Data Analysis

A. Data Analysis

  Data analysis was intended to find out whether the application of gallery walk strategy significantly affects the students’ ability at writing descriptive paragraph. The analysis was computed by applying the t test formula to discover the hypothesis of this research was accepted or rejected, before it the research should do requirement test at the first by using and homogeneity test.

A.1 The Description of Data

  The data on this research was quantitative data, they were taken from experiment research design, and there were thirty nine students who were taken as sample of his research. They were divided into two classes, namely experiment and control class. The students in experiment class were taught by using gallery walk strategy and the students in experiment class were taught by using lecturing method. The population of this research was the students of the eleventh grade at MAS Al-Ittihadiyah Medan. The experiment classs was XI-IPA and the control class was XI-IPS.

  The research was conducted in two weeks, within two treatments which include four meeting altogether. Both of experiment and control class were given essays writing test from on the pre-test and post-test.

  Table 4.1 Students’ Score in Experiment Class

  Num

  Initial Name

  Pre-test

  Post-test

  1 ARN

  2. AYT

  3. ES

  4. EM

  5. ER

  6. FH 70 88

  7. FOA

  8. HSP

  9. JHF

  10. KA

  11. LM

  12. MSBM

  13. NA

  14. RR

  15. RWH

  16. SAP

  17. SWH

  18. TMS

  19. TS

  Based on the table above, the students’ ability at writing descriptive paragraph that was taught by using Gallery Walk Strategy on showed the minimum score of pre-test was 40, the maximum score of pre-test was 78 and the mean of pre-test was 60. Meanwhile the minimum score of post-test was 60, the maximum score of post-test was 93, and the mean of post-test was 83.84.

  Table 4.2

  Students’ Score in Control Class

  Num

  Initial Name

  Pre-test

  Post-test

  1 AWP

  2. AMD

  3. BGD

  4. DF 60 65

  5. HW

  6. IM

  7. MFSK

  8. MTAGS

  9. MS

  10. NAR

  11. RPH

  12. SMP

  13. SD

  14. SVN

  15. SA

  16. SL

  17. SI

  18. TA

  The table above explained the students’ ability at writing descriptive paragraph that was taught by lecturing method showed the minimum score of pre-test was 40 and the maximum score was 76, the mean of pre-test 56.8. Meanwhile the minimum score of post-test was 40, and the maximum score was

  80, the mean of the post- test was 60.4

A.2. Normality Test

  Table 4.3 Normality Test of Pre-test in Experiment Class

  Num Score

  F Fcum

  (Zi)

  F(Zi)

  S(Zi) F(Zi)- S(Zi)

  To find Z score by using the formula :

  Zi =

  2. = -0.834

  3. = -0.167

  To find out S(Z i ) we use the formula : S(Z i ) =

  1. S(Z i )= =0.157

  2. S(Z i )= =0.263

  3. S(Z i )= =0.473

  4. S(Z i )= =0.631

  5. S(Z i )= =0.842

  6. S(Z i )= =0.947

  7. S(Z i )= =1.000

  Liliefors table L t =0.195.it can be concluded that the data distribution was normal,

  because L 0 (0.131 )

  Table 4.4 Normality Test of Post-test in Experiment Class

  Num Score

  F Fcum

  (Zi)

  F(Zi)

  S(Zi) F(Zi)- S(Zi)

  To find Z score by using the formula :

  Zi =

  2. = -0.632

  3. = -0.415

  4. = -0.090

  To find out S(Z i ) we use the formula : S(Z i ) =

  1. S(Z i )= = 0.105

  2. S(Z i )= = 0.157

  3. S(Z i )= = 0.263

  5. S(Z i )= = 0.368

  6. S(Z i )= = 0.473

  7. S(Z i )= = 0.736

  8. S(Z i )= = 0.842

  9. S(Z i )= = 0.894

  10. S(Z i )= = 0.947

  11. S(Z i )= = 1.000

  From the explanation above, it can be seen that the Liliefors Observation or L 0 = 0.171 with n= 19 and at real level α=0.05 from the list critical coefficient of Liliefors table L t = 0.195 . It can be concluded that the data distribution was

  normal, because L 0 (0.171) < L t (0.195 )

  Table 4.5 Normality Test of Pre-test in Control Class

  Num Score

  F Fcum

  (Zi)

  F(Zi)

  S(Zi) F(Zi)- S(Zi)

  1 40 2 3 -1.628

  2 45 1 4 -1.143

  3 50 2 6 -0.659

  4 53 1 7 -0.368

  5 55 1 8 -0.174

  To find Zi score by using the formula :

  Zi =

  1. = -1.628

  2. = -1.143

  3. = -0.659

  4. = -0.368

  5. = -0.174

  To find out S(Z i ) we use the formula : S(Z i ) =

  1. S(Z i )= = 0.15

  2. S(Z i )= = 0.2

  3. S(Z i )= = 0.3

  4. S(Z i )= = 0.35

  5. S(Z i )= = 0.4

  6. S(Z i )= = 0.45

  7. S(Z i )= = 0.65

  8. S(Z i )= = 0.7

  9. S(Z i )= = 0.75

  11. S(Z i )= = 0.9

  12. S(Z i )= = 0.95

  13. S(Z i )= = 1.000

  From the explanation above, it can be seen that the Liliefors Observation or L 0 =0.125with n=20 and at real level α=0.05from the list critical coefficient of Liliefors table L t =0.190 .it can be concluded that the data distribution was

  normal, because L 0 (0.125 ) < L t (0.190),

  Table 4.5 Normality Test of Post-test in Control Class

  Num Score

  F Fcum

  (Zi)

  F(Zi)

  S(Zi) F(Zi)-S(Zi)

  To find Zi score by using the formula :

  Zi =

  1. = -2.760

  2. = -1.678

  3. = -1.407

  4. = -0.730

  5. = -0.054

  To find out S(Z i ) we use the formula : S(Z i ) =

  1. S(Z i )= = 0.050

  2. S(Z i )= = 0.100

  3. S(Z i )= = 0.200

  4. S(Z i )= = 0.350

  5. S(Z i )= = 0.550

  7. S(Z i )= = 0.650

  8. S(Z i )= = 0.850

  9. S(Z i )= = 0.900

  10. S(Z i )= = 1.000

  From the explanation above, it can be seen that the Liliefors Observation or L 0 = 0.121 with n=20 and at real level α=0.05 from the list critical coefficient of Liliefors table L t =0.190 .It can be concluded that the data distribution was

  normal, because L 0 (0.121 ) < L t (0.19 )

A.3. Homogeneity Test

  Homogeneity test used F-test to know what the samples come from population that homogenous or not.

  Table 4.7 Homogeneity Test of Pre-test

  Num

  Data

  Variants F Observation

  F table Conclusion

  1. Pre-Test of

  experiment class

  1.35 1.78 Homogenous

  2 Pre-Test of

  Control Class

  F observation =

  = the smallest variant

  Bassed on the variants of both samples of pre-test found that :

  F observation =

  F observation =

  From the data above, it is found the coefficient of variant from students’ ability in writing descriptive paragraph in experimental and control class at real

  level α=0.05. So, by using the list critical value of F distribution is found F 0.05 Df=

  37 So that F observation (1.35)< F table (1.78), so it can be concluded that be variant

  from the data above is homogeneous.

  Table 4.8 Homogeneity Test of Post-test

  Num

  Data

  Variants F Observation

  F table Conclusion

  1. Pre-Test of

  experiment class

  2 Pre-Test of

  Control Class

  From the data above, it is found the coefficient of variant from students’ ability in writing descriptive paragraph in experimental and control class at real

  level α=0.05. So, by using the list critical value of F distribution is found F 0.05 Df=

  37 So that F observation (0.888)< F table (1.787), so it can be concluded that be variant

  from the data above is homogeneous.