CONCLUSION AND SUGGESTION
B. Suggestion
CHAPTER V
background knowledge and experience, which is needed for reading activity.
Teacher should be creative to decide kind of pre-reading activities that can be used such as picture or activating students’ background knowledge. Teacher is not only as the infomation giver, but also as facilitator. A good teacher has to give students guidance and direction how to comprehend a text. The tachers are suggested to use various techniques to teach the students.
41
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Sample Score
1 38
2 40
3 44
4 44
5 44
6 45
7 45
8 45
9 45
10 45
11 50
12 50
13 50
14 52
15 52
16 56
17 56
18 56
19 56
20 60
21 60
22 60
23 65
24 66
25 66
N 25
44 3 132
45 5 225
50 3 150
52 2 104
56 4 224
60 3 180
65 1 65
66 2 132
Total 25 1290
The formula of mean : 𝑋 =𝑓𝑥
𝑁 =1290
25 = 𝟓𝟏, 𝟔 3) Calculate the Standard Deviation
X F FX X x2 fx2
38 1 38 -13.6 184.96 1147.8544
40 1 40 -11.6 134.56 722. 5344
44 3 132 -7.6 57.76 570.2544
45 5 225 -6.6 43.56 118.3744
50 3 150 44.6 1989.16 62.0944
52 2 104 0.4 0.16 45.1632
56 4 224 4.4 19.36 0.7744
60 3 180 8.4 70.56 37.4976
65 1 65 13.4 179.56 224.7264
66 2 132 14.4 207.36 741.9264
Total 25 1290 3671.2
= √3671.2 25 = √146.848
= 12.1180856574 ≈ 𝟏𝟐. 𝟏𝟏 4) Determine the Z value of each data by the formula : Zi = Xi−X̅
S
Z1 =Xi−X̅
S = 38−𝟓𝟏,𝟔
𝟏𝟐.𝟏𝟏 = -1.22
To Calculate Z2 and so on, thfollow the method of calculating Z1
5) Determine the Ztable basedon the Zvalue
Z1 = round it into two numbers behind the comma become -1.22, then the minus value is released to be positive then in the table the critical value of the normal distribution is obtained by the value of Z1 is 0.3888
To find the Ztable from Z2 and so on can follow the method above.
6) Determine the F(Zi)basedon the Ztable
If Zi negative (-), so 0,5 - Ztable
If Zi positive (+), so 0,5 + Ztable
Z1 = -1,54 , because the value of Z1 is negative, so calculate the F(Zi) as follow :
F(Z1) = 0,5 – 0.3888= 0.1112
To find the F(Z2) and so on can follow the method above.
7) Determine the S(Zi) by the formula : S(Zi) = 𝐶𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝑛 = 1
25= 0,04 8) Calculate the Lcount by the formula : F(Z1) – S(Zi)
L1 = |F(Z1) – S(Zi)| = |0.1112– 0,04| = 0.0712 So that the table as below is found :
5 44 -0.68 0.2517 0.2483 5 0,20 0.0483
6 45 -0.59 0.2224 0.2776 6 0,24 0.0376
7 45 -0.59 0.2224 0.2776 7 0,28 0.0024
8 45 -0.59 0.2224 0.2776 8 0,32 0.0424
9 45 -0.59 0.2224 0.2776 9 0,36 0.0824
10 45 -0.59 0.2224 0.2776 10 0,40 0.1224
11 50 -0.14 0.0557 0.4443 11 0,44 0.0043
12 50 -0.14 0.0557 0.4443 12 0,48 0.0357
13 50 -0.14 0.0557 0.4443 13 0,52 0.0757
14 52 0.03 0.0120 0.488 14 0,56 0.072
15 52 0.03 0.0120 0.488 15 0,6 0.112
16 56 0.39 0.1517 0.6517 16 0,64 0.117
17 56 0.39 0.1517 0.6517 17 0,68 0.0283
18 56 0.39 0.1517 0.6517 18 0,72 0.0683
19 56 0.39 0.1517 0.6517 19 0,76 0.1083
20 60 0.75 0.2734 0.7734 20 0,80 0.0266
21 60 0.75 0.2734 0.7734 21 0,84 0.0666
22 60 0.75 0.2734 0.7734 22 0,88 0.1066
23 65 1.20 0.3849 0.8849 23 0,92 0.0351
24 66 1.29 0.4015 0.9015 24 0,96 0.0585
25 66 1.29 0.4015 0.9015 25 1 0.0985
9) The value of Ltable with the significance ơ = 0,05 and n = 25 , then found the Ltable of the critical value for the lilliefors test is 0,1730.
Based on the table, the Lmax value is 0.1283. Therefore, H0 is accepted because the result shows that Lmax is lower than Ltable or 0.1283≤ 0,1730. It means that the data in pre-test experimental class is normally distributed.
1 32
2 32
3 32
4 32
5 36
6 36
7 40
8 40
9 40
10 44
11 44
12 44
13 48
14 48
15 48
16 52
17 52
18 56
19 56
20 60
21 60
22 60
23 64
24 64
25 72
n 25
40 3 120
44 3 132
48 3 144
52 2 104
56 2 112
60 3 180
64 2 128
72 1 72
Total 25 1192
The formula of mean : 𝑋 =𝑓𝑥
𝑁 = 1192
25 = 𝟒𝟕, 𝟔𝟖 3) Calculate the Standard Deviation
X F FX X x2 fx2
32 4 128 -15,68 245,8624 983,4496
36 2 72 -11,68 136,4224 272,8448
40 3 120 -7,68 58,9824 176,9472
44 3 132 -3,68 13,5424 40,6272
48 3 144 0,32 0,1024 0,3072
52 2 104 4,32 18,6624 37,3248
56 2 112 8,32 69,2224 138,4448
60 3 180 12,32 151,7824 455,3472
64 2 128 16,32 266,3424 532,6848
72 1 72 24,32 591,4624 591,4624
Total 25 1192 27,2 1552,384 3229,44
= √3229,44 25 = √129,1776
= 11,365632407 ≈ 𝟏𝟏, 𝟑𝟕 4) Determine the Z value of each data by the formula : Zi = Xi−X̅
S
Z1 =Xi−X̅
S = 32−47,68
11,37 = -1,,38
To Calculate Z2 and so on, thfollow the method of calculating Z1
5) Determine the Ztable basedon the Zvalue
Z1 = round it into two numbers behind the comma become -1,38, then the minus value is released to be positive then in the table the critical value of the normal distribution is obtained by the value of Z1 is 0,4162
To find the Ztable from Z2 and so on can follow the method above.
6) Determine the F(Zi)basedon the Ztable
If Zi negative (-), so 0,5 - Ztable
If Zi positive (+), so 0,5 + Ztable
Z1 = -1,38 , because the value of Z1 is negative, so calculate the F(Zi) as follow :
F(Z1) = 0,5 – 0,4162 = 0,0838
To find the F(Z2) and so on can follow the method above.
7) Determine the S(Zi) by the formula : S(Zi) = 𝐶𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝑛 = 1
25= 0,04 8) Calculate the Lcount by the formula : F(Z1) – S(Zi)
L1 = |F(Z1) – S(Zi)| = |0,0838 – 0,04| = 0,0438
3 32 -1,38 0,4162 0,0838 3 0,12 0,0362
4 32 -1,38 0,4162 0,0838 4 0,16 0,0762
5 36 -1,02 0,3461 0,1539 5 0,20 0,0461
6 36 -1,02 0,3461 0,1539 6 0,24 0,0861
7 40 -0,68 0,2517 0,2483 7 0,28 0,0371
8 40 -0,68 0,2517 0,2483 8 0,32 0,0717
9 40 -0,68 0,2517 0,2483 9 0,36 0,1117
10 44 -0,32 0,1255 0,3745 10 0,40 0,0255
11 44 -0,32 0,1255 0,3745 11 0,44 0,0655
12 44 -0,32 0,1255 0,3745 12 0,48 0,1055
13 48 0,03 0,0120 0,5120 13 0,52 0,0080
14 48 0,03 0,0120 0,5120 14 0,56 0,0480
15 48 0,03 0,0120 0,5120 15 0,60 0.0880
16 52 0,38 0,1480 0,6480 16 0,64 0,0080
17 52 0,38 0,1480 0,6480 17 0,68 0,0320
18 56 0,73 0,2673 0,7673 18 0,72 0,0473
19 56 0,73 0,2673 0,7673 19 0,76 0,0073
20 60 1,08 0,3599 0,8599 20 0,80 0,0599
21 60 1,08 0,3599 0,8599 21 0,84 0,0199
22 60 1,08 0,3599 0,8599 22 0,88 0,0201
23 64 1,44 0,4251 0,9251 23 0,92 0,0051
24 64 1,44 0,4251 0,9251 24 0,96 0,0349
25 72 2,14 0,4838 0,9838 25 1 0,0162
9) The value of Ltable with the significance ơ = 0,05 and n = 25 , then found the Ltable of the critical value for the lilliefors test is 0,1730.
Based on the table, the Lmax value is 0,1117. Therefore, H0 is accepted because the result shows that Lmax is lower than Ltable or 0,1117 ≤ 0,1730. It means that the data in pre-test experimental class is normally distributed.
2 46
3 46
4 62
5 66
6 70
7 70
8 70
9 73
10 76
11 76
12 76
13 76
14 80
15 80
16 80
17 80
18 80
19 80
20 85
21 85
22 85
23 85
24 85
25 85
n 25
62 1 62
66 1 66
70 3 210
73 1 73
76 4 304
80 6 480
85 6 510
Total 25 1837
The formula of mean : 𝑋 =𝑓𝑥
𝑁 = 1837
25 = 𝟕𝟑, 𝟒𝟖 3) Calculate the Standard Deviation
X F FX X x2 fx2
40 1 40 -33,48 1120.9104 1120.9104
46 2 92 -27.48 755.1504 1510.308
62 1 62 -11.48 131.7904 131.7904
66 1 66 -7.48 55.9504 55.9504
70 3 210 -3.48 12.1104 36.3312
73 1 73 -0.48 0.2304 0.2304
76 4 704 2.52 6.3504 25.4016
80 6 480 6.52 42.5104 255.0624
85 6 510 11.52 132.7104 796.2624
Total 25 3932.24
= √3932.24 25 = √157.2896
= 12.54151506 ≈ 𝟏𝟐. 𝟓𝟒 4) Determine the Z value of each data by the formula : Zi = Xi−X̅
S
Z1 =Xi−X̅
S = 40−73,48
12.54 = -2.66
To Calculate Z2 and so on, thfollow the method of calculating Z1
5) Determine the Ztable basedon the Zvalue
Z1 = round it into two numbers behind the comma become -2.66, then the minus value is released to be positive then in the table the critical value of the normal distribution is obtained by the value of Z1 is 0.4961
To find the Ztable from Z2 and so on can follow the method above.
6) Determine the F(Zi)basedon the Ztable
If Zi negative (-), so 0,5 - Ztable
If Zi positive (+), so 0,5 + Ztable
Z1 = -2.66, because the value of Z1 is negative, so calculate the F(Zi) as follow :
F(Z1) = 0,5 – 0.4961= 0.0039
To find the F(Z2) and so on can follow the method above.
7) Determine the S(Zi) by the formula : S(Zi) = 𝐶𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝑛 = 1
25= 0,04 8) Calculate the Lcount by the formula : F(Z1) – S(Zi)
L1 = |F(Z1) – S(Zi)| = |0.0039 – 0,04| = 0.0361
3 46 -2.19 0.4857 0.0143 3 0,12 0.1057
4 62 -0.91 0.3186 0.1814 4 0,16 0.0214
5 66 -0.59 0.2224 0.2776 5 0,20 0.0776
6 70 -0.27 0.1064 0.3936 6 0,24 0.1536
7 70 -0.27 0.1064 0.3936 7 0,28 0.1136
8 70 -0.27 0.1064 0.3936 8 0,32 0.0736
9 73 0.03 0.0120 0.512 9 0,36 0.152
10 76 0.200 0.0793 0.5793 10 0,40 0.1430
11 76 0.200 0.0793 0.5793 11 0,44 0.1393
12 76 0.200 0.0793 0.5793 12 0,48 0.0993
13 76 0.200 0.0793 0.5793 13 0,52 0.0593
14 80 0.51 0.1950 0.695 14 0,56 0.135
15 80 0.51 0.1950 0.695 15 0,60 0.095
16 80 0.51 0.1950 0.695 16 0,64 0.055
17 80 0.51 0.1950 0.695 17 0,68 0.015
18 80 0.51 0.1950 0.695 18 0,72 0.025
19 80 0.51 0.1950 0.695 19 0,76 0.065
20 85 0.91 0.3186 0.8186 20 0,80 0.0186
21 85 0.91 0.3186 0.8186 21 0,84 0.0214
22 85 0.91 0.3186 0.8186 22 0,88 0.0614
23 85 0.91 0.3186 0.8186 23 0,92 0.1014
24 85 0.91 0.3186 0.8186 24 0,96 0.1414
25 85 0.91 0.3186 0.8186 25 1 0.1420
9) The value of Ltable with the significance ơ = 0,05 and n = 25 , then found the Ltable of the critical value for the lilliefors test is 0,1730.
Based on the table, the Lmax value is 0,1303. Therefore, H0 is accepted because the result shows that Lmax is lower than Ltable or 0.1536≤ 0,1730. It means that the data in pre-test experimental class is normally distributed.
2 32
3 36
4 40
5 40
6 44
7 44
8 48
9 48
10 48
11 52
12 52
13 56
14 56
15 60
16 60
17 64
18 68
19 72
20 76
21 76
22 80
23 80
24 84
25 84
N 25
40 2 80
44 2 88
48 3 144
52 2 104
56 2 112
60 2 120
64 1 64
68 1 68
72 1 72
76 2 152
80 2 160
84 2 168
Total 25 1432
The formula of mean : 𝑋 =𝑓𝑥
𝑁 =𝟏𝟒𝟑𝟐
25 = 𝟓𝟕. 𝟐𝟖 3) Calculate the Standard Deviation
X F FX X x2 fx2
32 2 64 -25.28 639.0784 1278.1568
36 1 36 -21.28 452.8384 452.8384
40 2 80 -17.28 298.5984 597.1968
44 2 88 -13.28 176.3584 352.7168
48 3 144 -9.28 86.1184 258.3552
52 2 104 -5.28 27.8784 55.7568
56 2 112 -1.28 1.6384 3.2768
60 2 120 2.72 7.3984 14.7968
64 1 64 6.72 45.1584 45.1584
68 1 68 10.72 114.9184 114.9184
72 1 72 14.72 216.6784 216.6784
The formula of Standard Deviation :
𝑆𝐷 = √∑ 𝑓𝑥2 𝑁
= √6551.04 25 = √262.0416
= 16.1876990335 ≈ 𝟏𝟔. 𝟏𝟗 4) Determine the Z value of each data by the formula : Zi = Xi−X̅
S
Z1 =Xi−X̅
S = 32−57.28
16.19 = -1.56
To Calculate Z2 and so on, thfollow the method of calculating Z1
5) Determine the Ztable basedon the Zvalue
Z1 = round it into two numbers behind the comma become -1.56, then the minus value is released to be positive then in the table the critical value of the normal distribution is obtained by the value of Z1 is 0.4406
To find the Ztable from Z2 and so on can follow the method above.
6) Determine the F(Zi)basedon the Ztable
If Zi negative (-), so 0,5 - Ztable
If Zi positive (+), so 0,5 + Ztable
Z1 = -1,56 , because the value of Z1 is negative, so calculate the F(Zi) as follow :
F(Z1) = 0,5 – 0.4406= 0.0594
To find the F(Z2) and so on can follow the method above.
7) Determine the S(Zi) by the formula : S(Zi) = 𝐶𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝑛 = 1
25= 0,04
No Xi Zi Ztable f(zi) f(kum) s(zi) Ifzi-szil
1 32 -1.56 0.4406 0.0594 1 0,04 0.0194
2 32 -1.56 0.4406 0.0594 2 0,08 0.0206
3 36 -1.31 0.4049 0.0951 3 0,12 0.0249
4 40 -1.07 0.3577 0.1423 4 0,16 0.0177
5 40 -1.07 0.3577 0.1423 5 0,20 0.0577
6 44 -0.82 0.2939 0.2061 6 0,24 0.0339
7 44 -0.82 0.2939 0.2061 7 0,28 0.0739
8 48 -0.57 0.2157 0.2843 8 0,32 0.0357
9 48 -0.57 0.2157 0.2843 9 0,36 0.0757
10 48 -0.57 0.2157 0.2843 10 0,40 0.1157
11 52 -0.33 0.1293 0.3707 11 0,44 0.0693
12 52 -0.33 0.1293 0.3707 12 0,48 0.1093
13 56 -0.08 0.0319 0.4681 13 0,52 0.0519
14 56 -0.08 0.0319 0.4681 14 0,56 0.0919
15 60 0.17 0.0675 0.5675 15 0,60 0.0325
16 60 0.17 0.0675 0.5675 16 0,64 0.0725
17 64 0.42 0.1628 0.6628 17 0,68 0.0172
18 68 0.66 0.2454 0.7454 18 0,72 0.0254
19 72 0.91 0.3186 0.8186 19 0,76 0.0586
20 76 1.16 0.3770 0.877 20 0,80 0.077
21 76 1.16 0.3770 0.877 21 0,84 0.037
22 80 1.40 0.4192 0.9192 22 0,88 0.0392
23 80 1.40 0.4192 0.9192 23 0,92 0.0008
24 84 1.65 0.4505 0.9505 24 0,96 0.0095
25 84 1.65 0.4505 0.9505 25 1 0.0495
9) The value of Ltable with the significance ơ = 0,05 and n = 25 , then found the Ltable of the critical value for the lilliefors test is 0,1730.
Based on the table, the Lmax value is 0.1157. Therefore, H0 is accepted because the result shows that Lmax is lower than Ltable or 0.1157≤ 0,1730. It means that the data in pre-test experimental class is normally distributed.
In order to know the homogeneity of the data, the writer did the homogeneity test. To do the homogeneity test, he tested the score of pretest and post-test in both experiment and control class using Fisher test.
Hypothesis:
H0 : The experiment class is homogenous to the controlled class Ha : The experiment class is not homogenous to the controlled class
The criteria of the test:
α = 0.05
H0 : 𝑓𝑐𝑜𝑢𝑛𝑡 ≤ 𝑓𝑡𝑎𝑏𝑙𝑒
Ha : 𝑓𝑐𝑜𝑢𝑛𝑡 ≥ 𝑓𝑡𝑎𝑏𝑙𝑒
The formula used in homogeneity test can be seen as follows : 𝐹ℎ𝑖𝑡𝑢𝑛𝑔 =𝑣𝑎𝑟𝑖𝑎𝑛𝑠𝑡𝑒𝑟𝑏𝑒𝑠𝑎𝑟
𝑣𝑎𝑟𝑖𝑎𝑛𝑠𝑡𝑒𝑟𝑘𝑒𝑐𝑖𝑙
With:
𝑆2 =∑(𝑥 − 𝑥)2 𝑛 − 1
A. Homogeneity Test of Experimental and Controll Class Pre-test