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Table 4.8

Distribution of frequency post-test control class

X F FX

46 1 46

55 1 55

60 1 60

62 1 62

65 2 130

70 3 210

74 2 148

75 3 225

76 4 304

78 2 156

20 1396

The formula of mean : X =

=

= 69,8

3 SD 10.11 8.01

4 0.09 0.11

5 0.19 0.19

We can seen based on table 4.9 shows the result of Locount

experiment class pre-test and post-test was (0,09 – 0,11) while the results are Locount with n = 20 and the level of significance ɑ = 0,05 = 0,19 of second Locount smaller than Lotable then the value Locountboth pre-test and post-test samples showed a smaller number than the Locount . Based on predetermined criteria, if the value Locount<

Lotablethan the sample can be said to be normality test.

b. Control class

Down below you will see the normality test result of pre-test and also post-test in control class.

Table 4.10

The results of the control class normality test No Statistic Pre-test Post-test

1 N 20 20

2 X 56.05 69.8

3 SD 11.62 8.35

4 0.11 0.14

5 0.19 0.19

The table above 4.10 shows us the result of Locount experiment class pre-test and post-test was (0,11 – 0,14) while the results are Locount with n = 20 and the level of significance ɑ = 0,05 = 0,19 of second Locount smaller than Lotable then the value Locountboth pre-test and post-test samples showed a smaller number than the Locount . Based on predetermined criteria, if the value Locount< Lotablethan the sample can be said to be normality test.

Loc nt Lot bel

Loc nt Lot bel

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2. Homogeneity test

a. pre-test Experimental and control Class

Down below you will see the result of homogeneity testfor pre-test experimental class and also control class.

Table 4.11

Homogeneous results pre-test experiment and control No Statistic pre-test

1 S2 135.1

2 S2 102.2

3 Fc 1.32

4 Fl 2.17

Description

S2 = Variance of data pre-test and post-test Fcount = Value count

Ftable = Value the table based on the calculated value

Based on the table above can be analyzed if the result of the samples class above that have homogeneous variance. From the results homogeneous test, and calculations get if Fcount1,32 And testing criteria has been determined that Fcount<Ftable then the second sample can be said homogeneous. Homogeneous test calculation process can be seen in appendix 5.

b. Post-test experimental and control class

Down below you will see the result of homogeneity testfor post-test experimental class and also control class.

Table 4.12

Homogeneous results post-test experimental and control class No Statistic P0st-test

1 S2 73.33

2 S2 64.17

3 Fc 1.14

4 Fl 2.17

Description

S2 = Variance of data pre-test and post-test Fcount = Value count

Ftable = Value the table based on the calculated value

Based on the table above can be concluded if the result of the two group of samples class have homogeneous variance. From the results homogeneous test calculations get if Fcount1,14. Based on testing criteria has been determined that Fcount<Ftable then the second sample can be said to be homogeneous. Homogeneous test calculation process can be seen in appendix 5.

3. Statistical hypothesis test a. Paired Sample t-test

To found a differences score between experimental group on after and before students given some of treatment therefore the researcher used paired sample t-test

a) HIf t-test (t0) ≥ t-table (t1) in the significance degree of 0,05, H0 (Null Hypothesis) is rejected

b) If t-test (t0) ≤ t-table (t1) in the significance degree of 0,05, H0 (Null Hypothesis) is accepted

Where :

a) Ha2 = There is a difference between the score of the student who are taught by using kahoot games application. and who are not taught by using kahoot games application.

b) Ho2 = There is no a difference between the score of the student who are taught by using kahoot games application. and who are not taught by using kahoot games application.

The researchers has conducted a calculation, and got the result of paired sample t-test and that result showed if H0 was rejected because the t-test score (t0) higher than t-table (tt) the score is4.1 ≥ 2,08 it means a significant difference between the pre-test and post-test in experimental group, the null hypothesis was rejected because there is a significant difference between the pre-test and post-test scores for experimental group. In other word, the treatment is success.

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b. Independent Sample t-test

The independent sample t-test was used for this research actually to test hypothesis that is was a significant effect for students‟

reading comprehension before and after used kahoot games.

The null hypothesis announces that there is no significant difference between the mean of experimental and control group‟s scores. Moreover, the alternative hypothesis reveals the means of score between the two groups that are significantly different.

The criteria below was used by researcher to analyze the test hypothesis:

a) If t-test (t0) ≥ t-table (t1) in the significance degree of 0.05, H0 (null hypothesis) is rejected

b) If t-test (t0) ≤ t-table (t1) in the significance degree of 0.05, H0 (null hypothesis) is accepted

Where :

a) Ha1:There is a difference between the score of the student who are taught by using kahoot games application and who are not taught by using kahoot games application.

b) Ho1 : There is no a difference between the score of the student who are taught by using kahoot games application and who are not taught by using kahoot games application.

Based on researchers‟ calculation, the result of paired sample t-test showed that H0 was rejected because the t-test (t0) was higher than t-table (tt) or 2,2 ≥ 2,09 it means that there is a significant difference between the pre-test and post-test in experimental group. Thus, the null hypothesis was rejected because there is a significant difference between the pre-test and post-test scores for experimental group. In other word, the treatment worked very well.

c. Conclusion

To found the conclusion from this research the researcher used some of formula, below :

Effect size

Based on the data, can be known :

20 20 20

The result of d is 0,66, so it show that the high effect size because d ≥ 0 and d ≥ 1,0 with the percentage 84%. It means that the effect of using Kahoot Games give high effect to use the class on students‟ reading comprehension

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