Description of the Data Analysis of the Data

34 − Teacher check students works by asking some of them to read out their answer. Now, Anita what is your answer for number one? Are you sure? Class, do you agree with Anitas answer? Etc. f. Closure − Teacher concludes the lesson. Teacher may say, So, what do you get from our lesson today? Do you enjoy your reading today? Etc. − Teacher closes the lesson. Teacher may say, I think time is up, see you next Tuesday, bye.

B. Findings of the Research

1. Description of the Data

+ , - - . - + . . . . 0 . + . . Table 4.1 1 35 1 1 1 2 1 1 1 2 1 1. Determining Mean of Pre-test Score of Experiment Class: M x 1 = x N = 1.290 20 = 64.5 2. Determining Mean of Post-test Score of Experiment Class: M y l = y N = 1.495 36 20 = 74.75 Table 4.2 The Test Score of the Controlled Class Students Pre-test x Post-test y 1 50 60 2 60 70 3 50 60 4 65 75 5 75 85 6 65 60 7 70 80 8 65 70 9 65 75 10 70 60 11 70 65 12 65 75 13 55 65 14 60 70 15 60 70 16 70 75 17 55 70 18 50 60 19 55 70 20 55 65 1. Determining Mean of Pre-test Score of Controlled Class: 37 M x2 = x N = 1.230 20 = 63.25 2. Determining Mean of Post-test Score of Controlled Class: M y2 = y N = 1.380 20 = 69 Table 4.3 The Calculation of the Test Score of the Experiment Class Students Pre-test x Post-test y D = x-y D2 = x-y 2 1 1 −10 100 2 −15 225 3 −10 100 4 −10 100 5 1 −5 25 6 −15 225 7 1 −20 400 8 −5 25 9 10 −15 225 11 −15 225 12 −10 100 13 1 −10 100 14 1 −20 400 15 −20 400 38 16 −15 225 17 −5 25 18 −10 100 19 1 −10 100 20 −5 25 = −215 D 2 = 3.125 Table 4.4 The Calculation 0f the Test Score of the Controlled Class Students Pre-test x Post-test y D = x-y D2 = x-y 2 1 50 60 −10 100 2 60 70 −10 100 3 50 60 −10 100 4 65 75 −10 100 5 75 85 −10 100 6 65 60 5 25 7 70 80 −10 100 8 65 70 −5 25 9 65 75 −10 100 10 70 60 10 100 11 70 65 5 25 12 65 75 −10 100 13 55 65 −10 100 14 60 70 −10 100 15 60 70 −10 100 16 70 75 −5 25 17 55 70 −15 225 18 50 60 −10 100 19 55 70 −15 225 39 20 55 65 −10 100 = −150 D 2 = 1.950

2. Analysis of the Data

From the data description above, the writer analyzed the score from the experiment and controlled class by integrating the results into the formula as follows: a. Determining Mean of Difference of Experiment Class: M D1 = D N = −215 20 = −10.75 b. Determining Mean of Difference of Controlled Class: M D 2 = D N = −150 20 = −7.5 c. Determining Deviation Standard of Difference of Experiment Class: SD D 1 = D 2 − D 2 N N = 3.125 − −215 20 20 = 156.25 − −2.31 = 156.25 + 2.31 = 158.56 = 12.6 d. Determining Deviation Standard of Difference of Controlled Class: SD D 2 = D 2 − D 2 N N 40 = 1.950 − −150 2 20 20 = 97.5 − −1.125 = 98.625 = 9.93 e. Determining Standard Error from Mean of Difference of Experiment Class: SE M D1 = SD D1 N − 1 = 12.6 20 − 1 = 12.6 19 = 12.6 4.3 = 2.9 f. Determining Standard Error from Mean of Difference of Controlled Class: SE M D2 = SD D2 N − 1 = 9.93 20 − 1 = 9.93 19 = 9.93 4.3 = 2.31 g. Determining t-observation t o : t o = M D1 – M D2 41 SE MD1 – SE MD2 = - 10.75 – -7.5 2.9 – 2.31 = -10.75 + 7.5 2.9 – 2.31 = -3.25 0.59 = -5.5 Note: Tanda – minus di sini bukanlah tanda aljabar; karena itu dengan t o sebesar −5.5 itu dapat kita baca: ada selisih derajat perbedaan sebesar 5.5. 35 h. Determining t-table t t in significance level 5 and 1 with degree of freedom df df = N1 + N2 – 2 = 20 + 20 – 2 = 40 – 2 = 38 See the table of t values of degree of freedom df = 38 at significance level 5 and 1 t table t t at significance level 5 = 2.02 t table t t at significance level 1 = 2.71 5 = t o t t = -5.5 2.02 1 = t o t t = -5.5 2.71

3. Test of Hypothesis