CHAPTER IV RESULT OF THE STUDY
In this chapter, the researcher presented the data which had been collected from the research in the field of study which consists of description of the data, result of data analysis, and interpretations.
A. Description of the Data
1. The result of Pre-Test of the Experiment Class
The Pre-Test at the experiment class had been conducted in class C with the number of student was 16 students. The Pre-test scores of class were presented in table 4.1.
Table 4.1 Pre-Test scores of Experimental class
NO Students’ Initial Names Pre-Test
1 A 46
2 BL 54
3 DR 53
4 DSW 40
5 ES 54
6 FF 31
7 LAR 47
8 MR 67
9 N 54
10 N 46
11 RH 55
12 RAA 49
13 SP 28
14 SW 38
15 UJ 36
16 YSS 32
Highest Score 74
Lowest Score 29
Based on the calculation result scores of pre test the highest score was 67 and the lowest one was 28. To determine the range of score, the class interval, and interval of temporary, the writer calculated using formula as follows:
The Highest Score (H) = 67 The Lowest Score (L) = 28
The Range of Score I = H – L + 1 = 67 – 28 + 1 = 40
The Class Interval (K) = 1 + (3.3) x Log n = 1 + (3.3) x Log 16
= 1 + (3.3) x 1.2041199827 = 1 + 3.9735959429
= 4.9735959429 = 5
Interval of Temporary (I) =
5 40
K R
= 8
3 3
The Frequency Distribution of the Pre test Score of the Experiment Class
Table 4.2 The Frequency Distribution of the Pre Test Scores of the Experiment Class
Figure 4.1 The Frequency Distribution of the Pre test Score of the Experiment Class
2. The result of Post-Test score of the Experiment Class
Table 4.3Post-Test scores of Experimental class
NO Students’ Initial Names Post-Test
1 A 64
2 BL 73
3 DR 80
4 DSW 70
5 ES 69
6 FF 69
7 LAR 73
8 MR 78
9 N 69
10 N 71
11 RH 62
12 RAA 72
13 SP 63
14 SW 69
15 UJ 78
16 YSS 66
Highest Score 80
Lowest Score 62
Based on the calculation result scores of post test, the highest score was 80 and the lowest score was 62. To determine the range of score, the class interval, and interval of temporary, the writer calculated using formula as follows:
The Highest Score (H) = 80 The Lowest Score (L) = 62
The Range of Score I = H – L + 1 = 80 – 62 + 1 = 19
= 1 + (3.3) x Log 16
So, the range of score was 18, the class interval was 5, and interval of temporary was 4. Then it was presented using frequency distribution in the following table:
Table 4.4 The Frequency Distribution of the Post Test Scores of the Experiment Class
3. Scores of students who are taught using Peer Correction Techniques
NO Students’ Initial Names PC_1 PC_2
1 A 69 72
The Frequency Distribution of the Post test Score of the Experiment Class
Mean 62.25 67.0625
Based on the calculation result of scores of the students who are taught using peer correction, the highest score achieved by students in the first meeting was 78 and the lowest score was 35. The range was 44, from the student’s number (N) 16. From the calculation result of statistic, the mean score achieved by students was 62.25. In the last meeting of peer correction, the highest score achieved by students was 78 and the lowest one was 39. The range was 40, from the student’s number (N) 16. From the calculation result of statistic, the mean score achieved by students was 67.0625.
4. Scores of students who are taught using Teacher Written Feedback Techniques
NO Students’ Initial Names TWF_1 TWF_2
1 A 68 74
2 BL 67 70
3 DR 75 75
4 DSW 63 66
5 ES 61 66
6 FF 55 55
7 LAR 70 76
8 MR 77 78
9 N 60 63
10 N 60 65
11 RH 57 61
12 RAA 63 68
13 SP 73 76
14 SW 74 76
15 UJ 71 80
16 YSS 50 77
Highest Score 77 80
Total 994 1049
Mean 62.25 70.375
Based on the calculation result of scores of the students who are taught using teacher written feedback, the highest score achieved by students in the first meeting was 77 and the lowest score was 50. The range was 28, from the student’s number (N) 16. From the calculation result of statistic, the mean
score achieved by students was 62.25. In the last meeting of teacher written feedback, the highest score achieved by students was 80 and the lowest one was 55. The range was 26, from the student’s number (N) 16. From the calculation result of statistic, the mean score achieved by students was 70.375.
B. Result of Data Analysis
1. The difference between the result of pre-test and post-test
16 YSS 32 66 34 6,0625 36,75390625
∑X1=
730
∑X2=
1126
∑d=
396 ∑ X2d = 1513
t =
√
N (N-1) Where:
Md : mean of the difference pre-test and post-test (post test – pre test) xd : deviation of each subject (d-Md)
∑ X2
d : sum of squared deviations N : total the subject on sample d.b : N-1
Md =
=
= 24. 75
t =
√
=
√
=
N (N-1) 16 x 15 tvalue =5.6911591786
Based on the manual calculating, the tvalue was 5.6911591786. Then it was consulted with ttable with df= (N-1) and level of significance ttable 0.05 = 2.13. Because tvalue = 5.6911591786was higher than ttable= 2.13 It meant that both of pre-test and post-test were significant.
2. Normality and Homogeneity Test
distribution or not.
a. Normality test of Pre-Test
One-Sample Kolmogorov-Smirnov Test
Pre_Test
N 16
Normal Parametersa,,b Mean 45.63
Std. Deviation 10.695 Most Extreme
Differences
Absolute .139
Positive .128
Negative -.139
Kolmogorov-Smirnov Z .556
Asymp. Sig. (2-tailed) .917
Based on the calculation used SPSS program, the dvalue of pre-test was 0.139. From the table of critical value of Kolmogrov-Smirnov test with the student’s number (N) = 16 at the significance level α= 0.05,
the score of dtable was 0.327. Because dvalue was lower than dtable (0.139 < 0.327), it could be concluded that the data was in normal distribution.
b. Normality test of Post- Test
One-Sample Kolmogorov-Smirnov Test
Post_Test
N 16
Normal Parametersa,,b Mean 70.38
Std. Deviation 5.277
Most Extreme Differences
Absolute .147
Positive .122
Negative -.147
Kolmogorov-Smirnov Z .589
Based on the calculation used SPSS program, the dvalue of post-test was 0.147. From the table of critical value of Kolmogrov-Smirnov test with the student’s number (N) = 16 at the significance level α= 0.05,
the score of dtable was 0.327. Because dvalue was lower than dtable (0.147< 0.327), it could be concluded that the data was in normal distribution.
c. Homogeneity Test
Homogeneity test was conducted to know whether data are homogeneous or not.
Levene's Test of Equality of Error Variances Dependent Variable:Post_Test
F df1 df2 Sig.
.979 12 3 .583
Tests the null hypothesis that the error variance of the dependent variable is equal across groups.
Based on the result of homogeneity test, the fvalue was 0.979 and the significant value was 0.583. The data are homogeneous if the significant value is higher than significance level α= 0.05. Because thesignificant value (0.583) was higher than significance level α= 0.05, it could be concluded that the data were homogeneous.
3. Testing Hypotheses
Table 4.5 Descriptive Statistics Descriptive Statistics
Mean Std. Deviation N
PC_1 62.25 10.063 16
TWF_1 65.25 7.853 16
TWF_2 70.38 7.191 16
PC_2 67.06 10.063 16
In table 4.5 showed the mean of peer correction were 62.25 in the first period and 67.06 in the second period and mean of teacher written feedback were 65.25 in the first period and 70.38 in the second period.
Table 4.6Multivariate Testsb
In Multivariate Tests showed the Fvalueof Peer Correction was 20.058 with the significant level was 0.000, the Fvalueof Teacher Written Feedback was 11.396 with the significant level was 0.004 and the FvalueofPeer
Effect Value F
Hypothe
sis df Error df Sig. PC Pillai's Trace .572 20.058a 1.000 15.000 .000
Wilks' Lambda .428 20.058a 1.000 15.000 .000 Hotelling's Trace 1.337 20.058a 1.000 15.000 .000 Roy's Largest Root 1.337 20.058a 1.000 15.000 .000 TWF Pillai's Trace .432 11.396a 1.000 15.000 .004 Wilks' Lambda .568 11.396a 1.000 15.000 .004 Hotelling's Trace .760 11.396a 1.000 15.000 .004 Roy's Largest Root .760 11.396a 1.000 15.000 .004 PC *
TWF
Correction and Teacher Written Feedback was 22.781 with the significant level was 0.000.
Table 4.7Mauchly's Test of Sphericityb
Within
Tests the null hypothesis that the error covariance matrix of the orthonormalized transformed dependent variables is proportional to an identity matrix.
a. To adjust the degrees of freedom for the averaged tests of significance. Corrected tests were displayed in the Tests of Within-Subjects Effects table. b. Design: Intercept
Within Subjects Design: PC + TWF + PC * TWF
The result of Mauchly’s test showed that the values of df was 0 and the significant values was -, so the researcher need to correct thevalues for this, but not for the treatments themselves.The researcher needed to adjust the degrees of freedom for the averaged tests of significance. Corrected tests were displayed in the Tests of Within-Subjects Effects table.
Table 4.8Tests of Within-Subjects Effects
Source
1122.250 1 1122.250 20.058 .000
Greenhouse-Geisser
1122.250 1.000 1122.250 20.058 .000 Huynh-Feldt 1122.250 1.000 1122.250 20.058 .000 Lower-bound 1122.250 1.000 1122.250 20.058 .000 Error(PC) Sphericity
Assumed
Greenhouse-Geisser
839.250 15.000 55.950 Huynh-Feldt 839.250 15.000 55.950 Lower-bound 839.250 15.000 55.950
TWF Sphericity
Assumed
390.063 1 390.063 11.396 .004
Greenhouse-Geisser
390.063 1.000 390.063 11.396 .004 Huynh-Feldt 390.063 1.000 390.063 11.396 .004 Lower-bound 390.063 1.000 390.063 11.396 .004 Error(TWF) Sphericity
Assumed
513.438 15 34.229
Greenhouse-Geisser
513.438 15.000 34.229 Huynh-Feldt 513.438 15.000 34.229 Lower-bound 513.438 15.000 34.229 PC * TWF Sphericity
Assumed
1105.563 1 1105.563 22.781 .000
Greenhouse-Geisser
1105.563 1.000 1105.563 22.781 .000 Huynh-Feldt 1105.563 1.000 1105.563 22.781 .000 Lower-bound 1105.563 1.000 1105.563 22.781 .000 Error(PC*TWF) Sphericity
Assumed
727.938 15 48.529
Greenhouse-Geisser
727.938 15.000 48.529 Huynh-Feldt 727.938 15.000 48.529 Lower-bound 727.938 15.000 48.529
researcher used the Greenhouse-Geisser corrected values as these were the most conservative.
Table 4.9 Tests of Within-Subjects Contrasts
Source
PC TWF
Type III Sum of Squares df
Mean
Square F Sig.
PC Linear 1122.250 1 1122.250 20.058 .000
Error(PC) Linear 839.250 15 55.950
TWF Linear 390.063 1 390.063 11.396 .004
Error(TWF) Linear 513.438 15 34.229
PC * TWF Linear Linear 1105.563 1 1105.563 22.781 .000 Error(PC*TWF) Linear Linear 727.938 15 48.529
Table 4.10Tests of Between-Subjects Effects
Source Type III Sum of Squares df Mean Square F Sig.
Intercept 445556.250 1 445556.250 1523.356 .000
Error 4387.250 15 292.483
C. Interpretation
From the calculation result of two ways repeated-measures ANOVA, it could be interpret that: