4. 62
86 24
5. 63
84 29
6. 51
77 26
7. 57
73 3
8. 66
60 6
9. 54
74 9
10. 65
60 6
11. 53
71 20
12. 57
68 14
13. 54
67 17
14. 68
71 9
15. 61
77 6
16. 51
66 8
17. 67
54 13
18. 57
80 12
19. 63
75 14
20. 62
77 18
21. 56
68 22
22. 54
60 7
23. 77
94 17
24. 59
63 20
25. 56
60 4
26. 60
73 13
27. 66
77 19
28. 68
73 13
29. 71
83 12
30. 75
84 14
31. 51
60 32.
78 83
5 Total
1965 2300
428 Average
61.40 71.88
13.38 Based on the table 4.2 above, it shows that the lowest score in pre-test
was 51 and the highest score was 78 with the average of pre-test score was 61.40. Furthermore, the highest score of post-test was 94 and the lowest
score was 54 with the average score was 71.88. After pre-test and post- test, the teacher got the average of the gained score was 13.38. it means
the gained score of control class is lower than experimental class.
B. Analysis of Data
1. Validity Test
The validity of the test is calculated using “ANATEST” software version 4.0.9 By using the software, the writer found 36 of 50 items are
valid and 14 items are not valid. It can be proved by the significance of each item. Besides, the validity or XY correlation of the instrument used
in this study was 0.72. it means the test was valid and categorized into high validity see appendix.
2. Reliabity Test
The writer calculated the reliability of the test using ANATEST. The reability score that the writer got is 0.84. it means the test is reliable.
Besides, the criterion of the test is categorized into high reliability see appendix.
3. Discriminating Power
The writer used ANATEST to analyze the discriminating power of test item used in this study see appendix. The result of the calculation using
ANATEST can be seen as follow:
Table 4.3 The Result of Discriminating Power
A-Multiple-Choice
Number of items DP index
Classification of DP
1. 25.00
Satisfactory 2.
37.50 Satisfactory
3. -25.00
Discarded
4. 0.00
Poor 5.
-12.50 Discarded
6. 62.50
Good 7.
12.50 Poor
8. 12.50
Poor 9.
87.50 Exellent
10. 12.50
Poor 11.
12.50 Poor
12. 12.50
Poor 13.
25.00 Satisfactory
14. 25.00
Satisfactory 15
12.50 Poor
16. 37.50
Satisfactory 17.
12.50 Poor
18. 50.00
Good 19.
50.00 Good
20. 37.50
Satisfactory 21.
62.50 Good
22. 0.00
Poor 23.
62.50 Good
24. 75.00
Exellent 25.
37.50 Satisfactory
26. 50.00
Good 27.
62.50 Good
28. 12.50
Poor 29.
75.00 Exellent
30. 75.00
Exellent 31.
37.50 Satisfactory
32. 12.50
Poor 33.
25.00 Satisfactory
34. 37.50
Satisfactory 35.
50.00 Good
36. 37.50
Satisfactory 37.
12.00 Poor
38. 12.00
Poor 39.
50.00 Good
40. 25.00
Satisfactory 41.
12.50 Poor
42. 25.00
Satisfactory 43.
12.00 Poor
44. 50.00
Good 45.
25.00 Satisfactory
46. 12.50
Poor 47.
50.00 Good
48. 25.00
Satisfactory 49.
0,00 Poor
50. 0.00
Poor
4. Item Difficulty
From the claculation using ANATEST, the writer categorized the items into five classification index. The writer got 18 items are very easy,
10 items are easy, 19 items are moderate, 3 items are difficult and 1 item is very difficult. The categorization of the items is on the table below:
Table 4.4 The Classification of Difficulty Item
Number of Items Classification of Difficulty
Items 4, 5,7, 10, 11, 12, 15, 16,
17, 20, 21, 37, 43, 44, 45, 46, 49, 50
Very Easy 1, 13, 14, 19, 22,
32, 34, 35, 39 Easy
2, 3, 6, 9, 23, 24, 25, 27, 28, 29, 30, 31, 34, 38,
41, 42, 47, 48 Moderate
8, 18, 40 Difficult
4 Very Difficult
5. Normality Test
a. Pre-test of Experimental Class Hypotheses:
H : Data of X normaly distributed
H
1
: Data of Y is not normally distributed
Table 4.5 Normality of Pre-test of Experimental Class
Tests of Normality
Kelas Kolmogorov-Smirnov
a
Shapiro-Wilk Statistic
df Sig.
Statistic Df
Sig. pretest
Experiment ,109
32 ,200
,962 32
,311 . This is a lower bound of the true significance.
a. Lilliefors Significance Correction
Based on the table above, it can be seen that the significant from normality test of Shapiro-Wilk shows 0.311. Therefore, the significant
score is higher than 0.05 0.311 0.05. it means that H is accepted so the
data is normally distributed. b. Post-test of Experimental Class
Table 4.6 Normality of Post-test of Experimental Class
Tests of Normality
Kelas Kolmogorov-Smirnov
a
Shapiro-Wilk Statistic
df Sig.
Statistic df
Sig. posttest
experiment ,125
32 ,200
,956 32
,207 . This is a lower bound of the true significance.
a. Lilliefors Significance Correction
Based on the table above, it can be seen that the significant from normality test of Shapiro-Wilk shows 0.207. therefore, the significant score
is higher than 0.05 0.207 . 0.05. it means that H is accepted so the data is
normally distributed. c. Pre-test of Controlled Class
Table 4.7 Normality of Pre-test of Controlled Class
Tests of Normality
kelas Kolmogorov-Smirnov
a
Shapiro-Wilk Statistic
Df Sig.
Statistic Df
Sig. pretest
control ,091
32 ,200
,972 32
,560 . This is a lower bound of the true significance.
a. Lilliefors Significance Correction
Based on the table above, it can be seen that the significant from normality test Shapiro-Wilk shows 0.560. Therefore, the significant score
is higher than 0.05 0.560 0.05. it means that H is accepted so the data
is normally dustributed. d. Post-test of Controlled Class
Table 4.8 Normality of Post-test Controlled Class
Tests of Normality
Kelas Kolmogorov-Smirnov
a
Shapiro-Wilk Statistic
Df Sig.
Statistic Df
Sig. Posttest
control ,124
32 ,200
,964 32
,356 . This is a lower bound of the true significance.
a. Lilliefors Significance Correction
Based on the table above, it can be seen that the significant from normality test ShapiroWilk shows 0.356. Therefore, the significant score is
higher than 0.05 0.356 0.05. it means that H is accepted so the data is
normally distributed.
6. Homogeneity Test
Based on the calculation of normality, the writer got the result that all data in pre-test and post-test of both experiment class and controlled class
have been normality distributed. The next step of the claculation was finding the homogeneity of tha data. The purpose of this claculation was
to see whether the data sample in both classes homogenous or heterogeneous.
Hypotheses:
H : The condition of experiment class is not different from controlled
class H
1
: The sample of experiment class is differentfrom controlled class. Based on the criteria, it can be concluded that H
is accepted. It means that the sample in experiment class and controlled clss were homogenous.
Moreover, the writer also used SPSS to calculated the homogeneity of the data. The result that the writer got can be seen on the table below:
Table 4.9 Test of Homogeneity of Variance
Test of Homogeneity of Variances
pretest Levene Statistic
df1 df2
Sig. ,130
1 62
,720
Based on the table above, it can be seen that the result of homogeneity test Lavene Statistic score shows 0.130 with the significant 0.720.
Therefore, the significant score is higher than 0.05 0.720 0.05. it means that the sample in experiment class and controlled class were homogenous.
Table 4.10 ANOVA Test
ANOVA
pretest Sum of Squares
df Mean Square
F Sig.
Between Groups 58,141
1 58,141 ,973
,328 Within Groups
3704,594 62
59,752 Total
3762,734 63
Based on the table above, F score from the result of calculation is 0.973 with the significant score 0.328. the writer found H
is accepted from
the comparison between F
o
F
observation
and F
table
which shows F
table
1.99 is higher than F
o
0.973. Therefore, the writer concluded that the data is homogenous.
C. Test of Hypothesis T-test
In this part, the researcher calculated the data to test hypohesis that whether there is significant different between stuents‟ vocabulary
understanding using crossword puzzle in experiment class and students‟
vocabulary understanding without crossword puzzle in controlled class. the researcher calculated the data using T-test formula. Two classes were
compared, the experimental class was X variable and the controlled class was Y variable. The next table is staistical calculation of the gain score both
experimental class using crossword puzzle and controlled class without crossword puzzle.
Table 4.11 The Statistical Calculaion of the Gain Score of Both
the Controlled and the Experimental Class
Students X
Y X=Mx-
X Y=My-
Y X
2
Y
2
1
17
7 -3,59
-6,38 289,00
289,00 2
26 29
26,00 29,00
676,00 676,00
3 33
9 11,00
9,00 1089,00
1089,00 4
17 24
17,00 24,00
289,00 289,00
5 12
29 12,00
29,00 144,00
144,00 6
10 26
10,00 26,00
100,00 100,00
7 26
3 26,00
3,00 676,00
676,00 8
34 6
34,00 6,00
1156,00 1156,00
9 9
9 9,00
9,00 81,00
81,00 10
30 6
30,00 6,00
900.00 900,00
11 22
20 22,00
20,00 484,00
484,00 12
20 14
20,00 14,00
400,00 400,00
13 11
17 11,00
17,00 121,00
121,00
14 11
9 11,00
9,00 121,00
121,00 15
10 6
8,00 6,00
64,00 64,00
16 18
8 18,00
8,00 324,00
324,00 17
17 13
17,00 13,00
289,00 269,00
18 29
12 29,00
12,00 841,00
144,00 19
28 14
28,00 14,00
784,00 196,00
20 16
18 16,00
1,00 256,00
324,00 21
23 22
23,00 22,00
529,00 484,00
22 17
7 17,00
7,00 289,00
49,00 23
37 17
37,00 17,00
1369,00 289,00
24 23
20 23,00
20,00 529,00
400,00 25
22 4
22,00 4,00
484,00 16,00
26 12
13 12,00
13,00 144,00
169,00 27
20 19
20,00 19,00
400,00 361,00
28 37
13 37,00
13,00 1369,00
169,00 29
13 12
13,00 12,00
169,00 144,00
30 20
17 20,00
17,00 400,00
289,00 31
17 17,00
0,00 289,00
0,00 32
24 5
24,00 5,00
576,00 25,00
Total 679
428 657,78
414,63 16189,00
7511,64 The table above described the result calculation of the gained score of
the experimental class X and the controlled class Y. Based on the table above, it can be concluded that the total score of the experimantal class X is
679 and the controlled class Y is 428. From the table aove, the writer calculated them based on the steps of t-
test formula, as follows: 1. Determining Mean of Variable X
M
x
= =
= 21.21 2. Determining Mean of Variable Y
M
y
= =
= 13.38 3. Determining Standar od Deviation Score of Variable X
SD
x
= √
= √
= √ = 22.49
4. Determining Standard of Deviation Score of Variable Y SD
y
= √
= √
= √ = 15.3
5. Determining Standard Error of Mean of Variable X SEM
x
=
√
=
√
=
√
= = 4.04
6. Determining Standard Error of Mean of Variable Y SEM
x
=
√
=
√
=
√
= = 2.74
7. SEM
x
– My = √ =
√ =
√ = √ = 3.68
Determining t t
= =
= 2.13 Dtermining t
table
in significance level 5 with df degree of freedom Df = N1+N2
– 2 = 32+32 – 2 = 62 The value of df 62 at the degree of significance 5 or t
table
is 1.670. It means Ha is accepted because T
2.13 is higher than t
table
1.999. 8. The Testing of Hypotheses
The statistical hypotheses of this research can be seen as: Ho
: There is no significant difference between students‟ vocabulary using crossword puzzle and without crossword
puzzle. Ha
: There is significant difference between students‟ vocabulary
using crossword puzzle and without crossword puzzle. And then the criteria used as follows:
1. If t-test t
o
t-table t
t
in significant degree of 0.05, Ho null hypothesis is rejected.
2. If t-test t
o
t-table t
t
in significant degree of 0.05, Ho null hypothesis is accepted.
In addition, the writer found To is 2.13 while t
table
is 1.999. After comparing between To and t
table
the writer found To is higher than t
table.
Therefore, it can be concluded that Ho is rejected. It means there is significance score gain between students‟ vocabulary using crossword
puzzle and without crossword puzzle.
D. Interpretation of the Data and Discussion
In the description of the data which was taken from 32 students of experimental class, the writer could explain briefly about the data got from the
students before they were analyzed. The description of the experimental class has the mean of pre-test 58.71 before using crossword puzzle. It means the
mean score is bad because it is lower than the standard minimum KKM. After giving 4 times treatments for experimental class using crossword
puzzle, the writer got the mean of post-test 74.56. So, the writer got the mean of the gain score 21.03. It is good because the mean score has reached KKM.
The smallest score in the pre-test was 50 and the highest score was 76. The data showed in the post-test that the smallest score was 70 and the highest
score was 94. It can be summarized that the lowest and the highest scores in post-test were higher than pre-test.
Meanwhile, from the description of score in controlled class which was the writer got the mean of pre-test 61.40. It means the mean score is bad.
In this class, the writer did not use crossword puzzle, but the writer only asked the students to memorize the vocabulary. After giving 4 times treatment
without crossword puzzle, the writer got the mean of post-test 71.25. it is bad because the mean score is still lower than KKM. The writer got the mean of
gain score was 15.09. it means that the gain score of experimental class was higher than controlled class. The smallest score in the pre-test was 51 and the
highest score was 78. The data showed in post-test that smallest score was 54 and the highest score was 94, it can be summarized that the lowest and the
highest were also higher than pre-test. Before testing the hypothesis, the writer analyzed the normality and
homogeneity of the data, the purpose of analyzing the normality was to see whether the data got in the research has been normally distributed or not. The
result of normality can be seen by comparing the value of significant score to 0.05. Meanwhile, the purpose of analyzing the homogeneity was to see
whether the data sample in both experimental and controlled class were homogenous or heterogonous.
In analyzing the normality, the result showed that both the data of pre- test and post-test in controlled class were distributed normally. According to
criteria of the test, it can be seen in the result that the significant score of pre- test 0.311 and post-test 0.207 of experimental class was higher than 0.05
0.311 and 0.207 0.05. Both the data of pre-test and post-test in experimental class also showed that they were distributed normally.
According to criteria of the test, it can be seen in the result that the significant score pre-test and post-test 0.05 0.560 and 0.356 0.05. It means that all
the data in both pre-test and post-test of experimental and control class were distributed normally. The next result that the writer got was from the
calculation of homogeneity, the result showed that 0.720 0.05. Based on the criteria, it can be concluded that H
is accepted. It means that the sample in experiment class and controlled class were homogenous.
The final calculation was testing the hypothesis. This was the main calculation to answer the problem of this research that whether there is
significant difference between students‟ vocabulary using crossword puzzle and without crossword puzzle. The writer used T-test formula in the
significance degree 0f 5. The result showed that t-test t
o
t-table 2.13 1.999. It means that t-test was higher than t-table. So, the null hypothesis
Ho is rejected. It means that alternative hypothesis Ha is accepted that there is significant difference between students‟ vocabulary using crossword
puzzle and without crossword puzzle.
44
CHAPTER V CONCLUSION AND SUGGESTION
This chapter presents the conclusion and the suggestion. In this chapter, the researcher would like to give some conclusions that may relate to
the subject.
A. Conclusion
Many games and media that become tool for teaching and learning activity effective such as cards, word wall, etc. One of games is crossword
puzzle that the writer expected effective for teaching and learning activity. So, the writer did the research using crossword puzzle in teaching vocabulary in
MTs. Muhammadiyah 1 Ciputat. Based on the anylisis result of this research, showed that the result of T-test formula to test the hypothesis of this research
in the significance degree of 5 is t-test t
t-table t
t
2.13 1.999. Therefore, the null hypothesis H
is rejected and alternative hypothesis H
a
is accepted. Besides, it can be seen from the comparison between the gained score
average of experimental class and the gained score average of controlled class on the table of the students‟ gained score, the gained score average in
experimental class 21.22 is higher than the gained score average in control class 13.38. it means that crossword puzzle is effective in improving
studen ts‟ vocabulary. Therefore, it can be concluded that the answer of
research question was proven that there is effectiveness of crosswrod puzzle on students‟ vocabulary at first of Islamic Junior High School
Muhammadiyah Ciputat in academic year 20152016.
