1. Scores of the students’ pre-test of experimental and control classes a. Scores of the students’ pre-test of experimental class - The Use Of Authentic Materials In Writing Recount Paragraphs At The Eighth Grade Students Of MTs Islamiyah Palangka Raya -
CHAPTER IV RESULT OF THE STUDY A. Description of the Data In this case, the writer divided the data of the students scores taken from on the
students’ English writing scores between those who taught using authentic materials and who taught using non-authentic materials of Eighth Grade Students of MTs Islamiyah Palangka Raya.
1. Scores of the students’ pre-test of experimental and control classes a. Scores of the students’ pre-test of experimental class
NO CODES SCORES
60
Table 4.1 Students’ Scores of Pre-Test of Experimental Class50 2500 3600 1600 4900 3600 1600 3600 2500 4900 3600 2500
60
70
50
60
40
60
70
40
50
X2
11 A01 A02 A03 A04 A05 A06 A07 A08 A09 A10 A11
Based on the test, the writer constructed the result are analyzed in following ways:
9
8
7
6
5
4
3
2
1
10
9.09
60
Furthermore, the writer arranged the data of the students’ scores as can be seen in the following table:
Table 4.2 The Distribution of Frequency of the students’ scores of pre-test of Experimental ClassNO SCORES F %
1
2
3
4
70
50
From the data above it is known highest score is 70, and the lowest score is 40.
40
4
10
6
2
18.18
45.46
27.27
The writer got the data from the result of test. It can be known: High score: 70, low score: 40 Range of score: R = H
Total 1260 73800
13
22 A13 A14 A15 A16 A17 A18 A19 A20 A21 A22
14
15
16
17
18
19
20
21
60
60 3600 4900 3600 2500 3600 2500 3600 4900 2500 3600
70
60
50
60
50
60
70
50
- – L + 1 = 70
- – 40 + 1 = 31
Total
22 100 Note: p = f / n × 100%
From the table above, it can be explained that on number 1 (one) there are 4 (four) students or about (18.18%) who obtained score 70. On number 2 (two) there are 10 (ten) students or about (45.46%) who obtained score 60. On number 3 (three) there are 6 (six) students or about (27.27%) who obtained score 50. On number 4 (four) there are 2 (two) students or about (9.09%) who obtained score
40. From the distribution of frequency above, the writer constructed the histogram as follow:
12
10
8
6
4
2
40
50
60
70 Figure 4.1 Histogram of Frequency Distribution of Students’ Scores of Pre-
Test of Experimental Class These are calculation of mean, median and modus can be seen at the following table: 1)
Mean The description and calculation of mean are presented in the following table:
Table 4.3 The Calculation of MeanNO X (SCORES) F fX
1
70 4 280
2
60 10 600
3
50 6 300
4
40
2
80 Total Ʃf = 22 ƩfX = 1260
From the data above, it is known: ∑fX
M =
x
N 1260 =
22 = 57.27
From the result of calculation above, it can be known that the mean score which have been obtained 57.27
2) Median
The description and calculation of median are presented as follows:
Table 4.4 The Calculation of MedianNO X (SCORES) F fX fk b fk a
1
70 4 280
22
4
2
60 10 600
18
14
3
50 6 300
8
20
4
40
2
80
2
22 Total Ʃf = 22 ƩfX = 1260
From the data above, it is known: N = 22, ½ N = 11 Mdn = 60 l = 59.5 fk b = 8 fi = 10 yields:
½N b
- – fk Median = l + f i
11
- – 8 = 59.5 +
10
= 59.5 + 0,3 = 59.8
10
60 2500 3600
50
2 B01 B02
1
X2
NO CODES SCORES
Table 4.6 Students’ Scores of pre-test of Control ClassBased on the test, the writer constructed the result are analyzed in the following ways:
From the data above, it is known that Modus is 60. It is known from score which has highest frequency.
2 Total Ʃf = 22
6
4
3) Modus
40
50
60
70
4
3
2
1
NO SCORES (X) F
Table 4.5 The Calculation of ModusThe description and calculation of modus are presented in the following table:
b. Scores of the students’ pre-test of control class
Furthermore, the writer arranged the data of the students’ scores as can be seen in the following table:
60
60
40
70
40
60
60
50
50
50
60
40
60
50 3600 1600 2500 2500 3600 1600 4900 1600 3600 3600 2500 3600 2500 3600 1600 3600 2500
Total
1050 56700 From the data above it is known highest score is 70, and the lowest score is 40.
The writer got the data from the result of test. It can be known: High score: 70 Low score: 40 Range of score: R = H
50
40
4
12
5
6
7
8
9
10
11
13
60
14
15
16
17
18
19
20 B04 B05 B06 B07 B08 B09 B10 B11 B12 B13 B14 B15 B16 B17 B18 B19 B20
- – L + 1 = 70
- – 40 + 1 = 31
NO SCORES F %
1
70
1
5
2
60
8
40
3
50
6
30
4
40
5
25 Total 20 100 Note: p = f/n x 100%
From the table above, it can be explained that on number 1 (one) there is 1 (one) students or about (5%) who obtained score 70. On number 2 (two) there are 8 (eight) students or about (40%) who obtained score 60. On number 3 (three) there are 6 (six) students or about (30%) who obtained score 50. On number 4 (four) there are 5 (five) students or about (25%) who obtained score 40.
From the distribution of frequency above, the writer constructed the histogram as follow:
9
8
7
6
5
4
3
2
1
40
50
60
70 Figure 4.2 Histogram of Frequency Distribution of Students’ Scores of Pre- test of Control Class
These are calculation of mean, median and modus can be seen at the following table: 1)
Mean The description and calculation of mean are presented in the following table:
Table 4.8 The Calculation of MeanNO X (SCORES) F fX
1
70
1
70
2
60 8 480
3
50 6 300
4
40 5 200
Total
Ʃf = ƩfX = 20 1050 From the data above, it is known:
∑fX M x = N
1050 =
20 = 52.5
From the result of calculation above, it can be known that the mean score which have been obtained 52.5 2)
Median The description and calculation of median are presented as follows:
Table 4.9 The Calculation of MedianNO X (SCORES) f fX fk b fk a
1
70
1
70
20
1
2
60 8 480
19
9
3
50 6 300
11
15
4
40 5 200
5
20 Total Ʃf = 20 ƩfX = 1050
From the data above, it is known: Mdn = 60 l = 59.5 fk b = 11 fi = 8 yields:
½N b
- – fk Median = l + f i
10
- – 11 = 59.5 +
8 = 59.5
− 0.125 = 59.375
3) Modus
The description and calculation of modus are presented in the following table:
Table 4.10 The Calculation of ModusSCORES (X) F NO
1
70
1
2
60
8
3
50
6
4
40
5 Total Ʃf = 20 From the data above, it is known that Modus is 60. It is known from score which has highest frequency.
2. Scores of the students’ post-test of experimental and control class a. Scores of the students’ post-test of experimental class
60
70
60
80
70
60
70
60
80
70
80
21 A01 A02 A03 A04 A05 A06 A07 A08 A09 A10 A11 A12 A13 A14 A15 A16 A17 A18 A19 A20 A21
70
80
70
60
80
60
70
80
70 4900 4900 3600 6400 4900 3600 4900 3600 6400 4900 3600 6400 4900 6400 4900 3600 6400 3600 4900 6400 4900
70
Based on the test, the writer constructed the result are analyzed in following ways:
Table 4.11 Students’ Scores of Post-Test of Experimental Class8
NO CODES SCORES (X)
X2
1
2
3
4
5
6
7
9
19
10
11
12
13
14
15
16
17
18
20
1550 110500
Total
From the data above it is known that highest score is 80, and the lowest score is 60.
The writer got the data from the result of test. It can be known: High score: 80, low score: 60 Range of score: R = H
- – L + 1 = 80
- – 60 + 1 = 21
Furthermore, the writer arranged the data of the students scores as can be seen in the following table:
Table 4.12 The Distribution of Frequency of the students’ score of post-test of Experimental ClassNO SCORES F %
1
80
7
31.82
2
70
9
40.91
3
60
6
27.27 Total 22 100 Note: p = f / n × 100%
From the table above, it can be explained that on number 1 (one) there are 7 (seven) students or about (31.82%) who obtained score 80. On number 2 (two) there are 9 (nine) students or about (40.91%) who obtained score 70. On number 3 (three) there are 6 (six) students or about (27.27%) who obtained score 60.
From the distribution of frequency above, the writer constructed the histogram as follow:
10
9 40,91
8
7 31,82 6 27,27
5
4
3
2
1
60
70
80 Figure 4.3 Histogram of Frequency Distribution of Students’ Scores of post-test of Experimental Class These are calculation of mean, median and modus can be seen at the following table: 1)
Mean The description and calculation of mean are presented in the following table:
Table 4.13 The Calculation of MeanNO X (SCORES) F fX
1
80 7 560
2
70 9 630
3
60 6 360
Total
Ʃf = 22 ƩfX = 1550 From the data above, it is known:
∑fX M x = N
1550 =
22 = 70.45
From the result of calculation above, it can be known that the mean score which have been obtained 70.45
2) Median
The description and calculation of median are presented as follows:
Table 4.14 The Calculation of MedianNO X (SCORES) F fX fk b fk a
1
80 7 560
22
7
2
70 9 630
15
16
3
60 6 360
6
22 Total Ʃf = 22 ƩfX = 1550
From the data above, it is known: N = 22, ½ N = 11 Mdn = 70 l = 69.5 fk b = 6 fi = 9 yields:
½N b
- – fk Median = l + f
i
11
- – 6 = 69.5 +
9
= 70.056 3)
Modus The description and calculation of modus are presented in the following table:
Table 4.15 The Calculation of ModusNO SCORES (X) F
1
2
3
80
70
60
7
9
6 Total Ʃf = 22 From the data above, it is known that Modus is 70. It is known from score which has highest frequency.
b. Scores of the students’ post-test of control class
Based on the test, the writer constructed the result are analyzed in the following ways:
NO CODES SCORES (X)
X2
1
2
3
4
5 B01 B02 B03 B04 B05
60
70
50
60
50 3600 4900 2500 3600 2500
Table 4.16 Students’ Scores of post-test of Control Class The writer got the data from the result of test. It can be known: High score: 70 Low score: 50 Range of score: R = HFrom the data above it is known highest score is 70, and the lowest score is 50.
60
Total 1180 70800
60 3600 3600 2500 4900 2500 3600 4900 3600 4900 2500 4900 2500 3600 3000
60
50
70
50
70
60
70
60
50
70
50
60
7
20 B07 B08 B09 B10 B11 B12 B13 B14 B15 B16 B17 B18 B19 B20
19
18
17
16
15
14
13
12
11
10
9
8
- – L + 1 = 70
- – 50 + 1 = 21
NO SCORES F %
1
70
5
25
2
60
8
40
3
50
7
35 Total 20 100 Note: p = f/n x 100%
From the table above, it can be explained that on number 1 (one) there are 5 (five) students or about (25%) who obtained score 70. On number 2 (two) there are 8 (eight) students or about (40) who obtained score 60. On number 3 (three) there are 7 (seven) students or about (35%) who obtained score 50.
From the distribution of frequency above, the writer constructed the histogram as follow:
9
8
40
7
35
6
5
25
4
3
2
1
50
60
70 Figure 4.4 Histogram of Frequency Distribution of Students’ Scores of post-test of Control Class
These are calculation of mean, median and modus can be seen at the following table: 1)
Mean The description and calculation of mean are presented in the following table:
Table 4.18 The Calculation of MeanNO X (SCORES) F fX
1
70 5 350
2
60 8 480
3
50 7 350
Total
Ʃf = 20 ƩfX = 1180 From the data above, it is known: ∑fX
M
50
b
From the data above, it is known: N = 20, ½ N = 10 Mdn = 60 l = 59.5 fk
20 Total Ʃf = 20 ƩfX = 1180
13
5
7
15
20
7 350 480 350
8
5
60
x
70
3
2
1
NO X (SCORES) f fX fk b fk a
Table 4.19 The Calculation of MedianMedian The description and calculation of median are presented as follows:
From the result of calculation above, it can be known that the mean score which have been obtained 59 2)
20 = 59
1180 =
= N
= 7 fi = 8 yields: ½N
b
- – fk Median = l + f i
10
- – 7 = 59.5 +
8 = 59.5 + 0.375 = 59.875
3) Modus
The description and calculation of modus are presented in the following table:
Table 4.20 The Calculation of ModusNO SCORES (X) F
1
70
5
2
60
8
3
50
7 Total Ʃf = 20
From the data above, it is known that Modus is 60. It is known from score which has highest frequency.
B.Result of the Data Analysis
1. Deviation standard of the students’ post-test of experimental class of eighth grade students of MTs Islamiyah Palangka Raya.
Ʃf = 22 ƩfX = 1550 Ʃfx
22 SD = 8.394
SD = √ 1550
SD = √ ∑fx
= 1295.455 To know the deviation standard where:
2
60 6 360 -10.45 109.2025 655.215 Total
The calculation of deviation standard is presented in the following table:
70 9 630 -0.45 0.2025 1.8225
80 7 560 9.55 91.2025 638.4175
2
fx
2
Scores (X) f fX X x
Table 4.21 The Calculation of Deviation Standard of Experimental Class2 N
2.Deviation standard of the students’ post-test of Control Class of eighth grade students of MTs Islamiyah Palangka Raya.
The calculation of deviation standard is presented in the following table:
SD = √ ∑fx
60 70 -10 100
2
50 70 -20 400
1
2
(Y) D D
Students’ scores after taught using non- authentic materials
NO Students’ scores before taught using authentic materials (X)
The Calculation of T-test Table 4.23 The Calculation of T-test
20 SD = 7.681 3.
SD = √ 1180
= 1180 To know the deviation standard where:
Table 4.22 The Calculation of Deviation Standard of Control Class2
Ʃf = 20 ƩfX = 1180 Ʃfx
50 7 350 -9 81 567 Total
8
1
1
60 8 480
70 5 350 11 121 605
2
fx
2
Scores (X) F fX X x
2 N
60 70 -10 100
17
14
70 80 -10 100
15
60 80 -20 400
16
50 60 -10 100
60 80 -20 400
13
18
50 60 -10 100
19
60 70 -10 100
20
70 80 -10 100
21
60 70 -10 100
60 80 -20 400
22
60 70 -10 100
4
70 80 -10 100
5
60 70 -10 100
6
40 60 -20 400
7
8
12
50 60 -10 100
9
70 80 -10 100
10
60 70 -10 100
11
50 60 -10 100
50 70 -20 400
- 290 4300 To know mean of difference, the writer used formula:
M D = ƩD
N M
D
= -290
22 M D = -13.18 To know SD
D
(Standard of Deviation of difference between score variable I and Score variable II), the writer used formula: SD D
2 N N
= √ ∑D
2
- – ∑D
2 SD D
= √4300 – -290 22 22 SD D =
√ 195.45454545 – 173.765124 SD D = 4.6571902957
To Calculate SE MD (Standard Error of Mean of Difference), the writer used formula: SE MD = SD D
√N-1 SE MD = 4.6571902957
√22-1 SE MD = 4.6571902957
4.5825756949 SE = 1.0162822408
MD
To know t o (t observed ), the writer used formula: t = M D SE
MD
t = -13.18 1.0162822408 t = -12.96883838253 t = -12.969 To know df (degree of freedom), the writer used formula: df = Nx + Ny
- – 2 = 22 + 22
- – 2 = 42
With the criteria: If t test (t ) > t table, Ha is accepted and Ho is rejected
If t test (t ) < t table, Ha is rejected and Ho is accepted Based on the data obtained, the result showed that the mean of students’
English writing scores of who taught using authentic materials was 70.45, while the mean of students’ English writing scores of who taught using non-authentic material was 59. From both means, there was different value that was 11.45. It meant there is different result of them in writing recount paragraphs. Furthermore, the writer arranged the
Mean Scores of students’ scores pre-test and students’ scores post-test of both classes as can be seen in the following table:
Table 4.24 Mean Scores of Students’ Scores Pre-test and Post-test of Experimental and
Control Classes
Mean Scores of Pre-test Mean Scores of Post-test Different Values
Experiment Control Experiment Control Pre-test Post-test
57.27
52.5
70.45
59
4.77
11.45 Based on the calculation above, it can be known the value from the result of calculation (t observed ) was -12,969. Then, it is consulted with t table (t t ) which db or df = (N
1 + N 2 ) table (t t ) = 2.02
- – 2 was (22 + 22) – 2 = 42. Significant standard 5% t and significant standard 1% t table (t t ) = 2.71. So, 2.02 < 12.969 > 2.71. It can be said that since the value of t (-12.969) was higher than t in the 5% (2.02)
observed table
and 1% (2.71) level of significance, it could be interpreted that Ha stating that there is a significant difference between who taught using authentic materials and who taught using non-authentic material of eighth grade students of MTs Islamiyah of Palangka Raya was accepted and Ho stating that there is no significant difference between who taught using authentic materials and who taught using non-authentic material of eighth grade students of MTs Islamiyah of Palangka Raya was rejected. It meant that there is a significant difference between who taught using authentic materials and who taught using non-authentic material.
Meanwhile, the writer also applied SPSS program to calculate t-test:
Table 4.25
The Calculation of the Result T-test using SPSS 16
Paired Samples Test
Paired Differences T df Sig.
(2- Mean tailed) Std. Devia tion
Std. Error
Mean 95% Confidence
Interval of the Difference
Lower Upper Pair
1 Pre-test Scores of Experimental Class - Post- test Scores of Experimental Class
- 13.182 4.767 1.016 -15.296 -11.068 -12.969 42 .000 The result of the t-test using SPSS also supported the interpretation above that was found the t observed (-12.969). Significant standard 5% t table (t t ) = 2.02 and significant standard 1% t table (t t ) = 2.71. So, 2.02 < 12.969 > 2.71. It can be said that since the value of t observed (-12.969) was higher than t table in the 5% (2.02) and 1% (2.71) level of significance, it could be interpreted that Ha stating that there is a significant difference between who taught using authentic materials and who taught using non-authentic material of eighth grade students of MTs Islamiyah of Palangka Raya was accepted and Ho stating that there is no significant difference between who taught using authentic materials and who taught using non-authentic