2. Data Analysis a. Analysis of the Linearity of Tests
The writer analyzed the linearity of the tests by using SPSS version 22.0. The linearity of the tests was checked in order to see whether the regression of
relationship between two variables in linear. The result of analyzing the linearity of the tests is presented in Anova table as follows:
From the table below, it reveals that vocabulary mastery and writing achievement have linear regression. The result of linearity is 0.004 and deviation
from linearity is 0.734, which means that the data distribution has a good linear regression. This because the significant of linearity is smaller than the significant
of 5 0.0040.05 and deviation of linearity is bigger than 0.05 0.7340.05.
Table 4.5 The Linearity Test Result of the Data
ANOVA Table Sum of
Squares Df Mean
Square F
Sig. Writing
Achievement Vocabulary
Mastery
Between Groups
Combined 233.952
11 21.268 1.486
.185
Linearity 136.451
1 136.451 9.533
.004 Deviation
from Linearity
97.501 10
9.750 .681
.734
Within Groups 458.048
32 14.314
Total 692.000
43
b. Analysis of the Normality of Tests
The normality of the tests was analyzed using SPSS version 22.0. The purpose of checking the normality is to see whether the data is normally
distributed. As the total number of the samples is less than 50, it is more suitable
to analyze the data using Saphiro-Wilk. The result of the analysis is presented in the following table:
Table 4.6 The Normality of the Test
Variable Kolmogorov-Smirnov
a
Shapiro-Wilk Statistic
df Sig.
Statistic df
Sig.
Score Vocabulary
.127 44
.073 .965
44 .197
Writing
.138 44
.034 .958
44 .110
From the result above, both vocabulary mastery and writing achievement score are normally distributed because the values of both scores are higher than
5 or 0.05. The test result reveal that the significant value of vocabulary mastery is 0.197, in which 0.197 0.05. Moreover, the significant value of writing
achievement is 0.110, in which 0.110 0.05.
c. Analysis of the Correlation Coefficient
After analyzing the linearity and the normality of the data, to measure the correlation coefficient, Pearson Moment Formula was used. The formula was used
because the data distribution is normal and linear. Before doing the calculation, the data are described such as below:
Table 4.7 The Data Analysis Table of Vocabulary Mastery and Writing Achievement
Participants X
Y XY
X
2
Y
2
Student 1 60
71 4260
3600 5041
Student 2 63
73 4599
3969 5329
Student 3 76
72 5472
5776 5184
Student 4 70
70 4900
4900 4900
Student 5 76
77 5852
5776 5929
Student 6 66
70 4620
4356 4900
Student 7 80
82 6560
6400 6724
Student 8 73
65 4745
5329 4225
Participants X
Y XY
X
2
Y
2
Student 9 60
66 3960
3600 4356
Student 10 66
73 4818
4356 5329
Student 11 83
76 6308
6889 5776
Student 12 60
68 4080
3600 4624
Student 13 83
70 5810
6889 4900
Student 14 70
79 5530
4900 6241
Student 15 83
70 5810
6889 4900
Student 16 66
76 5016
4356 5776
Student 17 83
82 6806
6889 6724
Student 18 63
69 4347
3969 4761
Student 19 83
72 5976
6889 5184
Student 20 66
68 4488
4356 4624
Student 21 73
73 5329
5329 5329
Student 22 70
76 5320
4900 5776
Student 23 83
80 6640
6889 6400
Student 24 70
74 5180
4900 5476
Student 25 56
75 4200
3136 5625
Student 26 76
73 5548
5776 5329
Student 27 86
79 6794
7396 6241
Student 28 76
75 5700
5776 5625
Student 29 80
74 5920
6400 5476
Student 30 70
75 5250
4900 5625
Student 31 76
73 5548
5776 5329
Student 32 80
70 5600
6400 4900
Student 33 63
70 4410
3969 4900
Student 34 86
77 6622
7396 5929
Student 35 80
76 6080
6400 5776
Student 36 66
80 5280
4356 6400
Student 36 80
78 6240
6400 6084
Student 38 73
75 5475
5329 5625
Student 39 90
77 6930
8100 5929
Student 40 76
76 5776
5776 5776
Student 41 76
76 5776
5776 5776
Student 42 73
73 5329
5329 5329
Student 43 86
75 6450
7396 5625
Student 44 93
77 7161
8649 5929
N=44 ΣX =
3268 ΣY =
3256 ΣXY =
242515 ΣX
2
= 246142
ΣY
2
= 241636
After getting the result above, the calculation of the data to Pearson Product Moment Formula is presented as follows:
Formula:
� = � ΣXY − ΣX ΣY
√[�ΣX
2
− ΣX
2
][�ΣY
2
− ΣY
2
]
Calculation:
N = 44 ΣX =
3268
ΣY = 3256 ΣX
2
= 246142 ΣY
2
= 241636 ΣX
2
= 10679824 ΣY
2
= 10601536 ΣXY =
242515
� = � ΣXY − ΣX ΣY
√[�ΣX
2
− ΣX
2
][�ΣY
2
− ΣY
2
] � =
− √[
− ][
− ]
� = −
√[ −
][ −
] � =
√[ ][
] � =
√ � =
. � = .
To make sure the result of the calculation above, the Pearson Product Moment in SPSS statistic program version 22.0 was used to know whether the
calculation that has been calculated manually is correct or not and to make sure that there is no mismatching calculation between score that the writer counted.
The results of those calculations; manual calculation and calculation using SPSS statistic program version 22.0 are equal, in which the value of r
xy
or r
o
is 0.444. It means that there is no mismatch in the process of calculating the data by
calculating manually or using the SPSS formula. The calculation of Pearson Product Moment is described as follows:
Table 4.8 Pearson Product Moment Table
Correlations Vocabulary
Mastery Writing
Achievement Vocabulary
Mastery Pearson
Correlation 1
.444
Sig. 2-tailed .003
N 44
44
Writing Achievement
Pearson Correlation
.444 1
Sig. 2-tailed
.003
N 44
44
d. Analysis of Determination Coefficient
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