Technique of Analyzing the Data

Based on the table above,, in pre-test there was onestudentwho got 40, one student who got 45, four students who got 50, one student who got 52, six students who got 55, one student who got 57, four students who got 60, one student who got 62, and the last one student who got 65. After post- test, their scores were higher than pre- test. It can be seen that the highest score in post- test was 72two students who got it and the lowest score was 55there were two students who got it. Whereas their average score for the pre-test is 54.55, and the average score for the post-test is 62.3. Moreover, the average gain score is 7.35. From the table above the writer can summarize the experiment class learning imperative sentences using TPR method got the higher score than the control class without TPR method when they learnt imperative sentences. The next table shows the result calculation of the post-test both Experiment Class using Total Physical Response and Control Class without using Total Physical Response.As mentioned before, that in analyzing the data from the result of pre- test and post- test, the writer used statistic calculation of the t- test formula with the degree of significance 5 and 1 . Table 4.3 Standard Deviation Table No X Y x y x 2 y 2 1 7 5 -3.2 -0.35 10.24 0.1225 2 17 2 6.8 -5.35 46.24 28.6225 3 17 5 6.8 -2.35 46.24 5.5225 4 7 7 -3.2 -0.35 10.24 0.1225 5 8 5 -2.2 -2.35 4.84 5.5225 6 7 7 -3.2 -0.35 10.24 0.1225 7 8 8 -2.2 0.65 4.84 0.4225 8 8 5 -2.2 -2.35 4.84 5.5225 Note: X = the students’ gained score in experimental class Y = the students’ gained score in control class x = X – MX y = Y – MY A. Data Analysis and Test of Hypothesis 1. Data Analysis The procedures of calculation were as follows. a. Determining Mean of Variable X, with formula: M 1 = X N 1 = Σ 204 20 = 10.2 No X Y x y x 2 y 2 9 7 7 -3.2 -0.35 10.24 0.1225 10 2 15 -0.2 7.65 0.04 58.5225 11 7 5 -3.2 -2.35 10.24 5.5225 12 13 7 2.8 -0.35 7.84 0.1225 13 10 5 -0.2 -2.35 0.04 5.5225 14 15 12 4.8 4.65 23.04 21.6225 15 13 5 2.8 -2.35 7.84 5.5225 16 9 5 -1.2 -2.35 1.44 5.5225 17 17 10 6.8 2.65 46.24 7.0225 18 15 20 4.8 12.65 23.04 160.0225 19 10 7 -0.2 -0.35 0.04 0.1225 20 7 5 -3.2 -2.35 10.24 5.5225 261 192 ∑x= 0 ∑y= 0 ∑x 2 = 278 ∑y 2 =321.15 b. Determining Mean of Variable Y, with formula: M 2 = Y N 2 = Σ 147 20 = 7.35 c. Determining Standard of Deviation Score of Variable X, with formula: SD 1 = X 2 N 1 = Σ 102 2 20 = 10.404 20 = 52.02 = 7.21 d. Determining Standard of Deviation Score of Variable Y, with formula: SD 2 = Y 2 N 2 = Σ 7.35 2 20 = 54.0225 20 = 2.701125 = 1.64 e. Determining Standard Error of Mean of Variable X, with formula: SE M1 = SD 1 N 1 −1 = 7.21 20 −1 = 7.21 19 = 0.37 = 0.61 f. Determining Standard Error of Mean of variable Y, with formula: SE M2 = SD 2 N 2 −1 = 1.64 20 −1 = 1.64 19 = 0.08 = 0.28 g. Determining Standard Error of Mean of Mean of variable X and variable Y, with formula: SE M1-M2 = SE M 1 1 + SE M 2 2 = 0.61 2 + 0.28 2 0.3721 + 0.0784 = 0.45 = 0.67 h. Determining t o with formula: �� = M 1 −M 2 SE M 1 − SE M 2 = 10.2 −7.35 0.67 = 2.85 0.67 = 4.25 i. Determining t-table in significance level 5 and 1 with degree of Freedom df: