Testing Hypothesis

2. Testing Hypothesis

The researcher uses t-test statistical calculation with significant level of the refusal null hypothesis a= 0.05. The researcher uses manual calculation and SPSS

a. Testing Hypothesis Using Manual Calculation

To test the hypothesis of the study, the researcher used t-test statistical calculation. Firstly, the researcher calculated the standard deviation and the standard error of X1 and X2. It was found the standard deviation and the standard error of posttest of X1 and X2 at the previous data presentation. The criteria of Ha is accepted when t obseved >t table , and Ho is refused when t observed <t table . It could be seen on this following table:

Table 4.17 The Standard Deviation and the Standard Error of X1and X2

Variable

Standard Deviation

Standard Error

Where: X1 = Experimental Group X2 = Control Group

The table showed the result of the standard deviation calculation of X1 was 12.035 and the result of the standard error mean calculation was 1.814. The result of the standard deviation calculation of X2 was 9.676 and the result of the standard error of mean calculation was 1.459. The next step, the researcher calculates the standard error of the differences mean between X1 and X2:

SE M1 – SE M2 = √ SE M1 – SE M2 = √ SE M1 – SE M2 = √ SE M1 – SE M2 = √

SE M1 – SE M2 = The calculation above shows the standard error of the differences mean between X1 and X 2 is 2.178. Then, it is inserted to t o formula to get the value of t observed as follows: SE M1 – SE M2 = The calculation above shows the standard error of the differences mean between X1 and X 2 is 2.178. Then, it is inserted to t o formula to get the value of t observed as follows:

With the criteria:

If t-test (t observed )≥t table , it means Ha is accepted and H0 is rejected.

If t-test (t observed ) <t table , it means Ha is rejected and H0 is accepted.

Then, the researcher interprets the result of t-test. Previously, the researcher accounts the degree of freedom (df) with the formula:

t table at df 86 at 5% significant level = 1.9 8793≈ 1.988 The calculation above shows the result of t-test calculation as in the table follows:

Table 4.18 The Result of T-test

t table

Variable t observed Df

Where: X1 = Experimental Class X2 = Control Class t observed = The calculated Value

t table = The distribution of t value

df = Degree of Freedom

The result of hypothesis test calculation (Table 4.22) proves that the value of t observed is higher than the value of t table at the level of significant in 5% or 1% that is 1.988<3.456>2.632. It shows that H a is accepted and H o is rejected. From the result of hypothesis test can be described that students who taught by using grammar discovery techniquehave significant effect on the passive voice mastery‟ score at the eleventh grade students of MA Muslimat NU Palangka Raya. On the other hand, students who taught by non-grammar discovery technique did not have better passive voice mastery than those taught by grammar discovery technique. Simply, it can be interpreted that independent variable gave significant effect on dependent variable.

b. Testing Hypothesis Using SPSS Program

The researcher also applied SPSS 18.0 program to calculate t-test in testing hypothesis of the study which supports the result of manual calculation. The result of the test using SPSS 18.0 program can be seen as follows:

Table 4.19 The Calculation of T-test Using SPSS 18.0

Independent Samples Test

Levene's Test for Equality of Variances

t-test for Equality of Means Std.

95% Confidence

Error Interval of the Sig. (2- Diffe Differe

Mean

Difference

nce Lower Upper SC Equal

F Sig.

df tailed)

rence

2.328 3.417 12.674 OR variances E assumed

2.328 3.414 12.677 variances not

assumed

The table showed the result of t-test calculation using SPSS 18.0 program. Since the result of post-test between experiment and control group had difference score of variance, it meant the t-test calculation used at the equal variances not assumed. It found that the result of t observed was 3.456, the result of mean difference between experiment and control group was 8.045, and the standard error difference between experiment and control group was 2.328.

To examine the truth or the false of null hypothesis stating that grammar discovery did not give significant effect on the students‟ passive voice mastery, the result of t-test was interpreted on the result of degree of freedom to get the table. The result of degree of freedom (df) was 86, it found from the total number of the students in both group minus 2. The following table was the result of t observed and table from 86 df at 5% and 1% significance level.

The interpretation of the result of t-test using SPSS 18.0 program, it was found the t observe was greater than the t table at 1 % and 5 % significance level or 1.988 <3.456> 2.632. It means, H a was accepted and H o was rejected. It could be interpreted based on the result of calculation that H a stating that grammar discovery technique gives significant effect on the students‟ passive voice mastery was accepted and H o stating that grammar discovery technique does not give significant effect on the students‟ passive voice mastery was rejected.

In conclusion, teaching English grammar (passive voice) using grammar discovery technique gave significant effect on the st udents‟ passive voice mastery at the eleventh grade students of MA Muslimat NU Palangka Raya.