1 Difficulty index around 0 , 0 – 0 , 3 shows the test is difficult
2 Difficulty index around 0 , 3 – 0 , 7 shows the test is enough difficulteasy
3 Difficulty index around 0 , 7 – 1 , 0 shows the test is easy
b. Testing discrimination index of a test item
Discrimination index of a test was used to know the ability of a test to differentiate which student who had high ability and low ability. To get such
information, firstly, the students were divided into two groups. The first group was a group for students who had higher score while the second group was a
group for students who had smaller score. Next, the discrimination index of the test was then calculated as follow:
9
D = -
= -
Note: D = Discrimination index
= Total number of students in the first group = Total number of students in the second group
= Total number of students in the first group who answer correctly = Total number of students in the second group who answer correctly
= Proportion of students in the first group who answer correctly = Proportion of students in the second group who answer correctly
Finally, such calculation was then interpreted by using the following classification
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: D: 0, 0 -0, 2: poor
D: 0, 2 -0, 4: satisfactory D: 0, 4 -0, 7: good
D: 0, 7 -1, 0: excellent D: negative: bad
9
Ibid, p.213
10
Ibid., p.218
From the above classification, a test item which was considered as a good item to be used in this study was a test item which had discrimination index
around 0, 4 – 1, 0.
E. The Technique of Data Collecting
After appropriate test items for this study had been gotten, the test as well as interview could be used to collect data for this study. The technique for collecting
data was through pretest, posttest, and interview. First, students were given a pretest before taking a treatment. The test was
arranged in the form of multiple choice items consisting of 25 items and in the each item there were four possible choices for students.
So, the students‘ task was just choosing one correct answer from the four possible choices in the each item.
The use of the test wa s to check students‘ past tense knowledge before treatment.
Further, students were given a posttest after taking a treatment. The test was arranged similarly like the pretest items in order the created posttest items could
bear a same difficulty index as the pretest items. Thus, by delivering such test, it was hoped that the real
information about students‘ progress in learning past tense could be gotten.
Finally, some students were given interview questions to get additional information about the use of CALL. The interview was delivered to 12 students in
the experimental group. It was used to find information whether CALL was effective for them or not. The result of the interview was used to make a
conclusion and suggestion of this study.
F. The Technique of Data Analyzing
After the data of the two samples had been gotten, the mean scores of the two samples were analyzed to determine whether the use of CALL in teaching past
tense was effective or not. At the end, this analysis could be useful to examine the truth or false of this study hypothesis. In this study, the parameter statistic was
used to calculate the data. So, as the requirement in the parameter statistic, the normality and homogeneity of the data had to be examined first.
1. Testing Normality of the Data
Testing normality was used to check whether the population had normal distribution or not. The formula was as follow
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:
= p-
ɸ
= value of p = sum value of data probability
ɸ = value of Kolmogorov table
To get the ɸ value, the standard score of the data z had to be calculated first
with formula: z =
̅
z = standard score x = students‘ score
̅ = the mean score s = standard deviation
After the value of was gotten, the value of normality table with
significance 5 was sought. After the value of normality table was found, it had to be compared with the value of
to find whether the data had a normal distribution or not. If the data had a normal distribution, the value of
would be same as or lower than the value of normality table. Conversely, if the data did
not have a normal distribution, the value of would be higher than the value
of normality table.
2. Testing Homogeneity of the Data
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Budi Susetyo, Statistika untuk analysis data penelitian, Bandung: Refika Aditama, 2010, p. 148-150
