In this post, I would like to share the Findings and Discussion in the journal that I've been found.
In findings and discussions, researchers used the quantitative method as follows.
As can be seen in Table 1, the highest score of pre-test of the experimental class was 7.7 and the lowest score was 5.5 with an average of 6.5. And then, the highest score of post-test of the experimental class was 9.7 and the lowest score was 6.7 with an average of 8.9.
Table 2 shows that the score of pre-test and post-test of the control class showed the highest score on the pre-test was 7.4 and the lowest score was 5.6 with an average of 6.3. And then, the highest score on post-test was 9.2 and the lowest score was 6.5 with an average of 8.4.
Interview Result
Regarding the students’ response toward film which is used in teaching, it can be concluded that students’ response is excellent. Most of the students feel more motivated and enthusiastic about it. It could be seen on how they focus on the film during the learning session. At that moment, the students kept on watching, repeating the words and expression that happened from the film.
Hypothesis Test
There are two research questions in this research: (1) is the use of film effective in improving students’ pronunciation ability?, and (2) what is the students’ response toward film in improving their pronunciation? To answer the first question, the statistical hypothesis can
be seen as follows:
Ha: there is a significant difference between students’ achievement in improving pronunciation which is taught by using film.
H0: there is no significant difference between students’ achievement in improving pronunciation which is taught by using film
The criteria is used as follows:
- If t-test (t0) > t-table (tt) in significant degree of .05, H0 (null hypothesis) is rejected.
- If t-test (t0) < t-table (tt) in significant degree of .05, H0 (null hypothesis) is accepted.
From the result of the calculation, it shows that the value of tt with df 58 in significance 5% is 1,673, while the value of t0 is 1,5. Since t0 score is lower than tt score obtain from the result of calculating, so the alternative hypothesis (Ha) is rejected and the null hypothesis (H0) is accepted.