2 times meeting, in the first meeting the researcher provided an overview and material about narrative text in accordance with syllabus learning materials. In this case, the researcher

gives the assignment after the material was explained and the students also assign an assignment to determine the three elements in the narrative text and they explain in front of the class with their groups. The students of experimental class use narrative storytelling to find and learn speaking and for control class are taught by using conventional method like discussion and teacher center. It can be seen in the first treatment for the experimental class the students began to be enthusiastic because they used learning aids and many of them feel more helped.

c. Posttest Score

Table 4.3

The Students’ Posttest Score f Control and Experiment Class Name Score Control Class Score Experiment Class

S-1 36 76

S-2 44 64

S-3 44 76

S-4 44 64

S-5 36 68

S-6 36 64

S-7 32 68

S-8 32 68

S-9 36 72

S-10 48 68

S-11 36 72

S-12 44 68

S-13 36 72

S-14 44 64

S-15 40 68

S-16 32 64

S-17 40 72

S-18 36 76

S-19 40 72

S-20 40 76

S-21 36 64

S-22 44 72

S-23 40 64

S-24 40 64

Total 936 1656

Mean 39 69

Max 48 76

Min 32 64

The explanation of the data can be seen in the appendix.

Table 4.4

The Data Description of Posttest No Class The Total Of

Students

Mean

1 Control Class 24 39

2 Experiment Class 24 69

Related to data posttest above, the higher score of control class was 48 and the lower score was 32 with the mean score was39, while the standard deviation was 4.45265. In addition, the higher score of experimental class was 76 and the lower score was 64 with the mean was 69 and the standard deviation was 4.4526494.

The mean value is obtained by using the formula = AVERAGE (data range) in the Microsoft Excel box then press ENTER.

According to the table, inform the scores of posttest has a significant score between control and experimental. The control class was increased in

pretest session than in post-test. For this session, the experimental class got higher score than control group.

d. The Calculated Score of Controlled and Experimental Class

Table 4.5

Calculated Table of Control Class45

Students Pre-test Post-test Difference

Y1 Y2 Y

S-1 36 36 0

S-2 32 44 12

S-3 32 44 12

S-4 40 44 4

S-5 36 36 0

S-6 44 36 -8

S-7 36 32 -4

S-8 36 32 -4

S-9 36 36 0

S-10 36 48 12

S-11 44 36 -8

S-12 32 44 12

S-13 44 36 -8

S-14 32 44 12

S-15 40 40 0

S-16 36 32 -4

S-17 40 40 0

S-18 44 36 -8

S-19 40 40 0

S-20 40 40 0

S-21 36 36 0

S-22 32 44 12

S-23 40 40 0

S-24 40 40 0

=24 =904 =936

45 , Test, Montong-Gamang, 04-18 November 2019

Table 4.6

Calculated Table of Experiment Class46 Students Pre-test Post-test Difference

X1 X2 Y

S-1 48 76 28

S-2 40 64 24

S-3 48 76 28

S-4 40 64 24

S-5 36 68 32

S-6 40 64 24

S-7 36 68 32

S-8 44 68 24

S-9 48 72 24

S-10 36 68 32

S-11 48 72 24

S-12 36 68 32

S-13 48 72 24

S-14 36 64 28

S-15 36 68 32

S-16 44 64 20

S-17 48 72 24

S-18 40 76 36

S-19 40 72 32

S-20 44 76 32

S-21 40 64 24

S-22 48 72 24

S-23 44 64 20

S-24 44 64 20

=24 =1012 =1656 =644

Well according to the data in table 12.1 that can be found likes: ƩN = 24, ƩY1 = 904, ƩY2 = 936, ƩY = 32 and ƩX1 =1012 ,ƩX2 =1656 and ƩX =644.

In addition, the researcher finds out the mean deviation and thesquare deviation of experiment and control classes using formula:

1) The mean deviation and the square deviation of experiment class

46 , Test, Montong-Gamang, 04-18 November 2019

=

= 26.8 Then :

= =17776-414736 =-396960

2) The mean deviation and the square deviation of the control class

=

=-1.33

Then :

=

= 1184-1024 =860

Where:

M: The mean score of two groups

X: The student final score for experimental group Y: The student final score for control group N: Is the number of sample

Is the sum

2. Test of Normality Data

Normality is used to determine whether sample data has been drawn from a normal distribution population. Test of normality of the control and experimental classes were conducted to examine whether all variables were normally distributed or not. The test of normality uses the Kolmogorov- Smirnov formula with calculated using SPSS 16. Furthermore, to find the

normality or not if the sig> 0.05. The results are described as follows:

Table 4.7 Test of Normality

No Class Sig Result

1 Posttest Control Class 0,009 Normal

2 Posttest Experiment Class 0,012 Normal

Based on the explanation above, it can be seen that the data of posttest related to the control and experimental has value was sig > 0,05. It can be concluded that the data of posttest is normalAnd the explanation can be seen in the appendix.

3. Test of Homogeneity

After knowing the level of normality data, then a test ofhomogeneity is used to know the level of similarity of variancesbetween the two classes,

control and experimental classes. Theresearcher compared the sig on Levena Statistic with sig > 0.05.

Table 4.8 Test of Homogeneity

Class Sig Statement

Control 0,655 Homogeneity

Experiment 0.581 Homogeneity

Based on the data of test of homogeneity of variance above, the significant value was 0,655 and 0,581. It mean higher than 0,05. So, the data in this research has homogeneity ofvarianceAnd the explanation can be seen in the appendix.

4. Test of Hypothesis

The last is the researcher find out the t-test with a summaryas follow:

Table 4.9 Test of Hypothesis

Paired Differences

T

Df Sig. (2-tailed) Mean

Std.

Deviation

Std. Error Mean

95%

Confide nce Interval of the Differen ce

Lower Upp er

Pair 1 pre test speaking skill control group - post test speaking skill control group

-1.33333 7.04437 1.43793 - 4.3079 1

1.641

24 -.927 23 .363

Pair 2 pre test speaking skill experime ntal group - post test speaking skill experime ntal group

-

2.68333E 1

4.64071 .94728 - 28.792 94

- 24.87 373

- 28.32 7

23 .000

Based on the table 4.9 above, the t-calculated was 28,327 with thedegree of freedom was 46(df =N1+N2 -2 24+24-2=46). In thedegree freedom of 46 was found in the t-tablewas 2,012 in significance value (sig 2-tailed) was 0.00. Consequently, the value of t- calculated > t-table (28,327> 2,012) and its significance value is less than 0.05 (P =0.00<

0.05).It can be concludedthat if the t-calculated>t-table (28,327> 2,012) and the significance value was less than 0.05 (0.00<0.05) it means alternative hypothesis (Ha) was accepted and the null hypothesis (Ho) was rejected.

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