Lesson Planning Validity and Reliability of the Test

2 Analyzing the normality distribution with Kolmogorov-Smirnov test. Comparing the asymp.sig 0.05, the null hypothesis is not rejected and alternative hypothesis is rejected, and the distribution of data is normal. Hence, if the asymp.sig 0.05, the null hypothesis is rejected and alternative hypothesis is not rejected, and it means the data is not normally distributed. Table 3.1 Normality test of pre-test Kolmogorov-Smirnov Shapiro-Wilk Statistic Df Sig. Statistic Df Sig. EXPERIMENTAL GROUP CONTROL GROUP .148 .119 24 24 .162 .200 .949 .949 24 24 .239 .234 The normality test above used Kolmogorov-Smirnov critical points table in determining the ttable. According to the consideration of n=45 and significance level 0.05, the ttable of this normality test was 0.198. Furthermore, table showed that the t value of experimental class was 0.115 and the tvalue of controlled. Class was 0.128. The values were smaller than ttable, in other words, tvalue ttable 0.1150.198 for experimental class, and 0.1280.198 for controlled class. Therefore, Ho was accepted and Hi was rejected. It meant that the distribution of the data was normal. Also, the significance of experimental class was 0.184 and the significance of controlled class was 0.070. The result shown that the significance of both classes are above 0.05. Therefore, the distribution of posttest score was normal. Table 3.2 Normality test of post-test Kolmogorov-Smirnov Shapiro-Wilk Statistic Df Sig. Statistic df Sig. VIIIA VIIIB .159 .149 24 24 .102 .160 .936 .936 24 24 .122 .117 The normality test in table used Kolmogorov-Smirnov critical points table in determining the ttable. According to the consideration of n=45 and significance level 0.05, the ttable of this normality test was 0.198. Furthermore, table 4.3 shown that the tvalue of experimental class was 0.102 and the tvalue of controlled class was 0.083. The values were smaller than ttable, in other words, tvalue ttable 0.1020.198 for experimental class, and 0.0830.198 for controlled class. Therefore, Ho was accepted and Hi was rejected. It meant that the distribution of the data was normal. Also, the significance of experimental class was 0.200 and the significance of controlled class was 0.200. The result show that the significance of both classes was above 0.05. Therefore, the distribution of post-test score was normal.

b. Homogeneity of Variance

The homogeneity of variance test used Levene test in SPSS program. The steps are follows: 1 Setting the level of significance p at 0.05 and establishing the alternative hypothesis as follows: H : the variances of the experimental and the controlled group are homogenous H 1 : the variances of the experimental and the controlled group are not homogenous 2 Analyzing the homogeneity of variance by using Levene test in SPSS. Comparing the asymp.sig with the level of significance to test the hypothesis. If the asymp.sig 0.05, the null hypothesis is not rejected and alternative hypothesis is rejected. It suggests that the variances of data are homogenous. However, it the asymp.sig ≤ 0.05, the null hypothesis is rejected and alternative hypothesis is not rejected. It clarifies that the variances are significantly different. Table 3.3 Homogeneity test of pre-test Levene Statistic df1 df2 Sig.