CHAPTER IV RESEARCH FINDINGS AND ANALYSIS

Transcription

1 CHAPTER IV RESEARCH FINDINGS AND ANALYSIS A. Description of the Result of Research The research was conducted in MA Sunan Kalijaga Bawang. This research is found that there were different result between experimental class which was taught using think pair share technique and control class which was taught without using think pair share technique in teaching speaking. Test to experimental class and control class were given to obtain the data. It was given before and after teaching and learning process. The subjects of this research were experimental class X.1 and control class X. of MA Sunan Kalijaga Bawang. There were test for pre-test that was given to experimental class and control class before teaching and learning process, whereas post-test was given after teaching and learning process. The researcher prepared the materials, technique and lesson plan before conducted teaching and learning process. The researcher analyzed the data after it was collected. The first analysis of the data analysis was conducted analysis of pre-test score both experimental class and control class. It was applied to know the normality, homogeneity and similarity of pre-test of experimental class and control class. It is used to know what two groups were 46

2 normal and had same variant. The second analysis of data analysis was taken from post-test score. It was applied to know the normality, homogeneity and similarity of post-test of experimental class and control class. It used to know what two groups were normal and had same variant. Both test used to prove the truth of hypothesis that has been planned. B. The Data Analysis and Test of Hypothesis 1. The Data Analysis of Pre-test Score of The Experimental Class and Control Class a. The Normality Experimental and Control Class of Pre-test Table 4.1 The List of Experimental and Control Class Pretest Score EXPERIMENT CONTROL NO CODE SCORE NO CODE SCORE 1 E C-1 60 E- 60 C E C E C E C E C E C E C

3 9 E C E C E C E C E C E C E C E C E C E C E C E C E C-1 55 E- 50 C E C E C E C E C E C E C E C E C Sum n

4 average 63, ,66667 Variant (s ) 59, ,50575 Standard of Deviation (S) 7, , ) The normality of the experimental class of the pre-test The normality test was used to know whether the data that obtained was normally distributed or not. Based on the table above, the normality test was: Hypothesis: Ha: The distribution list was normal Ho: The Distribution list was not normal Test of Hypothesis: The formula was used: k ( O E ) i i E i 1 The computation of normality test: Maximum score = 80 Minimum score = 50 K/ Number of class = 5 S = 74, 3 n = 30 Range = 30 i 49

5 Length of class = 5 = 6, 90 Table 4. The Frequency Distribution of Experimental Class Pretest Class f i X i X i f i.x i f i.x i ,5 756, ,5 34,3 409, ,5 4160, ,5 4970,3 35, ,5 5859,3 76,5 585, ,5 6806, Sum Table 4.3 The Frequency Observation of Experimental Class Pretest Class Bk Z i P(Z i ) Wide Area Ei Oi 49,5-1,55-0, ,60 0,1353 4,1 7, ,5-0,86-0, ,66 0,40 7, 7 0, ,5-0,16-0,

6 6 67 0,73 0,677 8,0 8 0, ,5 0,53 0, ,79 0,1874 5,6 5 0, ,5 1,3 0, ,86 0,083,5 1 0,875 85,5 1,93 0, ,93 0,07 0,7, ,5,6 0,4956 0,9 90 #REF! ² = 5,64 ² count = 5, 64 for a= 5%, dk = 6-3 = 3 was gotten ² table = 7, 81 Figure 4.1 Normality test of Experimental Class of Pre-test Ho accepted area 5,64 7,81 With α = 5% and dk = 6-3 = 3, from the chisquare distribution table, it was obtained ² table = 7, 81. Because ² count = 5, 64 was lower than ² table = 7, 81 (5, 64< 7, 81). So the distribution list was normal. 51

7 ) The Normality of the Control Class of Pre-test Hypothesis: Ho: The distribution list was normal. Ha: The distribution list was not normal. Test of hypothesis: The formula was used: k ( O E ) i i E i 1 i The computation of normality test: Maximum score = 80 Minimum score = 50 K/ Number of class = 6 S = 8, 87 n = 30 Range = 35 Length of class = 6 X = 63, 10 5

8 Table 4.4 Table Frequency Distribution of the Control Class Pre-test Class f i X i X i f i.x i f i.x i ,5 756, , ,5 34, , ,5 4160,5 3,5 0801, ,5 4970,5 35,5 4851, ,5 585, , ,5 6806, ,5 Sum Table 4.5 Table Frequency Observation of the Control Class Pre-test Class Bk Z i P(Z i ) Wide Area Ei Oi 49,5-1,53-0, ,60 0,133 4,0 6 1, ,5-0,86-0, ,66 0,37 7,0 10 1, ,5-0,18-0, ,73 0,617 7,9 5 1, ,5 0,50 0, ,79 0,1894 5,7 5 0,081 73,5 1,17 0,

9 ,86 0,0883,6 0, ,5 1,85 0, ,93 0,064 0,8 1, ,5,53 0,494 0,9 90 #REF! ² = 5,4 ² count = 5, 4 for a= 5%, dk = 6-3 = 3 was gotten ² table = 7,81 Figure 4. Normality test of Control Class of Pre-test Ho accepted area 5,4 7,81 With α = 5% and dk = 6-3 = 3, from the chisquare distribution table, it was obtained ² table = 7, 81. Because ² count = 5, 4 was lower than ² table = 7, 81 (5, 4 < 7, 81). So the distribution list was normal. b. Homogeneity Experimental and Control Class of Pre-test Hypothesis H o : σ 1 = σ 54

10 H a : σ 1 σ The Calculation Formula: The biggest Variants F The smallest Variants Ho is accepted if F F (1-a) (nb-1): (nk-1) F (1-a) (nb-1):(nk-1) Table 4.6 Homogeneity test of Experimental and Control Class Pre-test Variation Source Experiment Control Sum N X 63,333 63,667 Variant (s ) 59, ,506 Standard of deviation (s) 7,694 7,649 55

11 F = 59,195 58,506 = 1,01 For a = 5% with: df1 = nb 1 = 30-1 = 9 Df = nk -1 = 30-1 = 9 F (0.05)(9:9) = 1,861 ` Figure 4.3 Homogeneity Experimental and Control Class of Pre-test 1, 01 1,861 Because F count < F table, the experimental and control group had the same variance with α = 5% and dk = (30-1=9) : (30-1=9), it obtained F table = 1, 861. Because F count was lower than F table (1, 01 < 1, 861). So, Ho was accepted and two groups had same variant/ homogenous. 56

12 c. The Hypothesis Test of Pre-test In this research, because (has same variant), the t-test formula was as follows: t s x 1 1 n x 1 n 1 s n 1 s n Table 4.7 n 1 n s The Average Similarity Test of Experimental and Control Class Pre-test Variation Source Experiment Control Sum n X 63,333 63,667 Variants (s ) 59,195 58, 506 Standard of deviation (s) 7,694 7,649 ( ) ( ) = -0,168 57

13 With a = 5% with dk = = 58, it obtained t (0, 05)(58) =, 00 Figure 4.4 The Average Similarity Test of Pre-test -, 00-0,168, 00 Because t count was lower than t table (-0, 168<,00). So, Ho was accepted and there was no difference of pre-test average score from both of experimental and control groups.. The Data Analysis of Post-test Score of the Experimental Class and Control Class a. The Normality experimental and control class of post-test Table 4.8 The list of experimental and control class post-test score EXPERIMENTAL CONTROL NO CODE SCORE N0 CODE SCORE 1 E C-1 75 E- 90 C

14 3 E C E C E C E C E C E C E C E C E C E C E C E C E C E C E C E C E C E C E C-1 55 E- 80 C E C E C E C

15 6 E C E C E C E C E C Sum n average 7,5 67,667 Variants (s ) 73,707 8,99 Standard of deviation (S) 8,5857 9, ) The normality of the of Experimental Class Post-test The normality test was used to know whether the data that obtained was normally distributed or not. Based on the table above, the normality test was: Hypothesis: Ha: The distribution list was normal Ho: The distribution list was not normal 60

16 Test of Hypothesis: The formula was used: k ( O E ) i i E i 1 The computation of normality test: Maximum score = 90 Minimum score = 55 K/ Number of class = 6 s = 11, 14 n = 30 Range = 35 Length of class = 6 = 75, 50 Table 4.9 The Frequency Distribution of the Experimental Class Post Test Class f i X i X i f i.x i f i.x i Sum i 61

17 Table 4.10 Table Frequency Observation of the Experimental Class Post-Test Class Bk Z i P(Z i ) Wide Area Ei Oi 54,5-1,88-0, ,60 0,0748, 4 1,377 61,5-1,6-0, ,66 0,1604 4,8 4 0, ,5-0,63-0, ,73 0,350 7,1 8 0,176 75,5 0,00 0, ,79 0,350 7,1 5 0,5967 8,5 0,63 0, ,86 0,1604 4,8 5 0,007 89,5 1,6 0, ,93 0,0748, 4 1,377 96,5 96,5 0,470 0,9 90 #REF! ² = 3,6 ² count = 3, 6 for a= 5%, dk = 6-3 = 3 was gotten ² table = 7,81 6

18 Figure 4.5 The normality Test of experimental class of post-test Ho accepted area 3,6 7,81 With α = 5% and dk = 6-3 = 3, from the chi-square distribution table, it was obtained ² table = 7, 81. Because ² count = 3, 6 was lower than ² count = 7, 81 (3, 6< 7, 81). So the distribution list was normal. ) The Normality of the Control Class of Post-test Hypothesis: Ho: The distribution list was normal. Ha: The distribution list was not normal. Test of hypothesis: The formula was used: k ( O E ) i i E i 1 i 63

19 The computation of normality test: Maximum score = 85 Minimum score = 50 K/ Number of class = 6 S = 8, 99 n = 30 Range = 35 Length of class = 6 X = 67, 47 Table 4.11 The Frequency Distribution of Control Class Post-test Class f i X i X i f i.x i f i.x i Sum

20 Table 4.1 The Frequency Observation of Control Class Post-test Class Bk Z i P(Z i ) Wide Area Ei Oi 49,5 -,00-0, ,77 0,0885,7 4 0,685 56,5-1, -0, ,86 0,18 6,5 5 0, ,5-0,44-0, ,96 0,305 9,1 11 0, ,5 0,34 0, ,05 0,357 7,1 6 0, ,5 1, 1 0, ,14 0,103 3,1 3 0,009 84,5 1,89 0, ,3 0,053 0,8 1 0, ,5,67 0,496 0,1 9 #REF! ² = 1,70 ² count = 1, 70 for a= 5%, dk = 6-3 = 3 was gotten ² table = 7,81 65

21 Figure 4.6 The Normality of the Control Class of Post-test Ho accepted area 1,70 7,81 With α = 5% and dk = 6-3 = 3, from the chi-square distribution table, it was obtained ² count = 7, 81. Because ² count = 1, 70 was lower than ² table = 7, 81 (1, 70< 7, 81). So the distribution list was normal. b. Homogeneity Experimental and control class of Post-test Hypothesis H O : σ 1 = σ H a : σ 1 σ The Calculation Formula: The biggest Variants F The smallest Variants 66

22 Ho is accepted if F F (1-a) (nb-1): (nk-1) Ho accepted area F (1-a) (nb-1):(nk-1) Table 4.13 Homogeneity test of Experimental and Control Class of Post-test Variation Source Experiment Control Sum N X 7,500 67,667 Variant (s ) 73,707 8, 99 Standard of deviation (s) 8,585 9,07 F = 8, , 7069 = 1,

23 For a = 5% with: df1 = nb 1 = 30-1 = 9 df = nk -1 = 30-1 = 9 F (0.05)(9:9) = 1,861 Figure 4.7 Homogeneity Experimental and Control Class of Post-test Ho accepted area 1,117 1,861 Because F count < F table, the experimental and control group had the same variance with α = 5% and dk = (30-1=9) : (30-1=9), it obtained F table = 1, 861. Because F count was lower than F table (1, 117< 1, 861). So, Ho was accepted and two groups had same variant/ homogenous. c. The Hypothesis Test of Post-test In this research, because (has same variant), the t-test formula was as follows: t s x 1 1 n x 1 n 1 68

24 s n 1 s n 1 1 n 1 1 n s Ha is accepted if t count > t (1-a)(n1+n-) Ho accepted area Table 4.14 The Average Similarity Test of Experimental and Control Class of Post-test Variation Source Experiment Control Sum n X 7,500 67,667 Variant (s ) 73,707 8, 99 Standard of deviation (s) 8,5850 9, 07 ( ) ( ) 69

25 =,10 With a = 5% with dk = = 58, it obtained t (0, 05)(58) = 1,67 Figure 4.8 The Average Similarity Test of Post-test Ho accepted area 1,67, 10 Since t count >t table meant that there was a significant difference between experimental and control class on the post test. The experimental class was higher than the control class. Based on the computation above, by α = 5% of significance and dk = = 58. It was obtained t table = 1, 67 while t count =, 10. So, it can be concluded that Ho was rejected, Ha was accepted because t count was higher than the critical score on the t table (, 10>1, 67). Based on the result, the hypothesis in this research could be concluded that there was a significance difference in the result score of 70

26 descriptive text between experimental class and control class which was taught speaking descriptive text by using think pair share technique in experimental class and control class taught without using think pair share technique. C. Discussion of the Research Finding 1. The score of initial ability (pre-test) Based on the calculations of normality and homogeneity test from class X.1 as the experimental class and X. as the control class was normal distribution and homogeneous.. The score of final ability (Post-test) The result of this research is obtained the average score of experimental class was 7,500 which were higher than the result of control class was 67,667. The average score of experimental class was 7, 500 and (S) Standard of deviation was 8, 585. Teaching speaking of descriptive text in experimental class by using think pair share technique can encourage the students to be more active and interactive. They were also easy to understand the material when teaching and learning process conducted by using think pair share technique. It can be seen on the result of average score of experimental class which better than control class. 71

27 The average score of control class was 67, 667 and (S) Standard of deviation was 9, 07. The teaching speaking of descriptive text in control class by using lecturing method made students felt bored with the text that was presented because the teaching-learning process was monotonous. So, the students couldn t understand the teaching-learning process optimally. Based on the result of calculation t-test was obtained t count =, 10 and t table = 1, 67. This showed that t count >t table (t count higher than t table ). Thus, it meant that there was a significant difference between descriptive result score of students who taught speaking descriptive text by using think pair share technique and without think pair share technique in the teaching speaking of descriptive text. D. Limitations of the Research The researcher realized that this research had not been conducted optimally. There were constraints and obstacles faced during the research process. Some limitations of this research were: 1. The researcher was still lack of experience and knowledge of teaching-learning. It made the implementation process of this research was less smooth. But the researcher tried as maximal as possible to do this research. 7

28 . The researcher was limited at MA Sunan Kalijaga Bawang in the academic year 015/016. When the same research is conducted in others school, it was still possible that will be gained different score. Considering all those limitations, there was a need to do more researches about teaching speaking of descriptive text using think pair share technique or different technique. Hopefully, there will be better and has an optimal result. 73