The effect of context clues in reading comprehension of the eleventh grade student at SMA Negeri 3 Palangka Raya - Digital Library IAIN Palangka Raya
48
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 effect of using Context Clues
in teaching reading Narrative 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
CORRECT
Control Group
PREDICATE
CODE
SCORE
ANSWER
E-01
E-02
E-03
E-04
E-05
E-06
E-07
E-08
E-09
E-10
E-11
E-12
40
70
40
43,3
40
50
40
40
46,7
40
40
43,3
12
21
12
13
12
15
12
12
14
12
12
13
CORRECT
PREDICATE
ANSWER
FAIL
GOOD
FAIL
FAIL
FAIL
LESS
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
48
C-01
C-02
C-03
C-04
C-05
C-06
C-07
C-08
C-09
C-10
C-11
C-12
40
53,3
53,3
56,7
40
43,3
40
56,7
56,7
50
53,3
46,7
12
16
16
17
12
13
12
17
17
15
16
14
FAIL
LESS
LESS
LESS
FAIL
FAIL
FAIL
LESS
LESS
LESS
LESS
FAIL
49
Experimental Group
Code
Score
E-13
70
E-14
56,7
E-15
40
E-16
70
E-17
53,3
E-18
53,3
E-19
40
E-20
40
E-21
46,7
E-22
50
E-23
46,7
E-24
43,3
E-25
40
E-26
43,3
E-27
56,7
E-28
53,3
E-29
46,7
E-30
40
E-31
50
E-32
40
E-33
46,7
E-34
43,3
E-35
43,3
E-36
43,3
E-37
40
E-38
40
E-39
43,3
E-40
53,3
TOTAL
AVERAGE
Lowest Score
Highest Score
CORRECT
ANSWER
Control Group
PREDICATE
21
17
12
21
16
16
12
12
14
15
14
13
12
13
17
GOOD
LESS
FAIL
GOOD
LESS
LESS
FAIL
FAIL
FAIL
LESS
FAIL
FAIL
FAIL
FAIL
LESS
16
LESS
14
12
15
12
14
13
13
13
12
12
13
16
FAIL
FAIL
LESS
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
LESS
1866,5
46,7
40
70
Code
Score
40
40
40
40
40
40
40
53,3
70
40
40
40
40
40
40
43,3
40
43,3
46,7
43,3
40
50
46,7
40
TOTAL
AVERAGE
Lowest Score
Highest Score
C-13
C-14
C-15
C-16
C-17
C-18
C19
C-20
C-21
C-22
C-23
C-24
C-25
C-26
C-27
C-28
C-29
C-30
C-31
C-32
C-33
C-34
C-35
C-36
CORRECT
ANSWER
12
12
12
12
12
12
12
16
15
12
12
12
12
12
12
13
12
13
14
13
12
15
14
12
PREDICATE
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
LESS
GOOD
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
LESS
FAIL
FAIL
1626,6
45,2
40
70
50
Experimental Group
Control Group
Predicate
Percentages
Predicate
Percentages
Fail
28
Fail
26
Less
9
Less
9
Enough
0
Enough
0
Good
3
Good
1
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 (XI IPS-1)
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
The Range of Score (R)
=H–L+1
= 70 – 40+ 1
= 31
The Class Interval (K)
= 1 + (3.3) x Log n
51
= 1 + (3.3) x Log 40
= 1 + 5.2867979
= 6.2867979
=6
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 4.2 Frequency Distribution of the Pretest Score
Interval (I)
Frequency
(F)
Mid
Point
(x)
The
Limitation
of each
group
Frequency
Relative (%)
Frequency
Cumulative
(%)
1
2
66-70
61-65
3
0
68
63
66.5-70.5
61.5-65.5
7.5
0
100
92.5
3
56-60
2
58
56.5-60.5
5
92.5
4
51-55
5
53
51.5-55.5
12.5
87.5
5
46-50
7
48
46.5-50.5
17.5
75
6
40-45
23
42.5
40.5-45.5
Class
(K)
∑F=40
57.5
∑P = 100
57.5
The distribution of students’ predicate in pretest score of Experimental
group can also be seen in the following figure.
52
The Distribution of Student's Predicate
30
25
20
15
10
5
0
FAIL
LESS
ENOUGH
GOOD
Figure 4.1 The distribution frequency of students’ pretest score for
Experimental Group
Based on the figure above, it can be seen about the students’ predicate in
pretest score. There were twenty eight students who got Fail predicate. There were
ninestudents who gotLess Predicate.
There wasno students who gotEnough
predicate. There werethree 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:
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
66-70
3
2
61-65
0
F.X
FX2
Fka
Fkb
68
204
13872
3
40
63
0
0
3
37
53
3
56-60
2
58
116
6728
5
34
4
51-55
5
53
265
14045
10
29
5
46-50
7
48
336
16128
17
19
6
40-45
23
42.5
977.5
41543.75
40
2
∑=1898.5
∑=92316,75
∑F = 40
∑ Total
1) Calculating Mean
Mx=
∑𝐹𝑋 𝑖
𝑛
=
1898.5
= 46.6530
40
2) Median
1
= ℓ +2
Mdn
𝑁−𝑓𝑘𝑏
𝑓𝑖
1
40−2
2
= 39.5 +
= 39.5 +
𝑋𝑖
23
18
23
X5
𝑋5
= 201.41
3) Modus
Mo
=u+
𝑓𝑎
𝑓𝑎 +𝑓𝑏
= 39.5 +
16
𝑥𝑖
16+23
= 39.5 + 2.05
= 41.55
4) Standard Deviation
S=
S=
𝑛.∑𝐹𝑋 𝑖 2 − ∑𝐹𝑋 𝑖 2
𝑛 𝑛 −1
40.92316 .75− 92316 .75 2
40 40−1
𝑥5
54
S = 8.51868
5) Standard Error
SEmd =
𝑆
𝑁−1
=
8.51868
40−1
=
8.51868
6.24
= 1.36517
After Calculating, it was found that the standard deviation and the
standard error of pretest score were 8.51868 and 1.36517
b. The Result of Pretest Score of ControlGroup (XI IPS-2)
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 36
= 1 + 5.1357966
= 6.1357966
=6
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:
55
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
1
0
68
63
66.5-70.5
61.5-65.5
2.7
0
100
97.3
3
4
56-60
51-55
3
6
58
53
56.5-60.5
51.5-55.5
8.3
16.6
97.3
89
5
46-50
4
48
46.5-50.5
11.3
72.4
6
40-45
22
42.5
40.5-45.5
Class
(K)
∑F=36
61.1
∑P = 100
61.1
The distribution of students’ predicate in pretest score of Control group
The Distribution of Student's Predicate
30
25
20
15
10
5
0
FAIL
LESS
ENOUGH
GOOD
Figure 4.2 The distribution of students’ predicate in pretest score
of Control Group
Based on the figure above, it can be seen about the students’ predicate in
pretest score. There twenty sixstudents who got Fail predicate. There were
56
ninestudents who gotLess Predicate. There was no students who got Enough
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, 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
1
2
61-65
0
Class
(K)
Interval
(I)
3
56-60
3
4
51-55
5
46-50
6
40-45
∑ Total
F.X
FX2
Fka
Fkb
68
4624
0
14
0
0
1
14
F.X
FX2
Fka
Fkb
58
174
10092
1
13
6
53
318
16854
4
12
4
48
192
9216
10
8
22
∑F = 36
42.5
935
39737.5
14
2
68
63
Mid
Frequency
Point
(F)
(x)
∑=1687
1) Calculating Mean
Mx=
2) Median
Mdn
∑𝐹𝑋 𝑖
𝑛
=
1687
36
= 45.8611
1
𝑁−𝑓𝑘𝑏
2
=ℓ+
𝑓𝑖
1
= 39.5 + 2
𝑋𝑖
36−2
22
𝑋5
∑=80523.5
57
= 39.5 +
16
22
X5
= 43.13
3) Modus
Mo
𝑓𝑎
=u+
𝑥𝑖
𝑓𝑎 +𝑓𝑏
= 39.5 +
14
14+2
𝑥5
= 39.5 + 4.375
= 43.87
4) Standard Deviation
S=
S=
𝑛.∑𝐹𝑋 𝑖 2 − ∑𝐹𝑋 𝑖 2
𝑛 𝑛 −1
36.80523 .5− 80523 .5 2
36 36−1
S = 7.54464
5) Standard Error
SEmd =
𝑆
𝑁−1
=
7.54464
36−1
=
7.54464
5.91
= 1.27658
After Calculating, it was found that the standard deviation and the
standard error of pretest score were 7.54464 and 1.27658.
58
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
E-23
E-24
E-25
E-26
E-27
56,7
73,3
73,3
60
70
66,7
56,7
60
76,7
60
63,3
60
86,7
70
60
63,3
60
56,7
70
76,7
66,7
70
56,7
66,7
73,3
60
56,7
CORRECT
ANSWER
17
22
22
18
21
20
17
18
23
18
19
18
26
21
18
19
18
17
21
23
20
21
17
20
22
18
17
Control Group
PREDICATE
CODE
SCORE
LESS
GOOD
GOOD
ENOUGH
GOOD
ENOUGH
LESS
ENOUGH
GOOD
ENOUGH
ENOUGH
ENOUGH
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
C-25
C-26
C-27
53,3
60
60
60
56,7
53,3
63,3
70
56,7
53,3
66,7
66,7
53,3
60
56,7
66,7
70
76,7
60
66,7
53,3
56,7
53,3
56,7
60
66,7
70
EXCELLENT
GOOD
ENOUGH
ENOUGH
ENOUGH
LESS
GOOD
GOOD
ENOUGH
GOOD
LESS
ENOUGH
GOOD
ENOUGH
LESS
CORRECT
ANSWER
PREDICATE
16
18
18
18
17
16
19
21
17
16
20
20
16
18
17
20
21
23
18
20
16
17
16
17
LESS
ENOUGH
ENOUGH
ENOUGH
LESS
LESS
ENOUGH
GOOD
LESS
LESS
ENOUGH
ENOUGH
LESS
ENOUGH
LESS
ENOUGH
GOOD
GOOD
ENOUGH
ENOUGH
LESS
LESS
LESS
LESS
18
20
21
ENOUGH
ENOUGH
GOOD
59
E-28
60
E-29
56,7
E-30
70
E-31
70
E-32
66,7
E-33
60
E-34
70
E-35
56,7
E-36
63,3
E-37
56,7
E-38
60
E-39
73,3
E-40
56,7
TOTAL
AVERAGE
Lowest Score
Highest Score
18
ENOUGH
17
21
21
20
18
21
17
19
17
18
22
17
LESS
GOOD
GOOD
ENOUGH
ENOUGH
GOOD
LESS
ENOUGH
LESS
ENOUGH
GOOD
LESS
2590,3
64,8
56,7
86,7
53,3
60
63,3
53,3
60
53,3
56,7
66,7
53,3
TOTAL
AVERAGE
Lowest Score
Highest Score
C-28
C-29
C-30
C-31
C-32
C-33
C-34
C-35
C-36
Experimental Group
16
18
19
16
18
16
17
20
16
LESS
ENOUGH
ENOUGH
LESS
ENOUGH
LESS
LESS
ENOUGH
LESS
2166,7
60,2
53,3
76,7
Control Group
Predicate
Percentages
Predicate
Percentages
Fail
0
Fail
26
Less
9
Less
9
Enough
17
Enough
0
Good
13
Good
1
Excellent
1
Excellent
0
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
60
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 (XI IPS-1)
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 40
= 1 + 5.2867979
= 6.2867979
=6
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:
61
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
86.7-91.7
80.7-85.7
1
0
89.2
83.2
86.2-91.2
80.2-85.2
2.5
0
100
97.5
3
4
74.7-79.7
68.7-73.7
2
11
77.2
71.2
74.2-79.2
68.2-73.2
5
27.5
0
92.5
5
62.7-67.7
7
65.2
62.2-67.2
17.5
75
6
56.7-61.7
19
59.2
56.2-61.2
Class
(K)
∑F=40
47.5
∑P = 100
27.5
The distribution of students’ predicate in post-test score of Experimental
group.
The Distribution of Student's Predicate
18
16
14
12
10
8
6
4
2
0
LESS
ENOUGH
GOOD
EXCELLENT
Figure 4.3 The distribution of students’ predicate in post-test score
of Experimental Group
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
wereninestudentswho got Less predicate. There were seventeenstudents who
62
gotEnoughPredicate. There were thirteen 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
86.7-91.7
1
2
80.7-85.7
3
F.X
FX2
Fka
Fkb
89.2
89.2
7956.64
10
40
0
83.2
0
0
10
30
74.7-79.7
2
77.2
154.4
11919.68
12
20
4
68.7-73.7
11
71.2
783.2
55763.84
14
8
5
62.7-67.7
7
65.2
456.4
29757.28
21
4
6
56.7-61.7
∑ Total
19
∑F = 40
59.2
1124.8
66588.16
40
2
∑=2608
1) Calculating Mean
Mx=
2) Median
∑𝐹𝑋 𝑖
𝑛
=
2608 .5
40
1
Mdn
= ℓ +2
=64.2125
𝑁−𝑓𝑘𝑏
𝑓𝑖
= 56.2 +
= 56.2 +
= 4.7368
𝑋𝑖
1
40−2
2
19
18
19
X5
𝑋5
∑=171985.6
63
3) Modus
Mo
𝑓𝑎
=u+
𝑓𝑎 +𝑓𝑏
= 56.2+
40
𝑥𝑖
40+2
𝑥5
= 56.2 + 4.7619
= 60.9619
4) Standard Deviation
S=
S=
𝑛.∑𝐹𝑋 𝑖 2 − ∑𝐹𝑋 𝑖 2
𝑛 𝑛 −1
40.6801664 .5− 2608 2
40 40−1
S = 7.64816
5) Standard Error
SEmd =
𝑆
𝑁−1
=
7.64816
40−1
=
7.64816
6.24
= 1.22566
After Calculating, it was found that the standard deviation and the standard
error of pretest score were7.64816 and 1.22566
b. The Result of Post-test Score of Control Group (XI IPS-2)
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:
The Highest Score (H)
= 76.7
The Lowest Score (L)
= 53.3
The Range of Score (R)
=H–L+1
64
= 76.7 – 53.3 + 1
= 24.4
The Class Interval (K)
= 1 + (3.3) x Log n
= 1 + (3.3) x Log 36
= 1 + 5.1357928
= 6.1357928
=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
1
80.3
75.3
77.8-82.8
72.8-77.8
0
2.7
100
100
3
4
68.3-72.3
63.3-67.3
4
8
70.3
65.3
67.8-72.8
62.8-67.8
11.1
22.2
97.3
86.2
5
58.3-62.3
7
60.3
57.8-62.8
19.5
64
6
53.3-57.3
16
55.3
52.8-57.8
44.5
∑P = 100
44.5
Class
(K)
∑F=36
65
The distribution of students’ predicate in post-test score of Control Group.
The Distribution of Student's Predicate
18
16
14
12
10
8
6
4
2
0
LESS
ENOUGH
GOOD
EXCELLENT
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 sixteen
studentswho got Less predicate. There were sixteenstudents who gotEnough
Predicate. There were four 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:
66
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
1
36
1
75.3
75.3
5670.09
1
35
68.3-72.3
4
70.3
281.2
19768.36
5
34
4
63.3-67.3
8
65.3
522.4
34112.72
13
29
5
58.3-62.3
7
60.3
422.1
25452.63
20
16
6
53.3-57.3
∑ Total
16
∑F = 36
55.3
884.8
48929.44
36
4
∑=2185.8 ∑=133933.24
1) Calculating Mean
Mx=
2) Median
∑𝐹𝑋 𝑖
𝑛
=
2185 .8.16
36
1
Mdn
= ℓ +2
=60.7166
𝑁−𝑓𝑘𝑏
𝑓𝑖
= 52.8 +
= 52.8 +
𝑋𝑖
1
36−4
2
𝑋4
16
14
16
X4
= 56.3
3) Modus
Mo
=u+
𝑓𝑎
𝑓𝑎 +𝑓𝑏
= 52.8 +
36
36+4
= 52.8 + 3.6
= 56.4
𝑥𝑖
𝑥4
67
4) Standard Deviation
S=
S=
𝑛.∑𝐹𝑋 𝑖 2 − ∑𝐹𝑋 𝑖 2
𝑛 𝑛 −1
36.4777721 .64.4− 2185 .8 2
36 36−1
S = 6.48771
5) Standard Error
SEmd =
𝑆
𝑁−1
=
6.48771
36−1
=
6.48771
5.91
= 1.09775
After Calculating, it was found that the standard deviation and the
standard error of pretest score were 6.48771 and 1.09775
3. The Comparison result of Pre-test and Post-test of Experimental and
Control Group
NO
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
EXPERIMENTAL CLASS
SCORE
CODE
PRE- POST- DIFFE
TEST TEST RENCE
E-01
40
56,7
16,7
E-02
70
73,3
3,3
E-03
40
73,3
33,3
E-04
43,3
60
16,7
E-05
40
70
30
E-06
50
66,7
16,7
E-07
40
56,7
16,7
E-08
40
60
20
E-09
46,7
76,7
30
E-10
40
60
20
E-11
40
63,3
23,3
E-12
43,3
60
16,7
E-13
70
86,7
16,7
E-14
56,7
70
13,3
E-15
40
60
20
NO
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
CONTROL CLASS
SCORE
CODE
PREPOST- DIFFE
TEST
TEST RENCE
C-01
40
53,3
13,3
C-02
53,3
60
6,7
C-03
53,3
60
6,7
C-04
56,7
60
3,3
C-05
40
56,7
16,7
C-06
43,3
53,3
10
C-07
40
63,3
23,3
C-08
56,7
70
13,3
C-09
56,7
56,7
0
C-10
50
53,3
3,3
C-11
53,3
66,7
13,4
C-12
46,7
66,7
20
C-13
40
53,3
13,3
C-14
40
60
20
C-15
40
56,7
16,7
68
E-16
E-17
E-18
E-19
E-20
E-21
E-22
E-23
E-24
E-25
E-26
E-27
E-28
E-29
E-30
E-31
E-32
E-33
E-34
E-35
E-36
E-37
E-38
E-39
E-40
TOTAL
MEAN
LOWEST
HIGHEST
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
70
53,3
53,3
40
40
46,7
50
46,7
43,3
40
43,3
56,7
53,3
46,7
40
50
40
46,7
43,3
43,3
43,3
40
40
43,3
53,3
1866,5
46,7
40
70
63,3
60
56,7
70
76,7
66,7
70
56,7
66,7
73,3
60
56,7
60
56,7
70
70
66,7
60
70
56,7
63,3
56,7
60
73,3
56,7
2590,3
64,8
56,7
86,7
-6,7
6,7
3,4
30
36,7
20
20
10
23,4
33,3
16,7
0
6,7
10
30
20
26,7
13,3
26,7
13,4
20
16,7
20
30
3,4
723,8
18,1
C-16
16
C-17
17
C-18
18
C-19
19
C-20
20
C-21
21
C-22
22
C-23
23
C-24
24
C-25
25
C-26
26
C-27
27
C-28
28
C-29
29
C-30
30
C-31
31
C-32
32
C-33
33
C-34
34
C-35
35
C-36
36
TOTAL
MEAN
LOWEST
HIGHEST
40
40
40
40
53,3
70
40
40
40
40
40
40
43,3
40
43,3
46,7
43,3
40
50
46,7
40
1626,6
45,2
40
70
66,7
70
76,7
60
66,7
53,3
56,7
53,3
56,7
60
66,7
70
53,3
60
63,3
53,3
60
53,3
56,7
66,7
53,3
2166,7
60,2
53,3
76,7
From the table above the mean score of pre test and post test of the
experimental group were 46,7 and 64,8. 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 45,2 and 60,2. Meanwhile,
26,7
30
36,7
20
13,4
-16,7
16,7
13,3
16,7
20
26,7
30
10
20
20
6,6
16,7
13,3
6,7
20
13,3
540,1
15
69
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 3.1
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
studentsname
a
Kolmogorov-Smirnov
Statistic
Df
Shapiro-Wilk
Sig.
Statistic
Df
Sig.
experiment group
,229
40
,240
,765
40
,567
control group
,290
36
,176
,745
36
,115
Students score
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 76> 50, so to analyzed normality
data was used Kolmogorov-Smirnov. 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
70
Based on data above, significant data of experiment and control group used
Kolmogorov-Smirnov was 0.240> 0.05 and 0.176> 0.05. It could be concluded that
the data was normal distribution.
b. Testing normality of post-test experimental and control group
Table 4.12Testing normality of post-test experimental and control group
Tests of Normality
studentsname
a
Kolmogorov-Smirnov
Statistic
Df
Shapiro-Wilk
Sig.
Statistic
df
Sig.
experiment group
,217
40
,112
,890
40
,054
control group
,178
36
,076
,892
36
,064
Students score
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 76 > 50, so to analyzed normality
data was usedKolmogorov-Smirnov. 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
Kolmogorov-Smirnov was 0.112> 0.05 and 0.076> 0.05. It could be concluded that
the data was normal distribution.
2. Homogeneity Test
a. Testing Homogeneity of pre-test experimental and control group
71
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
Equal
,153
Sig.
,697
T
,808
Df
Sig.
Mean
Std. Error
99% Confidence Interval
(2-
Differen
Difference
of the Difference
tailed)
ce
Lower
Upper
74
,422
1,47167
1,82164
-3,34459
6,28793
,814 73,855
,418
1,47167
1,80711
-3,30643
6,24977
variances
Studentss assumed
core
Equal
variances 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 0,697. The result was 0,697> 0,01,
it meant the t-test calculation used at the equal variances assumed or data was
Homogeny distribution.
72
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
t-test for Equality of Means
Equality of
Variances
F
Sig.
T
Df
Sig. (2-
Mean
Std. Error
99% Confidence
tailed)
Difference
Difference
Interval of the
Difference
Lower
Equal
1,888
,174
Upper
2,910
74
,005
4,57167
1,57100
,41808
8,72525
2,933
73,855
,004
4,57167
1,55847
,45101
8,69233
variances
Students
assumed
score
Equal
variances 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 0,174. The result was 0,174> 0,01it
meant the t-test calculation used at the equal variances assumed or data was
Homogeny distribution.
73
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
Mean
Standard Error
Experimental Group
64.2125
1.22566
Control Group
60.7166
1.09775
The table showed the result of the standard deviation calculation of
Experiment group was 7.64816and the result of the standard error was 1.22566 .
The result of thestandard deviation calculation of Control group was 6.48771and
the result of standard error was 1.09775. To examine the hypothesis, the writer
used the formula as follow:
tobserved=
=
=
df
𝑀1−𝑀2
𝑆𝐸𝑚 1−𝑆𝐸𝑚 2
64.2125 −60.7166
1.22566 −1.09775
3.4959
0.12791
= 2.733
= (N1 + N2 – 2)
= 40+36-2
= 74
74
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 74. The following table was the
result of tobserved and ttable from 74 df at 5% and 1% significance level.
Table 4.16 The Result of T-Test Using Manual Calculation
t-table
t-observe
2.733
Df
5 % (0,05)
1 % (0,01)
2.000
2.660
74
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 2.733> 2.000,
2.733> 2.660.It meant Ha was accepted and Ho was rejected. It could be interpreted
based on the result of calculation that Ha stating that Context Clues was effective
for Teaching Reading Comprehension of the elevent grade students at SMA
Negeri 3 Palangka Raya was accepted and Ho stating that Context Clues was not
effective for Teaching Reading Comprehension of the eleventh grade students
at SMA Negeri 3 Palangka Raya was rejected. It meant that teaching reading with
Context Clues in Reading Comprehension of the eleventh grade students at
SMA Negeri 3 Palangka Raya gave significant effect at 5% and 1% significance
level.
75
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
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
Studentsname
N
Mean
Std. Deviation
Std. Error Mean
40
64,7550
7,30426
1,15491
36
60,1833
6,27856
1,04643
experiment group
Student score
control group
The table showed the result of mean calculation of experiment groupwas
64.7550, standard deviation calculation was 7.30426, and standard error of mean
calculation was 1.15491. The result of mean calculation of control
groupwas 60.1833, standard deviation calculation was 6.27856, and standard error
of mean was1.04643.
76
Table 4.18 The Calculation of T – Test Using SPSS 21.0
Independent Samples Test
Levene's Test for
t-test for Equality of Means
Equality of Variances
F
Sig.
T
Df
Sig. (2-
Mean
Std.
99% Confidence
tailed)
Difference
Error
Interval of the
Differen
Difference
ce
Equal
1,888
,174
2,910
74
,005
4,57167
1,57100
Lower
,41808 8,7252
variances
Students
assumed
score
Equal
Upper
5
2,933 73,855
,004
4,57167
1,55847
,45101 8,6923
variances
not assumed
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 0,174. The result was 0,174> 0,01,
it meant the t-test calculation used at the equal variances assumed. It found that the
result of tobserved was 2.910, the result of mean difference between experiment and
control group was 4.57167, and thestandard error difference between experiment and
control group was 1.57100.
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 74. The following table was the
result of tobserved and ttable from 74 df at 5% and 1% significance level.
3
77
Table 4.19 The Result of T-Test Using SPSS 21.0 Program
t-table
t-observe
2.910
Df
5 % (0,05)
1 % (0,01)
2.000
2.660
74
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 2.910> 2.000,
2.910> 2.660.It meant Ha was accepted and Ho was rejected. It could be interpreted
based on the result of calculation that Ha stating that Context Clues was effective
for Teaching Reading Comprehension of the eleventh grade students at SMA
Negeri 3 Palangka Raya was accepted and Ho stating that Context Clues was not
effective for Teaching Reading Comprehension of the eleventh grade students
at SMA Negeri 3 Palangka Raya was rejected. It meant that teaching reading with
Context Clues in Reading Comprehension of the eleventh grade students at
SMA Negeri 3 Palangka Raya gave significant effect at 5% and 1% significance
level.
D. Discussion
The result of analysis showed that there was significant effect of Context
Clues in Reading Comprehension forthe eleventh grade students at SMA
Negeri 3Palangka Raya.The students who were taught used Context Clues
reached higher score than those who were taught without used Context
Clues.
78
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
2.733> 2.000, 2.733> 2.660. 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 2.910> 2.000, 2.910> 2.660. 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 Context Clues in
Reading Comprehension for the eleventh grade students at SMA Negeri 3
Palangka Raya was accepted.On the contrary,the Null hypothesis (Ho) stating
that there was no any significant effect of Context Clues in Reading
Comprehension for the eleventh grade students at SMA Negeri 3 Palangka
Raya was rejected.Based on the result the data analysis showed that using
Context Clues gave significance effect for the students’ reading comprehension
scores of eleventh grade students at SMA Negeri 3 Palangka Raya.
After the students have been taught by using Context Clues, the reading
score were higher than before implementing Context Clues 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.67). This finding indicated that Context Clues was
effective and supports the previous research done by Seyed Jalal Abdolmanafi
Rokni and Hamid Reza Niknaqshthat also stated learning reading by using
Context Clues was effective.
There were some reason why using Context Clues gave significance effect
for the students’ reading comprehension scores of eleventh grade students at SMA
79
Negeri 3 Palangka Raya.First, Context Clues 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 64.8 and 60.2. (See p.68).
It was suitable withthe result of pre-test and post test for Experiment and
control Group. (See p.48). In the pre-test of experiment group there were twenty
eightstudents that got fail predicate. They were E-01, E-03, E-04, E-05, E-07, E08, E-09, E-10, E-11, E-12, E-15,E-19, E-20, E-21, E-23, E-24, E-25, E-26, E-29,
E-30, E-32, E-33, E-34, E-35, E-36, E-37, E-38, and E-39. There were nine
students that got less predicate. They were E-06, E-14, E-17, E-18, E-22, E-27, E28, E-31, and E-40. There was no student that got enough predicate. There were
three students that got good predicate. Theywere E-02,E-13, E-16. Then, in the
pre-test score of control group there were twenty six students that got fail
predicate. They were C-01, C-05, C-06, C-07, C-12, C-13, C-14, C-15, C-16, C17, C-18, C-19, C-22, C-23, C-24, C-25, C-26, C-27, C-28, C-29, C-30, C-31, C32, C-33, C-35, and C-36. There were nine students that got less predicate. They
were C-02, C-03, C-04, C-08, C-09, C-10, C-11, C-20, and C-34. There was no
student that got enough predicate. There was one student that got good predicate.
He was C-21.
Based on the result of post-test for experimental and control group, (See
p.58). In the experimental group, there was no student that got in fail predicate.
There were nine students that got less predicate, they were E-01, E-07, E-18, E23, E-27, E-29, E-35, E-37, and E-40. There were seventeen students that got
80
enough predicate. They were E-04, E-06, E-08, E-10, E-11, E-12, E-15, E-16, E17, E-21, E24, E-26, E-28, E-32, E-33, E-36, and E-38. There were thirteen
students that got good predicate. They were E-02, E-03, E-05, E-09, E-14, E-19,
E-20, E-22, E-25, E-30, E-31, E-34, and E-39. There was one student that got
excellent predicate, she was E-13. In the control group, there was no student that
got in fail predicate. There were sixteen students that got less predicate. They
were C-01, C-05, C-06, C-09, C-10, C-13, C-15, C-21, C-22, C-23, C-24, C-28,
C-31, C-33, C-34, and C-36. There were sixteen students that got enough
predicate. They were C-02, C-03, C-04, C-07, C-11, C-12, C-14, C-16, C-19, C20, C-25, C-26, C-29, C-30, C-32, and C-35. There were four students that got
good predicate. They were C-08, C-17, C-18, and C-27.
Those are the result of pre-test compared with post-test for experimental
group and control group of students at SMA Negeri 3 Palangka Raya. Based on
the theories and the writer’s result, Context Clues gave significance effect for the
students’ reading comprehension scores of Eleventh grade students at SMA
Negeri 3 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 effect of using Context Clues
in teaching reading Narrative 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
CORRECT
Control Group
PREDICATE
CODE
SCORE
ANSWER
E-01
E-02
E-03
E-04
E-05
E-06
E-07
E-08
E-09
E-10
E-11
E-12
40
70
40
43,3
40
50
40
40
46,7
40
40
43,3
12
21
12
13
12
15
12
12
14
12
12
13
CORRECT
PREDICATE
ANSWER
FAIL
GOOD
FAIL
FAIL
FAIL
LESS
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
48
C-01
C-02
C-03
C-04
C-05
C-06
C-07
C-08
C-09
C-10
C-11
C-12
40
53,3
53,3
56,7
40
43,3
40
56,7
56,7
50
53,3
46,7
12
16
16
17
12
13
12
17
17
15
16
14
FAIL
LESS
LESS
LESS
FAIL
FAIL
FAIL
LESS
LESS
LESS
LESS
FAIL
49
Experimental Group
Code
Score
E-13
70
E-14
56,7
E-15
40
E-16
70
E-17
53,3
E-18
53,3
E-19
40
E-20
40
E-21
46,7
E-22
50
E-23
46,7
E-24
43,3
E-25
40
E-26
43,3
E-27
56,7
E-28
53,3
E-29
46,7
E-30
40
E-31
50
E-32
40
E-33
46,7
E-34
43,3
E-35
43,3
E-36
43,3
E-37
40
E-38
40
E-39
43,3
E-40
53,3
TOTAL
AVERAGE
Lowest Score
Highest Score
CORRECT
ANSWER
Control Group
PREDICATE
21
17
12
21
16
16
12
12
14
15
14
13
12
13
17
GOOD
LESS
FAIL
GOOD
LESS
LESS
FAIL
FAIL
FAIL
LESS
FAIL
FAIL
FAIL
FAIL
LESS
16
LESS
14
12
15
12
14
13
13
13
12
12
13
16
FAIL
FAIL
LESS
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
LESS
1866,5
46,7
40
70
Code
Score
40
40
40
40
40
40
40
53,3
70
40
40
40
40
40
40
43,3
40
43,3
46,7
43,3
40
50
46,7
40
TOTAL
AVERAGE
Lowest Score
Highest Score
C-13
C-14
C-15
C-16
C-17
C-18
C19
C-20
C-21
C-22
C-23
C-24
C-25
C-26
C-27
C-28
C-29
C-30
C-31
C-32
C-33
C-34
C-35
C-36
CORRECT
ANSWER
12
12
12
12
12
12
12
16
15
12
12
12
12
12
12
13
12
13
14
13
12
15
14
12
PREDICATE
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
LESS
GOOD
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
LESS
FAIL
FAIL
1626,6
45,2
40
70
50
Experimental Group
Control Group
Predicate
Percentages
Predicate
Percentages
Fail
28
Fail
26
Less
9
Less
9
Enough
0
Enough
0
Good
3
Good
1
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 (XI IPS-1)
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
The Range of Score (R)
=H–L+1
= 70 – 40+ 1
= 31
The Class Interval (K)
= 1 + (3.3) x Log n
51
= 1 + (3.3) x Log 40
= 1 + 5.2867979
= 6.2867979
=6
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 4.2 Frequency Distribution of the Pretest Score
Interval (I)
Frequency
(F)
Mid
Point
(x)
The
Limitation
of each
group
Frequency
Relative (%)
Frequency
Cumulative
(%)
1
2
66-70
61-65
3
0
68
63
66.5-70.5
61.5-65.5
7.5
0
100
92.5
3
56-60
2
58
56.5-60.5
5
92.5
4
51-55
5
53
51.5-55.5
12.5
87.5
5
46-50
7
48
46.5-50.5
17.5
75
6
40-45
23
42.5
40.5-45.5
Class
(K)
∑F=40
57.5
∑P = 100
57.5
The distribution of students’ predicate in pretest score of Experimental
group can also be seen in the following figure.
52
The Distribution of Student's Predicate
30
25
20
15
10
5
0
FAIL
LESS
ENOUGH
GOOD
Figure 4.1 The distribution frequency of students’ pretest score for
Experimental Group
Based on the figure above, it can be seen about the students’ predicate in
pretest score. There were twenty eight students who got Fail predicate. There were
ninestudents who gotLess Predicate.
There wasno students who gotEnough
predicate. There werethree 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:
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
66-70
3
2
61-65
0
F.X
FX2
Fka
Fkb
68
204
13872
3
40
63
0
0
3
37
53
3
56-60
2
58
116
6728
5
34
4
51-55
5
53
265
14045
10
29
5
46-50
7
48
336
16128
17
19
6
40-45
23
42.5
977.5
41543.75
40
2
∑=1898.5
∑=92316,75
∑F = 40
∑ Total
1) Calculating Mean
Mx=
∑𝐹𝑋 𝑖
𝑛
=
1898.5
= 46.6530
40
2) Median
1
= ℓ +2
Mdn
𝑁−𝑓𝑘𝑏
𝑓𝑖
1
40−2
2
= 39.5 +
= 39.5 +
𝑋𝑖
23
18
23
X5
𝑋5
= 201.41
3) Modus
Mo
=u+
𝑓𝑎
𝑓𝑎 +𝑓𝑏
= 39.5 +
16
𝑥𝑖
16+23
= 39.5 + 2.05
= 41.55
4) Standard Deviation
S=
S=
𝑛.∑𝐹𝑋 𝑖 2 − ∑𝐹𝑋 𝑖 2
𝑛 𝑛 −1
40.92316 .75− 92316 .75 2
40 40−1
𝑥5
54
S = 8.51868
5) Standard Error
SEmd =
𝑆
𝑁−1
=
8.51868
40−1
=
8.51868
6.24
= 1.36517
After Calculating, it was found that the standard deviation and the
standard error of pretest score were 8.51868 and 1.36517
b. The Result of Pretest Score of ControlGroup (XI IPS-2)
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 36
= 1 + 5.1357966
= 6.1357966
=6
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:
55
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
1
0
68
63
66.5-70.5
61.5-65.5
2.7
0
100
97.3
3
4
56-60
51-55
3
6
58
53
56.5-60.5
51.5-55.5
8.3
16.6
97.3
89
5
46-50
4
48
46.5-50.5
11.3
72.4
6
40-45
22
42.5
40.5-45.5
Class
(K)
∑F=36
61.1
∑P = 100
61.1
The distribution of students’ predicate in pretest score of Control group
The Distribution of Student's Predicate
30
25
20
15
10
5
0
FAIL
LESS
ENOUGH
GOOD
Figure 4.2 The distribution of students’ predicate in pretest score
of Control Group
Based on the figure above, it can be seen about the students’ predicate in
pretest score. There twenty sixstudents who got Fail predicate. There were
56
ninestudents who gotLess Predicate. There was no students who got Enough
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, 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
1
2
61-65
0
Class
(K)
Interval
(I)
3
56-60
3
4
51-55
5
46-50
6
40-45
∑ Total
F.X
FX2
Fka
Fkb
68
4624
0
14
0
0
1
14
F.X
FX2
Fka
Fkb
58
174
10092
1
13
6
53
318
16854
4
12
4
48
192
9216
10
8
22
∑F = 36
42.5
935
39737.5
14
2
68
63
Mid
Frequency
Point
(F)
(x)
∑=1687
1) Calculating Mean
Mx=
2) Median
Mdn
∑𝐹𝑋 𝑖
𝑛
=
1687
36
= 45.8611
1
𝑁−𝑓𝑘𝑏
2
=ℓ+
𝑓𝑖
1
= 39.5 + 2
𝑋𝑖
36−2
22
𝑋5
∑=80523.5
57
= 39.5 +
16
22
X5
= 43.13
3) Modus
Mo
𝑓𝑎
=u+
𝑥𝑖
𝑓𝑎 +𝑓𝑏
= 39.5 +
14
14+2
𝑥5
= 39.5 + 4.375
= 43.87
4) Standard Deviation
S=
S=
𝑛.∑𝐹𝑋 𝑖 2 − ∑𝐹𝑋 𝑖 2
𝑛 𝑛 −1
36.80523 .5− 80523 .5 2
36 36−1
S = 7.54464
5) Standard Error
SEmd =
𝑆
𝑁−1
=
7.54464
36−1
=
7.54464
5.91
= 1.27658
After Calculating, it was found that the standard deviation and the
standard error of pretest score were 7.54464 and 1.27658.
58
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
E-23
E-24
E-25
E-26
E-27
56,7
73,3
73,3
60
70
66,7
56,7
60
76,7
60
63,3
60
86,7
70
60
63,3
60
56,7
70
76,7
66,7
70
56,7
66,7
73,3
60
56,7
CORRECT
ANSWER
17
22
22
18
21
20
17
18
23
18
19
18
26
21
18
19
18
17
21
23
20
21
17
20
22
18
17
Control Group
PREDICATE
CODE
SCORE
LESS
GOOD
GOOD
ENOUGH
GOOD
ENOUGH
LESS
ENOUGH
GOOD
ENOUGH
ENOUGH
ENOUGH
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
C-25
C-26
C-27
53,3
60
60
60
56,7
53,3
63,3
70
56,7
53,3
66,7
66,7
53,3
60
56,7
66,7
70
76,7
60
66,7
53,3
56,7
53,3
56,7
60
66,7
70
EXCELLENT
GOOD
ENOUGH
ENOUGH
ENOUGH
LESS
GOOD
GOOD
ENOUGH
GOOD
LESS
ENOUGH
GOOD
ENOUGH
LESS
CORRECT
ANSWER
PREDICATE
16
18
18
18
17
16
19
21
17
16
20
20
16
18
17
20
21
23
18
20
16
17
16
17
LESS
ENOUGH
ENOUGH
ENOUGH
LESS
LESS
ENOUGH
GOOD
LESS
LESS
ENOUGH
ENOUGH
LESS
ENOUGH
LESS
ENOUGH
GOOD
GOOD
ENOUGH
ENOUGH
LESS
LESS
LESS
LESS
18
20
21
ENOUGH
ENOUGH
GOOD
59
E-28
60
E-29
56,7
E-30
70
E-31
70
E-32
66,7
E-33
60
E-34
70
E-35
56,7
E-36
63,3
E-37
56,7
E-38
60
E-39
73,3
E-40
56,7
TOTAL
AVERAGE
Lowest Score
Highest Score
18
ENOUGH
17
21
21
20
18
21
17
19
17
18
22
17
LESS
GOOD
GOOD
ENOUGH
ENOUGH
GOOD
LESS
ENOUGH
LESS
ENOUGH
GOOD
LESS
2590,3
64,8
56,7
86,7
53,3
60
63,3
53,3
60
53,3
56,7
66,7
53,3
TOTAL
AVERAGE
Lowest Score
Highest Score
C-28
C-29
C-30
C-31
C-32
C-33
C-34
C-35
C-36
Experimental Group
16
18
19
16
18
16
17
20
16
LESS
ENOUGH
ENOUGH
LESS
ENOUGH
LESS
LESS
ENOUGH
LESS
2166,7
60,2
53,3
76,7
Control Group
Predicate
Percentages
Predicate
Percentages
Fail
0
Fail
26
Less
9
Less
9
Enough
17
Enough
0
Good
13
Good
1
Excellent
1
Excellent
0
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
60
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 (XI IPS-1)
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 40
= 1 + 5.2867979
= 6.2867979
=6
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:
61
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
86.7-91.7
80.7-85.7
1
0
89.2
83.2
86.2-91.2
80.2-85.2
2.5
0
100
97.5
3
4
74.7-79.7
68.7-73.7
2
11
77.2
71.2
74.2-79.2
68.2-73.2
5
27.5
0
92.5
5
62.7-67.7
7
65.2
62.2-67.2
17.5
75
6
56.7-61.7
19
59.2
56.2-61.2
Class
(K)
∑F=40
47.5
∑P = 100
27.5
The distribution of students’ predicate in post-test score of Experimental
group.
The Distribution of Student's Predicate
18
16
14
12
10
8
6
4
2
0
LESS
ENOUGH
GOOD
EXCELLENT
Figure 4.3 The distribution of students’ predicate in post-test score
of Experimental Group
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
wereninestudentswho got Less predicate. There were seventeenstudents who
62
gotEnoughPredicate. There were thirteen 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
86.7-91.7
1
2
80.7-85.7
3
F.X
FX2
Fka
Fkb
89.2
89.2
7956.64
10
40
0
83.2
0
0
10
30
74.7-79.7
2
77.2
154.4
11919.68
12
20
4
68.7-73.7
11
71.2
783.2
55763.84
14
8
5
62.7-67.7
7
65.2
456.4
29757.28
21
4
6
56.7-61.7
∑ Total
19
∑F = 40
59.2
1124.8
66588.16
40
2
∑=2608
1) Calculating Mean
Mx=
2) Median
∑𝐹𝑋 𝑖
𝑛
=
2608 .5
40
1
Mdn
= ℓ +2
=64.2125
𝑁−𝑓𝑘𝑏
𝑓𝑖
= 56.2 +
= 56.2 +
= 4.7368
𝑋𝑖
1
40−2
2
19
18
19
X5
𝑋5
∑=171985.6
63
3) Modus
Mo
𝑓𝑎
=u+
𝑓𝑎 +𝑓𝑏
= 56.2+
40
𝑥𝑖
40+2
𝑥5
= 56.2 + 4.7619
= 60.9619
4) Standard Deviation
S=
S=
𝑛.∑𝐹𝑋 𝑖 2 − ∑𝐹𝑋 𝑖 2
𝑛 𝑛 −1
40.6801664 .5− 2608 2
40 40−1
S = 7.64816
5) Standard Error
SEmd =
𝑆
𝑁−1
=
7.64816
40−1
=
7.64816
6.24
= 1.22566
After Calculating, it was found that the standard deviation and the standard
error of pretest score were7.64816 and 1.22566
b. The Result of Post-test Score of Control Group (XI IPS-2)
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:
The Highest Score (H)
= 76.7
The Lowest Score (L)
= 53.3
The Range of Score (R)
=H–L+1
64
= 76.7 – 53.3 + 1
= 24.4
The Class Interval (K)
= 1 + (3.3) x Log n
= 1 + (3.3) x Log 36
= 1 + 5.1357928
= 6.1357928
=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
1
80.3
75.3
77.8-82.8
72.8-77.8
0
2.7
100
100
3
4
68.3-72.3
63.3-67.3
4
8
70.3
65.3
67.8-72.8
62.8-67.8
11.1
22.2
97.3
86.2
5
58.3-62.3
7
60.3
57.8-62.8
19.5
64
6
53.3-57.3
16
55.3
52.8-57.8
44.5
∑P = 100
44.5
Class
(K)
∑F=36
65
The distribution of students’ predicate in post-test score of Control Group.
The Distribution of Student's Predicate
18
16
14
12
10
8
6
4
2
0
LESS
ENOUGH
GOOD
EXCELLENT
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 sixteen
studentswho got Less predicate. There were sixteenstudents who gotEnough
Predicate. There were four 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:
66
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
1
36
1
75.3
75.3
5670.09
1
35
68.3-72.3
4
70.3
281.2
19768.36
5
34
4
63.3-67.3
8
65.3
522.4
34112.72
13
29
5
58.3-62.3
7
60.3
422.1
25452.63
20
16
6
53.3-57.3
∑ Total
16
∑F = 36
55.3
884.8
48929.44
36
4
∑=2185.8 ∑=133933.24
1) Calculating Mean
Mx=
2) Median
∑𝐹𝑋 𝑖
𝑛
=
2185 .8.16
36
1
Mdn
= ℓ +2
=60.7166
𝑁−𝑓𝑘𝑏
𝑓𝑖
= 52.8 +
= 52.8 +
𝑋𝑖
1
36−4
2
𝑋4
16
14
16
X4
= 56.3
3) Modus
Mo
=u+
𝑓𝑎
𝑓𝑎 +𝑓𝑏
= 52.8 +
36
36+4
= 52.8 + 3.6
= 56.4
𝑥𝑖
𝑥4
67
4) Standard Deviation
S=
S=
𝑛.∑𝐹𝑋 𝑖 2 − ∑𝐹𝑋 𝑖 2
𝑛 𝑛 −1
36.4777721 .64.4− 2185 .8 2
36 36−1
S = 6.48771
5) Standard Error
SEmd =
𝑆
𝑁−1
=
6.48771
36−1
=
6.48771
5.91
= 1.09775
After Calculating, it was found that the standard deviation and the
standard error of pretest score were 6.48771 and 1.09775
3. The Comparison result of Pre-test and Post-test of Experimental and
Control Group
NO
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
EXPERIMENTAL CLASS
SCORE
CODE
PRE- POST- DIFFE
TEST TEST RENCE
E-01
40
56,7
16,7
E-02
70
73,3
3,3
E-03
40
73,3
33,3
E-04
43,3
60
16,7
E-05
40
70
30
E-06
50
66,7
16,7
E-07
40
56,7
16,7
E-08
40
60
20
E-09
46,7
76,7
30
E-10
40
60
20
E-11
40
63,3
23,3
E-12
43,3
60
16,7
E-13
70
86,7
16,7
E-14
56,7
70
13,3
E-15
40
60
20
NO
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
CONTROL CLASS
SCORE
CODE
PREPOST- DIFFE
TEST
TEST RENCE
C-01
40
53,3
13,3
C-02
53,3
60
6,7
C-03
53,3
60
6,7
C-04
56,7
60
3,3
C-05
40
56,7
16,7
C-06
43,3
53,3
10
C-07
40
63,3
23,3
C-08
56,7
70
13,3
C-09
56,7
56,7
0
C-10
50
53,3
3,3
C-11
53,3
66,7
13,4
C-12
46,7
66,7
20
C-13
40
53,3
13,3
C-14
40
60
20
C-15
40
56,7
16,7
68
E-16
E-17
E-18
E-19
E-20
E-21
E-22
E-23
E-24
E-25
E-26
E-27
E-28
E-29
E-30
E-31
E-32
E-33
E-34
E-35
E-36
E-37
E-38
E-39
E-40
TOTAL
MEAN
LOWEST
HIGHEST
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
70
53,3
53,3
40
40
46,7
50
46,7
43,3
40
43,3
56,7
53,3
46,7
40
50
40
46,7
43,3
43,3
43,3
40
40
43,3
53,3
1866,5
46,7
40
70
63,3
60
56,7
70
76,7
66,7
70
56,7
66,7
73,3
60
56,7
60
56,7
70
70
66,7
60
70
56,7
63,3
56,7
60
73,3
56,7
2590,3
64,8
56,7
86,7
-6,7
6,7
3,4
30
36,7
20
20
10
23,4
33,3
16,7
0
6,7
10
30
20
26,7
13,3
26,7
13,4
20
16,7
20
30
3,4
723,8
18,1
C-16
16
C-17
17
C-18
18
C-19
19
C-20
20
C-21
21
C-22
22
C-23
23
C-24
24
C-25
25
C-26
26
C-27
27
C-28
28
C-29
29
C-30
30
C-31
31
C-32
32
C-33
33
C-34
34
C-35
35
C-36
36
TOTAL
MEAN
LOWEST
HIGHEST
40
40
40
40
53,3
70
40
40
40
40
40
40
43,3
40
43,3
46,7
43,3
40
50
46,7
40
1626,6
45,2
40
70
66,7
70
76,7
60
66,7
53,3
56,7
53,3
56,7
60
66,7
70
53,3
60
63,3
53,3
60
53,3
56,7
66,7
53,3
2166,7
60,2
53,3
76,7
From the table above the mean score of pre test and post test of the
experimental group were 46,7 and 64,8. 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 45,2 and 60,2. Meanwhile,
26,7
30
36,7
20
13,4
-16,7
16,7
13,3
16,7
20
26,7
30
10
20
20
6,6
16,7
13,3
6,7
20
13,3
540,1
15
69
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 3.1
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
studentsname
a
Kolmogorov-Smirnov
Statistic
Df
Shapiro-Wilk
Sig.
Statistic
Df
Sig.
experiment group
,229
40
,240
,765
40
,567
control group
,290
36
,176
,745
36
,115
Students score
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 76> 50, so to analyzed normality
data was used Kolmogorov-Smirnov. 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
70
Based on data above, significant data of experiment and control group used
Kolmogorov-Smirnov was 0.240> 0.05 and 0.176> 0.05. It could be concluded that
the data was normal distribution.
b. Testing normality of post-test experimental and control group
Table 4.12Testing normality of post-test experimental and control group
Tests of Normality
studentsname
a
Kolmogorov-Smirnov
Statistic
Df
Shapiro-Wilk
Sig.
Statistic
df
Sig.
experiment group
,217
40
,112
,890
40
,054
control group
,178
36
,076
,892
36
,064
Students score
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 76 > 50, so to analyzed normality
data was usedKolmogorov-Smirnov. 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
Kolmogorov-Smirnov was 0.112> 0.05 and 0.076> 0.05. It could be concluded that
the data was normal distribution.
2. Homogeneity Test
a. Testing Homogeneity of pre-test experimental and control group
71
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
Equal
,153
Sig.
,697
T
,808
Df
Sig.
Mean
Std. Error
99% Confidence Interval
(2-
Differen
Difference
of the Difference
tailed)
ce
Lower
Upper
74
,422
1,47167
1,82164
-3,34459
6,28793
,814 73,855
,418
1,47167
1,80711
-3,30643
6,24977
variances
Studentss assumed
core
Equal
variances 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 0,697. The result was 0,697> 0,01,
it meant the t-test calculation used at the equal variances assumed or data was
Homogeny distribution.
72
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
t-test for Equality of Means
Equality of
Variances
F
Sig.
T
Df
Sig. (2-
Mean
Std. Error
99% Confidence
tailed)
Difference
Difference
Interval of the
Difference
Lower
Equal
1,888
,174
Upper
2,910
74
,005
4,57167
1,57100
,41808
8,72525
2,933
73,855
,004
4,57167
1,55847
,45101
8,69233
variances
Students
assumed
score
Equal
variances 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 0,174. The result was 0,174> 0,01it
meant the t-test calculation used at the equal variances assumed or data was
Homogeny distribution.
73
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
Mean
Standard Error
Experimental Group
64.2125
1.22566
Control Group
60.7166
1.09775
The table showed the result of the standard deviation calculation of
Experiment group was 7.64816and the result of the standard error was 1.22566 .
The result of thestandard deviation calculation of Control group was 6.48771and
the result of standard error was 1.09775. To examine the hypothesis, the writer
used the formula as follow:
tobserved=
=
=
df
𝑀1−𝑀2
𝑆𝐸𝑚 1−𝑆𝐸𝑚 2
64.2125 −60.7166
1.22566 −1.09775
3.4959
0.12791
= 2.733
= (N1 + N2 – 2)
= 40+36-2
= 74
74
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 74. The following table was the
result of tobserved and ttable from 74 df at 5% and 1% significance level.
Table 4.16 The Result of T-Test Using Manual Calculation
t-table
t-observe
2.733
Df
5 % (0,05)
1 % (0,01)
2.000
2.660
74
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 2.733> 2.000,
2.733> 2.660.It meant Ha was accepted and Ho was rejected. It could be interpreted
based on the result of calculation that Ha stating that Context Clues was effective
for Teaching Reading Comprehension of the elevent grade students at SMA
Negeri 3 Palangka Raya was accepted and Ho stating that Context Clues was not
effective for Teaching Reading Comprehension of the eleventh grade students
at SMA Negeri 3 Palangka Raya was rejected. It meant that teaching reading with
Context Clues in Reading Comprehension of the eleventh grade students at
SMA Negeri 3 Palangka Raya gave significant effect at 5% and 1% significance
level.
75
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
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
Studentsname
N
Mean
Std. Deviation
Std. Error Mean
40
64,7550
7,30426
1,15491
36
60,1833
6,27856
1,04643
experiment group
Student score
control group
The table showed the result of mean calculation of experiment groupwas
64.7550, standard deviation calculation was 7.30426, and standard error of mean
calculation was 1.15491. The result of mean calculation of control
groupwas 60.1833, standard deviation calculation was 6.27856, and standard error
of mean was1.04643.
76
Table 4.18 The Calculation of T – Test Using SPSS 21.0
Independent Samples Test
Levene's Test for
t-test for Equality of Means
Equality of Variances
F
Sig.
T
Df
Sig. (2-
Mean
Std.
99% Confidence
tailed)
Difference
Error
Interval of the
Differen
Difference
ce
Equal
1,888
,174
2,910
74
,005
4,57167
1,57100
Lower
,41808 8,7252
variances
Students
assumed
score
Equal
Upper
5
2,933 73,855
,004
4,57167
1,55847
,45101 8,6923
variances
not assumed
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 0,174. The result was 0,174> 0,01,
it meant the t-test calculation used at the equal variances assumed. It found that the
result of tobserved was 2.910, the result of mean difference between experiment and
control group was 4.57167, and thestandard error difference between experiment and
control group was 1.57100.
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 74. The following table was the
result of tobserved and ttable from 74 df at 5% and 1% significance level.
3
77
Table 4.19 The Result of T-Test Using SPSS 21.0 Program
t-table
t-observe
2.910
Df
5 % (0,05)
1 % (0,01)
2.000
2.660
74
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 2.910> 2.000,
2.910> 2.660.It meant Ha was accepted and Ho was rejected. It could be interpreted
based on the result of calculation that Ha stating that Context Clues was effective
for Teaching Reading Comprehension of the eleventh grade students at SMA
Negeri 3 Palangka Raya was accepted and Ho stating that Context Clues was not
effective for Teaching Reading Comprehension of the eleventh grade students
at SMA Negeri 3 Palangka Raya was rejected. It meant that teaching reading with
Context Clues in Reading Comprehension of the eleventh grade students at
SMA Negeri 3 Palangka Raya gave significant effect at 5% and 1% significance
level.
D. Discussion
The result of analysis showed that there was significant effect of Context
Clues in Reading Comprehension forthe eleventh grade students at SMA
Negeri 3Palangka Raya.The students who were taught used Context Clues
reached higher score than those who were taught without used Context
Clues.
78
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
2.733> 2.000, 2.733> 2.660. 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 2.910> 2.000, 2.910> 2.660. 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 Context Clues in
Reading Comprehension for the eleventh grade students at SMA Negeri 3
Palangka Raya was accepted.On the contrary,the Null hypothesis (Ho) stating
that there was no any significant effect of Context Clues in Reading
Comprehension for the eleventh grade students at SMA Negeri 3 Palangka
Raya was rejected.Based on the result the data analysis showed that using
Context Clues gave significance effect for the students’ reading comprehension
scores of eleventh grade students at SMA Negeri 3 Palangka Raya.
After the students have been taught by using Context Clues, the reading
score were higher than before implementing Context Clues 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.67). This finding indicated that Context Clues was
effective and supports the previous research done by Seyed Jalal Abdolmanafi
Rokni and Hamid Reza Niknaqshthat also stated learning reading by using
Context Clues was effective.
There were some reason why using Context Clues gave significance effect
for the students’ reading comprehension scores of eleventh grade students at SMA
79
Negeri 3 Palangka Raya.First, Context Clues 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 64.8 and 60.2. (See p.68).
It was suitable withthe result of pre-test and post test for Experiment and
control Group. (See p.48). In the pre-test of experiment group there were twenty
eightstudents that got fail predicate. They were E-01, E-03, E-04, E-05, E-07, E08, E-09, E-10, E-11, E-12, E-15,E-19, E-20, E-21, E-23, E-24, E-25, E-26, E-29,
E-30, E-32, E-33, E-34, E-35, E-36, E-37, E-38, and E-39. There were nine
students that got less predicate. They were E-06, E-14, E-17, E-18, E-22, E-27, E28, E-31, and E-40. There was no student that got enough predicate. There were
three students that got good predicate. Theywere E-02,E-13, E-16. Then, in the
pre-test score of control group there were twenty six students that got fail
predicate. They were C-01, C-05, C-06, C-07, C-12, C-13, C-14, C-15, C-16, C17, C-18, C-19, C-22, C-23, C-24, C-25, C-26, C-27, C-28, C-29, C-30, C-31, C32, C-33, C-35, and C-36. There were nine students that got less predicate. They
were C-02, C-03, C-04, C-08, C-09, C-10, C-11, C-20, and C-34. There was no
student that got enough predicate. There was one student that got good predicate.
He was C-21.
Based on the result of post-test for experimental and control group, (See
p.58). In the experimental group, there was no student that got in fail predicate.
There were nine students that got less predicate, they were E-01, E-07, E-18, E23, E-27, E-29, E-35, E-37, and E-40. There were seventeen students that got
80
enough predicate. They were E-04, E-06, E-08, E-10, E-11, E-12, E-15, E-16, E17, E-21, E24, E-26, E-28, E-32, E-33, E-36, and E-38. There were thirteen
students that got good predicate. They were E-02, E-03, E-05, E-09, E-14, E-19,
E-20, E-22, E-25, E-30, E-31, E-34, and E-39. There was one student that got
excellent predicate, she was E-13. In the control group, there was no student that
got in fail predicate. There were sixteen students that got less predicate. They
were C-01, C-05, C-06, C-09, C-10, C-13, C-15, C-21, C-22, C-23, C-24, C-28,
C-31, C-33, C-34, and C-36. There were sixteen students that got enough
predicate. They were C-02, C-03, C-04, C-07, C-11, C-12, C-14, C-16, C-19, C20, C-25, C-26, C-29, C-30, C-32, and C-35. There were four students that got
good predicate. They were C-08, C-17, C-18, and C-27.
Those are the result of pre-test compared with post-test for experimental
group and control group of students at SMA Negeri 3 Palangka Raya. Based on
the theories and the writer’s result, Context Clues gave significance effect for the
students’ reading comprehension scores of Eleventh grade students at SMA
Negeri 3 Palangka Raya.