The value of df 60 at the degree of significance 5 or t
table
5 of df 60 = 2. 00
The value of df 60 at the degree of significance 1 or t
table
of df 1 the = 2. 65
3. Hypothesis of Data
The research to answer the question “Does jigsaw technique effect students’ reading achievement?” Moreover, the writer questioning “Do
the students who taught by using jigsaw technique show better achievement in their reading comprehension than the students who
taught by
using conventional
technique in
their reading
comprehension?” To get the answer from the question above, the writer proposes
Alternative Hypothesis Ha and Null Hypothesis Ha as follows: Ho : “There is no significant difference between students’ reading
comprehension achievement who taught by using jigsaw technique experiment class and those who taught without
using jigsaw technique”. Ha : “There is a significant difference between the students’
reading comprehension achievement who taught by using jigsaw technique experiment class and those who taught
without using jigsaw technique”. To prove the hypothesis, the data obtained from experiment class
and controlled class are calculated by using t-test formula with assumption as follows:
If t
o
≥ t
table
, the Null Hypothesis H
o
is rejected. It means there is a significant difference between the students’ reading comprehension
achievement who taught by using jigsaw technique and those who taught without using jigsaw technique”.
If t
o
≤ t
table,
the Null Hypothesis H
o
accepted. It means there is no significant difference between the students’ reading comprehension
achievement who taught by using jigsaw technique and those who taught without using jigsaw technique”.
From the result of the calculation, is obtained the value of t
o
= 3.332. The degree of freedom is 58. The writer uses the closer df of 58
mentioned in t
table
, that is 60. Then the writer uses the degree of significance 5 and 1. Based on both of significance, it can be seen
that on df = 60 in significant 5 and 1, the value of the degree significance are 2. 00 and 2. 65.
By comparing the result of t
t
and t
o
, that in significance 5, t
t
: t
o
= 3.332 2.00, and in significance1, t
t
: t
o
= 3.332 2. 65. According to those results, the writer get conclusion that H
o
is rejected and H
a
is accepted.
4. Interpretation of Data
