Suggestions

Based on the findings and conclusions above, the researcher provides suggestions as follow:

1. This research finding indicates that Collaborative Writing was one of the effective way to help the students in writing at grade VIII of MTsN Kamang. Therefore, it is suggested that English teachers at MTsN Kamang apply Collaborative Writing as a technique of teaching writing.

2. In teaching writing, English teachers need to find the appropriate strategy for the students, by considering that the students become the center of learning. The teacher also to consider the students‟ reading habits.

3. The moderator variable in this research is reading habits. It is suggested to the other researcher to conduct a research on other moderator variable like motivation, participation, and so on.

4. It is suggested for further researcher to develop this research on larger population and sample in order to get the knowledge and the empirical data. Besides that, they also suggested to conduct the same research for other skill and other kind of text.

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NORMALITY TESTING OF POPULATION

Normality Testing (Liliefors testing) of VIII.1 writing score The formula of Normality testing of VIII.1 writing score :

1. 𝑥 = ^𝑥𝑖

𝑛 =¹⁴⁸⁸

24 = 62 2. 𝑆₁² =^{𝑛 𝑥2− 𝑥 2}

𝑛 𝑛−1

𝑆₁² = 24 93064 − 1488 ²

24 23 = 35.13 𝑆₁ = 35.13 = 5.93

3. S

x z_i xⁱ





For 𝑥 = 52 so:

𝑧₁ =⁵²⁻⁶²

5.93 = −1.69

4. Compute the prequency F

 

zi by using z table, so 𝐹 𝑧₁ = 𝐹 −1.69 = 0.0455

5. n

z yang z z Banyaknyaz z

S _i ⁿ  ⁱ

 , ,...,

)

( ^{1 2}

𝑆 𝑧₁ = ¹

24 = 0.04167 6. Compute F(z_i)S(z_i)

So, 𝐹 𝑧₁ − 𝑆 𝑧₁ = 0.0455 − 0.04167 =0.00383

7. The formula of each classess are the same and the result can be seen by the tables below:

NO NILAI X² Zi F(Zi) S(Zi) F(Zi)- S(Zi) | F(Zi)- S(Zi)|

1 52 ₂₇₀₄ -1.6863406 0.0455 0.04167 0.00383 0.00383

2 54 ₂₉₁₆ -1.3490725 0.0885 0.125 -0.0365 0.0365

3 54 ₂₉₁₆ -1.3490725 0.0885 0.125 -0.0365 0.0365

4 56 ₃₁₃₆ -1.0118044 0.1562 0.1667 -0.0105 0.0105

5 58 ₃₃₆₄ -0.6745363 0.2514 0.2917 -0.0403 0.0403

6 58 3364 -0.6745363 0.2514 0.2917 -0.0403 0.0403

7 58 3364 -0.6745363 0.2514 0.2917 -0.0403 0.0403

8 60 3600 -0.3372681 0.3669 0.4583 -0.0914 0.0914

9 60 3600 -0.3372681 0.3669 0.4583 -0.0914 0.0914

10 60 ₃₆₀₀ -0.3372681 0.3669 0.4583 -0.0914 0.0914

11 60 ₃₆₀₀ -0.3372681 0.3669 0.4583 -0.0914 0.0914

12 62 _{3844 0 0.5 0.625 -0.125 0.125}

13 62 3844 0 0.5 0.625 -0.125 0.125

14 62 3844 0 0.5 0.625 -0.125 0.125

15 62 3844 0 0.5 0.625 -0.125 0.125

16 64 4096 0.33726813 0.6331 0.75 -0.1169 0.1169

17 64 4096 0.33726813 0.6331 0.75 -0.1169 0.1169

18 64 4096 0.33726813 0.6331 0.75 -0.1169 0.1169

19 66 4356 0.67453626 0.7486 0.875 -0.1264 0.1264

20 66 4356 0.67453626 0.7486 0.875 -0.1264 0.1264

21 66 4356 0.67453626 0.7486 0.875 -0.1264 0.1264

22 70 4900 1.34907251 0.9099 0.9617 -0.0518 0.0518

23 72 5184 1.68634064 0.9545 0.9583 -0.0038 0.0038

24 78 6084 2.69814503 0.9964 1 -0.0036 0.0036

ΣX 1488 93064

X ̅ 62

𝐿_{𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑} = 0.1264 ; 𝑛 = 24; ∝= 0.05 ; 𝐿_{𝑡𝑎𝑏𝑙𝑒} = 0.173

So Lobserved < Ltable 0.1264 < 0.173

Conclusion : if Lobserved < Ltable, it means H0 was accepted or the data was normally distributed.

Lobserved 0.1264

Ltable 0.173

NO NILAI X² Zi F(Zi) S(Zi) F(Zi)- S(Zi) | F(Zi)- S(Zi)|

1 52 ₂₇₀₄ -1.70126874 0.0446 0.125 -0.0804 0.0804

2 54 ₂₉₁₆ -1.35524798 0.0869 0.125 -0.0381 0.0381

3 54 ₂₉₁₆ -1.35524798 0.0869 0.125 -0.0381 0.0381

4 54 ₂₉₁₆ -1.35524798 0.0869 0.2083 -0.1214 0.1214

5 56 ₃₁₃₆ -1.00922722 0.1562 0.2083 -0.0521 0.0521

6 58 ₃₃₆₄ -0.66320646 0.2546 0.2917 -0.0371 0.0371

7 58 ₃₃₆₄ -0.66320646 0.2546 0.2917 -0.0371 0.0371

8 58 ₃₃₆₄ -0.66320646 0.2546 0.375 -0.1204 0.1204

9 60 ₃₆₀₀ -0.3171857 0.3745 0.375 -0.0005 0.0005

10 60 ₃₆₀₀ -0.3171857 0.3745 0.4583 -0.0838 0.0838

11 60 ₃₆₀₀ -0.3171857 0.3745 0.4583 -0.0838 0.0838

12 60 ₃₆₀₀ -0.3171857 0.3745 0.5 -0.1255 0.1255

13 62 ₃₈₄₄ 0.028835063 0.512 0.5833 -0.0713 0.0713

14 62 ₃₈₄₄ 0.028835063 0.512 0.5833 -0.0713 0.0713

15 64 ₄₀₉₆ 0.374855825 0.6443 0.6667 -0.0224 0.0224

16 64 ₄₀₉₆ 0.374855825 0.6443 0.6667 -0.0224 0.0224

17 66 ₄₃₅₆ 0.720876586 0.7642 0.8333 -0.0691 0.0691

18 66 ₄₃₅₆ 0.720876586 0.7642 0.8333 -0.0691 0.0691

19 68 ₄₆₂₄ 1.066897347 0.8577 0.8333 0.0244 0.0244

20 68 ₄₆₂₄ 1.066897347 0.8577 0.8333 0.0244 0.0244

21 68 ₄₆₂₄ 1.066897347 0.8577 0.9583 -0.1006 0.1006

22 70 ₄₉₀₀ 1.412918108 0.9207 0.9583 -0.0376 0.0376

23 70 ₄₉₀₀ 1.412918108 0.9207 0.9583 -0.0376 0.0376

24 72 ₅₁₈₄ 1.75893887 0.9608 1 -0.0392 0.0392

ΣX 1484 92528

X ̅ 61.833333

Lhitung 0.1255 Ltable 0.173

𝐿𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 = 0.1255 ; 𝑛 = 24; ∝= 0.05 ; 𝐿𝑡𝑎𝑏𝑙𝑒 = 0.173

So Lobserved < Ltable 0.1255 < 0.173

Conclusion : if Lobserved < Ltable, it means H0 was accepted or the data was normally distributed.

NO NILAI X² Zi F(Zi) S(Zi) F(Zi)- S(Zi) | F(Zi)- S(Zi)|

1 50 ₂₅₀₀ -1.978114478 0.0239 0.0417 -0.0178 0.0178

2 54 ₂₉₁₆ -1.304713805 0.0968 0.0833 0.0135 0.0135

3 56 ₃₁₃₆ -0.968013468 0.166 0.1667 -0.0007 0.0007

4 56 ₃₁₃₆ -0.968013468 0.166 0.1667 -0.0007 0.0007

5 58 ₃₃₆₄ -0.631313131 0.2643 0.2917 -0.0274 0.0274

6 58 ₃₃₆₄ -0.631313131 0.2643 0.2917 -0.0274 0.0274

7 58 ₃₃₆₄ -0.631313131 0.2643 0.2917 -0.0274 0.0274

8 60 ₃₆₀₀ -0.294612795 0.3859 0.5417 -0.1558 0.1558

9 60 ₃₆₀₀ -0.294612795 0.3859 0.5417 -0.1558 0.1558

10 60 ₃₆₀₀ -0.294612795 0.3859 0.5417 -0.1558 0.1558

11 60 ₃₆₀₀ -0.294612795 0.3859 0.5417 -0.1558 0.1558

12 60 ₃₆₀₀ -0.294612795 0.3859 0.5417 -0.1558 0.1558

13 60 ₃₆₀₀ -0.294612795 0.3859 0.5417 -0.1558 0.1558

14 62 ₃₈₄₄ 0.042087542 0.516 0.625 -0.109 0.109

15 62 ₃₈₄₄ 0.042087542 0.516 0.625 -0.109 0.109

16 64 ₄₀₉₆ 0.378787879 0.648 0.7917 -0.1437 0.1437

17 64 ₄₀₉₆ 0.378787879 0.648 0.7917 -0.1437 0.1437

18 64 ₄₀₉₆ 0.378787879 0.648 0.7917 -0.1437 0.1437

19 64 ₄₀₉₆ 0.378787879 0.648 0.7917 -0.1437 0.1437

20 66 ₄₃₅₆ 0.715488215 0.7642 0.8333 -0.0691 0.0691

21 68 ₄₆₂₄ 1.052188552 0.8531 0.9167 -0.0636 0.0636

22 68 ₄₆₂₄ 1.052188552 0.8531 0.9167 -0.0636 0.0636

23 72 ₅₁₈₄ 1.725589226 0.9582 0.9583 -1E-04 1E-04

24 78 ₆₀₈₄ 2.735690236 0.9969 1 -0.0031 0.0031

ΣX 1482 92324

X ̅ 61.75

Lhitung 0.0135 Ltable 0.173

𝐿𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 = 0.0135 ; 𝑛 = 24; ∝= 0.05 ; 𝐿𝑡𝑎𝑏𝑙𝑒 = 0.173

So Lobserved < Ltable 0.0135 < 0.173

Conclusion : if Lobserved < Ltable, it means H0 was accepted or the data was normally distributed.

NO NILAI X² Zi F(Zi) S(Zi) F(Zi)- S(Zi) | F(Zi)- S(Zi)|

1 50 ₂₅₀₀ -1.74796748 0.0409 0.0833 -0.0424 0.0424

2 50 ₂₅₀₀ -1.74796748 0.0409 0.0833 -0.0424 0.0424

3 52 ₂₇₀₄ -1.422764228 0.0778 0.2083 -0.1305 0.1305

4 52 ₂₇₀₄ -1.422764228 0.0778 0.2083 -0.1305 0.1305

5 52 ₂₇₀₄ -1.422764228 0.0778 0.2083 -0.1305 0.1305

6 56 ₃₁₃₆ -0.772357724 0.2206 0.25 -0.0294 0.0294

7 58 ₃₃₆₄ -0.447154472 0.3264 0.3333 -0.0069 0.0069

8 58 ₃₃₆₄ -0.447154472 0.3264 0.3333 -0.0069 0.0069

9 60 ₃₆₀₀ -0.12195122 0.4522 0.4167 0.0355 0.0355

10 60 ₃₆₀₀ -0.12195122 0.4522 0.4167 0.0355 0.0355

11 62 ₃₈₄₄ 0.203252033 0.5793 0.625 -0.0457 0.0457

12 62 ₃₈₄₄ 0.203252033 0.5793 0.625 -0.0457 0.0457

13 62 ₃₈₄₄ 0.203252033 0.5793 0.625 -0.0457 0.0457

14 62 ₃₈₄₄ 0.203252033 0.5793 0.625 -0.0457 0.0457

15 62 ₃₈₄₄ 0.203252033 0.5793 0.625 -0.0457 0.0457

16 64 ₄₀₉₆ 0.528455285 0.7019 0.75 -0.0481 0.0481

17 64 ₄₀₉₆ 0.528455285 0.7019 0.75 -0.0481 0.0481

18 64 ₄₀₉₆ 0.528455285 0.7019 0.75 -0.0481 0.0481

19 66 ₄₃₅₆ 0.853658537 0.8023 0.8333 -0.031 0.031

20 66 ₄₃₅₆ 0.853658537 0.8023 0.8333 -0.031 0.031

21 68 ₄₆₂₄ 1.178861789 0.881 0.9167 -0.0357 0.0357

22 68 ₄₆₂₄ 1.178861789 0.881 0.9167 -0.0357 0.0357

23 70 ₄₉₀₀ 1.504065041 0.9332 1 -0.0668 0.0668

24 70 ₄₉₀₀ 1.504065041 0.9332 1 -0.0668 0.0668

ΣX 1458 89444

X ̅ 60.75

Lhitung 0.1305 Ltable 0.173

𝐿_{𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑} = 0.1305 ; 𝑛 = 24; ∝= 0.05 ; 𝐿_{𝑡𝑎𝑏𝑙𝑒} = 0.173

So Lobserved < Ltable 0.1305 < 0.173

Conclusion : if Lobserved < Ltable, it means H0 was accepted or the data was normally distributed.

NO NILAI X² Zi F(Zi) S(Zi) F(Zi)- S(Zi) | F(Zi)- S(Zi)|

1 54 ₂₉₁₆ -1.529338327 0.063 0.125 -0.062 0.062

2 54 ₂₉₁₆ -1.529338327 0.063 0.125 -0.062 0.062

3 54 ₂₉₁₆ -1.529338327 0.063 0.125 -0.062 0.062

4 56 ₃₁₃₆ -1.154806492 0.1251 0.2083 -0.0832 0.0832

5 56 ₃₁₃₆ -1.154806492 0.1251 0.2083 -0.0832 0.0832

6 58 3364 -0.780274657 0.2177 0.25 -0.0323 0.0323

7 60 3600 -0.405742821 0.3409 0.4167 -0.0758 0.0758

8 60 3600 -0.405742821 0.3409 0.4167 -0.0758 0.0758

9 60 3600 -0.405742821 0.3409 0.4167 -0.0758 0.0758

10 60 ₃₆₀₀ -0.405742821 0.3409 0.4167 -0.0758 0.0758

11 62 ₃₈₄₄ -0.031210986 0.488 0.5833 -0.0953 0.0953

12 62 ₃₈₄₄ -0.031210986 0.488 0.5833 -0.0953 0.0953

13 62 3844 -0.031210986 0.488 0.5833 -0.0953 0.0953

14 62 3844 -0.031210986 0.488 0.5833 -0.0953 0.0953

15 64 4096 0.343320849 0.6331 0.6667 -0.0336 0.0336

16 64 4096 0.343320849 0.6331 0.6667 -0.0336 0.0336

17 66 4356 0.717852684 0.7642 0.875 -0.1108 0.1108

18 66 4356 0.717852684 0.7642 0.875 -0.1108 0.1108

19 66 4356 0.717852684 0.7642 0.875 -0.1108 0.1108

20 66 4356 0.717852684 0.7642 0.875 -0.1108 0.1108

21 66 4356 0.717852684 0.7642 0.875 -0.1108 0.1108

22 70 4900 1.466916355 0.9292 0.9583 -0.0291 0.0291

23 70 4900 1.466916355 0.9292 0.9583 -0.0291 0.0291

24 74 5476 2.215980025 0.9868 1 -0.0132 0.0132

ΣX 1492 93408

X ̅ 62.166667

Lhitung 0.1108 Ltable 0.173 𝐿_{𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑} = 0.1108 ; 𝑛 = 24; ∝= 0.05 ; 𝐿_{𝑡𝑎𝑏𝑙𝑒} = 0.173

So Lobserved < Ltable 0.1108 < 0.173

Conclusion : if Lobserved < Ltable, it means H0 was accepted or the data was normally distributed.

HOMOGENEITY TESTING OF POPULATION

The homogenity testing was analyzed by Barlett (Walpole). The hypothesis before testing the homogenity:

 H0 = All of the population have homogenity

 H1 = Not all of the population have homogenity The steps were:

1. Collect the reading comprehension score of all population 2. The formula for testing variance of sample:

𝑆₁² = ^{𝑛 𝑥2− 𝑥 2}

𝑛 𝑛−1

𝑆₁² = 24 93064 − 1488 ²

24 23

= 35.13

𝑆₂² = 33.36 ; 𝑆₃² = 35.24 ; 𝑆₄² = 37.85 ; 𝑆₅² = 28.49

3. Barlett Testing Table

No N Dk = N -1 Si^2 Dk.Si^2 Dk/N-k (Si^2)^(Dk/N-k) bk N*bk

1 24 23 33.36 767.28 0.191666667 1.958626644 0.9195 22.068

2 24 23 35.24 810.52 0.191666667 1.979316359 0.9195 22.068

3 24 23 37.85 870.55 0.191666667 2.006608408 0.9195 22.068

4 24 23 28.49 655.27 0.191666667 1.900273241 0.9195 22.068

5 24 23 35.13 807.99 0.191666667 1.97813068 0.9195 22.068

jml 120 115 170.07 3911.61 0.958333333 9.822955332 4.5975 110.34

4. Compute the variance of all sample by the formula belows:

𝑆_𝑝² = ^𝑘_𝑖=1 𝑛_𝑖 − 1 𝑆_𝑖²

𝑁 − 𝑘 = 3911.61

115 = 34.014

𝑏 = 𝑆₁^{2 𝑛1−1} 𝑆₂^{2 𝑛2−1}… … 𝑆_𝑘^{2 𝑛𝑘−1 𝑁−𝑘}

𝑆_𝑝² = 0.9195

6. Compute btabel by the formula below:

𝑏_𝑘 á; 𝑛₁, 𝑛₂, … , 𝑛_𝑘 = 𝑛₁𝑏_𝑘 á; 𝑛₁ + 𝑛₂𝑏_𝑘 á; 𝑛₂ + ⋯ + 𝑛_𝑘𝑏_𝑘 á; 𝑛_𝑘 𝑁

𝑏₅ 0.05; 24,24,24,24,24 = 0.859692

𝑏 > 𝑏₅ 0.05; ,24,24,24, 24,24 it indicated that H⁰ was accepted. It means the variance of all classes was homogen.

LESSON PLAN OF EXPERIMENTAL CLASS

School : MTsN Kamang

Subject : English Grade/semester : VIII/1

Text type : Recount text

Theme : Holiday

Language skill : Writing Time Allocation : 2 x 40 minutes

A. Standard of Competence : 6. Expressing the meaning in written text functional and simple short essay in descriptive form, and recount for interact surroundings environment.

B. Basic Competence : 6.1 Expressing the meaning in simple short written text functional