Range Number of Classes Class length Average Median Me Modus Mo Standard Deviation S

Appendix 2 Pre-test Results of Experimental Class Pre-test results of Experimental Class is as follows : 30 35 30 25 30 25 25 40 25 30 35 25 30 20 20 25 25 30 30 30 35 40 35 25 35 40 20 25 35 20 35 35 Based on tha data above that the maximum score X max is 50 and score minimum X min is 30. So that it can be made of a frequency distribution table after first determining the value range R, the number of classes K, and length class P.Three values obtained based on the following calculation.

a. Range

R 20 20 40 min      X X R mx

b. Number of Classes

K 6 97 , 5 97 , 4 1 50 , 1 3 , 3 1 32 log 3 , 3 1 log 3 , 3 1            n K So the number of classes is 6

c. Class length

P 4 33 , 3 6 20     K R P So that the length class is 4. . Distribution table is as follows . Class Class Limit Middle Class Values x i Frekuensi f i f i . x i f i . x i 2 20 - 23 19.5 21.5 4 86 1849 24 - 27 23.5 25.5 9 229.5 5852.25 28 - 31 27.5 29.5 8 236 6962 32 - 35 31.5 33.5 8 268 8978 36 - 39 35.5 37.5 40 - 43 39.5 41.5 3 124.5 5166.75 Jumlah ∑ 177 189 32 944 28808 Based on the frequency Distribution Table, it can be determined average value X , median Me, mode Mo, and standard deviation S value of this pre- test. Here is a calculation to determine those values.

a. Average

X 5 , 29 32 944       i i i f x f X

b. Median Me

The median value is determined by the following statistical formula.                f F n P b Me 2 1 Where: b = lower limit of the class median = 23.5 P = length class = 4 n = number of data = 32 F = value of cumulative frequency before the median class = 4 f = frequency of median class value = 9 Based on these data, the median value can be determined from the results of this pre-test are as follows.   83 , 28 33 , 5 5 , 23 33 , 1 4 5 , 23 9 4 32 . 2 1 4 5 , 23                      Me

c. Modus Mo

Mode value is determined using the following statistical formula.          2 1 1 b b b P b Mo Where : b = lower limit of the class median = 23,5 P = length class = 4 b 1 = class frequency mode frequency is reduced previous class = 9 – 4 = 5 b 2 = class frequency mode frequency is reduced subsequent class = 9 – 8 = 1 Based on these data, it can be determined from the pre-test value of this mode is as follows.   83 , 26 33 , 3 5 , 23 83 , 4 5 , 23 1 5 5 4 5 , 23                Mo

d. Standard Deviation S

Standard deviation value determined with the following statistical formula.     56 , 5 97 , 30 31 960 31 27848 28808 31 32 891136 28808 1 32 32 944 28808 1 . . 2 2 2                  i i i i i i f f x f x f S i Appendix 3 Pre-test Results of Control Class Pre-test results of Control Class is as follows : 30 40 25 20 40 15 30 25 20 30 25 25 20 30 25 35 30 30 30 15 20 25 25 40 15 30 20 30 30 40 20 30 From there it is obtained that the maximum val X max is 40 and minimum value X min is 15. So that it can be made of a frequency distribution table after first determining the value range R, the number of classes K, and length class P. Three values obtained based on the following calculation.

a. Range R

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