Action Action Research Cycle Two a. Revised Plan

48 To compare the result between pre-test and post-test of each cycle, the writer uses some steps. First, the writer calculates the students‟ mean score for each test. Second, she computes the percentage of students‟ improvement score from pre-test to post-test cycle 1 and cycle 2. In analyzing the data of pre-test, the step is to get the mean score of the class. It is calculated as following: x = ∑x n x = 2660 39 x = 68.21 The mean score of the class in pre-test is 68.21. It means that the students‟ pronunciation mean score before using minimal pair drill or before implementing CAR is 68.21. Then, the writer calculates the post-tests of CAR for each cycle to measure the students‟ score improvement from the pre-test to post-test 1 result and then the post-test 1 result will be compared to the post-test 2 result of CAR cycle 2. The first step, the writer calculates the post-test 1 to know the students‟ improvement after implementing CAR in cycle 1. 33 60 85 100 34 80 85 95 35 45 75 85 36 90 90 100 37 80 90 100 38 70 85 90 39 80 90 95 TOTAL 2660 3200 3565 Mean: x = ∑x N

68.21 82.05

91.41 49 The calculating of post-test 1 is as follows: x = ∑x n x = 3200 39 x = 82.05 From that calculation, the mean score of the class in post-test 1 is 82.05. It means that the students‟ pronunciation mean score after using minimal pair drill or after implementing CAR showed the improvement, it is 82.05. If we compare with the previous test pre-test score 82.05, the difference is 13.84 points by calculation as follows 82.05-68.21 = 13.84. So, the writer could say that the improvemen t students‟ pronunciation score is 13.84. The next phase is to know the percentage of pre-test and post-test difference, it is calculated as follows: P= y1-y x 100 y P= 82.05-68.21 x 100 68.21 P= 13.84 x 100 68.21 P= 20.29 So , the percentages of students‟ mean score improvement from pre- test to post-test 1 is 20.29. The improvement did not valuably increase. Next, after calculating the students‟ post-test 1 score, the next step is to calculate the post-test 2; it can be seen as follows: x = ∑x n x = 3565 39 x = 91.41

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