Normality of the Data

2. Normality of post-test in Experiment Class Hypotheses H o = Data of X is normally distributed H 1 = Data of X is not normally distributed Criteria of the test: In the significant degree of 0,05, the value in the table of Lilyfors shows: T 0,0532 = 0, 157 H = T 0,157 H 1 = T 0,157 The result showed that T max T table 0,091 0,157, it means that the data is normally distributed. Table 4.5 Calculation of Post-test Normality in Experimental Class xi f fx x2 fx2 p=fn z=xi-Ms Φ Σp T=Φ-Σp 72 2 144 5184 10368 0.063 -1.843 0.033 0.063 -0.030 73 2 146 5329 10658 0.063 -1.707 0.044 0.125 0.081 74 1 74 5476 5476 0.031 -1.572 0.058 0.156 -0.098 77 1 77 5929 5929 0.031 -1.167 0.122 0.188 0.066 79 1 79 6241 6241 0.031 -0.896 0.185 0.219 -0.034 81 1 81 6561 6561 0.031 -0.626 0.266 0.250 -0.016 82 1 82 6724 6724 0.031 -0.491 0.312 0.281 0.031 83 1 83 6889 6889 0.031 -0.356 0.361 0.313 -0.049 85 2 170 7225 14450 0.063 -0.085 0.466 0.375 0.091 87 3 261 7569 22707 0.094 0.185 0.573 0.469 -0.105 88 3 264 7744 23232 0.094 0.320 0.626 0.563 0.063 89 2 178 7921 15842 0.063 0.455 0.676 0.625 -0.051 90 1 90 8100 8100 0.031 0.590 0.723 0.656 0.066 91 2 182 8281 16562 0.063 0.726 0.766 0.719 -0.047 93 2 186 8649 17298 0.063 0.996 0.840 0.781 0.059 94 2 188 8836 17672 0.063 1.131 0.871 0.844 -0.027 95 2 190 9025 18050 0.063 1.266 0.897 0.906 -0.009 96 1 96 9216 9216 0.031 1.402 0.919 0.938 0.018 98 2 196 9604 19208 0.063 1.672 0.953 1 -0.047 1627 32 2767 140503 241183 1 0.091 85.63 145.63 7394.89 12693.84 � 2 = ∑ �� 2 � − � ∑ �� � � 2 = 241183 32 − � 2767 32 � 2 = 7536.98 − [86.5] 2 = 7536.98 − 7482.25 = 54.73 = √54.73 = 7,4 S = 7.4 S 2 = 54.73 M = 85.6 T max = 0,091 T table = 0.157 3. Normality of pre-test in Control Class Hypotheses H o = Data of X is normally distributed H 1 = Data of X is not normally distributed Criteria of the test: In the significant degree of 0,05, the value in the table of Lilyfors shows: T 0,0532 = 0, 157 H = T 0,157 H 1 = T 0,157 The result showed that T max T table 0.104 0.157, it means that the data is normally distributed. Table 4.6 Calculation of Pre-test Normality in Control Class xi f fx x2 fx2 p=fn z=xi-Ms Φ Σp T=Φ-Σp 45 1 45 2025 2025 0.031 -1.884 0.030 0.031 -0.001 48 1 48 2304 2304 0.031 -1.549 0.061 0.063 -0.002 50 1 50 2500 2500 0.031 -1.326 0.092 0.094 -0.001 51 2 102 2601 5202 0.063 -1.215 0.112 0.156 -0.044 53 1 53 2809 2809 0.031 -0.992 0.161 0.188 -0.027 55 1 55 3025 3025 0.031 -0.769 0.221 0.219 0.002 56 3 168 3136 9408 0.094 -0.658 0.255 0.313 -0.057 57 1 57 3249 3249 0.031 -0.546 0.292 0.344 -0.051 58 1 58 3364 3364 0.031 -0.435 0.332 0.375 -0.043 60 1 60 3600 3600 0.031 -0.212 0.416 0.406 0.010 61 1 61 3721 3721 0.031 -0.101 0.460 0.438 0.022 64 2 128 4096 8192 0.063 0.233 0.592 0.500 0.092 65 1 65 4225 4225 0.031 0.345 0.635 0.531 0.104 66 3 198 4356 13068 0.094 0.456 0.676 0.625 0.051 69 2 138 4761 9522 0.063 0.791 0.785 0.688 0.098 70 1 70 4900 4900 0.031 0.902 0.816 0.719 0.098 72 4 288 5184 20736 0.125 1.125 0.870 0.844 0.026 73 1 73 5329 5329 0.031 1.236 0.892 0.875 0.017 74 1 74 5476 5476 0.031 1.348 0.911 0.906 0.005 75 2 150 5625 11250 0.063 1.459 0.928 0.969 -0.041 78 1 78 6084 6084 0.031 1.793 0.964 1 -0.036 1300 32 2019 82370 129989 1 0.104 61.90

96.14 3922.38 6190

� 2 = ∑ �� 2 � − � ∑ �� � � 2 = 129989 32 − � 2019 32 � 2 = 4062.16 − [63,1] 2 = 4062.16 − 3981.61 = 80.55 = √80.55 = 8.97 S = 8.97 S 2 = 80.55 M = 61.9 T max = 0.104 T table = 0.157 4. Normality of post-test in Control Class Hypotheses H o = Data of X is normally distributed H 1 = Data of X is not normally distributed Criteria of the test: In the significant degree of 0,05, the value in the table of Lilyfors shows: T 0,0532 = 0, 157 H = T 0,157 H 1 = T 0,157 The result showed that T max T table 0,109 0,157, it means that the data is normally distributed. Table 4.7 Calculation of Post-test Normality in Control Class xi f fx x2 fx2 p=fn z=xi-Ms Φ Σp T=Φ-Σp 51 1 51 2601 2601 0.031 -2.686 0.004 0.031 -0.028 61 1 61 3721 3721 0.031 -1.210 0.113 0.063 0.051 62 2 124 3844 7688 0.063 -1.063 0.144 0.125 0.019 63 3 189 3969 11907 0.094 -0.915 0.180 0.219 -0.039 64 3 192 4096 12288 0.094 -0.768 0.221 0.313 -0.091 66 2 132 4356 8712 0.063 -0.472 0.318 0.375 -0.057 68 1 68 4624 4624 0.031 -0.177 0.430 0.406 0.023 71 4 284 5041 20164 0.125 0.266 0.605 0.406 0.109 72 2 144 5184 10368 0.063 0.413 0.660 0.594 0.067 73 3 219 5329 15987 0.094 0.561 0.713 0.688 0.025 74 1 74 5476 5476 0.031 0.708 0.761 0.719 0.042 76 4 304 5776 23104 0.125 1.004 0.842 0.844 -0.002 78 2 156 6084 12168 0.063 1.299 0.903 0.906 -0.003 79 1 79 6241 6241 0.031 1.447 0.926 0.938 -0.012 80 2 160 6400 12800 0.063 1.594 0.945 1 -0.055 1038 32 2237 72742 157849 1 0.109

69.2 149.13 4849.5 10523.3

� 2 = ∑ �� 2 � − � ∑ �� � � 2 = 157849 32 − � 2237 32 � 2 = 4932.78 − [69.9] 2 = 4932.78 − 4886.88 = 45,9 = √45.9 = 6.8 S = 6.8 S 2 = 45.9 M = 69.2 T max = 0.109 T table = 0.157 ∑ � � = 674 32 = ��, �� ∑ � � = 218 32 = �, �� From the result of statistical calculation, it can be seen that both of pre-test and post-test in both experiment and control class is normally distributed because the result of T-table is bigger than T-max. If the data is normal, the sample is also normally distributed for the research.

C. Data Analysis

After comparing the score between experiment class 8-H and control class 8-G, the writer made an analysis of data from the result. Afterwards, the writer calculated them based on the step of the t-test formula, as follow: 1. Determining Mean X Mean variable X= 2. Determining Mean Y Mean variable Y = 3. Determining of Standard Deviation of variable X 4. Determining of Standard Deviation of variable Y 5. Determining Standard of Error Mean of variable X �� � = � ∑ � 2 � = � ����.�� �� = √���. �� = 10,44 �� � = � ∑ � 2 � = � ���.�� �� = √��, �� = 4,16 ��� � = �� � √� − 1 = 10,44 √32 − 1 = 10,44 √31 = 10,44 5,57 = �, �� ��� � = �� � √� − 1 = 4,16 √32 − 1 = 4,16 √31 = 4,16 5,57 = �, �� = �3,49 + 0,56 = �4,05 = 2,01 ��� � − ��� � = ���� � 2 + ��� � 2 = �1,87 2 + 0,75 2 6. Determining Standard of Error Mean of variable Y 7. Determining Standard of Error Mean difference of M x dan M y 8. Determining t o : 9. Determining t-table in significant level 5 and 1 with df �� = �1 + �2 − 2 = 32 + 32 − 2 = �� At the degree of significant of 5 = 1.7 At the degree of significant of 1 = 2.4 10. The comparison between t-score with t-table t-score = 1.77,092.4 � � = � � − � � ��� � − ��� � = 21,06 − 6,81 2,01 = 14,25 2,01 = �, ��

D. The Testing of Hypothesis

The research was held to answer the question wheter using guided questions in MTs Pembangunan UIN Jakarta is effective in teaching students’ writing of narrative text at the eighth grade. In order to answer the question about it, the writer make two hypothesis, the Alternative Hypothesis H a and Null Hypothesis H were proposed as follows: 1. Null Hypothesis H : guided question is not effective in teaching students’ writing of narrative text. 2. Alternative Hypothesis H a : guided question is effective in teaching students’ writing of narrative text To prove the hypothesis, the obtained data from experiment class and control class were calculated by using t test formula with assumption as follows: 1. If t o ≤ R t table in significant degree of 1, the Null Hypothesis H is accepted and the Hypothesis Alternative H a is rejected. Moreover, in significant degree of 5 is to support the significant degree of 1 and make sure that the H is accepted. It means that there is no significant effect of guided questions on students writing of narrative text. 2. If t o ≥ R t table in significant degree of 1, the Null Hypothesis H is rejected and the Hypothesis Alternative H a is accepted. Furthermore, in significant degree of 5 is to support the significant degree of 1 and make sure that the H a is accepted. It means that there is significant effect of guided questions on students’ writing of narrative text. The hypothesis criteria above states that: if t o t t = Ha is accepted and H is rejected, and if t t t = H a is rejected and H is accepted. Where: H a = Hypothesis Alternative H = Null Hypothesis t o = t observation t t = t test. The result of the statistic calculation indicates that the value of t is 7,09 which is higher that t-table t t at significance level 5 = 1,7 and t-table t t at significance level 1 2,4 it means that the Null Hypothesis H is rejected and the Hypothesis Alternative H a is accepted.

E. Data Interpretation

From the data above, seeing by the Mx in post-test of experiment class that higher than My in post-test of control class, 86.1970.06, is already prove that the technique in this research is effective. Furthermore, in testing of hypothesis, the result indicate that Hypothesis Alternative is accepted, it means that there is significant effect in using guided questions in teaching students’ narrative writing, especially for eighth grade students of MTs Pembangunan Jakarta. Teaching learning process in eighth grade is purposed to make students write narrative story in well organized and well grammatically. The data had shown that with guided questions, students can write in better structure and grammatically. This result could be the consideration in English language teaching which require students to write in particular rule such as the generic structure of narrative text. By this consideration, this technique can be applied in English language teaching. This final result, answer the question whether guided questions effective or not in teaching narrative writing. This is related to the title of the research, “The Effectiveness of Guided Questions in Teaching Students’ Narrative Writing”.