From the description above, the writer get the students’ pretest score of controlled class with minimum score is 30 and maximum
score 65, mean is 49, median is 49.2, mode is 49. As described it below:
Table 3.5
The table of frequency distribution of pre-test controlled class No Interval
f x
fx x2
fx2 Limit
1 30-35
2 32,5 65
1056,25 2112,5
29,5-35,5 2
36-41 2 38,5
77 1482,25
2964,5 35,5-41,5
3 42-47
6 44,5 267
1980,25 11881,5 41,5-47,5
4 48-53
7 50,5 353,5 2550,25 17851,75 47,5-53,5 5
54-59 4 56,5
226 3192,25
12769 53,5-59,5
6 60-65
3 62,5 187,5 3906,25 11718,75 59,5-65,5 24
1176 14167,5 59298
After they are taught by using conventional method, the writer get the students’ postest score of controlled class with minimum score
is 50 and maximum score 75, mean is 60.66, median is 59.5, and mode is 58.8. As described it follows:
Table 3.6
The table of frequency distribution of post-test controlled class No Interval
f x
fx x2
fx2 Limit
1 50-53
2 51,5 103
2652,25 5304,5
49,5-53,5 2
54-57 6 55,5
333 3080,25
18481,5 53,5-57,5
3 58-61
8 59,5 476
3540,25 28322
57,5-61,5 4
62-65 4 63,5
254 4032,25
16129 61,5-65,5
5 66-69
0 67,5 4556,25
65,5-69,5 6
70-73 3 71,5 214,5 5112,25
15336,75 69,5-73,5 7
74-77 1 75,5 75,5
5700,25 5700,25
73,5-77,5 24
1456 28673,75 89274
B. Data Analysis
As stated above that the writer uses “t-test” in analyzing the data by this formula:
. 2
2 .
1 2
1 2
1 2
2 2
1 2
1
N N
N N
N N
x x
M M
t
o
+ −
+ +
− =
Table 3.7 The result calculation of test in both experimental and controlled class
STUDENT X
1
X
2 1
2 1
2 2
2
1 70
55 3,75
5,625 14,0625 31,640625
2 75
55 -1,25
5,625 1,5625
31,640625 3
70 60
3,75 0,625
14,0625 0,390625
4 75
55 -1,25
5,625 1,5625
31,640625 5
55 60
18,75 0,625
351,5625 0,390625 6
80 50
-6,25 10,625
39,0625 112,89063 7
70 60
3,75 0,625
14,0625 0,390625
8 80
60 -6,25
0,625 39,0625
0,390625 9
75 60
-1,25 0,625
1,5625 0,390625
10 65
50 8,75
10,625 76,5625 112,89063
11 70
65 3,75
-4,375 14,0625 19,140625
12 75
70 -1,25
-9,375 1,5625
87,890625 13
75 55
-1,25 5,625
1,5625 31,640625
14 60
65 13,75
-4,375 189,0625 19,140625 15
80 65
-6,25 -4,375
39,0625 19,140625 16
80 55
-6,25 5,625
39,0625 31,640625 17
65 75
8,75 -14,375 76,5625 206,64063
18 75
60 -1,25
0,625 1,5625
0,390625 19
85 65
-11,25 -4,375 126,5625 19,140625
20 70
60 3,75
0,625 14,0625
0,390625 21
85 70
-11,25 -9,375 126,5625 87,890625
22 75
55 -1,25
5,625 1,5625
31,640625 23
70 70
3,75 -9,375
14,0625 87,890625 24
90 60
-16,25 0,625
264,0625 0,390625
1770 1455
1462,5 965,625
74 24
1776 .
1 1
1
= =
= N
X f
M 66
. 60
24 1456
.
2 2
2
= =
= N
X f
M
. 2
2 .
1 2
1 2
1 2
2 2
1 2
1
N N
N N
N N
x x
M M
t
o
+ −
+ +
− =
24 .
24 24
24 .
2 24
24 625
. 965
5 .
1462 66
. 60
74 +
− +
+ −
=
576 48
. 46
125 .
2428 34
. 13
=
083 .
78 .
52 34
. 13
x =
092 .
2 34
. 13
38 .
4 34
. 13
= =
37 .
6 =
o
t
C. Test of Hypothesis
The writer states the hypothesis as follows: 1 Ha: “There is a significant difference in students’ score of reading comprehension with
or without using jigsaw technique”, 2 Ho: “There is no significant difference in students’ score of reading comprehension with or without
using jigsaw technique”. According to Anas Sudijono, if the result of calculation of t
o
T observation is higher than t
t
T table; so the null hypothesis Ho is rejected, and if the result of calculation of t
o
T observation is lower than t
t
T table, so the alternative hypothesis Ha is accepted.
31
As stated above, the result of statistic calculation indicates that the value of t
o
is 6.37 and the value of df 24 on degree of significance of 5 and 1 are 2.069 and 2.807. Comparing the calculation of t
o
with each values of degree of significance, the result is 2.069 6.37 2.807.
Since t
o
which is obtained from the result of calculating is higher than t
t
,
31
Prof.DrsAnas Sudijono. Pengantar Statistik Pendidikan. Jakarta: PT.Raja Grafindo Persada, 2005, p.328
so the alternative hypothesis Ha is accepted and the null hypothesis Ho is rejected.
D. Data Interpretation
