Saturday, October 17, 2015

CHAPTER 4

CAPTER IV
DATA ANALYISIS, FINDING and DISCUSSION

4.1 The Scores of Pre – Test and Post – Test
NO
The Names of Student
Pre – Test (x)
Post – Test (y)
1
Ai Elis Hilmi
5
5
2
Ai Jamilah
4
6
3
Andrian
4
6,5
4
Ahmad Hapid
5
5,5
5
Asep Rahmat
5
5,5
6
Cecep Rian
7
7
7
Dede Joharudin
4
4,5
8
Devi Azhari
6
8,5
9
Dede siti Maryam
7
8,5
10
Encep Muhammad
6
7
11
Firmansyah
6
6
12
Fitriyani
6
7
13
Herdiana
6
6,5
14
Ihsan Ramdani
5
6,5
15
Lisna Wati
6
6,5
16
Irfan Maulana
7
8,5
17
Mi’raj
5
6
18
Neng Nia
6
6,5
19
Nenden Martiasari
4
6
20
Nurul Hiudayat
6,5
7
21
Rina Andrian
7
8
22
Risda Nuriah
6
6,5
23
Royani
5
8
24
Revi Pandina
5
6
25
Ridwan Jaelani
7
8,5
26
Siti Maryama
4
4,5
27
Siti Nur Azizah
5
8,5
28
Siti Aisyah
6
6
29
Silvi Susanti
4
7
30
Shopi Yanti
4
6,5
31
Sulaeman
4
6,5
32
Saepul Rohmat
4
4,5
33
Weni Hindayani
5
6,5
34
Yani Nuryanti
7
7,5
35
Yusup Maulana
7
8
36
Rina
6
7,5
37
Windi
6,5
8,5
38
Sandi Muhammad
5
6,5
39
Yana Suryana
5
6,5
40
Yogi
4
8,5











NO
Pree – Test (x)
Post – Test (y)
D
D2
1
5
5
0
0
2
4
6
2
4
3
4
6,5
2,5
6,25
4
5
5,5
0,5
0,25
5
5
5,5
0,5
0,25
6
7
7
0
0
7
6
8,5
2,5
6,25
8
4
4,5
0,2
0,25
9
7
8,5
1,5
2,25
10
6
7
0,5
1
11
6
6
0
0
12
6
7
1
1
13
6
6,5
0,5
0,25
14
5
6,5
1,5
2,25
15
6
6,5
0,5
0,25
16
7
8,5
1,5
2,25
17
5
6
1
1
18
6
6,5
0,5
0,25
19
4
6
2
4
20
5
6,5
1,5
2,25
21
6,5
7
1,5
2,25
22
7
8
1
1
23
6
6,5
0,5
0,25
24
5
8
3
9
25
5
6
1
1
26
7
8,5
1,5
2,25
27
4
4,5
0,5
0,25
28
5
8,5
3,5
12, 25
29
6
6
0
0
30
4
7
3
9
31
4
6,5
2,5
6,25
32
4
6,5
2,5
6,25
33
4
4,5
0,5
0,25
34
3
6,5
1,5
2,25
35
7
7,5
0,5
0,25
36
7
8
1
1
37
6
7,5
1,5
2,25
38
6,5
8,5
2
4
39
5
6,5
1,5
2,25
40
4
8,5
4,5
20,25

∑ x = 217
∑ y = 270,5
∑ D = 54,5
∑ D2  = 116,25

  X  = 5,43
   Y = 6,77
    D = 1,36










4.2 Statistical Computation.
a. n = 40
b. D = ∑ D      
             n
      = 54,5
           40
      = 1,36
c.  ∑ D2 = 116,25
d. (∑ D )2  = 54,52       
    e. ∑ D = 2970.25
                      40
                = 74,26
f. n – 1 = 40 – 1
              = 39
g. Count the Standard deviation
S=                           
                                N
                     n-1
     
                        39
          
                 39
          
               =1.03

h.
      = 6.32

i.      Count t  - Value
T observe =
                    S
= 1.36 x 6,3
        1.03
 = 8.59
          1.03
       = 8.34

j. Degree of freedom
    d.f = (n – 1) = 39
k. t critical; value at 5% significance level for these degree of freedom is 2.021, but the t observe = 8.34, the t observe  t critical


4.3      Interpreting the result of the computation
After the formulas and counting the result of the test, the data of the research as follow:
a.       N = 40
b.      The – man of the pre test   and post test  )
c.       Standard deviations
S = 1.03 (see the appendix)
d.      T observe = 8.34
e.      Df = 39
f.        T critical for the first test on the level significance 0,5 with degree of freedom 39 is 2.021
g.       We can observe now that t observe is more than  t critical, this indicated that the differences between x and y is significant.



4.4   Testing Hypothesis
From the data computation above, the writer has found that the value of the t observe is, 8.34. Thus, based on the significance 5 % and n = 40 the t is 2.021, the t observe t critical, its means that the hypothesis is accepted. So, the kinds of activities can develop students’ speaking ability.



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