Post-test means' comparison

CHAPTER III: DATA ANALYSIS AND FINDINGS

3.2. Post-test means' comparison

Table 3.6 Number of students scored less from X i

Class Number of students

Test takers Number of students scored less from X i

0 1 2 3 4 5 6 7 8 9

Experimental 31 31 0 0 0 0 1 7 9 25 30 31

Control 31 31 0 0 0 0 6 11 24 28 31 31

Table 3.7 Statistics of % of students scored less from X i

Class

Number of students

Test takers

Statistics of % of students scored less from X i

0 1 2 3 4 5 6 7 8 9 10

Experimental 31 31 0 0 0 0 3,2 22,6 29 80 96,7 100 100

Control 31 31 0 0 0 0 9,3 35,5 77,4 90,3 100 100 100 Table 3.8 The statistics parameters of the two groups

Class Cases Number of

test X S 2 S.d V

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Experimental 31 1 6,5 1,66 1,29 0,19

Control 31 1 5,5 1,69 1,30 0,23

There appeared from the table that means scores of the experimental class is higher than that of the control class. However, it is arguable that the observed differences in the scores of both classes may have been brought about by some other factors than the different vocabulary checking techniques. In order to confirm the differences on the test are due to the experimental treatment, the researcher continues analyzing the data using the correct statistic

"t-test".

3.2.1 Choosing the correct statistic "t-test"

The means of two groups were compared using t-test because all assumptions for the t- test were found to be met in the study. There are two assumptions underlying the t-test which are stated as follows:

1. the scores in each group are normally distributed 2. the variances for the scores of the two groups are equal

The first assumption requires that the distribution of scored for each group must be approximately normal. The results from data analysis provided me with information to plot the scores of both groups as follows:

10 20 30 40 50 60 70 80 90 100

1 2 3 4 5 6 7 8 9 10

0 N 0 % o f st u d en ts s co re d l es s fr o m X i

Score X i

control class experimental class

Figure 3.2: Cumulative frequency curves of the two groups

Cumulative frequency graph of the two groups shows the statistics of % students scored less from X i . These figures appeared to be clear that experimental group got better results of test than control one due to different vocabulary checking techniques.

The second assumption of t-test is referred to as the "homogeneity of variances". It means that the squared value of standard deviation (sd 2 ) should be about the same.

Fortunately, the squared value of standard deviation of both groups in the study were nearly about the same (1.29) 2 = (1.30) 2 . This result clearly showed that the second assumption of the t-test was not violated.

3.2.2. Statistical hypotheses

The statistical hypotheses of this study were stated as follows:

H 0: X E = X C

H 1 = X E > X C

H 0 is called null hypothesis and H 1 is called right-tailed hypothesis (or positive hypothesis). And the subscript E is used here for the experimental class mean and subscript C is used for the control class.

In order to complete t-test, the following parameters need to be set or calculated in advance.

Choosing significant level α = 0.05 to test the hypothesis H 1 , the researcher uses the random quantity Z (the observed statistics) with

X E X C

Z =

S E 2

N E +

S C

N C

6.5 5.5

1.66 31 +

1.69 31

0.05 +

0.05

1 0.3287

= = 3.042

This study set the alpha level at α = 0.05, we could find out critical value Z t : ϕ ( Z t ) = ( 1 - 2x α)/2 = ( 1 - 2x 0.05)/2 = 0.45

According to the Laplace table, we could find out Z t = 1.65

Compared Z to Z t , we can see that Z > Z t with the significant level α = 0.05. It means that the null hypothesis (H 0 ) is rejected and H 1 is accepted. There is only a 5% probability that the observed means deference, X E > X C , occurred by chance alone or 95% probability that was due to factors other than chance. It means that using vocabulary checking techniques improves students' vocabulary retention.

3.2.3. Interpretation of results Steps

1 Look at the hypothesis H 0: X E = X C

H 1 = X E > X C

2 Look at the α level α < 0.05, one-tailed decision 3 Compare the observed and critical

statistics

Z = 3.042 Z t = 1.65

a. If the observed statistic is less than the critical statistic, accept the null hypothesis and stop.

b. If the observed statistic is greater than the critical hypothesis, reject the null hypothesis and continue.

Mentioned above

5 Decide which alternative hypothesis is more logical

Because X E was expected to be greater if there was any mean difference, there was only one alternative hypothesis:

H 1 = X E > X C

6 Interpret the results in terms of the p level

(H 0 ) is rejected and H 1 is accepted. There is only a 5% probability that the observed means deference, X E > X C , occurred by chance alone or 95% probability that was due to factors other than chance.

V E = 0.19; V C = 0.23. V E < V C . This can be explained that students of experimental

7 Look and compare coefficient of variation

class can retain vocabulary better so differences between students in the class are less than the control class.

Using communicative activities to check vocabulary

Vocabulary checking techniques in combination with communicative activities