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c. The Normality Test
The researcher used SPSS Statistic 20 to count the normality of each instruments. The instruments would be normal distribution if the result of
calculation from SPSS Statistic 20 showed less than the result from Shapiro- Wilk table with the number of sample n is 35.
These tests are valued using SPSS Statistic 20. The normality Using One-Sample Shapiro-Wilk test results can be seen as bellow:
Table 4.4 The Normality Data of Variable X and Y
Tests of Normality
Kolmogorov-Smirnov
a
Shapiro-Wilk Statistic
df Sig.
Statistic Df
Sig. Parents Involvement
.108 35
.200 .950
35 .114
Students English Learning Achievement
.103 35
.200 .957
35 .183
. This is a lower bound of the true significance. a. Lilliefors Significance Correction
From the result above, both parents’ involvement score and student’s
English learning achievement score data are categorized normally distributed because the value of significant are higher than 0.05. The significant value of
X is 0.114, which means that 0.114 0.05 and the significant value of Y is 0.200, which means that 0.200 0.05. Hence, it can be concluded that all the
data are normally distributed. As supported by Siregar statement about the degree of normality test:
Hypothesis: H
o
: Population are normally distributed H
1
: Population are not normally distributed randomly distributed Criteria of normality test are:
H
o
is rejected if probability of significant value 0.05
39
H is accepted if probability of significant value 0.05
1
Additionally, to visualize the normality of data, the writer use Q-Q plot in SPSS 20 as the description. The normality result presented in the
picture below:
Figure 4.3 The Normal Q-
Q plot of Parents’ Involvement
Figure 4.4 The Normal Q-
Q Plot of Student’s English Learning Achievement
1
Syofian Siregar, Statistika Deskriptif untuk Penelitian, Jakarta: Rajawali Press, 2011, p. 256.
40
d. The Correlational Result
Table 4.5 The Result of Correlation Calculation
Correlations
Parents Involvement
Students English
Learning Achievement
Parents Involvement Pearson Correlation
1 .996
Sig. 2-tailed .000
N 35
35 Students English Learning
Achievement Pearson Correlation
.996 1
Sig. 2-tailed .000
N 35
35 . Correlation is significant at the 0.01 level 2-tailed.
From the table, it is described that r coefficient is 0.996 and the sig, 2-tailed is 0.000. This result is equivalent with the calculation that the
writer did manually before in which = 0.996. It also means that there is
no mismatching calculation in the process of calculating that the researcher did. The next step is to find the significance of variables by calculating
. It is tested by significance test formula:
√ √
In which: = t value
= 0.996 n
= 35 Therefore, it is calculated that:
√ √
=
√ √
=
√ √
=
√
= =
64.1
41
The researcher made two hypotheses of significance before testing the as follow:
Ha : There is significant correlation between two variables
Ho : There is no significant correlation between two variables
The formulation of test: 1. If
it means that the null hypotheses is rejected and there is significant correlation.
2. If it means that the null hypotheses is accepted and there
is no significant correlation. Based on the calculation, the result is compared by
in the significant 0.05 and 0.01 and n = 35, the researcher found the Degree of
Freedom Df with the formula: Df = n- nr
= 35 – 2
= 33 From Df = 33, it is obtained
of 0.05 = 2.035 and of 0.01 =
2.733. It means that is higher than
64.1 2.035 and 64.1 2.733. Therefore, the null hypothesis Ho is rejected. On the other words,
there is significant correlation between parents’ involvement and student’s English learning achievement.
3. The Interpretation of the Data
Based on the analysis drawn, the result of normality test of questionnaire showed that the data are normally distributed and it is found
that the correlation value is = 0.996. After knowing the correlation value,
then the researcher needs to classify the category of relationship as the table follow:
Kategori