30 16
8 -11.7
-6.2 136.89
38.44 Total
∑X=831 ∑Y=426 ∑x=0 ∑y=0
∑x
2
=4448.3 ∑y
2
=784.8 It shows that the score of both classes are there a difference in minimum
and maximum standard of each class. The table shows that the experimental class has higher score.
2. Data Analysis
In analyzing the data, the writer uses the comparative technique. The writer compares the gain score of both experimental class and controlled class.
This technique is useful to prove statically whether there is any effectiveness of the two variables, in this case, between using pictures and without using
pictures in reinforcing degrees of comparison of adjective. To find out the difference of students‟ score in using pictures in reinforcing
degrees of comparison of adjective will be compared to the students‟ score that
without using pictures in reinforcing degrees of comparison of adjective. The writer calculated them based on the steps of the t
– test formula, as follow: 1. Determining Mean I with formula:
7 ,
27 30
831 N
X I
Variable Mean
2. Determining Mean II with formula:
2 ,
14 30
426 N
Y II
Variable Mean
3. Determining of Standard Deviation of variable I: 18
. 12
17690 .
12 377
. 148
30 4448.3
N X
SD
2 X
4. Determining standard of error mean of variable II: 115
. 5
1146847 .
5 16
. 26
30 8
. 784
SD
2 y
N Y
5. Determining standard of error mean of variable I:
262 .
2 26183
. 2
385 .
5 18
. 12
29 18
. 12
1 30
18 .
12 1
- N
SD SEM
X X
6. Determining standard of error mean of variable II:
950 .
949861 .
385 .
5 115
. 5
29 115
. 5
1 -
30 115
. 5
1 -
N SD
SEM
X Y
7. Determining standard of error mean difference of M
x
and M
y
:
2 2
2 Y
2 X
Y X
950 .
262 .
2 SEM
SEM M
SEM
9025 .
116644 .
5
019144
. 6
453394 .
2
453 .
2
8. Determining t
o
with formula:
503 .
5 503465
. 5
453 .
2 5
. 13
453 .
2 2
. 14
7 ,
27 M
SEM M
M o
Y X
Y X
t
9. Determining t – table in significance level 5 and 1 with df:
df = N 2 + N 1 – 2 = 30 + 30 – 2 = 60 – 2 = 58
df = 58 see the table of t-scores at the degree of significance of 5 and 1. The writer gained t-table:
t-table t
t
at significance level of 5 = 2,03 t-table t
t
at significance level of 1 = 1.67
10. The comparison between t – score with t – table:
SL 5 = t
o
t
t
= 5.503 2.00 SL 1 = t
o
t
t
= 5.503 1.67
Form the comparison above, it shows that t-score or t-observation t
o
is
higher on the degree of significance of 5 than t-table t
t
. It shows that the treatment given using pictures is accepted.
3. Data Interpretation
