37 The Description of Data Students’ Improvement Using TPR Method in
= 3.97
2
+ 2.84
2
= 15.76 + 8.07
=
23.83
= 4.88
h. Determining t − observation�
i.
=
19.60 4.88
= 4.01
j. Statistic Hypothesis
The hypothesis criterion sates that: if �
�
�
There is significant difference between variable X and variable Y. It means the null hypothesis
is rejected and the alternative hypothesis
�
is accepted. If �
�
�
there is no significant difference between variable X and variable Y. it means the null hypothesis
is accepted and the alternative hypothesis
�
is rejected.
k. Determining t-table �
�
in significant level 5 and 1 with degree of freedom.
df = N
– 1 = 30
– 1 = 29 see the table of “t” values at the degree of
significance of 5 and 1
Because the value of 29 is not mentioned in the table, the writer used the closest value to 29 is 30 as degree of freedom df. Based on the table, df at
significance level of
5 and 1 are: t. table
�
�
at the significance level of 5 = 2.04 t. table
�
�
at the significance level of 1 = 2.75 Based on the formula above, the result of the statistic calculation indicated
that value of �
�
is 4.01and the value of df degree of freedom was 29 on degree of significance of 5 is 2.04. Comparing the
�
�
with each values of the degree of
significance, the result is 2.04 4.01 2.75.
Since �
�
score in the table is higher than �
�
score obtained from the result of calculating, so the alternative hypothesis
�
is accepted and null hypothesis is rejected.