The effectiveness of K-W-L strategy toward reading comprehension for the seventh graders at SMP al-Amin Palangka Raya - Digital Library IAIN Palangka Raya
44
CHAPTER IV
RESULT OF THE STUDY
This chapter covers Description of the data, test of normality and
homogeneity, result of the data analyses and discussion.
A. Description of The Data
This section described the obtained data of the effectiveness of using K-WL Strategy in teaching reading Descriptive text. The presented data consisted of
Mean, Median, Modus, Standard Deviation and Standard Error.
1. The descriptiondata of Pre-Test Score
The students’ pre test score are distributed in the following table in order
toanalyze the students’ knowledge before conducting the treatment.
Table 4.1Pre test score of experimental and control group
Experimental Group
Code
Score
E-01
E-02
E-03
E-04
E-05
E-06
E-07
E-08
E-09
E-10
E-11
E-12
E-13
40
43.3
66.7
70
56.7
40
53.3
50
56.7
53.3
60
50
53.3
CORRECT
ANSWER
12
13
20
21
17
12
16
15
17
16
18
15
16
Control Group
PREDICATE
CODE
SCORE
FAIL
FAIL
ENOUGH
GOOD
LESS
FAIL
LESS
LESS
LESS
LESS
ENOUGH
LESS
LESS
C-01
C-02
C-03
C-04
C-05
C-06
C-07
C-08
C-09
C-10
C-11
C-12
C-13
46.7
43.3
40
40
40
40
60
46.7
66.7
70
56.7
70
53.3
44
CORRECT
ANSWER
14
13
12
12
12
12
18
14
20
21
17
21
16
PREDICATE
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
ENOUGH
FAIL
ENOUGH
GOOD
LESS
GOOD
LESS
45
E-14
E-15
E-16
E-17
E-18
E-19
E-20
E-21
E-22
53.3
50
43.3
66.7
50
40
43.3
40
50
16
15
13
20
15
12
13
12
15
TOTAL
AVERAGE
Lowest Score
Highest Score
LESS
LESS
FAIL
ENOUGH
LESS
FAIL
FAIL
FAIL
LESS
1129,9
51,4
40
70
C-14
C-15
C-16
C-17
C-18
C19
C-20
C-21
C-22
C-23
C-24
40
53.3
50
53.3
40
60
43.3
46.7
40
40
53.3
12
16
15
16
12
18
13
14
12
12
16
TOTAL
AVERAGE
Lowest Score
Highest Score
FAIL
LESS
LESS
LESS
FAIL
ENOUGH
FAIL
FAIL
FAIL
FAIL
LESS
1193,3
49,7
40
70
The table above showed us the comparison of pre-test score achieved by
experimental and control group students, both class’ achievement are at the same
level. It can be seen that from the students’ score. The highest score 70 and the
lowest score 40, both experimental and control group. It meant that the
experimental and control group have the same level in reading comprehension
before getting the treatment.
a. The Result of Pretest Score of ExperimentalGroup (VII-A)
Based on the data above, it was known the highest score was 70 and the
lowest score was 40. To determine the range of score, the class interval, and
interval of temporary,the writer calculated using formula as follows:
The Highest Score (H)
= 70
The Lowest Score (L)
= 40
46
The Range of Score (R)
=H–L+1
= 70 – 40+ 1
= 31
The Class Interval (K)
= 1 + (3.3) x Log n
= 1 + (3.3) x Log 22
= 1 + 4.4299948
= 5.4299948
=5
Interval of Temporary (I) =
𝑅
𝐾
=
31
5
= 6,2 = 6 or 2
So, the range of score was 31, the class interval was 5, and interval of
temporary was 6 or 2. Then, it was presented using frequency distribution in
the following table:
Table 4.2 Frequency Distribution of the Pretest Score
Interval (I)
Frequency
(F)
Mid
Point
(x)
The
Limitation
of each
group
Frequency
Relative (%)
Frequency
Cumulative
(%)
1
2
65-70
59-64
3
1
67.5
61.5
64.5-70.5
58.5-64.5
13.6363
4.5454
100
86.3636
3
53-58
6
52.7
52.5-58.5
27.2727
81.8182
4
47-52
5
55.5
46.5-52.5
22.7272
54.5455
5
40-46
7
43
39.5-46.5
31.8181
∑P = 100
31.8183
Class
(K)
∑F=22
47
The distribution of students’ predicate in pretest score of Experimental
group can also be seen in the following figure.
The Distribution of Student's Predicate
12
10
8
6
4
2
Figure 4.1 The distribution frequency of students’ pretest score for
Experimental Group
0
Fail
Less
Enough
Good
Figure 4.1 The distribution of students’ predicate in pretest score
for Experimental Group
Based on the figure above, it can be seen about the students’ predicate in
pretest score. There were seven students who got Fail predicate. There were
elevenstudents who gotLess Predicate. There were three students who gotEnough
predicate. There was one student who got Good predicate.
The next step, the writer tabulated the scores into the table for the
calculation of mean, standard deviation, and standard error as follows:
48
Table 4.3the Table for Calculating Mean, median, modus, Standard
deviation. and standard error of Pretest Score.
Frequency Mid
Point
(F)
(x)
Class
(K)
Interval
(I)
1
65-70
3
2
59-64
3
F.X
FX2
Fka
67.5
202.5
41006.25
3
1
61.5
61.5
3782.25
4
53-58
6
52.7
316.2
99982.44
10
4
47-52
5
55.5
277.5
77006.25
15
5
40-46
∑ Total
7
∑F = 22
43
301
90601
22
∑=1158.7 ∑=312378.19
1) Calculating Mean
Mx=
∑𝐹𝑋 𝑖
𝑛
=
1158 .7
22
=52.67
2) Median
Mdn
1
𝑁−𝑓𝑘𝑏
2
=ℓ+
𝑓𝑖
1
= 39.5 + 2
𝑋𝑖
22−7
7
4
= 39.5 + X 6
7
𝑋6
= 45.93
3) Modus
Mo
=u+
𝑓𝑎
𝑓𝑎 +𝑓𝑏
= 39.5 +
2
2+7
= 39.5 + 1.11
𝑥𝑖
𝑥5
Fkb
22
19
18
12
7
49
= 40.61
4) Standard Deviation
S=
S=
𝑛.∑𝐹𝑋 𝑖 2 − ∑𝐹𝑋 𝑖 2
𝑛 𝑛 −1
22.1158 .7− 312378 .19 2
22 22−1
S = 6.29
5) Standard Error
SEmd =
𝑆
𝑁−1
=
6.29
22−1
=
6.29
4.58
= 1.37
After Calculating, it was found that the standard deviation and the
standard error of pretest score were 6.29 and 1.37
b. The Result of Pretest Score of ControlGroup (VII-B)
Based on the data pretest score of control group, it was known the
highest score was 70 and the lowest score was 40. To determine the range of
score, the class interval, and interval of temporary,the writer calculated using
formula as follows:
The Highest Score (H)
= 70
The Lowest Score (L)
= 40
The Range of Score (R)
=H–L+1
= 70 – 40+ 1
= 31
The Class Interval (K)
= 1 + (3.3) x Log n
= 1 + (3.3) x Log 24
= 1 + 4.554697
= 5.554697
=6
50
Interval of Temporary (I) =
𝑅
𝐾
=
31
6
= 5.1 = 5
So, the range of score was 31, the class interval was 6, and interval of
temporary was 5. Then, it was presented using frequency distribution in the
following table:
Table 4.4 Frequency Distribution of the Pretest Score
Interval (I)
Frequency
(F)
Mid
Point
(x)
The
Limitation
of each
group
Frequency
Relative (%)
Frequency
Cumulative
(%)
1
2
66-70
61-65
3
0
68
63
65.5-70.5
60.5-65.5
12.5
0
100
87.5
3
4
56-60
51-55
3
4
58
53
55.5-60.5
50.5-55.5
12.5
16.6667
87.5
75
5
46-50
4
48
45.5-50.5
16.6667
58.3333
6
40-45
10
42.5
39.5-45.5
41.6666
∑P = 100
41.6666
Class
(K)
∑F=24
The distribution of students’ predicate in pretest score of Control group
The distribution of student's predicate
14
12
10
8
6
4
2
0
Fail
Less
Enough
Good
Figure 4.2 The distribution of students’ predicate in pretest score
of Control Group
51
Based on the figure above, it can be seen about the students’ predicate in
pretest score. There thirteenstudents who got Fail predicate. There were
sixstudents who gotLess Predicate. There were three students who got Enough
predicate. There weretwo student who got Good predicate.
The next step, the writer tabulated the scores into the table for the
calculation of mean, median, modus, standard deviation, and standard error as
follows:
Table 4.5the Table for Calculating Mean, median, modus, Standard
deviation. and standard error of Pretest Score of Control group.
Frequency Mid
Point
(F)
(x)
Class
(K)
Interval
(I)
1
66-70
3
2
61-65
3
F.X
FX2
Fka
Fkb
68
204
41616
3
24
0
63
0
0
3
21
56-60
3
58
174
30276
6
21
4
51-55
4
53
212
44944
10
18
5
46-50
4
48
192
36864
14
14
6
40-45
∑ Total
10
∑F = 24
42.5
425
180625
24
10
∑=1207
1) Calculating Mean
Mx=
2) Median
Mdn
∑𝐹𝑋 𝑖
𝑛
=
1207
24
=50.29
1
𝑁−𝑓𝑘𝑏
2
=ℓ+
𝑓𝑖
1
= 39.5 + 2
𝑋𝑖
24−10
10
𝑋5
∑=334325
52
= 39.5 +
7
10
X5
= 43
3) Modus
Mo
𝑓𝑎
=u+
𝑓𝑎 +𝑓𝑏
= 39.5 +
𝑥𝑖
6
6+10
𝑥5
= 39.5 + 1.87
= 41.37
4) Standard Deviation
S=
𝑛.∑𝐹𝑋 𝑖 2 − ∑𝐹𝑋 𝑖 2
S=
24.1207 − 334325 2
𝑛 𝑛 −1
24 24−1
S = 5.38
5) Standard Error
SEmd =
𝑆
𝑁−1
=
5.38
24−1
=
5.38
4.79
= 1.18
After Calculating, it was found that the standard deviation and the
standard error of pretest score were 5.38 and 1.18
53
2.
The description data of Post-Test Score
The students’ score are distributed in the following table in order toanalyze
the students’ knowledge after conducting the treatment.
Table 4.6Post-test score of Experimental and Control Group
Experimental Group
Code
Score
E-01
E-02
E-03
E-04
E-05
E-06
E-07
E-08
E-09
E-10
E-11
E-12
E-13
E-14
E-15
E-16
E-17
E-18
E-19
E-20
E-21
E-22
66.7
70
80
86.7
76.7
56.7
60
76.7
73.3
80
70
66.7
80
83.3
76,7
70
73.3
60
63.3
73.3
70
76.7
TOTAL
AVERAGE
Lowest Score
Highest Score
CORRECT
ANSWER
20
21
24
26
23
17
18
23
22
24
21
20
24
25
23
21
22
18
19
22
21
23
PREDICATE
ENOUGH
GOOD
GOOD
EXCELLENT
GOOD
LESS
ENOUGH
GOOD
GOOD
GOOD
GOOD
ENOUGH
GOOD
GOOD
GOOD
GOOD
GOOD
ENOUGH
ENOUGH
GOOD
GOOD
GOOD
1590.1
72.3
56.7
86.7
Control Group
CODE
SCORE
C-01
C-02
C-03
C-04
C-05
C-06
C-07
C-08
C-09
C-10
C-11
C-12
C-13
C-14
C-15
C-16
C-17
C-18
C19
C-20
C-21
C-22
C-23
C-24
56.7
60
66.7
60
60
63.3
60
70
66.7
76,7
63.3
63.3
73.3
70
73.3
70
63.3
60
73.3
60
70
73.3
60
53.3
TOTAL
AVERAGE
Lowest Score
Highest Score
CORRECT
ANSWER
17
18
20
18
18
19
18
21
20
23
19
19
22
21
22
21
19
18
22
18
21
22
18
16
PREDICATE
LESS
ENOUGH
ENOUGH
ENOUGH
ENOUGH
ENOUGH
ENOUGH
GOOD
ENOUGH
GOOD
ENOUGH
ENOUGH
GOOD
GOOD
GOOD
GOOD
ENOUGH
ENOUGH
GOOD
ENOUGH
GOOD
GOOD
ENOUGH
LESS
1566.5
65.3
53.3
76.7
54
The table above showed us the comparison of post-test score achieved by
experimental and control group students. Both class’ achievement have different
score. It can be seen from the highest score 86.7 and 76.7 and the lowest score
56.7 and 53.3. It meant that the experimental and control group have the different
level in reading comprehension after getting the treatment.
a. The Result of Post-test Score of Experimental Group (VII-A)
Based on the data Post-test score of Experimental group, it was known the
highest score was 86.7 and the lowest score was 56.7. To determine the range of
score, the class interval, and interval of temporary, the writer calculated using
formula as follows:
The Highest Score (H)
= 86.7
The Lowest Score (L)
= 56.7
The Range of Score (R)
=H–L+1
= 86.7 – 56.7 + 1
= 31
The Class Interval (K)
= 1 + (3.3) x Log n
= 1 + (3.3) x Log 22
= 1 + 4.4299948
= 5.4299948
=5
Interval of Temporary (I) =
𝑅
𝐾
=
31
5
= 6,2 = 6
55
So, the range of score was 31, the class interval was 5, and interval of
temporary was 6. Then, it was presented using frequency distribution in the
following table:
Table 4.7 Frequency Distribution of the Post-test Score
Interval (I)
Frequency
(F)
Mid
Point
(x)
The
Limitation
of each
group
Frequency
Relative (%)
Frequency
Cumulative
(%)
1
2
84.7-90.7
77.7-83.7
1
4
87.7
80.7
83.2-91.2
73.2-83.2
4.5454
18.1818
100
95.4545
3
4
70.7-76.7
63.7-69.7
11
3
73.7
66.7
69.2-73.2
62.2-69.2
50
13.6364
45.4546
31.8182
5
56.7-62.7
3
59.7
55.2-62.2
13.6364
∑P = 100
18.1818
Class
(K)
∑F=22
The distribution of students’ predicate in post-test score of Experimental
group.
The distribution of student's predicate pos-test
16
14
12
10
8
6
4
2
0
Fail
Less
Enough
Good
Exellent
Figure 4.3 The distribution of students’ predicate in post-test score
of Experimental Group
56
Based on the figure above, it can be seen about the students’ predicate in
pretest score. There was no student who got Fail predicate. There wasone
studentswho got Less predicate. There were fivestudents who gotEnough
Predicate. There were fifteen students who got Good predicate. There was one
student who got Excellent predicate.
The next step, the writer tabulated the scores into the table for the
calculation of mean,median, modus, standard deviation, and standard error as
follows:
Table 4.8the Table for Calculating Mean, median, modus, Standard
deviation. and standard error of Post- test Score.
Frequency Mid
Point
(F)
(x)
Class
(K)
Interval (I)
1
84.7-90.7
1
2
77.7-83.7
3
F.X
FX2
Fka
Fkb
87.7
87.7
7691.29
1
22
4
80.7
322.8
104199.84
5
23
70.7-76.7
11
73.7
810.7
657234.49
16
17
4
63.7-69.7
3
66.7
200.1
40040.01
19
6
5
56.7-62.7
∑ Total
3
∑F = 22
59.7
179.1
32076.81
22
3
∑=1600.4 ∑=841242.44
1) Calculating Mean
Mx=
∑𝐹𝑋 𝑖
𝑛
=
1600 .4
22
=72.7455
2) Median
1
Mdn
= ℓ +2
𝑁−𝑓𝑘𝑏
𝑓𝑖
𝑋𝑖
57
= 70.2 +
= 70.2 +
1
22−17
2
𝑋6
11
7.5
11
X6
= 74.2909
3) Modus
Mo
𝑓𝑎
=u+
𝑥𝑖
𝑓𝑎 +𝑓𝑏
= 70.5 +
7
7+6
𝑥6
= 70.5 + 3.2307
= 73.7307
4) Standard Deviation
S=
S=
𝑛.∑𝐹𝑋 𝑖 2 − ∑𝐹𝑋 𝑖 2
𝑛 𝑛 −1
22.1600 .4− 1600 .4 2
22 22−1
S = 7.74545
5) Standard Error
SEmd =
𝑆
=
𝑁−1
7.74545
22−1
=
7.74545
4.58
= 1.69115
After Calculating, it was found that the standard deviation and the standard
error of pretest score were 7.74545 and 1.69115
b. The Result of Post-test Score of Control Group (VII-B)
Based on the data Post-test score of control group, it was known the highest
score was 76.7 and the lowest score was 53.3. To determine the range of score,
the class interval, and interval of temporary, the writer calculated using formula as
follows:
58
The Highest Score (H)
= 76.7
The Lowest Score (L)
= 53.3
The Range of Score (R)
=H–L+1
= 76.7 – 53.3 + 1
= 24.4
The Class Interval (K)
= 1 + (3.3) x Log n
= 1 + (3.3) x Log 24
= 1 + 4.554697
= 5.554697
=6
Interval of Temporary (I) =
𝑅
𝐾
=
24.4
6
= 4,06 = 4
So, the range of score was 24.4, the class interval was 6, and interval of
temporary was 4. Then, it was presented using frequency distribution in the
following table:
Table 4.9 Frequency Distribution of the Post-test Score
Interval (I)
Frequency
(F)
Mid
Point
(x)
The
Limitation
of each
group
Frequency
Relative (%)
Frequency
Cumulative
(%)
1
2
78.3-82.3
73.3-77.3
0
5
80.3
75.3
77.8-82.8
72.8-77.8
0
20.8333
100
79.1667
3
4
68.3-72.3
63.3-67.3
4
6
70.3
65.3
67.8-72.8
62.8-67.8
16.6667
25
62.5
37.5
5
58.3-62.3
7
60.3
57.8-62.8
29.1667
8.3333
6
53.3-57.3
2
55.3
52.8-57.8
8.3333
∑P = 100
Class
(K)
∑F=24
0
59
The distribution of students’ predicate in post-test score of Control Group.
The distribution of students' predicate in post-test
14
12
10
8
6
4
2
0
Fail
Less
Enough
Good
Exellent
Figure 4.4 The distribution of students’ predicate in post-test score
of Control Group
Based on the figure above, it can be seen about the students’ predicate in
post-test score. There was no student who got Fail predicate. There were two
studentswho got Less predicate. There were thirteenstudents who gotEnough
Predicate. There were nine students who got Good predicate. There was no
student who got Excellent predicate.
The next step, the writer tabulated the scores into the table for the
calculation of mean,median, modus, standard deviation, and standard error as
follows:
60
Table 4.10the Table for Calculating Mean, median, modus, Standard
deviation. and standard error of Post- test Score.
Frequency Mid
Point
(F)
(x)
Class
(K)
Interval (I)
1
78.3-82.3
0
2
73.3-77.3
3
F.X
FX2
Fka
Fkb
80.3
0
0
0
24
5
75.3
376,5
141752,25
5
19
68.3-72.3
4
70.3
281,2
79073,44
9
15
4
63.3-67.3
6
65.3
391,8
153507,24
15
9
5
58.3-62.3
7
60.3
422,1
178168,41
22
2
6
53.3-57.3
∑ Total
2
∑F = 24
55.3
110,6
12232,36
24
0
∑=1582.2
∑=564733.7
1) Calculating Mean
Mx=
∑𝐹𝑋 𝑖
𝑛
=
1582 .2
24
= 65.925
2) Median
1
Mdn
= ℓ +2
𝑁−𝑓𝑘𝑏
𝑓𝑖
1
= 57.8 + 2
= 57.8 +
𝑋𝑖
24−2
7
11
7
X4
𝑋4
= 64.0857
3) Modus
Mo
=u+
𝑓𝑎
𝑓𝑎 +𝑓𝑏
= 58.3 +
1
1+5
𝑥𝑖
𝑥4
= 58.3 + 0.6667
61
= 58.9667
4) Standard Deviation
S=
S=
𝑛.∑𝐹𝑋 𝑖 2 − ∑𝐹𝑋 𝑖 2
𝑛 𝑛 −1
24.1582 .2− 1582 .2 2
24 24−1
S = 6.5925
5) Standard Error
SEmd =
𝑆
𝑁−1
=
6.5925
24−1
=
6.5925
4.79
= 1.3764
After Calculating, it was found that the standard deviation and the
standard error of pretest score were 6.5925 and 1.37634.
3. The Comparison result of Pre-test and Post-test of Experimental and
Control Group
EXPERIMENTAL GROUP
CONTROL GROUP
SCORE
SCORE
NO CODE
PREPOST- DIFFE NO CODE
PREPOST- DIFFE
TEST
TEST RENCE
TEST
TEST RENCE
E-01
40
66,7
26,7
C-01
46,7
56,7
10
1
1
E-02
43,3
70
26,7
C-02
43,3
60
16,7
2
2
E-03
66,7
80
13,3
C-03
40
66,7
26,7
3
3
E-04
70
86,7
16,7
C-04
40
60
20
4
4
E-05
56,7
76,7
20
C-05
40
60
20
5
5
E-06
40
56,7
16,7
C-06
40
63,3
23,3
6
6
E-07
53,3
60
6,7
C-07
60
60
0
7
7
E-08
50
76,7
26,7
C-08
46,7
70
23,3
8
8
E-09
56,7
73,3
16,6
C-09
66,7
66,7
0
9
9
E-10
53,3
80
26,7
C-10
70
76,7
6,7
10
10
E-11
60
70
10
C-11
56,7
63,3
6,6
11
11
E-12
50
66,7
16,7
C-12
70
63,3
-6,7
12
12
E-13
53,3
80
26,7
C-13
53,3
73,3
20
13
13
E-14
53,3
83,3
30
C-14
40
70
30
14
14
62
15
16
17
18
19
20
21
22
E-15
E-16
E-17
E-18
E-19
E-20
E-21
E-22
TOTAL
MEAN
LOWEST
HIGHEST
50
43,3
66,7
50
40
43,3
40
50
1129,9
51,4
40
70
76,7
70
73,3
60
63,3
73,3
70
76,7
1590,1
72,3
56,7
86,7
26,7
26,7
6,6
10
23,3
30
30
26,7
460,2
20,9
15
16
17
18
19
20
21
22
23
24
C-15
C-16
C-17
C-18
C-19
C-20
C-21
C-22
C-23
C-24
TOTAL
MEAN
LOWEST
HIGHEST
53,3
50
53,3
40
60
43,3
46,7
40
40
53,3
1193,3
49,7
40
70
73,3
70
63,3
60
73,3
60
70
73,3
60
53,3
1566,5
65,3
53,3
76,7
20
20
10
20
13,3
16,7
23,3
33,3
20
0
373,2
15,6
From the table above the mean score of pre test and post test of the
experimental group were 51,4 and 72,3. Meanwhile, the highest score pre test and
post test of the experimental group were 70 and 86,7, the lowest scores pre test
and post test of the experimental group were 40 and 56,7. In addition, the mean
score pre test and post test of the control group were 49,7 and 65,3. Meanwhile,
the highest score pre test and post test of the control group were 70 and 76,7. The
lowest scores pre test and post test of the control group were 40 and 53,3. Based
on the data above, the difference of mean score between experimental and control
group score were 5.3.
63
B. Testing of Normality and Homogeinity
1. Normality Test
a. Testing normality of pre-test experimental and control group
Table 4.11 Testing normality of pre-test experimental and control group
Tests of Normality
a
Kolmogorov-Smirnov
Shapiro-Wilk
Group
Statistic
experiment
Scores
Df
Sig.
Statistic
df
Sig.
*
,920
22
,676
,863
24
,114
,142
22
,200
,168
24
,076
group
control group
The table showed the result of test normality calculation using SPSS 21.0
program. To know the normality of data, the formula could be seen as follows:
If the number of sample. > 50 = Kolmogorov-Smirnov
If the number of sample. < 50 = Shapiro-Wilk
Based on the number of data the writer was 46 < 50, so to analyzed normality
data was used Shapiro-Wilk. The next step, the writer analyzed normality of data
used formula as follows:
If Significance > 0.05 = data is normal distribution
If Significance < 0.05 = data is not normal distribution
Based on data above, significant data of experiment and control group used
Shapiro-Wilk was 0.676 > 0.05 and 0.114 > 0.05. It could be concluded that the data
was normal distribution.
b. Testing normality of pre-test experimental and control group
64
Table 4.12Testing normality of post-test experimental and control group
Tests of Normality
a
Kolmogorov-Smirnov
Group
Scores
Statistic
df
Shapiro-Wilk
Sig.
Statistic
df
Sig.
experiment group
.122
22
.200
*
.971
22
.744
control group
.174
24
.058
.933
24
.115
The table showed the result of test normality calculation using SPSS 21.0
program. To know the normality of data, the formula could be seen as follows:
If the number of sample. > 50 = Kolmogorov-Smirnov
If the number of sample. < 50 = Shapiro-Wilk
Based on the number of data the writer was 46 < 50, so to analyzed normality
data was used Shapiro-Wilk. The next step, the writer analyzed normality of data
used formula as follows:
If Significance > 0.05 = data is normal distribution
If Significance < 0.05 = data is not normal distribution
Based on data above, significant data of experiment and control group used
Shapiro-Wilk was 0.744 > 0.05 and 0.115 > 0.05. It could be concluded that the data
was normal distribution.
65
2. Homogeneity Test
a. Testing Homogeneity of pre-test experimental and control group
Table 4.13Testing Homogeneity of pre-test experimental and control
group
Homogeneity Test
Levene's Test for
t-test for Equality of Means
Equality of
Variances
F
Sig.
T
Df
Sig. (2-
Mean
Std. Error
99% Confidence
tailed)
Differen
Difference
Interval of the
ce
Difference
Lower
Equal
,743
,393
,584
44
,562
1,63826
2,80478
variances
Upper
-
9,18951
5,91299
assumed
Scores
Equal
,587 43,985
,560
1,63826
2,79147
variances not
5,87727
assumed
The table showed the result of Homogeneity test calculation using SPSS 21.0
program. To know the Homogeneity of data, the formula could be seen as follows:
If Sig. > 0,01 = Equal variances assumed or Homogeny distribution
If Sig. < 0,01 = Equal variances not assumedor not Homogeny distribution
Based on data above, significant data was 0,393. The result was 0,393 > 0,01,
it meant the t-test calculation used at the equal variances assumed or data was
Homogeny distribution.
9,15378
66
b. Testing Homogeneity of post-test experimental and control group
Table 4.14Testing Homogeneity of post-test experimental and control
group
Homogeneity Test
Levene's Test for
Equality of
Variances
t-test for Equality of Means
99% Confidence
Interval of the
Difference
Sig. (2-
F
Scores
Equal variances
.607
Sig.
.440
t
3.352
df
tailed)
44
.002
Mean
Std. Error
Difference Difference
7.00644
2.09010
Lower
1.37931 12.633
assumed
Equal variances
57
3.320 40.219
.002
7.00644
2.11064
1.29985 12.713
not assumed
The table showed the result of Homogeneity test calculation using SPSS 21.0
program. To know the Homogeneity of data, the formula could be seen as follows:
If Sig. > 0,01 = Equal variances assumed or Homogeny distribution
If Sig. < 0,01 = Equal variances not assumedor not Homogeny distribution
Based on data above, significant data was 0,440. The result was 0,440 > 0,01,
it meant the t-test calculation used at the equal variances assumed or data was
Homogeny distribution.
Upper
03
67
C. The Result of Data Analysis
1. Testing Hypothesis Using Manual Calculation
Table 4.15 The Standard Deviation and the Standard Error of Experiment
and Control Group
Group
Standard Deviation
Standard Error
Experimental Group
7.74545
1.69115
Control Group
6.5925
1.37634
The table showed the result of the standard deviation calculation of
Experiment group was 7.74545and the result of the standard error was 1.69115.
The result of thestandard deviation calculation of Control group was 6.5925 and
the result of standard error was 1.37634. To examine the hypothesis, the writer
used the formula as follow:
tobserved=
=
=
df
𝑀1−𝑀2
𝑆𝐸𝑚 1−𝑆𝐸𝑚 2
7.74545 −6.5925
1.69115 −1.37634
1.15295
0.31481
= 3.663
= (N1 + N2 – 2)
= 22+24-2
= 44
68
a. Interpretation
The result of t – test was interpreted on the result of degree of freedom to get
the ttable. The result of degree of freedom (df) was 44. The following table was the
result of tobserved and ttable from 44 df at 5% and 1% significance level.
Table 4.16 The Result of T-Test Using Manual Calculation
t-table
t-observe
3.663
Df
5 % (0,05)
1 % (0,01)
2.021
2.704
44
The interpretation of the result of t-test using manual calculation, it was found
the tobserved was higher than the ttable at 5% and 1% significance level or 3.663 > 2.021,
3.663 > 2.704. It meant Ha was accepted and Ho was rejected. It could be interpreted
based on the result of calculation that Ha stating that K-W-L strategy was effective
for Teaching Reading Comprehension of the seventh grade students at SMP
Al-Amin Palangka Raya was accepted and Ho stating thatK-W-L strategy was
not effective for Teaching Reading Comprehension of the seventh grade
students at SMP Al-Amin Palangka Raya was rejected. It meant that teaching
reading with K-W-L Strategy Toward Reading Comprehension for the seventh
grade students at SMP Al-Amin Palangka Raya gave significant effect at 5% and
1% significance level.
2. Testing Hypothesis Using SPSS 21.0 Program
The writer also applied SPSS 21.0 program to calculate t – test in testing
hypothesis of the study. The result of t – test using SPSS 21.0 was used to support
69
the manual calculation of t – test. The result of t – test using SPSS 21.0 program
could be seen as follows:
Table 4.17 Mean, Standard Deviation and Standard Error of experiment
group and control groupusing SPSS 21.0 Program
Group Statistics
Group
Scores
N
Mean
Std. Deviation
Std. Error Mean
experiment group
22
72.2773
7.86032
1.67583
control group
24
65.2708
6.28597
1.28312
The table showed the result of mean calculation of experiment groupwas
72.2773, standard deviation calculation was 7.86032, and standard error of mean
calculation was 1.67583. The result of mean calculation of control groupwas
65.2708, standard deviation calculation was 6.28597, and standard error of mean
was 1.28312.
Table 4.18 The Calculation of T – Test Using SPSS 21.0
Independent Samples Test
Levene's Test for
Equality of
Variances
t-test for Equality of Means
99% Confidence
Interval of the
Difference
Sig. (2-
F
Scores
Equal variances
.607
Sig.
.440
T
3.352
Df
tailed)
44
.002
Mean
Difference Difference
7.00644
assumed
Equal variances
not assumed
Std. Error
Lower
Upper
2.09010 1.37931 12.6335
7
3.320 40.219
.002
7.00644
2.11064 1.29985 12.7130
3
70
The table showed the result of t – test calculation using SPSS 21.0 program.
To know the variances score of data, the formula could be seen as follows:
If Sig. > 0,01 = Equal variances assumed
If Sig. < 0,01 = Equal variances not assumed
Based on data above, significant data was 0,440. The result was 0,440 > 0,01,
it meant the t-test calculation used at the equal variances assumed. It found that the
result of tobserved was 3.352, the result of mean difference between experiment and
control group was 7.00644, and thestandard error difference between experiment and
control group was 2.09010.
a. Interpretation
The result of t – test was interpreted on the result of degree of freedom to get
the ttable. The result of degree of freedom (df) was 44. The following table was the
result of tobserved and ttable from 44 df at 5% and 1% significance level.
Table 4.19 The Result of T-Test Using SPSS 21.0 Program
t-table
t-observe
3.352
Df
5 % (0,05)
1 % (0,01)
2.021
2.704
44
The interpretation of the result of t-test using SPSS 21.0 program, it was
found the tobserved was higher than the ttable at 5% and 1% significance level or 3.352 >
2.021, 3.352 > 2.704. It meant Ha was accepted and Ho was rejected. It could be
interpreted based on the result of calculation that Ha stating that K-W-L strategy
was effective for Teaching Reading Comprehension of the seventh grade
71
students at SMP Al-Amin Palangka Raya was accepted and Ho stating thatKW-L strategy was not effective for Teaching Reading Comprehension of the
seventh grade students at SMP Al-Amin Palangka Raya was rejected. It meant
that teaching reading with K-W-L Strategy Toward Reading Comprehension for
the seventh grade students at SMP Al-Amin Palangka Raya gave significant
effect at 5% and 1% significance level.
D. Discussion
The result of analysis showed that there was significant effect of K-W-L
strategy Toward Reading Comprehension forthe seventh grade students at
SMP Al-Amin Palangka Raya.The students who were taught used K-W-L
strategyreached higher score than those who were taught without used K-WL strategy.
Meanwhile, after the data was calculated using manual calculation of ttest.It
was found the tobserved was higher than the ttable at 5% and 1% significance level or
3.663 > 2.021, 3.663 > 2.704. It meant Ha was accepted and Ho was rejected. And the
data calculated using SPSS 21.0 program, it was found the tobserved was higher than
the ttable at 5% and 1% significance level or 3.352 > 2.021, 3.352 > 2.704. It meant Ha
was accepted and Ho was rejected. This finding indicated that the alternative
hypothesis (Ha) stating that there was any significant effect of K-W-L strategy
Toward Reading Comprehension for the seventh grade students at SMP AlAmin Palangka Raya was accepted.On the contrary,the Null hypothesis (Ho)
stating that there was no any significant effect of K-W-L strategy Toward
Reading Comprehension for the seventh grade students at SMP Al-Amin
72
Palangka Raya was rejected.Based on the result the data analysis showed that
using K-W-L Strategy gave significance effect for the students’ reading
comprehension scores of Seventh grade students at SMP Al-amin Palangka Raya.
After the students have been taught by using K-W-L Strategy, the reading
score were higher than before implementing K-W-L Strategy as a learning
strategy. It can be seen in the comparison of pre test and post test score of
experimental group and control group (See p.61). This finding indicated that KW-L strategy was effective and supports the previous research done by Zhang
Fengjuan, Eviani Damastuti and Anne Crout Shelleythat also stated teaching
reading by using K-W-L strategy was effective.
There were some reason why using K-W-L Strategy gave significance effect
for the students’ reading comprehension scores of Seventh grade students at SMP
Al-amin Palangka Raya.First, K-W-L Strategy was effective in terms of
improving the students’ English reading score. It can be seen from the
improvement of the students’ score average in the post-test. From the mean
score of control and experiment were 72.3 and 65.3. (See p.62).It supports the
previous study by Eviani Damastuti and Sugini states that using K-W-L strategy
increase in reading ability significantly intensified.
It was suitable withthe result of pre-test and post test for Experiment and
control Group. (See p.44). In the pre-test of experiment group there wereseven
students that got fail predicate. They were E-01, E-02, E-06, E-16, E-19, E-20 and
E-21. There were eleven students that got less predicate. They were E-05, E-07,
E-08, E-09, E-10, E-12, E-13, E-14, E-15, E-18 and E-22. There were three
73
students that got enough predicate. They are E-03, E-11 and E-17. There was one
student that got good predicate. She was E-04. Then, in the pre-test score of
control group there were thirteen students that got fail predicate. They were C-01,
C-02, C-03, C-04, C-05, C-06, C-08, C-14, C-18, C-20, C-21, C-22 and C-23.
There were six students that got less predicate. They were C-11, C-13, C-15, C16, C-17 and C-24. There were three students that got enough predicate. They
were C-07, C-09 and C-19. There were two students that got good predicate. They
were C-10 and C-12.
Based on the result of post-test for experimental and control group, (See
p.53). In the experimental group, there was no student that got in fail predicate.
There was one student that got less predicate, he was E-06. There were five
students that got enough predicate. They were E-01, E-07, E-12, E-18 and E-18.
There were fifteen students that got good predicate. They were E-02, E-03, E-05,
E-08, E-09, E-10, E-11, E-13, E-14, E-15, E-16, E-17, E-20, E-21 and E-22.
There was one student that got excellent predicate, she was E-04. In the control
group, there was no student that got in fail predicate. There were two students that
got less predicate. They were C-01 and C-24. There were thirteen students that got
enough predicate. They were C-02, C-03, C-04, C-05, C-06, C-07, C-09, C-11, C12, C-17, C-18, C-20 and C-23. There were nine students that got good predicate.
They were C-08, C-10, C-13, C-14, C-15, C-16, C-19, C-21 and C-22.
The next reason wasK-W-L strategycan motivate students in teaching
learning process. It was suitable with the students response when learning process
is going, they enthusiasm to wrote in coloum (K) for what they Know about the
74
topic with their background knowledge. It was necessary to keep responses inside
the topic. It indicated that using K-W-L strategy was effective in enhance reading
motivation and encouragement. (See appendix 6). It supports with Ferdinand
Nicholas Boonde states that K-W-L strategycan motivate the students to take a
part in the teaching learning process and Filling the columns is effective to help
the students understand the reading text.
The last reason was K-W-L strategy gave the students can answer both
literal and inferential reading comprehension types. It indicated the test was
suitable for junior high school student (see appendix4). It supports with Zhang
FengjuanK-W-L strategy is an instructional scheme that develops active reading of
descriptive texts by activating learners’ background knowledge.
Those are the result of pre-test compared with post-test for experimental
group and control group of students at SMP Al-Amin Palangka Raya. Based on
the theories and the writer’s result,K-W-L Strategy gave significance effect for the
students’ reading comprehension scores of Seventh grade students at SMP Alamin Palangka Raya.
CHAPTER IV
RESULT OF THE STUDY
This chapter covers Description of the data, test of normality and
homogeneity, result of the data analyses and discussion.
A. Description of The Data
This section described the obtained data of the effectiveness of using K-WL Strategy in teaching reading Descriptive text. The presented data consisted of
Mean, Median, Modus, Standard Deviation and Standard Error.
1. The descriptiondata of Pre-Test Score
The students’ pre test score are distributed in the following table in order
toanalyze the students’ knowledge before conducting the treatment.
Table 4.1Pre test score of experimental and control group
Experimental Group
Code
Score
E-01
E-02
E-03
E-04
E-05
E-06
E-07
E-08
E-09
E-10
E-11
E-12
E-13
40
43.3
66.7
70
56.7
40
53.3
50
56.7
53.3
60
50
53.3
CORRECT
ANSWER
12
13
20
21
17
12
16
15
17
16
18
15
16
Control Group
PREDICATE
CODE
SCORE
FAIL
FAIL
ENOUGH
GOOD
LESS
FAIL
LESS
LESS
LESS
LESS
ENOUGH
LESS
LESS
C-01
C-02
C-03
C-04
C-05
C-06
C-07
C-08
C-09
C-10
C-11
C-12
C-13
46.7
43.3
40
40
40
40
60
46.7
66.7
70
56.7
70
53.3
44
CORRECT
ANSWER
14
13
12
12
12
12
18
14
20
21
17
21
16
PREDICATE
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
ENOUGH
FAIL
ENOUGH
GOOD
LESS
GOOD
LESS
45
E-14
E-15
E-16
E-17
E-18
E-19
E-20
E-21
E-22
53.3
50
43.3
66.7
50
40
43.3
40
50
16
15
13
20
15
12
13
12
15
TOTAL
AVERAGE
Lowest Score
Highest Score
LESS
LESS
FAIL
ENOUGH
LESS
FAIL
FAIL
FAIL
LESS
1129,9
51,4
40
70
C-14
C-15
C-16
C-17
C-18
C19
C-20
C-21
C-22
C-23
C-24
40
53.3
50
53.3
40
60
43.3
46.7
40
40
53.3
12
16
15
16
12
18
13
14
12
12
16
TOTAL
AVERAGE
Lowest Score
Highest Score
FAIL
LESS
LESS
LESS
FAIL
ENOUGH
FAIL
FAIL
FAIL
FAIL
LESS
1193,3
49,7
40
70
The table above showed us the comparison of pre-test score achieved by
experimental and control group students, both class’ achievement are at the same
level. It can be seen that from the students’ score. The highest score 70 and the
lowest score 40, both experimental and control group. It meant that the
experimental and control group have the same level in reading comprehension
before getting the treatment.
a. The Result of Pretest Score of ExperimentalGroup (VII-A)
Based on the data above, it was known the highest score was 70 and the
lowest score was 40. To determine the range of score, the class interval, and
interval of temporary,the writer calculated using formula as follows:
The Highest Score (H)
= 70
The Lowest Score (L)
= 40
46
The Range of Score (R)
=H–L+1
= 70 – 40+ 1
= 31
The Class Interval (K)
= 1 + (3.3) x Log n
= 1 + (3.3) x Log 22
= 1 + 4.4299948
= 5.4299948
=5
Interval of Temporary (I) =
𝑅
𝐾
=
31
5
= 6,2 = 6 or 2
So, the range of score was 31, the class interval was 5, and interval of
temporary was 6 or 2. Then, it was presented using frequency distribution in
the following table:
Table 4.2 Frequency Distribution of the Pretest Score
Interval (I)
Frequency
(F)
Mid
Point
(x)
The
Limitation
of each
group
Frequency
Relative (%)
Frequency
Cumulative
(%)
1
2
65-70
59-64
3
1
67.5
61.5
64.5-70.5
58.5-64.5
13.6363
4.5454
100
86.3636
3
53-58
6
52.7
52.5-58.5
27.2727
81.8182
4
47-52
5
55.5
46.5-52.5
22.7272
54.5455
5
40-46
7
43
39.5-46.5
31.8181
∑P = 100
31.8183
Class
(K)
∑F=22
47
The distribution of students’ predicate in pretest score of Experimental
group can also be seen in the following figure.
The Distribution of Student's Predicate
12
10
8
6
4
2
Figure 4.1 The distribution frequency of students’ pretest score for
Experimental Group
0
Fail
Less
Enough
Good
Figure 4.1 The distribution of students’ predicate in pretest score
for Experimental Group
Based on the figure above, it can be seen about the students’ predicate in
pretest score. There were seven students who got Fail predicate. There were
elevenstudents who gotLess Predicate. There were three students who gotEnough
predicate. There was one student who got Good predicate.
The next step, the writer tabulated the scores into the table for the
calculation of mean, standard deviation, and standard error as follows:
48
Table 4.3the Table for Calculating Mean, median, modus, Standard
deviation. and standard error of Pretest Score.
Frequency Mid
Point
(F)
(x)
Class
(K)
Interval
(I)
1
65-70
3
2
59-64
3
F.X
FX2
Fka
67.5
202.5
41006.25
3
1
61.5
61.5
3782.25
4
53-58
6
52.7
316.2
99982.44
10
4
47-52
5
55.5
277.5
77006.25
15
5
40-46
∑ Total
7
∑F = 22
43
301
90601
22
∑=1158.7 ∑=312378.19
1) Calculating Mean
Mx=
∑𝐹𝑋 𝑖
𝑛
=
1158 .7
22
=52.67
2) Median
Mdn
1
𝑁−𝑓𝑘𝑏
2
=ℓ+
𝑓𝑖
1
= 39.5 + 2
𝑋𝑖
22−7
7
4
= 39.5 + X 6
7
𝑋6
= 45.93
3) Modus
Mo
=u+
𝑓𝑎
𝑓𝑎 +𝑓𝑏
= 39.5 +
2
2+7
= 39.5 + 1.11
𝑥𝑖
𝑥5
Fkb
22
19
18
12
7
49
= 40.61
4) Standard Deviation
S=
S=
𝑛.∑𝐹𝑋 𝑖 2 − ∑𝐹𝑋 𝑖 2
𝑛 𝑛 −1
22.1158 .7− 312378 .19 2
22 22−1
S = 6.29
5) Standard Error
SEmd =
𝑆
𝑁−1
=
6.29
22−1
=
6.29
4.58
= 1.37
After Calculating, it was found that the standard deviation and the
standard error of pretest score were 6.29 and 1.37
b. The Result of Pretest Score of ControlGroup (VII-B)
Based on the data pretest score of control group, it was known the
highest score was 70 and the lowest score was 40. To determine the range of
score, the class interval, and interval of temporary,the writer calculated using
formula as follows:
The Highest Score (H)
= 70
The Lowest Score (L)
= 40
The Range of Score (R)
=H–L+1
= 70 – 40+ 1
= 31
The Class Interval (K)
= 1 + (3.3) x Log n
= 1 + (3.3) x Log 24
= 1 + 4.554697
= 5.554697
=6
50
Interval of Temporary (I) =
𝑅
𝐾
=
31
6
= 5.1 = 5
So, the range of score was 31, the class interval was 6, and interval of
temporary was 5. Then, it was presented using frequency distribution in the
following table:
Table 4.4 Frequency Distribution of the Pretest Score
Interval (I)
Frequency
(F)
Mid
Point
(x)
The
Limitation
of each
group
Frequency
Relative (%)
Frequency
Cumulative
(%)
1
2
66-70
61-65
3
0
68
63
65.5-70.5
60.5-65.5
12.5
0
100
87.5
3
4
56-60
51-55
3
4
58
53
55.5-60.5
50.5-55.5
12.5
16.6667
87.5
75
5
46-50
4
48
45.5-50.5
16.6667
58.3333
6
40-45
10
42.5
39.5-45.5
41.6666
∑P = 100
41.6666
Class
(K)
∑F=24
The distribution of students’ predicate in pretest score of Control group
The distribution of student's predicate
14
12
10
8
6
4
2
0
Fail
Less
Enough
Good
Figure 4.2 The distribution of students’ predicate in pretest score
of Control Group
51
Based on the figure above, it can be seen about the students’ predicate in
pretest score. There thirteenstudents who got Fail predicate. There were
sixstudents who gotLess Predicate. There were three students who got Enough
predicate. There weretwo student who got Good predicate.
The next step, the writer tabulated the scores into the table for the
calculation of mean, median, modus, standard deviation, and standard error as
follows:
Table 4.5the Table for Calculating Mean, median, modus, Standard
deviation. and standard error of Pretest Score of Control group.
Frequency Mid
Point
(F)
(x)
Class
(K)
Interval
(I)
1
66-70
3
2
61-65
3
F.X
FX2
Fka
Fkb
68
204
41616
3
24
0
63
0
0
3
21
56-60
3
58
174
30276
6
21
4
51-55
4
53
212
44944
10
18
5
46-50
4
48
192
36864
14
14
6
40-45
∑ Total
10
∑F = 24
42.5
425
180625
24
10
∑=1207
1) Calculating Mean
Mx=
2) Median
Mdn
∑𝐹𝑋 𝑖
𝑛
=
1207
24
=50.29
1
𝑁−𝑓𝑘𝑏
2
=ℓ+
𝑓𝑖
1
= 39.5 + 2
𝑋𝑖
24−10
10
𝑋5
∑=334325
52
= 39.5 +
7
10
X5
= 43
3) Modus
Mo
𝑓𝑎
=u+
𝑓𝑎 +𝑓𝑏
= 39.5 +
𝑥𝑖
6
6+10
𝑥5
= 39.5 + 1.87
= 41.37
4) Standard Deviation
S=
𝑛.∑𝐹𝑋 𝑖 2 − ∑𝐹𝑋 𝑖 2
S=
24.1207 − 334325 2
𝑛 𝑛 −1
24 24−1
S = 5.38
5) Standard Error
SEmd =
𝑆
𝑁−1
=
5.38
24−1
=
5.38
4.79
= 1.18
After Calculating, it was found that the standard deviation and the
standard error of pretest score were 5.38 and 1.18
53
2.
The description data of Post-Test Score
The students’ score are distributed in the following table in order toanalyze
the students’ knowledge after conducting the treatment.
Table 4.6Post-test score of Experimental and Control Group
Experimental Group
Code
Score
E-01
E-02
E-03
E-04
E-05
E-06
E-07
E-08
E-09
E-10
E-11
E-12
E-13
E-14
E-15
E-16
E-17
E-18
E-19
E-20
E-21
E-22
66.7
70
80
86.7
76.7
56.7
60
76.7
73.3
80
70
66.7
80
83.3
76,7
70
73.3
60
63.3
73.3
70
76.7
TOTAL
AVERAGE
Lowest Score
Highest Score
CORRECT
ANSWER
20
21
24
26
23
17
18
23
22
24
21
20
24
25
23
21
22
18
19
22
21
23
PREDICATE
ENOUGH
GOOD
GOOD
EXCELLENT
GOOD
LESS
ENOUGH
GOOD
GOOD
GOOD
GOOD
ENOUGH
GOOD
GOOD
GOOD
GOOD
GOOD
ENOUGH
ENOUGH
GOOD
GOOD
GOOD
1590.1
72.3
56.7
86.7
Control Group
CODE
SCORE
C-01
C-02
C-03
C-04
C-05
C-06
C-07
C-08
C-09
C-10
C-11
C-12
C-13
C-14
C-15
C-16
C-17
C-18
C19
C-20
C-21
C-22
C-23
C-24
56.7
60
66.7
60
60
63.3
60
70
66.7
76,7
63.3
63.3
73.3
70
73.3
70
63.3
60
73.3
60
70
73.3
60
53.3
TOTAL
AVERAGE
Lowest Score
Highest Score
CORRECT
ANSWER
17
18
20
18
18
19
18
21
20
23
19
19
22
21
22
21
19
18
22
18
21
22
18
16
PREDICATE
LESS
ENOUGH
ENOUGH
ENOUGH
ENOUGH
ENOUGH
ENOUGH
GOOD
ENOUGH
GOOD
ENOUGH
ENOUGH
GOOD
GOOD
GOOD
GOOD
ENOUGH
ENOUGH
GOOD
ENOUGH
GOOD
GOOD
ENOUGH
LESS
1566.5
65.3
53.3
76.7
54
The table above showed us the comparison of post-test score achieved by
experimental and control group students. Both class’ achievement have different
score. It can be seen from the highest score 86.7 and 76.7 and the lowest score
56.7 and 53.3. It meant that the experimental and control group have the different
level in reading comprehension after getting the treatment.
a. The Result of Post-test Score of Experimental Group (VII-A)
Based on the data Post-test score of Experimental group, it was known the
highest score was 86.7 and the lowest score was 56.7. To determine the range of
score, the class interval, and interval of temporary, the writer calculated using
formula as follows:
The Highest Score (H)
= 86.7
The Lowest Score (L)
= 56.7
The Range of Score (R)
=H–L+1
= 86.7 – 56.7 + 1
= 31
The Class Interval (K)
= 1 + (3.3) x Log n
= 1 + (3.3) x Log 22
= 1 + 4.4299948
= 5.4299948
=5
Interval of Temporary (I) =
𝑅
𝐾
=
31
5
= 6,2 = 6
55
So, the range of score was 31, the class interval was 5, and interval of
temporary was 6. Then, it was presented using frequency distribution in the
following table:
Table 4.7 Frequency Distribution of the Post-test Score
Interval (I)
Frequency
(F)
Mid
Point
(x)
The
Limitation
of each
group
Frequency
Relative (%)
Frequency
Cumulative
(%)
1
2
84.7-90.7
77.7-83.7
1
4
87.7
80.7
83.2-91.2
73.2-83.2
4.5454
18.1818
100
95.4545
3
4
70.7-76.7
63.7-69.7
11
3
73.7
66.7
69.2-73.2
62.2-69.2
50
13.6364
45.4546
31.8182
5
56.7-62.7
3
59.7
55.2-62.2
13.6364
∑P = 100
18.1818
Class
(K)
∑F=22
The distribution of students’ predicate in post-test score of Experimental
group.
The distribution of student's predicate pos-test
16
14
12
10
8
6
4
2
0
Fail
Less
Enough
Good
Exellent
Figure 4.3 The distribution of students’ predicate in post-test score
of Experimental Group
56
Based on the figure above, it can be seen about the students’ predicate in
pretest score. There was no student who got Fail predicate. There wasone
studentswho got Less predicate. There were fivestudents who gotEnough
Predicate. There were fifteen students who got Good predicate. There was one
student who got Excellent predicate.
The next step, the writer tabulated the scores into the table for the
calculation of mean,median, modus, standard deviation, and standard error as
follows:
Table 4.8the Table for Calculating Mean, median, modus, Standard
deviation. and standard error of Post- test Score.
Frequency Mid
Point
(F)
(x)
Class
(K)
Interval (I)
1
84.7-90.7
1
2
77.7-83.7
3
F.X
FX2
Fka
Fkb
87.7
87.7
7691.29
1
22
4
80.7
322.8
104199.84
5
23
70.7-76.7
11
73.7
810.7
657234.49
16
17
4
63.7-69.7
3
66.7
200.1
40040.01
19
6
5
56.7-62.7
∑ Total
3
∑F = 22
59.7
179.1
32076.81
22
3
∑=1600.4 ∑=841242.44
1) Calculating Mean
Mx=
∑𝐹𝑋 𝑖
𝑛
=
1600 .4
22
=72.7455
2) Median
1
Mdn
= ℓ +2
𝑁−𝑓𝑘𝑏
𝑓𝑖
𝑋𝑖
57
= 70.2 +
= 70.2 +
1
22−17
2
𝑋6
11
7.5
11
X6
= 74.2909
3) Modus
Mo
𝑓𝑎
=u+
𝑥𝑖
𝑓𝑎 +𝑓𝑏
= 70.5 +
7
7+6
𝑥6
= 70.5 + 3.2307
= 73.7307
4) Standard Deviation
S=
S=
𝑛.∑𝐹𝑋 𝑖 2 − ∑𝐹𝑋 𝑖 2
𝑛 𝑛 −1
22.1600 .4− 1600 .4 2
22 22−1
S = 7.74545
5) Standard Error
SEmd =
𝑆
=
𝑁−1
7.74545
22−1
=
7.74545
4.58
= 1.69115
After Calculating, it was found that the standard deviation and the standard
error of pretest score were 7.74545 and 1.69115
b. The Result of Post-test Score of Control Group (VII-B)
Based on the data Post-test score of control group, it was known the highest
score was 76.7 and the lowest score was 53.3. To determine the range of score,
the class interval, and interval of temporary, the writer calculated using formula as
follows:
58
The Highest Score (H)
= 76.7
The Lowest Score (L)
= 53.3
The Range of Score (R)
=H–L+1
= 76.7 – 53.3 + 1
= 24.4
The Class Interval (K)
= 1 + (3.3) x Log n
= 1 + (3.3) x Log 24
= 1 + 4.554697
= 5.554697
=6
Interval of Temporary (I) =
𝑅
𝐾
=
24.4
6
= 4,06 = 4
So, the range of score was 24.4, the class interval was 6, and interval of
temporary was 4. Then, it was presented using frequency distribution in the
following table:
Table 4.9 Frequency Distribution of the Post-test Score
Interval (I)
Frequency
(F)
Mid
Point
(x)
The
Limitation
of each
group
Frequency
Relative (%)
Frequency
Cumulative
(%)
1
2
78.3-82.3
73.3-77.3
0
5
80.3
75.3
77.8-82.8
72.8-77.8
0
20.8333
100
79.1667
3
4
68.3-72.3
63.3-67.3
4
6
70.3
65.3
67.8-72.8
62.8-67.8
16.6667
25
62.5
37.5
5
58.3-62.3
7
60.3
57.8-62.8
29.1667
8.3333
6
53.3-57.3
2
55.3
52.8-57.8
8.3333
∑P = 100
Class
(K)
∑F=24
0
59
The distribution of students’ predicate in post-test score of Control Group.
The distribution of students' predicate in post-test
14
12
10
8
6
4
2
0
Fail
Less
Enough
Good
Exellent
Figure 4.4 The distribution of students’ predicate in post-test score
of Control Group
Based on the figure above, it can be seen about the students’ predicate in
post-test score. There was no student who got Fail predicate. There were two
studentswho got Less predicate. There were thirteenstudents who gotEnough
Predicate. There were nine students who got Good predicate. There was no
student who got Excellent predicate.
The next step, the writer tabulated the scores into the table for the
calculation of mean,median, modus, standard deviation, and standard error as
follows:
60
Table 4.10the Table for Calculating Mean, median, modus, Standard
deviation. and standard error of Post- test Score.
Frequency Mid
Point
(F)
(x)
Class
(K)
Interval (I)
1
78.3-82.3
0
2
73.3-77.3
3
F.X
FX2
Fka
Fkb
80.3
0
0
0
24
5
75.3
376,5
141752,25
5
19
68.3-72.3
4
70.3
281,2
79073,44
9
15
4
63.3-67.3
6
65.3
391,8
153507,24
15
9
5
58.3-62.3
7
60.3
422,1
178168,41
22
2
6
53.3-57.3
∑ Total
2
∑F = 24
55.3
110,6
12232,36
24
0
∑=1582.2
∑=564733.7
1) Calculating Mean
Mx=
∑𝐹𝑋 𝑖
𝑛
=
1582 .2
24
= 65.925
2) Median
1
Mdn
= ℓ +2
𝑁−𝑓𝑘𝑏
𝑓𝑖
1
= 57.8 + 2
= 57.8 +
𝑋𝑖
24−2
7
11
7
X4
𝑋4
= 64.0857
3) Modus
Mo
=u+
𝑓𝑎
𝑓𝑎 +𝑓𝑏
= 58.3 +
1
1+5
𝑥𝑖
𝑥4
= 58.3 + 0.6667
61
= 58.9667
4) Standard Deviation
S=
S=
𝑛.∑𝐹𝑋 𝑖 2 − ∑𝐹𝑋 𝑖 2
𝑛 𝑛 −1
24.1582 .2− 1582 .2 2
24 24−1
S = 6.5925
5) Standard Error
SEmd =
𝑆
𝑁−1
=
6.5925
24−1
=
6.5925
4.79
= 1.3764
After Calculating, it was found that the standard deviation and the
standard error of pretest score were 6.5925 and 1.37634.
3. The Comparison result of Pre-test and Post-test of Experimental and
Control Group
EXPERIMENTAL GROUP
CONTROL GROUP
SCORE
SCORE
NO CODE
PREPOST- DIFFE NO CODE
PREPOST- DIFFE
TEST
TEST RENCE
TEST
TEST RENCE
E-01
40
66,7
26,7
C-01
46,7
56,7
10
1
1
E-02
43,3
70
26,7
C-02
43,3
60
16,7
2
2
E-03
66,7
80
13,3
C-03
40
66,7
26,7
3
3
E-04
70
86,7
16,7
C-04
40
60
20
4
4
E-05
56,7
76,7
20
C-05
40
60
20
5
5
E-06
40
56,7
16,7
C-06
40
63,3
23,3
6
6
E-07
53,3
60
6,7
C-07
60
60
0
7
7
E-08
50
76,7
26,7
C-08
46,7
70
23,3
8
8
E-09
56,7
73,3
16,6
C-09
66,7
66,7
0
9
9
E-10
53,3
80
26,7
C-10
70
76,7
6,7
10
10
E-11
60
70
10
C-11
56,7
63,3
6,6
11
11
E-12
50
66,7
16,7
C-12
70
63,3
-6,7
12
12
E-13
53,3
80
26,7
C-13
53,3
73,3
20
13
13
E-14
53,3
83,3
30
C-14
40
70
30
14
14
62
15
16
17
18
19
20
21
22
E-15
E-16
E-17
E-18
E-19
E-20
E-21
E-22
TOTAL
MEAN
LOWEST
HIGHEST
50
43,3
66,7
50
40
43,3
40
50
1129,9
51,4
40
70
76,7
70
73,3
60
63,3
73,3
70
76,7
1590,1
72,3
56,7
86,7
26,7
26,7
6,6
10
23,3
30
30
26,7
460,2
20,9
15
16
17
18
19
20
21
22
23
24
C-15
C-16
C-17
C-18
C-19
C-20
C-21
C-22
C-23
C-24
TOTAL
MEAN
LOWEST
HIGHEST
53,3
50
53,3
40
60
43,3
46,7
40
40
53,3
1193,3
49,7
40
70
73,3
70
63,3
60
73,3
60
70
73,3
60
53,3
1566,5
65,3
53,3
76,7
20
20
10
20
13,3
16,7
23,3
33,3
20
0
373,2
15,6
From the table above the mean score of pre test and post test of the
experimental group were 51,4 and 72,3. Meanwhile, the highest score pre test and
post test of the experimental group were 70 and 86,7, the lowest scores pre test
and post test of the experimental group were 40 and 56,7. In addition, the mean
score pre test and post test of the control group were 49,7 and 65,3. Meanwhile,
the highest score pre test and post test of the control group were 70 and 76,7. The
lowest scores pre test and post test of the control group were 40 and 53,3. Based
on the data above, the difference of mean score between experimental and control
group score were 5.3.
63
B. Testing of Normality and Homogeinity
1. Normality Test
a. Testing normality of pre-test experimental and control group
Table 4.11 Testing normality of pre-test experimental and control group
Tests of Normality
a
Kolmogorov-Smirnov
Shapiro-Wilk
Group
Statistic
experiment
Scores
Df
Sig.
Statistic
df
Sig.
*
,920
22
,676
,863
24
,114
,142
22
,200
,168
24
,076
group
control group
The table showed the result of test normality calculation using SPSS 21.0
program. To know the normality of data, the formula could be seen as follows:
If the number of sample. > 50 = Kolmogorov-Smirnov
If the number of sample. < 50 = Shapiro-Wilk
Based on the number of data the writer was 46 < 50, so to analyzed normality
data was used Shapiro-Wilk. The next step, the writer analyzed normality of data
used formula as follows:
If Significance > 0.05 = data is normal distribution
If Significance < 0.05 = data is not normal distribution
Based on data above, significant data of experiment and control group used
Shapiro-Wilk was 0.676 > 0.05 and 0.114 > 0.05. It could be concluded that the data
was normal distribution.
b. Testing normality of pre-test experimental and control group
64
Table 4.12Testing normality of post-test experimental and control group
Tests of Normality
a
Kolmogorov-Smirnov
Group
Scores
Statistic
df
Shapiro-Wilk
Sig.
Statistic
df
Sig.
experiment group
.122
22
.200
*
.971
22
.744
control group
.174
24
.058
.933
24
.115
The table showed the result of test normality calculation using SPSS 21.0
program. To know the normality of data, the formula could be seen as follows:
If the number of sample. > 50 = Kolmogorov-Smirnov
If the number of sample. < 50 = Shapiro-Wilk
Based on the number of data the writer was 46 < 50, so to analyzed normality
data was used Shapiro-Wilk. The next step, the writer analyzed normality of data
used formula as follows:
If Significance > 0.05 = data is normal distribution
If Significance < 0.05 = data is not normal distribution
Based on data above, significant data of experiment and control group used
Shapiro-Wilk was 0.744 > 0.05 and 0.115 > 0.05. It could be concluded that the data
was normal distribution.
65
2. Homogeneity Test
a. Testing Homogeneity of pre-test experimental and control group
Table 4.13Testing Homogeneity of pre-test experimental and control
group
Homogeneity Test
Levene's Test for
t-test for Equality of Means
Equality of
Variances
F
Sig.
T
Df
Sig. (2-
Mean
Std. Error
99% Confidence
tailed)
Differen
Difference
Interval of the
ce
Difference
Lower
Equal
,743
,393
,584
44
,562
1,63826
2,80478
variances
Upper
-
9,18951
5,91299
assumed
Scores
Equal
,587 43,985
,560
1,63826
2,79147
variances not
5,87727
assumed
The table showed the result of Homogeneity test calculation using SPSS 21.0
program. To know the Homogeneity of data, the formula could be seen as follows:
If Sig. > 0,01 = Equal variances assumed or Homogeny distribution
If Sig. < 0,01 = Equal variances not assumedor not Homogeny distribution
Based on data above, significant data was 0,393. The result was 0,393 > 0,01,
it meant the t-test calculation used at the equal variances assumed or data was
Homogeny distribution.
9,15378
66
b. Testing Homogeneity of post-test experimental and control group
Table 4.14Testing Homogeneity of post-test experimental and control
group
Homogeneity Test
Levene's Test for
Equality of
Variances
t-test for Equality of Means
99% Confidence
Interval of the
Difference
Sig. (2-
F
Scores
Equal variances
.607
Sig.
.440
t
3.352
df
tailed)
44
.002
Mean
Std. Error
Difference Difference
7.00644
2.09010
Lower
1.37931 12.633
assumed
Equal variances
57
3.320 40.219
.002
7.00644
2.11064
1.29985 12.713
not assumed
The table showed the result of Homogeneity test calculation using SPSS 21.0
program. To know the Homogeneity of data, the formula could be seen as follows:
If Sig. > 0,01 = Equal variances assumed or Homogeny distribution
If Sig. < 0,01 = Equal variances not assumedor not Homogeny distribution
Based on data above, significant data was 0,440. The result was 0,440 > 0,01,
it meant the t-test calculation used at the equal variances assumed or data was
Homogeny distribution.
Upper
03
67
C. The Result of Data Analysis
1. Testing Hypothesis Using Manual Calculation
Table 4.15 The Standard Deviation and the Standard Error of Experiment
and Control Group
Group
Standard Deviation
Standard Error
Experimental Group
7.74545
1.69115
Control Group
6.5925
1.37634
The table showed the result of the standard deviation calculation of
Experiment group was 7.74545and the result of the standard error was 1.69115.
The result of thestandard deviation calculation of Control group was 6.5925 and
the result of standard error was 1.37634. To examine the hypothesis, the writer
used the formula as follow:
tobserved=
=
=
df
𝑀1−𝑀2
𝑆𝐸𝑚 1−𝑆𝐸𝑚 2
7.74545 −6.5925
1.69115 −1.37634
1.15295
0.31481
= 3.663
= (N1 + N2 – 2)
= 22+24-2
= 44
68
a. Interpretation
The result of t – test was interpreted on the result of degree of freedom to get
the ttable. The result of degree of freedom (df) was 44. The following table was the
result of tobserved and ttable from 44 df at 5% and 1% significance level.
Table 4.16 The Result of T-Test Using Manual Calculation
t-table
t-observe
3.663
Df
5 % (0,05)
1 % (0,01)
2.021
2.704
44
The interpretation of the result of t-test using manual calculation, it was found
the tobserved was higher than the ttable at 5% and 1% significance level or 3.663 > 2.021,
3.663 > 2.704. It meant Ha was accepted and Ho was rejected. It could be interpreted
based on the result of calculation that Ha stating that K-W-L strategy was effective
for Teaching Reading Comprehension of the seventh grade students at SMP
Al-Amin Palangka Raya was accepted and Ho stating thatK-W-L strategy was
not effective for Teaching Reading Comprehension of the seventh grade
students at SMP Al-Amin Palangka Raya was rejected. It meant that teaching
reading with K-W-L Strategy Toward Reading Comprehension for the seventh
grade students at SMP Al-Amin Palangka Raya gave significant effect at 5% and
1% significance level.
2. Testing Hypothesis Using SPSS 21.0 Program
The writer also applied SPSS 21.0 program to calculate t – test in testing
hypothesis of the study. The result of t – test using SPSS 21.0 was used to support
69
the manual calculation of t – test. The result of t – test using SPSS 21.0 program
could be seen as follows:
Table 4.17 Mean, Standard Deviation and Standard Error of experiment
group and control groupusing SPSS 21.0 Program
Group Statistics
Group
Scores
N
Mean
Std. Deviation
Std. Error Mean
experiment group
22
72.2773
7.86032
1.67583
control group
24
65.2708
6.28597
1.28312
The table showed the result of mean calculation of experiment groupwas
72.2773, standard deviation calculation was 7.86032, and standard error of mean
calculation was 1.67583. The result of mean calculation of control groupwas
65.2708, standard deviation calculation was 6.28597, and standard error of mean
was 1.28312.
Table 4.18 The Calculation of T – Test Using SPSS 21.0
Independent Samples Test
Levene's Test for
Equality of
Variances
t-test for Equality of Means
99% Confidence
Interval of the
Difference
Sig. (2-
F
Scores
Equal variances
.607
Sig.
.440
T
3.352
Df
tailed)
44
.002
Mean
Difference Difference
7.00644
assumed
Equal variances
not assumed
Std. Error
Lower
Upper
2.09010 1.37931 12.6335
7
3.320 40.219
.002
7.00644
2.11064 1.29985 12.7130
3
70
The table showed the result of t – test calculation using SPSS 21.0 program.
To know the variances score of data, the formula could be seen as follows:
If Sig. > 0,01 = Equal variances assumed
If Sig. < 0,01 = Equal variances not assumed
Based on data above, significant data was 0,440. The result was 0,440 > 0,01,
it meant the t-test calculation used at the equal variances assumed. It found that the
result of tobserved was 3.352, the result of mean difference between experiment and
control group was 7.00644, and thestandard error difference between experiment and
control group was 2.09010.
a. Interpretation
The result of t – test was interpreted on the result of degree of freedom to get
the ttable. The result of degree of freedom (df) was 44. The following table was the
result of tobserved and ttable from 44 df at 5% and 1% significance level.
Table 4.19 The Result of T-Test Using SPSS 21.0 Program
t-table
t-observe
3.352
Df
5 % (0,05)
1 % (0,01)
2.021
2.704
44
The interpretation of the result of t-test using SPSS 21.0 program, it was
found the tobserved was higher than the ttable at 5% and 1% significance level or 3.352 >
2.021, 3.352 > 2.704. It meant Ha was accepted and Ho was rejected. It could be
interpreted based on the result of calculation that Ha stating that K-W-L strategy
was effective for Teaching Reading Comprehension of the seventh grade
71
students at SMP Al-Amin Palangka Raya was accepted and Ho stating thatKW-L strategy was not effective for Teaching Reading Comprehension of the
seventh grade students at SMP Al-Amin Palangka Raya was rejected. It meant
that teaching reading with K-W-L Strategy Toward Reading Comprehension for
the seventh grade students at SMP Al-Amin Palangka Raya gave significant
effect at 5% and 1% significance level.
D. Discussion
The result of analysis showed that there was significant effect of K-W-L
strategy Toward Reading Comprehension forthe seventh grade students at
SMP Al-Amin Palangka Raya.The students who were taught used K-W-L
strategyreached higher score than those who were taught without used K-WL strategy.
Meanwhile, after the data was calculated using manual calculation of ttest.It
was found the tobserved was higher than the ttable at 5% and 1% significance level or
3.663 > 2.021, 3.663 > 2.704. It meant Ha was accepted and Ho was rejected. And the
data calculated using SPSS 21.0 program, it was found the tobserved was higher than
the ttable at 5% and 1% significance level or 3.352 > 2.021, 3.352 > 2.704. It meant Ha
was accepted and Ho was rejected. This finding indicated that the alternative
hypothesis (Ha) stating that there was any significant effect of K-W-L strategy
Toward Reading Comprehension for the seventh grade students at SMP AlAmin Palangka Raya was accepted.On the contrary,the Null hypothesis (Ho)
stating that there was no any significant effect of K-W-L strategy Toward
Reading Comprehension for the seventh grade students at SMP Al-Amin
72
Palangka Raya was rejected.Based on the result the data analysis showed that
using K-W-L Strategy gave significance effect for the students’ reading
comprehension scores of Seventh grade students at SMP Al-amin Palangka Raya.
After the students have been taught by using K-W-L Strategy, the reading
score were higher than before implementing K-W-L Strategy as a learning
strategy. It can be seen in the comparison of pre test and post test score of
experimental group and control group (See p.61). This finding indicated that KW-L strategy was effective and supports the previous research done by Zhang
Fengjuan, Eviani Damastuti and Anne Crout Shelleythat also stated teaching
reading by using K-W-L strategy was effective.
There were some reason why using K-W-L Strategy gave significance effect
for the students’ reading comprehension scores of Seventh grade students at SMP
Al-amin Palangka Raya.First, K-W-L Strategy was effective in terms of
improving the students’ English reading score. It can be seen from the
improvement of the students’ score average in the post-test. From the mean
score of control and experiment were 72.3 and 65.3. (See p.62).It supports the
previous study by Eviani Damastuti and Sugini states that using K-W-L strategy
increase in reading ability significantly intensified.
It was suitable withthe result of pre-test and post test for Experiment and
control Group. (See p.44). In the pre-test of experiment group there wereseven
students that got fail predicate. They were E-01, E-02, E-06, E-16, E-19, E-20 and
E-21. There were eleven students that got less predicate. They were E-05, E-07,
E-08, E-09, E-10, E-12, E-13, E-14, E-15, E-18 and E-22. There were three
73
students that got enough predicate. They are E-03, E-11 and E-17. There was one
student that got good predicate. She was E-04. Then, in the pre-test score of
control group there were thirteen students that got fail predicate. They were C-01,
C-02, C-03, C-04, C-05, C-06, C-08, C-14, C-18, C-20, C-21, C-22 and C-23.
There were six students that got less predicate. They were C-11, C-13, C-15, C16, C-17 and C-24. There were three students that got enough predicate. They
were C-07, C-09 and C-19. There were two students that got good predicate. They
were C-10 and C-12.
Based on the result of post-test for experimental and control group, (See
p.53). In the experimental group, there was no student that got in fail predicate.
There was one student that got less predicate, he was E-06. There were five
students that got enough predicate. They were E-01, E-07, E-12, E-18 and E-18.
There were fifteen students that got good predicate. They were E-02, E-03, E-05,
E-08, E-09, E-10, E-11, E-13, E-14, E-15, E-16, E-17, E-20, E-21 and E-22.
There was one student that got excellent predicate, she was E-04. In the control
group, there was no student that got in fail predicate. There were two students that
got less predicate. They were C-01 and C-24. There were thirteen students that got
enough predicate. They were C-02, C-03, C-04, C-05, C-06, C-07, C-09, C-11, C12, C-17, C-18, C-20 and C-23. There were nine students that got good predicate.
They were C-08, C-10, C-13, C-14, C-15, C-16, C-19, C-21 and C-22.
The next reason wasK-W-L strategycan motivate students in teaching
learning process. It was suitable with the students response when learning process
is going, they enthusiasm to wrote in coloum (K) for what they Know about the
74
topic with their background knowledge. It was necessary to keep responses inside
the topic. It indicated that using K-W-L strategy was effective in enhance reading
motivation and encouragement. (See appendix 6). It supports with Ferdinand
Nicholas Boonde states that K-W-L strategycan motivate the students to take a
part in the teaching learning process and Filling the columns is effective to help
the students understand the reading text.
The last reason was K-W-L strategy gave the students can answer both
literal and inferential reading comprehension types. It indicated the test was
suitable for junior high school student (see appendix4). It supports with Zhang
FengjuanK-W-L strategy is an instructional scheme that develops active reading of
descriptive texts by activating learners’ background knowledge.
Those are the result of pre-test compared with post-test for experimental
group and control group of students at SMP Al-Amin Palangka Raya. Based on
the theories and the writer’s result,K-W-L Strategy gave significance effect for the
students’ reading comprehension scores of Seventh grade students at SMP Alamin Palangka Raya.