1. CHAPTER IV RESEARCH FINDINGS AND DISCUSSION This chapter presents the data that was collected during the experimental research. First analysis focuses on the homogeneity of the sample; the second analysis focuses on the validity, reliability, index difficulty, and discriminating power of instruments. And the third analysis represents the result of pre-test and post-test that was done both in experimental and control group. A. First Analysis The first analysis was homogeneity test of the sample. That was previous summative score of students XI IPS 4 as experimental group and students of XI IPS I as control group. The analysis was meant to get homogeneous class of XI IPS 4 and IPS 1. In this study, the homogeneity of the test was measured by comparing the obtained score (Fscore) with Ftable. Thus, if the obtained score (Fscore) was lower than Ftable or equal, it could be said that Ho was accepted. It meant those the classes were homogeneous. The analysis of homogeneity test could be seen in table I. Table .I. Test of Homogeneity Experimental G Control G 2244 2236 37 38 60.65 58.84 29.790 43.164 5.46 6.57 Variants (s2) Standart deviation (s) n X Variant Sources Sum By knowing the mean and the variance, the researcher was able to test the similarity of the two variants with the homogeneity test from students previous score between IPS 4 and IPS 1. The computation of the rest of homogeneity as follows: F = Biggest Variance Smallest Variance 43.164 29.790 F = F = 1.448938879 On a 5 % with df1 numerator =37, df2 =38, it was found Ftable = 1.721142152. Because Fscore < Ftable / 1.448938879 rtable Calculation: Below is the example of the item validity of number 1 3. 15 22 10 30 18 6016 121 100 100 29 T-02 T-03 T-06 T-07 T-08 72 42 24 27 24 72 81 30 24 56 39 54 42 36 45 18 196 144 81 144 324 225 81 4 100 144 196 169 324 196 144 225 32416 9 4 9 4 16 4 1 9 4 16 9 9 9 9 9 14 12 9 12 18 9 11 10 10 9 T-15 T-32 T-12 T-13 10 12 14 13 18 14T-28 T-35 T-36 T-37 T-33 T-04 31 25 26 27 28 T-11 T-16 T-24 T-26 T-27 2 2 1 3 2 21 22 23 24 30 4 3 2 3 2 4 4 19 20 3 2 4 3 3 3 3 3 13 14 15 16 17 18 8 30 16 51 27 39 56 12 15 18 9 9 16 9 4 100 196 81 9 4 No Kode X 9 Y X2 Y2 XY 51 2 T-20 3 15 9 1 T-09 3 16 9 256 3 17 9 289 9 289 225 45 48 51 9 225 3 T-25 4 16 16 256 3 174 T-29 6 T-10 3 15 289 45 5 T-05 3 17 9 289 51 64 80 64 51 7 T-14 8 T-21 3 17 9 289 9 T-30 169 4 20 16 400 17 13 326 7044 1482 T-17 T-22 4 3 3 9 14 36 37 T-33 10 11 32 33 34 35 12 2 12 T-34 3 T-01 T-18 T-19 3 2 10 T-31 1 8 106 496 144 24 1 64 8 4 Where N = 37 X2 = 326 X = 106 Y2 = 7044 Y = 496 XY = 1482 37 37 2 37 2 = = 1482 106 0.650 326 rxy 496 496 106 7044 Because of rxy > rtable, so item no 1 is valid. 2. Reliability of Try Out Test After validity items had been done, the next analysis was to test the reliability of instrument. It was done to find out whether a test had higher critical score and gave the stability or consistency of the test scores or not. From the computation of reliability of the try out instruments using advertisements video, it was obtained 0,743, for 5 % with N = 37. It was obtained 0,725. It could be concluded that the instruments that were used in his research was reliable. The complete analysis and the computation as follow : 4. The computation of reliability using advertisement video Formula : Criteria : Criteria : The Try out is reliable if > rtable Calculation: = + + = + + = + . . .+ 0.92490.6201 1.0751 0.7973 + . . .+ 4.4474 Variance 2 S2 = 7044 10.970 37 36 496 = Index Reliability 5 1 5 10.970 4.447 10.970 = 0.743 = The result shows that 0.743 is more than 0.725, it meant the items of instrument were valid. 3. Discriminating Power of Try Out Test The discriminating power of the five items analysis of speaking was satisfied. It showed that all speaking items had strong discrimination. The complete analysis and the sample of computation as follow. The Computation of Discriminating Power 5. Formula: Where: MH : The average upper class ML :The average lower class x1 2 : The number of individual quadrate deviation upper class x2 2 : The number of individual quadrate deviation lower class Ni : 27% x N, with N= the number of students Criteria: The result of tscore is compared with ttable ,df= (N1-1) + (N2-1) and alpha =5%. If tscore > ttable, so the discriminating power of the item is significant. Calculation: Below is the example of the computation of discriminating power on item number 1. Ni = 27% x 45=12 = = = t = + 10 10 1 1.60 8.40 1.80 8.401.60 2.603.20 The result obtained Discriminating power is = 1.80 Because ttable = 1.725 and tscore > ttable, so the item number 1 is significant. Because the result tscore > ttable. So the item number 1 is satisfactory. 4. Difficulty Level Of Try Out Test 6. From the computation of difficulty level of the five items analysis of speaking, it was found that the difficulty level is easy. So it could be concluded that the final total items analysis for the instruments were categorized satisfactory. The sample of computation is as follow. The Computation of Difficulty Index Formula : DL = N G x 100% Where: DL: Difficulty level G : The number of students who fail in answer N : The number of students Criteria: < < < < < < 0.27 Easy 0.27 DL 0.72 medium 0.00 Interval P Criteria DL 1.00 Difficult DL 0.72 Calculation : Below is the example of the computation of difficulty level on item number 1. = % DL = 37 10 x 100% 27.03 The result obtained DL = 27.03% Because of the result is between 0.27-0.72, so the item number 1 is medium. C. Third Analysis 7. The second analysis represents the result of pre-test and post-test that was done both in experimental and control group. This analysis will answer the research question Is advertisement video effective to teach speaking of advertisement? We can conclude advertisement video is effective when the result of post-test of the experimental class (using advertisement video) and control class (using conventional media) has significant differences or the assumption that those classes is equal is not fulfilled. Before the researcher tested the hypothesis that had been mentioned in the chapter two, the researcher analyzed and tested hypothesis prerequisites which contained of normality test and homogeneity test. Second analysis dealt with normality test, homogeneity test, and t-test (test difference two variants) in pre-test and post-test. 1. Analysis of Pre-test The experimental group (XI IPS 4) was given a pre test on Mei 21, 2012 and control group (XI IPS 1) was given a pre-test on Mei 22, 2012. They were asked to practiced became advertiser to advertise certain product. a. Test of Normality Test of normality was used to find out whether data of control and experimental group which had been collected from the research come from normal distribution normal or not. The result computation of Chi- quadrate ( X2 score ) then was compared with table of Chi-quadrate ( X2 table ) by using 5 % alpha of significance. If X2 score < X2 table meant that the data spread of research result distributed normally. Based on the research result of XI IPS 1 student in the control group before they were taught speaking of advertisement text without advertisement video, they reached the maximum score 72 and minimum score 48. The stretches of score were 24. So there were 6 classes with length with classes 4. From the computation of frequency distribution, it was found ( f i Xi ) = 2309, and ( fi Xi 2 ) = 141990.9. So the average score (X )was 72,25 and the standard deviation (S) was 6.75587. After counting the average sore and standard deviation, table of observation frequency was needed to measure Chi-quadrate ( X2 score ). Table 4. 1 Table of the Observation Frequency of Control Group 8. 47.50 -1.96 -0.475 48 52 0.0725 2.7544 4 0.5633 52.00 -1.30 -0.403 53 56 0.1667 6.3352 7 0.0698 56.50 -0.63 -0.236 57 61 0.2220 8.4363 9 0.0377 61.00 0.04 0.014 62 65 0.2444 9.2876 8 0.1785 65.50 0.70 0.258 66 70 0.1558 5.9215 6 0.0010 70.00 1.37 0.414 71 74 0.0648 2.4611 4 0.9622 74.50 2.03 0.479 hitung = 1.8125 large area Ei OiClass Class limit Zi P(Zi) Based on the Chi-quadrate table X2 table for 5% alpha significance with 6-3= 3, it was found X2 table = 7.81. Because of X2 score < X2 table, so the initial data of experimental group distributed normally. While from the result of XI IPS 4 students in experimental group, before they were taught speaking of advertisement by using advertisement video, was found that the maximum score was 76 and minimal score was 52. The stretches of score were 24. So, there were 6 classes (fi xi ) = 2434. 25, and ( f i xi 2 ) = 161853.8. So, the average score (X) was 65.7905 and the standard deviation ( S ) was 6.87829. After counting the average score and standard deviation, table of observation frequency was needed to measure Chi-quadrate ( X2 score ). Table IV.2 Table of the Observation Frequency of Experimental Group 51.50 -2.08 -0.481 52 56 0.0584 2.1622 3 0.3246 56.00 -1.42 -0.423 57 60 0.1436 5.3127 7 0.5359 60.50 -0.77 -0.279 61 65 0.2334 8.6340 6 0.8036 65.00 -0.11 -0.046 66 69 0.1594 5.8981 9 1.6313 69.50 0.54 0.205 70 74 0.1785 6.6049 7 0.0236 74.00 1.19 0.384 75 78 0.0840 3.1085 5 1.1510 78.50 1.85 0.468 X = 4.4700 Ld OiP(Zi) EiZiLcClass Based on the Chi-quadrate table ( X2 table ) for 5% alpha of significance with df 6-3=3, it was found X2 table = 7.81. Because of X2 score < X2 table, so the initial data of experimental group distributed normally. b. Test of Homogeneity 9. Test of Homogeneity was done to know whether sample in the research come from the population that had same variance or not. In this study, the homogeneity of the test was measured by comparing the obtained score ( Fscore ) with Ftable . Thus, if the obtained score ( Fscore ) was lower than the Ftable or equal, it could be said that the Ho was accepted. It meant that the variance was homogenous. The analysis of homogeneity test could be seen in table IV.3. Table IV.3. Test of Homogeneity (Pre-Test) Eksperiment G Control G 2244 2236 37 38 60.65 58.84 29.790 43.164 5.46 6.57 Varians (s 2 ) Standard deviation (s) n X Variant Sources Sum By knowing the mean and the variance, the writer was able to test the similarity of the two variants in the pre-test between experimental and control group. The computation of the test of homogeneity as follows: F = Biggest Variance Smallest Variance 43.164 29.790 F = F = 1.448938879 On a 5% with df 1 = 37, df 2 = 38, it was found Ftable = 1.721142152. Because of Fscore Ftable, so it concluded that both experimental and control group had no differences. The result showed both groups had similar variants (homogenous). c. Test difference two variants in pre-test between experiment and control group After counting standard deviation and variance, it could be concluded that both group have no differences in the test of similarity 10. between two variances in pre-test score. So, to differentiate whether the students result of speaking of advertisement in experimental and control group were significant or not, the writer used t-test to test the hypothesis that had been mentioned in the chapter two. The writer used formula: 21 21 11 nn s xx t + = Where: 2 )1()1( 21 2 22 2 11 + + = nn snsn s Based on table IV.3, first the writer had to find out S by using the formula above: (37-1) . 29.790 + (38-1) . 43.164 S 2 = 36.568 S = 6.047 =S 2 37 + 38 -2 = After S was found, the next step was to measure t-test: t = - 1 1 37 38 60.649 58.842 6.047 + 1.807 1.397 = = 1.293 After getting t-test result, then it would be consulted to the critical score of ttable to check whether the difference is significant or not. For a = 5% with df = 37+ 38-2= 73, it was found ttable (0.975)(80)= 1.99. Because of tscore < ttable, so it could be concluded that were no significance of difference between the experimental and control group. It meant that both experimental and control group had same condition before getting treatments. 2. Analysis of Post-Test The experimental group was given post-test on Mei 28, 2012 and control group was given a pos test Mei 29, 2012. Post test was conducted after all treatments were done. Advertisements video was used as medium in the teaching speaking of 11. advertisement text to students in experimental group. While for students in control group, they were given treatments without advertisements video. Post-test was aimed to measure students ability after they got treatments. They were asked to advertise certain products. a. Test of Normality Test of normality was used to find out whether data of control and experimental group, which had been collected after they got treatments, come from normal distribution normal or not. The formula, that was used, was Chi-Quadrate. The result computation of Chi-Quadrate ( X2 score ), then was compared with table of Chi-Quadrate ( X2 score ) then was compared with table of Chi-quadrate ( X2 table ) by using 5% alpha of significance. If X2 score < X2 table meant that the data spread of research result distributed normally. Based on the research result of XI IPS 1 students in the control group after they got usual treatments in the teaching speaking of advertisements they reached the maximum score 80 and minimum score 60. The stretches score 20. So there were 6 classes with length of 4 classes. From the computation of frequency distribution it was found fi xi = 2709 and ( fi xi 2 ) = 194457.5. So the average score ( X ) was 71.2895 and the standard deviation (S) was 6.00521. It meant that there was an improvement of students score after they got treatments. After counting the average score and standard deviation, table of observation frequency was needed to measure Chi quadrate ( X2 score ). Table IV.4 Table of the Observation Frequency of Control Group 59.50 -1.96 -0.475 60 63 0.0725 2.7544 4 0.5633 63.50 -1.30 -0.403 64 67 0.1667 6.3352 7 0.0698 67.50 -0.63 -0.236 68 71 0.2220 8.4363 9 0.0377 71.50 0.04 0.014 72 75 0.2444 9.2876 8 0.1785 75.50 0.70 0.258 76 79 0.1558 5.9215 6 0.0010 79.50 1.37 0.414 80 83 0.0648 2.4611 4 0.9622 83.50 2.03 0.479 hitung = 1.8125 Luas Daerah Ei OiClass Lc Zi P(Zi) 12. Based on the Chi-quadrate table ( X2 table ) for 5 % alpha of significance with df 6-3 = 3, it was found X2 table = 7.81. Because of X2 score < X2 table, so the data of experimental group after getting treatments distributed normally. While from the result of XI IPS 4 students in experimental group, after they were taught using advertisement video, was found that the maximum score was 92, and minimal score was 64. The stretches of score were 28. So, there were 6 classes with length of classes 5. From the computation of frequency distribution, it was found ( fiXi ) = 2939, and ( fiXi 2 ) = 235237, so the average score (X) was 79. 3784 and the standard deviation ( S ) was 7.64254. By seeing the average score of students in experimental group, it could be concluded that there an improvement of students score after they got treatments by using advertisement video. After counting the average score and standard deviation, table of observation frequency was needed to measure Chi-quadrate ( X2 score ). Table IV. 5 Table of the Observation Frequency of Experimental Group 63.50 -2.08 -0.481 64 68 0.0584 2.1622 3 0.3246 68.50 -1.42 -0.423 69 73 0.1436 5.3127 7 0.5359 73.50 -0.77 -0.279 74 78 0.2334 8.6340 6 0.8036 78.50 -0.11 -0.046 79 83 0.1594 5.8981 9 1.6313 83.50 0.54 0.205 84 88 0.1785 6.6049 7 0.0236 88.50 1.19 0.384 89 93 0.0840 3.1085 5 1.1510 93.50 1.85 0.468 X = 4.4700 P(Zi) Ei Luas Daerah OiZiLcClass Based on the Chi-quadrate table (X2 table) for 5 % alpha of significance with df = 6-3 = 3, it was found X2 table 7.81. Because of X2 score < X2 table, so the data of experimental group after getting treatments distributed normally. b. Test of Homogeneity 13. The writer determined the mean and variance of the students score either in experimental or control group. By knowing the mean and variance, the writer was able to test the similarity of the two variance in the post-test between experimental and control group. Table. IV.6 Test of Homogeneity (Post-test) Eksperimen Kontrol 2860 2700 37 38 77.30 71.05 53.381 32.376 7.31 5.69 Varians (s 2 ) Standart deviation (S) n X Varians Sources Jumlah The computation of the test of homogeneity as follows: F = Biggest Variance Smallest Variance 32.376 53.381 F = F = 0.60649486 On a 5 % with df 1 = 37, df 2 = 38,it was found Ftable = 1.721142152. Because of Fscore Ftable, so it could be concluded that both experimental and control group had no differences. The result showed both groups had similar variance (homogenous). c. Test of difference two variants in post-test between experiment and control group After counting standard deviation and variance, it could be concluded that both group have no differences in the test of similarity between two variances in the post-test score. So, to differentiate if the students result of speaking of advertisement text in experimental and control group after getting treatments were significant or not, the writer used t-test to test the hypothesis that had been mentioned in chapter two. To see the difference between the experimental and control group, the writer used formula : 14. 21 21 11 nn s xx t + = Where: 2 )1()1( 21 2 22 2 11 + + = nn snsn s Based on the table IV.6, first the writer had to find out S by using the formula above: (37-1) . 53.381 + (38-1) . 32.376 S 2 = 42.735 S = 6.537 =S 2 37 + 38 -2 = After S was found, the next step was to measure t-test: - 1 1 37 38 6.245 1.510 t = 4.136 t = = 77.297 71.053 6.537 + t = After getting t-test result, then it would be consulted to the critical score of ttable to check whether the difference is significant or not. For a = 5% with df = 37+38-2=73, it was found ttable (0.05)(95)= 1.67. Because of tsore > ttable,, so it could be concluded that there was significance of difference between the experimental and control group. It meant that experimental group was better than control group after getting treatments. Since the obtained t-score was higher than the critical score on the table, the difference was statistically significance. Therefore, based on the computation there was a significance difference between the teaching of speaking of advertisement using advertisement video and teaching speaking of advertisement without advertisement video for the eleventh grade students of SMA Negeri 1 Pegandon Kendal. Teaching speaking of advertisement using advertisement video to be more effective than teaching speaking of advertisement without using advertisement video. It can be seen from the result of the test where the students taught speaking 15. of advertisement by using advertisement video got higher scores than the students taught speaking of advertisement without advertisement video. D. Discussions The data were obtained from the students achievement scores of the test of speaking of advertisement. They were pre-test and post-test scores from the experimental and control group. The average score for experimental group was 60.65(pre-test) and 77.3 (post-test). The average score for control group was 58.84(pre-test) and 71.65(post-test). The following was the simple tables of pre and post-test students average score and students average score of each speaking components. Table IV. 7 The Pre-test and Pos-test Students Average Scores of the Experimental and Control Group No Group The Average Percentage of Pre-test The Average Percentage of Post-test 1 Experimental 60.65 77.3 2 Control 58.84 71.65 Table IV. 8 The Pre-test and Post-test StudentsAverage Scores of the Experimental and Control No Component of Speaking Group The Average score of Pre-test The Average score of Post-test 1 Pronunciation Experimental 2.9 3.8 Control 2.9 3.3 2 Grammar Experimental 2.8 3.8 Control 3 3.6 3 Vocabulary Experimental 3.1 3.8 Control 2.9 3.6 4 Fluency Experimental 3 3.8 Control 2.8 3.3 5 Comprehension Experimental 3.3 4.2 Control 3.1 3.9 16. a. Students Condition in Control Group In this study, source of data that become as control group was class IPS 1. In the control group, there was not a new treatment in a teaching learning process. They were given a usual treatment. They were taught speaking of advertisement in conventional media. By listen teacher explanation and use the realia media and pictures. Teacher only give little example in speaking how to advertise certain product. Students could not enjoy in practicing their skill in speaking because they only saw teachers way in advertise certain product. It could not effective to teach students in speaking of advertisement because the teacher seldom practice became good advertiser. It was proven with the control groups average in the post test (71.65) which was lower than experimental group (77.3). b. Students condition in experimental group 1. Analysis students speaking before treatment (pre-test) In the pre-test, students ability in speaking of advertisement was low. Pre- test was conducted before treatment. From the result of pre-test, it was known was students faced many difficulties in speaking of advertisement. Sentences which were used by students to convoy the idea, were influenced by Indonesian. Moreover they dont know what should they say when they want to convey their meaning. Students ability was in low level when they had to arrange words to be a good sentence that comprehensible by considering main function. It meant that the idea was not clearly stated and the sentences were not well-organized to support to transformation of meaning. Students word voice (Pronunciation and fluency) was also far from being perfect. Not only the way they convey their idea was not clear but also there were many difficulties in grammar and vocabulary; therefore, students ability of speaking of advertisement text was hard to be understood. To minimize the number of students mistakes of their speaking, the researcher collected students speaking in writing form after they do advertise certain product, gave correction and returned the paper to them. From the correction of their mistakes, students were supposed to learn more and improve and improve their ability in speaking of advertisement. 2. Analysis students speaking after treatment (post-test) 17. Based on the analysis of students ability, it was found that students ability after getting treatment was improved. In the treatment, students were doing speaking of advertisement that was in line with the function of some expressions they learn. The vocabulary choice, sentences arrangement, and the way they produce the word were good and relevance to the topic or (their meaning) so the meaning were easy to be understood. Their speaking was still comprehensible however; there were some mistakes in grammar and pronunciation. The finding that shows students ability in namely the increasing of students average score. There were still some mistakes that students had made like grammar and pronunciation. But it was very human. So, it could be concluded that the implementation of using advertisement video as media in the teaching speaking of advertisement was effective. It was proven with students average score in experimental group was higher than control group. By considering the students final score after getting treatment, the teaching of speaking of advertisement using advertisement video as media was better than without advertisement video. Based on t-test analysis was done, it was found that t-score (4.136) was higher than t-table using 5 % alpha of significance (1.67). Since tscore > ttable , it proved that there was significant difference between the improvement of students achievement that was given a new treatment (using advertisement video) and the improvement of students achievement that was given a usual treatment. c. The advantages and disadvantages of using advertisement video in teaching of speaking of advertisement 1) The advantages of using advertisement video in teaching of speaking of advertisement After conducting the research, there were some advantages of using advertisement video in teaching speaking of advertisement: a. Advertisement video gave students the real situations of a chronological action. It helped students to express their imagination to be transferred in to the reality. The use advertisement video was actually meant to help them expressing their ideas easily. b. Students boredom i learning speaking could be avoided. The treatment gave students different nuances of teaching and learning process so they were interested in following the lesson. Video advertisement that gives students chance to notice how to be good advertiser and how use the expression appropriately according the situation, so they will motivated to 18. imitate practiced to be good advertiser by using clear pronunciation and appropriate expression. It also could measure how far their speaking skill is. 2) The disadvantages of using advertisement video in teaching of speaking of advertisement The disadvantages were described below: a. It spent a lot of time, because the students skill was too low, they cant directly speaking in front of class to advertise certain product after teacher distributed certain product to be advertised. They need time to prepare their performance. b. It was not easy enough to manage the class because, sometime the students will be very hysteric when they see their friends practicing in front of them. Their voice can disturb another class. E. Limitation of research The writer realized that there were some hindrances and barriers in doing this research. The hindrances and barriers occurred was not caused by inability of the researcher but caused by limitation of the research like time, fund, and equipment of research. 19. .