Statistical Analysis .1 Computation between the Two Means
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c. Relative frequency of the result of the writing post-tests both groups.
Chart 3 The Polygon of the Result of the Post-Test of Both Groups
This polygon presents the achievement of both group in doing the post- test. The percentage of the students who got grade A, B, C, D for the experimental
group were 10.00, 36.67, 50.00, 3.33. While the percentage of the students who got grade B, C, D, E for the
control group were 10.00, 63.33, 20.00 and 6.67.
4.2 Statistical Analysis 4.2.1 Computation between the Two Means
Means is the average value of the scores. In order to know the significant difference of the experiment could be seen through the difference of the two
means. The following formula was used to get the means:
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Me =
N Xe
∑
Mc =
N Xc
∑
Where, Me = the mean score of the experimental group
∑
Xe = the sum of all scores of the experimental group
Mc = the mean scores of the control group
∑
Xc = the sum of all scores of the control group
N = the number of the subject sample The computation of the scores of the experimental group and control group
was calculated as follows:
Me =
N Xe
∑
=
30 2439
= 81.3 The mean score of the experimental group was 81.3
Mc =
N Xc
∑
=
30 2196
= 73.2 The mean score of the control group was 73.2
If we compared the two means it was clear that the mean of the experimental group was higher than that of the control group. The difference
between the two means was 8.1.
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To make the analysis more reliable, t-test formula was used.
⎟ ⎠
⎞ ⎜
⎝ ⎛
+ ⎟
⎠ ⎞
⎜ ⎝
⎛ −
+ +
− =
Nc Ne
Nc Ne
SSc SSe
Mc Me
t 1
1 2
Where, t = t-test
Me = the mean difference of the experimental group Mc
= the mean difference of the control group SSe= Sum of quadrate deviation of the experimental group
SSc= Sum of quadrate deviation of the control group Ne = the number of the experimental group
Nc = number of the control group
Me =
= =
Σ 30
2439 Ne
Xe
81.30
SSe =
Ne Xe
Xe
2
Σ −
Σ
= 200299 -
30 2439
2
= 200299 -
30 5948721
= 200299 – 198290.7 = 2008.30
Mc =
20 .
73 30
2196 = =
Σ Nc
Xc
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SSc =
Nc Xc
Xc
2
Σ −
Σ
= 161962 -
30 2196
2
= 161962 -
30 4822416
= 161962 – 160747.2 = 1214.80
⎟ ⎠
⎞ ⎜
⎝ ⎛
+ ⎟
⎠ ⎞
⎜ ⎝
⎛ −
+ +
− =
30 1
30 1
2 30
30 80
. 1214
30 .
2008 20
. 73
30 .
81 t
= ⎟
⎠ ⎞
⎜ ⎝
⎛ ⎟
⎠ ⎞
⎜ ⎝
⎛ 30
2 58
10 .
3223 10
. 8
= 067
. 57
. 55
10 .
8
= 70
. 3
10 .
8
=
92 .
1 10
. 8
= 4.208 Ne + Nc – 2 = 30 + 30 – 2 = 58, the result t
00 .
2
58 965
.
=
After getting t-value, I consulted the critical value of the t-table to check whether the difference was significant or not. Based on the computation the t-
value was 4.208 and it is higher than the critical value on the table 4.2082.00.
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It is concluded that there was significant difference between teaching writing using pictures and teaching writing without pictures.