A. Data Presentation - The effectiveness of picture series toward writing score of the eighth graders of MTs Muslimat NU Palangka Raya - Digital Library IAIN Palangka Raya

CHAPTER IV RESEARCH FINDING AND DISCUSSION In this chapter, the writer presents the data which had been collected from

  the research in the field of study. The data are the result of pre-test experiment and control class, the result of post-test experiment and control class, result of data analysis, interpretation, and discussion.

A. Data Presentation

  The Pre-test and Post-test at the experiment class had been conducted on

  th st

  July, 30 2015(Thursday, at time 08.00-09.20) for Pre-test and August, 21 2015 (Friday, at time 07.20

• –08.40) for Post- test in class VIII C of MTs Muslimat Nu Palangka Raya with the number of student was 36 students. Then the control class

  th

  had been conducted on August, 4 2015(Tuesday, at time 06.30

• –08.00) for Pre-

  st

  test and August, 1 2015 (Tuesday, at time 06.30

• –08.00) for Post- test in the class

  VIII A of MTs Muslimat Nu Palangka Raya with the number of student was 36 students.

  In this chapter, the writer presents the obtained data of the students’ writing score, experiment class who was taught with picture series and control class who was taught without picture series.

1. Distribution of the Pre-Test Scores of the Experiment Class

  The pre-test scores of the experiment class are presented in the following table.

Table 4.1 The Description of the Pre-Test Scores of the Experiment Class NO STUDENT CODE SCORE

  60

  75

  26 E26

  79

  25 E25

  70

  24 E24

  23 E23

  54

  70

  22 E22

  54

  21 E21

  56

  20 E20

  27 E27

  28 E28

  19 E19

  33 E33

  36 E36

  65

  35 E35

  60

  34 E34

  64

  58

  61

  32 E32

  69

  31 E31

  58

  30 E30

  63

  29 E29

  58

  54

  1 E1

  5 E5

  8 E8

  66

  7 E7

  61

  6 E6

  70

  60

  9 E9

  4 E4

  70

  3 E3

  65

  2 E2

  61

  74

  78

  18 E18

  70

  65

  17 E17

  70

  16 E16

  66

  15 E15

  14 E14

  10 E10

  65

  13 E13

  70

  12 E12

  65

  11 E11

  78

  66

  58.5

  1 79 - 83

  7 61 19.44444

  63.5

  25

  66

  9

  68.5

  8 71 22.22222

  73.5

  4 76 11.11111

  78.5

  1 81 2.77778

  Relative Frequency (%) The Limitation of Each Group

Table 4.1 highlights that the student’s highest score is 79 and the student’s

  Class (k) Interval (I) Frequency (F)

Midpoint

(X)

Table 4.2 The Frequency Distribution of the Pre-Test Score of the Experiment Class

     So, the range of score is 25, the class interval is 6, and interval of temporary is 5. Then, it is presented using frequency distribution in the following table:

  or K R

  25

  6

  4

  5 . 4 16667

  Interval of Temporary (I) =

  The Class Interval (K) = 1+ 3.3 log n = 1+ 3.3 log 36 = 1+ 3.3 (1.56) = 1+ 5.148 = 6.148 = 6

  = 79 - 54 = 25

  lowest score is 54. To determines the range of score, the class interval, and the interval temporary the writer calculates using formula as follows: The Highest Score (H) = 79 The Lowest Score (L) = 54 The Range of Score (R) = H - L

• – 83.5 2 74 - 78
• – 78.5 3 69 - 73
• – 73.5 4 64 - 68
• – 68.5 5 59 - 63
• – 63.5
• – 58.5

• – 58.5 58.5 – 63.5 63.5 – 68.5 68.5 – 73.5 73.5 – 78.5 78.5 – 83.5
^{7 7 9 8 4 1} F R E Q U E N C Y The Limitation of Each Group The Frequency Distribution of the Pre Test Scores of the Experiment Class

  1

  23 ^{2 4 6 8 10}

  22

  66 594

  9

  31 64 - 68

  13

  71 568

  8

  35 69 - 73

  5

  76 304

  4

  36 74 - 78

  81

  6 54 - 58

  81

  1

  79 - 83

  Interval (I) Frequency (F) Midpoint (X) FX fk (a) fk (b)

Table 4.3 The Calculation of Mean, Median, and Modus of the Pre-Test Scores of the Experiment Class

  The next step, the writer tabulates the scores into the table for the calculation of mean, median, and modus as follows:

  The students’ score which pass the standard value (67) are 23 students and there are 13 students whose score are below the standard value.

  below standard value. Whereas the lowest of number in the score 78.5-83.5 which is only 1 student.

Figure 4.1 shows that most of the students got score 63.5-68.5. It is proved that there are 9 students which is the greatest of number but their score are stillFigure 4.1. The Frequency Distribution of the Pre-Test Score of the Experiment Class

  TOTAL

36 100

  53.5

  7 56 19.44444

  53.5

  29

  14 59 - 63

  7

  61 427 54 - 58

  7

  56

  36

  7 392

  TOTAL

  36 2366 a.

  Mean

  X

  =

  2366

  =

  36

  = 65.72222 b. Median _{1 2} −

  Me

  = l + i ₁

  36 ₂ −14

  = 63.5 + 5

  9

  = 63.5 + 5 0.44444

  = 63.5 + (2.2222) = 65.7222 c. Modus

  Mo

  = l +

• 8

  = 63.5 +

  5

  8+7

  = 63.5 + 0.53333 5

  = 63.5 + 2.66665 = 66.16665

  From the calculation, the mean score is 65.72222, median score is 65.7222,

• 7
• 14

  5

  68

  5   

     

   

  SD _{2 1} ) 05556 . ( 88889 .

  1

  5   SD . 88889 00309 .

  1

  5 ₁  

  SD 8858 .

  1

  5 ₁  SD 37324 .

  1

  1

  2

   

  SD 8662 .

  6

  1

  

  SD b.

   Standard Error

  1 ₁ ^{1 1}  

  N SD SEM

  1

  36 8662 .

  6 ₁  

  SEM

  35 8662 .

  36

  36

  The last step, the writer tabulates the scores into the table for the calculation of standard deviation and the standard error as follows:

  16 69 - 73

Table 4.4 The Calculation of the Standard Deviation and the Standard Error of the Pre-Test Scores of the Experiment Class

  Interval (I) Frequency (F) Midpoint (X) x’ Fx’ F( x’

  2 )

  79 - 83

  1

  81

  3

  3

  9 74 - 78

  4

  76

  2

  8

  8

  1

  71

  1

  8

  8 64 - 68

  9

  66 59 - 63 7 61 -1

  7 54 - 58 7 56 -2

  28 TOTAL

  36 -2

  68 a. Standart Deviation  

  N Fx N Fx

  SD i ^{2 2} ₁ ' '

    

  2

  6 ₁  SEM

  6 . 8662

  SEM ₁  5 . 9161 SEM

   ₁ 1 . 16059

  The result of calculation reports that the standard deviation of pre test score of experiment class is 6.8662 and the standard error of pre test score of experiment class is 1.16059.

  The next step, the writer calculates the scores of the pre test in experiment class using SPSS as follows:

Table 4.5 The Table of Calculation of the Pre-Test Scores of the Experiment Class Using SPSS 16.0 Program

  Statistics

  VAR00001 N Valid

  36 Missing

  62 Mean 65.2222 Std. Error of Mean 1.13607 Median 65.0000 Mode

  70.00 Std. Deviation 6.81641 Variance 46.463 Range

  25.00 Minimum

  54.00 Maximum

  79.00 Sum 2348.00

2. Distribution of the Pre-Test of the Control Class The pre test scores of the control class are presented in the following table.

  24 C24

  27 C27

  60

  26 C26

  74

  25 C25

  70

  65

  28 C28

  23 C23

  56

  22 C22

  56

  21 C21

  56

  65

  65

  60

  65

  60

  36 C36

  66

  35 C35

  70

  34 C34

  33 C33

  29 C29

  61

  32 C32

  54

  31 C31

  63

  30 C30

  64

  20 C20

Table 4.6 The Description of the Pre-Test Scores of the Control Class NO STUDENT CODE SCORE

  1 C1

  65

  74

  8 C8

  70

  7 C7

  65

  6 C6

  5 C5

  60

  73

  4 C4

  74

  3 C3

  63

  2 C2

  70

  9 C9

  10 C10

  66

  15 C15

  18 C18

  70

  17 C17

  65

  16 C16

  61

  73

  65

  14 C14

  64

  13 C13

  58

  12 C12

  58

  11 C11

  19 C19

  57.5

  75.5 8,333333

  59.5 22,22222

  8

  61.5

  63.5 33,33333

  12

  65.5

  67.5 5,555556

  2

  69.5

  71.5 19,44444

  7

  73.5

  3

Table 4.6 highlights that the student’s highest score is 74 and the student’s

  1 74 - 77

  Class (k) Interval (I) Frequency (F) Midpoint (X) Relative Frequency (%) The Limitation of Each Group

Table 4.7 The Frequency Distribution of the Pre-Test Score of the Control Class

     So, the range of score is 20, the class interval is 6, and interval of temporary is 4. Then, it is presented using frequency distribution in the following table:

  or K R

  20

  6

  3

  4 . 3 33333

  Interval of Temporary (I) =

  The Class Interval (K) = 1+ 3.3 log n = 1+ 3.3 log 36 = 1+ 3.3 (1.56) = 1+ 5.148 = 6.148 = 6

  = 74 - 54 = 20

  lowest score is 54. To determines the range of score, the class interval, and the interval temporary the writer calculates using formula as follows: The Highest Score (H) = 74 The Lowest Score (L) = 54 The Range of Score (R) = H - L

• – 77.5 2 70 - 73
• – 73.5 3 66 - 69
• – 69.5 4 62 - 65
• – 65.5 5 58 - 61
• – 61.5

  6 54 - 57

  4

  55.5 11,11111

  53.5

• – 57.5

  TOTAL

36 100

Figure 4.2 The Frequency Distribution of the Pre-Test Score of the Control Class
• – 57.5 57.5 – 61.5 61.5 – 65.5 65.5 – 69.5 69.5 – 73.5 73.5 – 77.5
^{4 8 12 2 7 3} F R E Q U E N C Y The Limitation of Each Group The Frequency Distribution of the Pre-Test Scores of the Control Class
Figure 4.2 shows that most of the students got score 61.5-65.5. It is proved that there are 12 students which is the greatest of number but their score are still

  below standard value (67). Whereas the lowest of number in the score 65.5-69.5 which is only 2 students and both score are still below standard value (67). The students’ score which pass the standard value (67) are 10 students and there are 26 students whose score are below the standard value.

  The next step, the writer tabulates the score into the table for the calculation of mean, median, and modus as follows: ^{2 4 6 8 10 12}

  53.5

Table 4.8 The Calculation of Mean, Median, and Modus of the Pre-Test Scores of the Control Class

  Interval Frequency Midpoint FX fk (a) fk (b) (I) (F) (X)

  74 - 77

  3

  75.5

  3

  36 226,5

  10

  33 70 - 73

  7

  71.5 500,5 66 - 69

  2

  67.5

  12

  26 135 62 - 65

  12

  63.5

  24

  24 762 58 - 61

  8

  59.5

  32

  12 476 54 - 57

  4

  55.5

  36

  4 222

  TOTAL

  36 2322 a.

  Mean

  X

  =

  2322

  =

  36

  = 64.5 b. Median ₁

  −

  = ₁ ²

• Me

  36 ₂ −12

  = 61.5 + 4

  12

  = 61.5 + 4

  0.5 = 61.5 + 2 = 63.5 c. Modus

  Mo

  = l +

• 2
• 8
• 8

  2

  2

  2

  36

  9

  36

  81

  4   

    

   

  SD _{2 2} ) ( 25 . 25 .

  2  4  SD 0625 .

  25 .

  4 ₂   SD 1875 .

  SD i ^{2 2} ₂ ' '

  2

  4

  2

  

  SD 47901 .

  1

  4

   91604 .

  5

  2

  

  SD

    

  N Fx N Fx

  = 61.5 + 0.2 4

  71.5

  = 61.5 + 0.8 = 62.3

  From the calculation, the mean score is 64.5, median score is 63.5, and modus score is 62.3 of the pre-test of the control class.

  The last step, the writer tabulates the scores into the table for the calculation of standard deviation and the standard error as follows:

Table 4.9 The Calculation of the Standard Deviation and the Standard Error of the Pre-Test Scores of the Control Class

  Interval (I) Frequency (F) Midpoint (X) x’ Fx’ Fx’2

  74 - 77

  3

  75.5

  3

  9

  27 70 - 73

  7

  2

  81 a. Standard Deviation  

  14

  28 66 - 69

  2

  67.5

  1

  2

  2 62 - 65

  12

  63.5 58 - 61 8 59.5 -1

  8 54 - 57 4 55.5 -2

  16 TOTAL

  36

  9

2 SD x

b. Standard Error

  SD ₂ SEM ₂  N ₁ 

  1 5 . 91604 SEM ₂ 

  36

  1  5 . 91604

  SEM ₂ 

  35

  5 . 91604

  SEM ₂  5 . 91608 SEM ₂ 

  0.99999 The result of calculation reports that the standard deviation of pre test score of control class is 5.91604 and the standard error of pre test score of control class is 0.99999.

  The next step, the writer calculates the scores of pre-test in control class using SPSS as follows:

Table 4.10 The Table of Calculation of the Pre-Test Scores of the Control Class Using SPSS 16.0 Program

  

Statistics

  VAR00002 N Valid

  36 Missing

  62 Mean 64.5556

  Std. Error of Mean .93218 Median 65.0000 Mode

  65.00 Std. Deviation 5.59308 Variance 31.283 Range

  20.00 Minimum

  54.00 Maximum

  74.00 Sum 2324.00

3. Distribution of the Post-Test Scores of the Experiment Class

  80

  79

  25 E25

  75

  24 E24

  71

  23 E23

  22 E22

  79

  71

  21 E21

  71

  20 E20

  75

  19 E19

  26 E26

  27 E27

  18 E18

  75

  79

  34 E34

  80

  33 E33

  69

  32 E32

  31 E31

  70

  70

  30 E30

  80

  29 E29

  71

  28 E28

  70

  The post test scores of the experiment class are presented in the following table.

Table 4.11 The Description of the Post-Test Scores of the Experiment Class NO STUDENT CODE SCORE

  70

  73

  7 E7

  69

  6 E6

  71

  5 E5

  4 E4

  80

  80

  3 E3

  75

  2 E2

  73

  1 E1

  8 E8

  9 E9

  17 E17

  78

  75

  16 E16

  71

  15 E15

  74

  14 E14

  13 E13

  79

  75

  12 E12

  75

  11 E11

  80

  10 E10

  75

  76.5

  So, the range of score is 11, the class interval is 6, and interval of temporary is 2. Then, it is presented using frequency distribution in the following table:

  77.5

  1

  78.5

  25

  79.5

  9

  1 79 - 80

  Class (k) Interval (I) Frequency (F) Midpoint (X) Relative Frequency (%) The Limitation of Each Group

Table 4.12 The Frequency Distribution of the Post-Test Score of the Experiment Class

  K R

  35 E35

  11   

  6

  1

  Interval of Temporary (I) = . 2 83333

  The Class Interval (K) = 1+ 3.3 log n = 1+ 3.3 log 36 = 1+ 3.3 (1.56) = 1+ 5.148 = 6.148 = 6

  = 80 - 69 = 11

  73 Table 4.11 highlights that the student’s highest score is 88 and the student’s lowest score is 70. To determines the range of score, the class interval, and the interval temporary the writer calculates using formula as follows: The Highest Score (H) = 80 The Lowest Score (L) = 69 The Range of Score (R) = H - L

  36 E36

  75

• – 80.5 2 77 - 78

  3 75 - 76

  11

  75.5

  74.5 30,55556 – 76.5 4 73 - 74

  2

  73.5

  72.5

• – 74.5 5,555556 5 71 - 72

  6

  71.5

  70.5 16,66667 – 72.5 6 69 - 70

  7

  69.5

  68.5

• – 70.5 19,44444

  TOTAL

36 100

Figure 4.3. The Frequency Distribution of the Post-Test Scores of the Experiment Class

  The Frequency Distribution of the Post Test Scores of ₁₂ the Experiment Class ₁₁ F _{10 9} R E ₈ ⁷ ₆ Q ₆ U E ₄ N _{2 1} C ² Y 68.5 – 70.5 70.5 – 72.5 72.5 – 74.5 74.5 – 76.5 76.5 – 78.5 78.5 – 80.5

  The Limitation of Each Group

Figure 4.3 shows that most of the students got score 74.5-76.5. It is proved that there are 11 students which is the greatest of number. Whereas the lowest of

  number in the score 76.5-78.5 which is only 1 student. From the figure clarify that all of the students’ score are pass the standard value (67).

  The next step, the writer tabulates the score into the table for the calculation of mean, median, and modus as follows:

Table 4.13 The Calculation of Mean, Median, and Modus of the Post-Test Scores of the Experiment Class

  Interval Frequency Midpoint FX fk (a) fk (b) (I) (F) (X)

  79 - 80

  9

  79.5

  9

  36 715.5

  10

  27 77 - 78

  1

  77.5

  77.5 75 - 76

  11

  75.5

  21

  26 830.5 73 - 74

  2

  73.5

  23

  15 147 71 - 72

  6

  71.5

  29

  13 429 69 - 70

  7

  69.5

  36

  7 486.5

  TOTAL

  36 2686 a.

  Mean

  X

  =

  2686

  =

  36

  = 74.61111 b. Median _{1 2} −

  = ₁

• Me

  36 ₂ −13

  = 74.5 + 2

  11

  = 74.5 + 2 0.45455

  = 74.5 + 0.9091 = 75.4091 c. Modus

  =

• Mo
• 1
• 6
• 14

  SD _{2 1} ) 55556 . ( 61111 .

  SD i ^{2 2} ₁ ' '

    

  2

  1

  36

  20

  36 130

  2   

    

   

  3  2  SD . 61111 30865 .

  36 20 130 a. Standard Deviation  

  3

  2

  1

   

  SD 30246 .

  3

  2

  1

  

  SD 81727 .

  1

  N Fx N Fx

  28 TOTAL

  = 74.5 + 0.33333 x 2 = 74.5 + 0.66666 = 75.16666

  81 77 - 78

  From the calculation, the mean score is 74.61111, median score is 75.4091, and modus score is 75.16666 of the post-test of the experiment class.

  The last step, the writer tabulates the scores into the table for the calculation of standard deviation and the standard error as follows:

Table 4.14 The Calculation of the Standard Deviation and the Standard Error of the Post-Test Scores of Experiment Class

  Interval (I) Frequency (F) Midpoint (X) x’ Fx’ F( x’

  2 )

  79 - 80

  9

  79.5

  3

  27

  1

  6 69 - 70 7 69.5 -2

  77.5

  2

  2

  4 75 - 76

  11

  75.5

  1

  11

  11 73 - 74

  2

  73.5 71 - 72 6 71.5 -1

  2 ₁   SD

  SD ₁  3 . 63454 b.

   Standard Error SD ₁ SEM ₁ 

  N ₁ 

  1 3 . 63454 SEM ₁  36 

  1 3 . 63454 SEM ₁ 

  35

  3 . 63454

  SEM ₁  5 . 91608 SEM

   ₁ . 61435

  The result of calculation reports that the standard deviation of the post-test score of the experiment class is 3.63454 and the standard error of post-test score of experiment class is 0.61435.

  The next step, the writer calculates the scores of the post-test in experiment class using SPSS as follows:

Table 4.15 The Table of Calculation of the Post-Test Scores of the Experiment Class Using SPSS 16.0 Program

  

Statistics

  VAR00001 N Valid

  36 Missing

  36 Mean 74.5833 Std. Error of Mean .65146 Median 75.0000 Mode

  75.00 Std. Deviation 3.90878 Variance 15.279 Range

  11.00 Maximum

  80.00 Sum 2685.00 4.

  22 C22

  16 C16

  74

  17 C17

  70

  18 C18

  70

  19 C19

  70

  20 C20

  61

  21 C21

  70

  61

  15 C15

  23 C23

  74

  24 C24

  74

  25 C25

  66

  26 C26

  71

  27 C27

  75

  28 C28

  74

  29 C29

  63

  70

   Distribution of the Post-Test of the Control Class The post test scores of the control class are presented in the following table.

  65

Table 4.16 The Description of the Post-Test Scores of the Control Class NO STUDENT CODE SCORE

  1 C1

  74

  2 C2

  74

  3 C3

  78

  4 C4

  78

  5 C5

  78

  6 C6

  7 C7

  14 C14

  74

  8 C8

  78

  9 C9

  66

  10 C10

  71

  11 C11

  75

  12 C12

  70

  13 C13

  66

  74

  30 C30

  75

  31 C31

  78

  32 C32

  74

  33 C33

  70

  34 C34

  63

  35 C35

  74

  36 C36

  70 Table 4.16 highlights that the student’s highest score is 78 and the student’s lowest score is 61. To determines the range of score, the class interval, and the interval temporary the writer calculates using formula as follows: The Highest Score (H) = 78 The Lowest Score (L) = 61 The Range of Score (R) = H - L

  = 78

• – 61 = 17 The Class Interval (K) = 1+ 3.3 log n

  = 1+ 3.3 log 36 = 1+ 3.3 (1.56) = 1+ 5.148 = 6.148 = 6

  R

  17

  or

  Interval of Temporary (I) = 2 . 83333

  2

  3   

  K

  6 So, the range of score is 17, the class interval is 6, and interval of temporary is 3. Then, it is presented using frequency distribution in the following table:

Table 4.17 The Frequency Distribution of the Post-Test Score of the Control Class

  

Class Interval Frequency Midpoint Relative The Limitation

(k) (I) (F) (X) Frequency (%) of Each Group

  1 76 - 78

  5

  77

  75.5

• – 78.5 13,88889 2 73 - 75

  13

  74

  72.5

• – 75.5 36,11111 3 70 - 72

  10

  71

  69.5

• – 72.5 27,77778 4 67 - 69

  68

  66.5

• – 69.5 5 64 - 66

  4

  65

  63.5

• – 66.5 11,11111 6 61 - 63

  4

  62

  60.5 11,11111 – 63.5

  TOTAL

36 100

Figure 4.4 The Frequency Distribution of the Post-Test Score of the Control Class

  The Frequency Distribution of the Post-Test Scores ₁₄

of the Control Class

₁₃ F _{12 10} R ₁₀ E Q ₈ U _{6 4 4} ⁵ E ₄ N C ₂ Y

  60.5

• – 63.5 63.5 – 66.5 66.5 – 69.5 69.5 – 72.5 72.5 – 75.5 75.5 – 78.5

  The Limitation of Each Group

Figure 4.4 shows that most of the students got score 72.5-75.5. It is proved that there are 13 students which is the greatest of number. Whereas the lowest of

  number in the score 66.5-69.5 which there is no students got that score. The students’ score which pass the standard value (67) are 28 students and there are 8

  = 72.5 + 3 = 72.5 + 0

  Mean

  32

  8 61 - 63

  4

  62 248

  36

  4 TOTAL

  36 2565 a.

  X

  4

  = =

  2565

  36

  = 71.25 b. Median

  Me =

  = 72.5 + 3 ^{1 2}

  36 −18

  13

  65 260

  8 64 - 66

  students whose score are below the standard value. It means that the most of students got score which pass the standard value (67).

  36 73 - 75

  The next step, the writer tabulates the score into the table for the calculation of mean, median, and modus as follows:

Table 4.18 The Calculation of Mean, Median, and Modus of the Post-Test Scores of the Control Class

  Interval (I) Frequency (F) Midpoint (X) FX fk (a) fk (b)

  76 - 78

  5

  77 385

  5

  13

  28

  74 962

  18

  31 70 - 72

  10

  71 710

  28

  18 67 - 69

  68

• ¹
² −
• = 72.5 +

  SD i ^{2 2} ₂ ' '

  68

  4 65 -2

  16 61 - 63 4 62 -3

  36 TOTAL

  36

  3

  85 a.

   Standard Deviation  

  N Fx N Fx

    

  10

  2

  2

  36

  3

  36

  85

  3   

    

   

  71 67 - 69

  13 70 - 72

  SD

Table 4.19 The Calculation of the Standard Deviation and the Standard Error of the Post-Test Scores of the Control Class

  c.

  Modus

  Mo

  = l +

  5 5+10

  3 = 72.5 +

  0.33333 3 = 72.5 + 0.99999 = 73.49999

  From the calculation, the mean score is 71.25, median score is 72.5, and modus score is 73.49999 of the post-test of the control class.

  The last step, the writer tabulates the scores into the table for the calculation of standard deviation and the standard error as follows:

  Interval (I) Frequency (F) Midpoint (X) x’ Fx’ Fx’2

  13

  76 - 78

  5

  77

  2

  10

  20 73 - 75

  13

  74

  1

• 1 64 - 66
• 8
• 12

2 SD

  3 2 . 36111  (  . 08333 ) SD

  ₂ 

  

  3 2 . 36111  . 00694

  2 SD

  

  3 2 . 35417

  2 SD x

  

  3 1 . 53433

  2 SD 4 . 60299 ₂  b.

   Standard Error SD ₂ SEM ₂ 

  N ₁ 

  1 4 . 60299 SEM ₂  36 

  1 4 . 60299 SEM ₂ 

  35

  4 . 60299

  SEM ₂  5 . 91608 SEM

   . 77804

2 The result of calculation reports that the standard deviation of post-test

  score of control class is 4.60299 and the standard error of post-test score of control class is 0.77804.

  The next step, the writer calculates the scores of the post-test in control class using SPSS as follows:

Table 4.20 The Table of Calculation of the Post-Test Scores of the Control

  Class Using SPSS 16.0 Program

Statistics

  VAR00002 N Valid

  36 Missing Std. Error of Mean .80966 Median 72.5000 Mode

  74.00 Std. Deviation 4.85798 Variance 23.600 Range

  17.00 Minimum

  61.00 Maximum

  78.00 Sum 2568.00 B.

   Result of Data Analysis 1. Testing of Normality and Homogeneity

  The writer calculates the result of pre-test and post-test score of experiment and control class by using SPSS 16.0 program. It is used to know the normality of the data that is going to be analyzed whether both groups have normal distribution or not. Also, homogeneity is used to know whether experiment class and control class, that are decided, come from population that has relatively same variant or not.

  a.

  

Testing of Normality and Homogeneity of Pre-Test of Experiment and

Control Class
Table 4.21 Testing of Normality One-Sample Kolmogorov-Smirnov Test One-Sample Kolmogorov-Smirnov Test

  VAR00001 N

  72 Normal Parameters

  ab

  Mean 64.8889 Std. Deviation 6.19985

  Most Extreme Differences

  Absolute .109 Positive .109 Negative -.101 Kolmogorov-Smirnov Z .929 Asymp. Sig. (2-tailed) .354 a.

  Test distribution is Normal.

  b.

  Calculated from data Based on the calculation uses SPSS program, the asymptotic significant normality of experiment class and control class is 0.354. Then the normality both of class is consulted with table of Kolmogorov- Smirnov with the level of significant 5% (α=0.05). Since asymptotic significant of experiment and asymptotic significant of control= 0.354

  ≥ α = 0.05. It can be concluded that the data distribution is normal.

Table 4.22 Testing Homogeneity Levene's Test of Equality of Error

  a Variances Test of Homogeneity of Variances

  Dependent variable: Achievement Levene Statistic df1 df2 Sig.

  1.374

  1 70 .245 Based on the result of homogeneity test, the data are homogeneous if the significant

  value

  is higher than significant level α= 0.05. Table 4.22 proves that the significant value (0.245) i s higher than significant level α= 0.05, it can be concluded that the data are homogeneous. It means that both of classes have same variants.

  

b.Testing of Normality and Homogeneity for Post-Test of Experiment and

Control Class

  Test distribution is Normal.

  1.293

  Dependent variable: Achievement Levene Statistic df1 df2 Sig.

  a

Test of Homogeneity of Variances

Table 4.24 Testing of Homogeneity Levene's Test of Equality of Error Variances

  (α=0.05). Since asymptotic significant of experiment and asymptotic significant of control= 0.176 ≥ α = 0.05. It can be concluded that the data distribution is normal.

  Calculated from data Based on the calculation uses SPSS program, the asymptotic significant normality of experiment class and control class are 0.176. Then the normality both of class are consulted with table of Kolmogorov- Smirnov with the level of significant 5%

  b.

  Kolmogorov-Smirnov Z 1.102 Asymp. Sig. (2-tailed) .176 a.

Table 4.23 Testing of Normality One-Sample Kolmogorov-Smirnov Test

  Absolute .130 Positive .109 Negative -.130

  Most Extreme Differences

  Mean 72.9583 Std. Deviation 4.67368

  ab

  72 Normal Parameters

  VAR00002 N

  

One-Sample Kolmogorov-Smirnov Test

  1 70 .259 Based on the result of homogeneity test, the data are homogeneous if the significant value is higher than significant level α= 0.05. Table 4.24 proves that the significant value (0.259) i s higher than significant level α= 0.05, it can be concluded that the data are homogeneous. It means that both of classes have same variants.

2. Testing Hypothesis a. Testing Hypothesis Using T-test

  The writer uses t-test statistical calculation with significant level of the the refusal null hypothesis α= 0.05. The writer uses manual calculation and SPSS

  16.0. Program test the hypothesis using t-test statistical calculation. The criteria of Ha is accepted when t > t , and Ho is refused when t < t . The

  obseved table observed table result of testing hypothesis explained in the following table.

Table 4.25 The Standard Deviation and the Standard Error of X 1 and X

  2 Variable The Standard Deviation The Standard Error

  X 1 3.63454 0.61435

  X 2 4.60299 0.77804

  Where: X = Experimental Class

  1 X 2 = Control Class

  The table shows the result of the standard deviation calculation of X

  1 is

  3.63454 and the result of the standard error mean calculation is 0.61435. The result of the standard deviation calculation of X

  2 is 4.60299 and the result of the The next step, the writer calculates the standard error of the differences mean between X

  1 dan X 2 as follows: ^{2 2} SE M1 -SE M2 =

• 1
₂

  2

₂

• 0.77804 SE M1 -SE M2 =

  0.61435 SE M1 -SE M2 =

  0.37743 + 0.60534 SE M1 -SE M2 =

  0.99135 SE M1 -SE M2 = 0.99135

  The calculation above shows the standard error of the differences mean between X

  1 dan X 2 is 0.99135. Then, it is inserted to the t o formula to get the value

  of t as follows:

  observed M M t

   _o

  1

  2

  =

SE SE

  _{M M} 

  1

  2 t _o 74 . 61111  71 .

  25 =

  . 99135

  t _o 3 . 36111

  = . 99135

  t _o

  = 3.390 With the criteria: If t-test (t observed table , it means Ha is accepted and H0 is rejected.

  ) ≥ t If t-test (t observed ) < t table , it means Ha is rejected and H0 is accepted.

  Then, the writer interprets the result of t-test. Previously, the writer accounts the degree of freedom (df) with the formula:

  N (  N  _{1 2} 2 )

  df = = (

  36  36  2 )

  t _{table at df 70/60 at 5% significant level = 2.00}

  The calculation above shows the result of t-test calculation as in the table follows:

Table 4.26 The Result of T-test t table Variable t observed

  Df/db 5% 1%

  X - X 3.390

  2.00 2.66 70/60

  1

2 Where:

  X

  1 = Experimental Class

  X

  2 = Control Class

  t observed = The calculated Value t table = The distribution of t value df/db = Degree of Freedom

  The result of hypothesis test calculation (Table 4.26) proves that the value of t observed is higher than the value of t table at the level of significant in 5% or 1% that is 2.00 <3.390> 2.66. It shows that Ha is accepted and H0 is rejected. From the result of hypothesis test can be described, students who taught by using picture series gave significant effect on the students’ writing score of the eighth graders of MTs Muslimat Nu Palangka Raya. On the other hand, students who taught by non picture series do not have better writing achievement than those taught by picture series. Simply, it can be interpreted that null hypothesis is rejected.