Techniques of Collecting Data

21 40 65 -25 625 22 40 65 -25 625 23 60 55 5 25 24 45 75 -30 900 25 55 65 -10 100 26 55 65 -10 100 27 55 55 N= 36 ∑x = 1350 ∑x = 1680 ∑ D = -330 ∑ D 2 = 9900 Notes: N = Number of cases x = The score of pre-test y = The score of post-test D = Difference ∑x = The total score of x ∑y = The total score of y ∑D = The total of difference ∑D 2 = The total of quadrate difference Based on the data in the table , writer calculated the result of ∑ D = -330 and ∑ D 2 = 9900. Then, the writer tried to find out the standard deviation with the formula: SD D = ∑ D 2 ∑ D 2 N N = 9900 -330 2 27 27 = 366.67 -12.22 2 = 276.38 149.38 = 217.28 = 14.73 Based on that computation, the standard deviation is 14.73. the standard deviation is used to find out the standard error from mean of difference. Next the writer tried to know mean of difference. It is used by the formula: MD = ∑ D N = -330 27 = -12.22 From that calculation the MD is -12.22, this calculation is used for finding out t-test. Meanwhile to get SE MD Standard Error of the Mean of Difference, the writer used the formula as follows: SE MD = SD D N-1 = 14.73 27-1 = 13. 11 26 = 13. 11 5.09 = 2. 89 From that calculation it could be seen that the standard error of the mean difference is 2.89. This SE MD is used to find out the T-Test. Furthermore, to get t o t observation of the test, the writer used formula: to = MD SE MD = -12.22 2.89 = - 4.22 Based on that computation, it indicates that the T-test is -4.22, it means there was a difference of degree as much as – 4.22. Regardless the minus, it does not indicate negative score. The result of analyzing the data above it shows that the coofecient is 4.22. Then, to find critic “t” result in t-table, then the writer must calculate degree of freedom with formula as follows: df = N – 1 = 27 - 1 = 26 df = 26 See table of “t” value at degree of significance of 5 and 1 in appendix 4 Degree of Significance Degree of freedom 5 2.06 1 2.78 Statitiscal Hypothesis: a If t o t t : There is a significant deference and the alternative hypothesis Ha is accepted and null hypothesis Ho is rejected. b If t o t t : There is no significant deference and the alternative hypothesis Ha is rejected and null hypothesis Ho is accepted. The value of t o has been found with the amount 4.22. so, with formula The result is 2.06 4.22 2.78, it means that t- observation is higher than t- table. According to the data, the score of t o is higher than t t which means that the alternative hypothesis is accepted and null hypothesis is rejected. There is a significance increases that the simple present tense taught through contextual teaching and learning approach.

2. Interpretation of Data

Based on the data analysis, the writer would like to interpret the data result of to T-Test which is to know whether students get the improvement scores from the pre-test to post-test in learning the simple present tense by using CTL. To get the to T-Test, the writer tried to get the Mean of difference MD, and Standard Error of mean difference SE MD . First, to get Mean of Difference the writer tried to know the Difference between pre-test and post-test score and the number of students. The Difference between pre-test and post-test is -330. Meanwhile, the number of students is 27. Therefore, The Mean of Difference MD is -12.22. it means that there is the difference between pre-test and post-test and the result is 12.22. Second, to get Standard Error of Mean Difference the writer tried to know Standard Deviation from the Difference and the Number of students subtracted with 10ne. the Standard Deviation is 14.73 and the number of student subtracted with 1 is 26. Therefore, the Standard Error of Mean difference is 2.89 it means that the standard error shows to avoid the error in choosing sampling and the standard error 2.89 is used to find out the critic of “t” to determine whether the hypothesis is rejected or accepted. Furthermore, After calculation the writer get the result of Mean of Difference, the result of Mean Difference is-12.22 and Standard Error of Mean Difference is 2.89, then after MD and SE MD has been known the writer calculate to which gain from MD divided by SE MD and the result of to is 4.22, the to 4.22 will be compared by table ‘to’ to know whether teaching the Simple Present Tense using CTL accepted or rejected. Based on the statistical hypothesis that if to t-observation is higher than t-table than it means there is a significant deference or using CTL in teaching the Simple Present Tense is accepted. Meanwhile, if to t-observation is lower than t- table than it means there is no significant deference using CTL in teaching the Simple Present Tense is rejected. For to, the result is 4.22 and degree of freedom df is 26. Here we have to see t- table which df is 26, at df 26 the result of critical t-table for Degree of Significance of 5 is 2.06 and for Degree of Significance of 1 is 2.78. Based on the explanation it shows that to is higher than t-table at both degree of significant 5 and 1, or 4.22 is higher than 2.06 and 2.78. It shows that there is some significant increases in this research. It means that student after being taught the Simple Present Tense using CTL approach get some improvement score from pre- test and post-test.. Therefore, we can see that the teaching simple present tense through contextual teaching and learning approach is adequate success.