The Description of the Data The Analysis of the Data
Table 4.4 The Frequency and Percentage of the Students’ Errors in the
Meaning of Can Students’
Number Number of
Item Test Correct
Answer Frequency of
Errors Percentage
of Errors
1. 10
4 6
4.08 2.
10 7
3 2.04
3. 10
7 3
2.04 4.
10 9
1 0.68
5. 10
7 3
2.04 6.
10 6
4 2.27
7. 10
4 6
4.08 8.
10 8
2 1.36
9. 10
9 1
0.68 10.
10 9
1 0.68
11. 10
7 3
2.04 12.
10 7
3 2.04
13. 10
7 3
2.04 14.
10 8
2 1.36
15. 10
8 2
1.36 16.
10 8
2 1.36
17. 10
6 4
2.27 18.
10 7
3 2.04
19. 10
6 4
2.27 20.
10 8
2 1.36
21. 10
9 1
0.68 22.
10 9
1 0.68
23. 10
8 2
1.36 24.
10 8
2 1.36
To find out the percentage of the students’ errors in the meaning of can, the writer used the formula below:
P = Frequency of Errors x 100
Frequency of Errors + Correct Answers
= 147 x 100
147 + 253 =
14700 400
= 36.75
Students’ Number
Number of Item Test
Correct Answer
Frequency of Errors
Percentage of Errors
25. 10
7 3
2.04 26.
10 7
3 2.04
27. 10
7 3
2.04 28.
10 7
3 2.04
29. 10
9 1
0.68 30.
10 8
2 1.36
31. 10
8 2
1.36 32.
10 9
1 0.68
33. 10
6 4
2.27 34.
10 6
4 2.27
35. 10
8 2
1.36 36.
10 7
3 2.04
37. 10
7 3
2.04 38.
10 8
2 1.36
39. 10
7 3
2.04 40.
10 6
4 2.27
Total 253
147 100
From the data above, the writer can conclude that the average of the students’ errors in using can is 36.75, and it can be concluded that the rest is
63.25, which means that the majority of the students did not do errors on the test.
After the writer got the description of the meaning of can above, she
would like to analyze the data description of the students’ errors in the meaning of
may, as follows:
Table 4.5 The Frequency and Percentage of
the Students’ Errors in the Meaning of May
Students’ Number
Number of Item Test
Correct Answer
Frequency of Errors
Percentage of Errors
1. 10
5 5
2.69 2.
10 8
2 1.07
3. 10
7 3
1.61 4.
10 5
5 2.69
5. 10
6 4
2.15 6.
10 8
2 1.07
7. 10
7 3
1.61 8.
10 2
8 4.30
9. 10
5 5
2.69 10.
10 2
8 4.30
11. 10
7 3
1.61 12.
10 6
4 2.15
13. 10
7 3
1.61 14.
10 5
5 2.69
15. 10
7 3
1.61 16.
10 6
4 2.15
17. 10
6 4
2.15
To find out the percentage of the students’ errors in the meaning of may, the writer used the formula below:
Students’ Number
Number of Item Test
Correct Answer
Frequency of Errors
Percentage of Errors
18. 10
3 7
3.76 19.
10 4
4 2.15
20. 10
8 2
1.07 21.
10 3
7 3.76
22. 10
4 6
3.22 23.
10 5
5 2.69
24. 10
1 9
4.84 25.
10 3
7 3.76
26. 10
6 4
2.15 27.
10 6
4 2.15
28. 10
6 4
2.15 29.
10 4
6 3.22
30. 10
6 4
2.15 31.
10 6
4 2.15
32. 10
8 2
1.07 33.
10 6
4 2.15
34. 10
7 3
1.61 35.
10 7
3 1.61
36. 10
7 3
1.61 37.
10 8
2 1.07
38. 10
6 4
2.15 39.
10 7
3 1.61
40. 10
6 4
2.15
Total 214
186 100
P = Frequency of Errors x 100
Frequency of Errors + Correct Answers
= 186 x 100
186 + 214 =
18600 400
= 46.05 From the data above, the writer can conclude that the average of the
students’ errors in using may is 46.05, and it can be concluded that the rest is 53.95, which means that the majority of the students did not do errors on the
test. After the writer got the description of the meaning of
may above, she would like to analyze the data description of the students’ errors in the form of
can and may, as follows:
Table 4.6 The Frequency and Percentage of the Students’ Errors in the Form of
Can and May Students’
Number Number of
Item Test Correct
Answer Frequency of
Errors Percentage
of Errors
1. 10
6 4
5.06 2.
10 7
3 3.80
3. 10
7 3
3.80 4.
10 10
5. 10
8 2
2.53 6.
10 7
3 3.80
7. 10
8 2
2.53 8.
10 9
1 1.26
9. 10
8 2
2.53
Students’ Number
Number of Item Test
Correct Answer
Frequency of Errors
Percentage of Errors
10. 10
10 11.
10 7
3 3.80
12. 10
8 2
2.53 13.
10 9
1 1.26
14. 10
7 3
3.80 15.
10 7
3 3.80
16. 10
9 1
1.26 17.
10 7
3 3.80
18. 10
7 3
3.80 19.
10 7
3 3.80
20. 10
9 1
1.26 21.
10 9
1 1.26
22. 10
7 3
3.80 23.
10 6
4 5.06
24. 10
10 25.
10 8
2 2.53
26. 10
6 4
5.06 27.
10 9
1 1.26
28. 10
10 29.
10 8
2 2.53
30. 10
7 3
3.80 31.
10 6
4 5.06
32. 10
8 2
2.53 33.
10 9
1 1.26
34. 10
9 1
1.26 35.
10 8
2 2.53
36. 10
9 1
1.26 37.
10 9
1 1.26
To find out the percentage of the students’ errors in the form of can and may , the writer used the formula below:
P = Frequency of Errors x 100
Frequency of Errors + Correct Answers
= 79 x 100
79 + 321 =
7900 400
= 19.75 From the data above, the writer can conclude that the average of the
students’ errors in using the form of can and may is 19.75, and it can be concluded that the rest is 80.25, which means that the majority of the
students did not do errors on the test.