because this calculation functions as an answer for the research problem to know whether the inductive technique is effective or not on
students’ mastery of comparison degrees. Before the writer found the frequency distribution, she had
recapitulated the mean scores of both experimental and control classes.
Table 4.7 Brief Summary of Mean Scores Mean Scores
Experimental Control
Pre – Test
67.3 78.13
Post – Test
69.06 72.26
a. Experimental Class Frequency Distribution
Table 4.8 Frequency Distribution Experimental Class Pre-Test No. Interval Fi
Xi Xi
2
Fixi Fixi
2
Xi-X
2
fXi- X
2
1 52-58
8 55
3025 440
193600 151.29
1210.32 2
59-65 7
62 3844
434 188356
28.1 196.7
3 66-72
6 69
4761 414
171396 2.89
17.34 4
73-79 4
76 5776
304 92416
75.69 302.76
5 80-87
3 83.5 6972.25
250.5 62750.25
262.44 787.32
6 88-94
2 91
8281 182
33124 561.69
1123.38 Total
30 2024.5
3637.82
1 Mean X
M
x
=
�
= 2024,5 = 67.48 30
2 Variance
�
�
� = ∑� �� − x
∑�
= 3637.82 = 121.261 30
3 Standard Deviation S
S=
√
∑ �−
2
∑
�
=
√
.
= √
. = 11.01
Table 4.9 Frequency Distribution Experimental Class Post-Test No. Interval Fi
Xi Xi
2
Fixi Fixi
2
Xi-X
2
fXi- X
2
1 60-66
3 63
3969 189
35721 228.92
686.76 2
67-72 5
69.5 4830.25
347.5 120756.25
74.48 372.4
3 73-78
5 75.5
5700.25 377.5
142506.25 6.92
34.6 4
79-84 12
81.5 6642.25
978 956484
11.36 136.32
5 85-90
2 87.5
7656.25 175
30625 87.79
175.58 6
91-96 3
93.5 8742.25
280.5 78680.25
236.23 708.69
Total 30
2347.5 2114.35
4 Mean X
M
x
= = 2347,5 = 78.25
30
5 Variance
�
�
� = ∑� �� − x
∑�
= 2114.35 = 70.48 30
6 Standard Deviation S
S= √
∑ �−
2
∑
�
=
√
.
= √ . = 8.39
b. Control Class Frequency Distribution
Table 4.10 Frequency Distribution Control Class Pre-Test No. Interval Fi
Xi Xi
2
Fixi Fixi
2
Xi-X
2
fXi- X
2
1 52-57
3 54.5 2970.25
163.5 26732.25
211.99 635.97
2 58-63
6 60.5 3660.25
363 131769
73.27 439.62
3 64-69
6 66.5 4422.25
399 159201
6.55 39.3
4 70-75
4 72.5 5256.25
290 84100
11.83 47.32
5 76-82
10 79
6241 790
624100 98.8
988 6
83-88 1
85.5 7310.25 85.5
7310.25 270.27
270.27 Total
30 2091
2420.48
1 Mean X
M
x
= = 2091 = 69.7
30
2 Variance
�
�
� = ∑� �� − x
∑� = 2420.48 = 80.683
30
3 Standard Deviation S
S= √
∑ �−
2
∑
�
=
√
.
= √ .
= 8.98
Table 4.11 Frequency Distribution Control Class Post-Test No. Interval Fi
Xi Xi
2
Fixi Fixi
2
Xi-X
2
fXi- X
2
1 56-61
7 58.5
3422.25 409.5
167690.3 189.34
1325.4 2
62-67 64.5
4160.25 60.22
3 68-73
8 70.5
4970.25 564
318096 3.09
24.72 4
74-79 6
76.5 5852.25
459 210681
17.98 107.88
5 80-85
7 82.5
6806.25 577.5
333506,3 104.86
734.02 6
86-91 2
88.5 7832.25
177 31329
263.74 527.5
Total 30
2187 2719.52
4 Mean X
M
x
= = 2187 = 72.9
30
5 Variance
�
�
� = ∑� �� − x
∑� = 2719.52 = 90.65
30
6 Standard Deviation S
S= √
∑ �−
2
∑
�
=
√
.
= √ . = 9.52
From the calculation above, here is the table of brief summary from the
calculation above. Table 4.12 Brief Summary of the Calculation of Post-Test Data
Variable Total of
Students n
Mean X Variance
S
2
Standar Deviation S
Experimental Class
30 78.25
70.48 8.39
Control Class
30 72.9
90.65 9.52
After the writer found the calculation of mean, variance, and standar deviation score, she found the score of t-test which has the significance
α= 0.05. Here is the description of t-test calculation:
Descriptions: �
= the price of t value �̅
= average score of experimental class �̅
= average score of control class �
= variant data of experimental class �
= variant data of control class � = standard deviation of both classes
� = the total students of experimental class
� = the total students of control class
� = √
� − � + � − � � + � −
= √ −
. + −
. +
−
= √ . +
.
= √ .
= √ .
= 8.98 � =
�̅ − �̅ � √� + �
� = . − .
. √ +
� = .
. √
� = .
. √ .
� = .
. . .
� = .
.
� = .
Based on calculating above, the result of t-test from experimental and controlled classes is 2.47.
c. Determining the t-test significance level α= 0.05 by calculating the degree of freedom
