The Calculation of Taking Groups as Samples for The Experiments

(1)

APPENDIX ( 1 )

(2)

Number of

Students

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

Total

Mean

Range

=

66,83 - 46,25

=

20,58

A

57,8

61,8

54

66,2

70,2

65,4

56,8

55,8

59,8

58

56,8

58

62,2

59,2

59

68,8

51,6

57,4

61,6

54,6

1195

59,75

Kelas

=

1

+

3,323 log 5

=

3,3227

=

3

Range

Lebar Kelas

=

Kelas

20,58

=

3

=

6,86

B

54,4

52,4

59

52,2

61,6

61,8

54,4

52

52,8

53,2

52,8

58,6

--50

56

58,6

45,4

49,8

-

--925

46,25

.GROUPS

c

D

E

55,2

73,4

70,6

63,8

69,6

59,6

61,6

77,2

73

64

66,4

61,4

74,8

63,6

48,6

68,6

57,6

53,6

--

81,6

59,2

61,4

56,6

60,8

59,6

68

56,2

70,8

67,8

57,6

52,4

51,2

57,2

67,6

69,2

53,6

54,8

76,8

56

68

62

47,4

58,6

77,2

53,2

46

62,8

53

---

65

50,4

61,2

63,4

57,2

5_6,4

76,8

52,6

69,2

67,4

(3)

46,25

53,12

69,99

53,11

59,98

66,85

XA

=

59,75

XB

=

46,25

Xc

=

56,13

XD

=

57,80

XE

=

66,83

Bad

Good

Very good

So, the average classes are A, C, and D.

Analysis to prove that there is No Significant Difference among Groups.

HO :

J..lA

=

J..lc

=

J..lo : there is no significant difference among groups.

HA : J..lo

=

J..lc

J..lo : there is significant difference among groups

HO

is accepted if Fe< Ft

=

3,16

HO

is rejected if Fe> Ft

=

3,16

Fe=

0,39

So,

HO

is accepted because Fe< Ft

(4)

THE TRY-OUT RESULTS

"NATIVE OR NON-NATIVE TEACHERS"

Number of Number

Students I 2 3 4 5 6 7

I 4 4 4 4 6 4 6

2

2 2 4 3 6 6 3

3 2 2 I 2 6 4 3

4 4 2 I

2

6 3 4

5 2 1 2 3 3 4 3

6 4 3 4 4 3 3 4

7 4 3 3 4 6 4 3

8 4 3 3 4 6 4 4

9 4 3 4 2 6 3 2

10 4 4 3 4 3 3 3

11 4 4 4 3 6 3 4

12 4 1 3 4 6 3 3

13 4 4 4 4 6 4 5

14 4 3 4 4 6 4 6

15 4 4 4 3 6 4 4

16 4 3 3 4 6 4 6

17 4 3 3 4 6 4 5

18 4 3 4 4 6 4 4

19 4 4 3 4 6 4 4

20 4 3 4 4 6 6 6

Vi

0,53 0,89 0,93 0,57 1,20 0,72 1,46

Evi 42,43

Vt = 156,829

r = 0,786 (high )

Notes:

X

= total amount

Vi

= sum of variance

Evi

=

sum of total variance Vt = total sum of variance r

= reliability

of Items

8 9 10 II

6 10 5 8 6 5 10 5

I 5 5 5

6 10 5 5

2 5 5 9

3

5

10

5

6 9 10 10

6 10 10 10

6 10 5 5 3 5 10

5

6 10 10 9 6 5 5

5

6 10 5 10 3 10

.

10 10

6 10 10 10

3 5 9 10

3 lO 10 10

6 10 9 10

4 10

9

10

2,95 5,85 5,41 5,52

X

12 13 14

8 10 10 89 5 10 5 72

9 5 8 58

5 5 5 63 5 5 9 59

10 10

5

73

10 5 10 87

10 5 10 89

5 10 5 70

9

5

10 71

5 9 10 87

5

60

10 10 10 92

9 9 10 92

5 9 9 88

10

5

5 77

10

5

82

9 10 10

93

10 5 9 88

10 10 9 95

(5)

The Calculation of Discrimination Power and Difficulty Index

Number Of

Students l

2

3 4

20 l l l l

18 l l I l

14 l l l l

13 l l l I

8 I I l l

I I I I I

15 l 1 l 1

19 I I I I

[[ _l _I _l ₁

7 I I 1 1

Correct

Answer ( U) 10 10 10 10

17 1 I 1 1

16 1 1 1 I

6 I 1 1 I

2 0 1 1 I

lO I 1 I I

9 I 1 1 0

4 I 0 0 0

12

l 0 1 I

5 0 0 0 0

3 0 0 0 l

Correct

Answer (L) 7 6 7 7

DP 0,3 0,4 0,3 0,3

(U-L)

IN

Interpretation

s s

s

Correctly (C) 17 16 17 17

Dl= C I Total 0,85 0,8 0,85 0,85

Interpretation E E E E

Criterion of Discrimination Power

0,00 - 0,20 : Poor 0,20 - 0,40 : Satisfactory 0,40 - 0,70 :

Good

0,70 - 1,00 : Excellent

5 l l I l I I I l l I 10 1 1 0 I 0

1

l I I 0 7 0,3

s

17 0,85 E

Number Of Items

6 7 8

I I I

I I 0

l I l

I I I

l l l

l I l

0 I l

I 0 1

9 9 9

l 1 0

0

1

0

0 1 0

I 0 I

0 0 0

0 0 1

0 I I

0 0 I

I 0 0

4 4 4

0,5 0,5 0,5

G G G

13 13 13

0,65 0,65 0,65

M M M

9 10 [[ ₁₂ ₁₃ ₁₄

l l I I l l

l l l l l l

l l l I l l

l 0 l l l I

I l l l 0 l

I 0 I l l l

l 1 1 0 1 I

l l l I 0 l

I I I 0 I I

I I I I 0 I

10 8 10 8 7 10

1 l 1

1

0 0

0 l I I 0 0

0 I 0 I I 0

0 . I 0 0 I 0

0 1 0 I 0 I

I 0 0 0 l 0

I 0 0 0 0 0

0 0 0 0 0 0

0 0 0 I 0 I

0 0 I 0 0 I

3 5 3 5 3 3

0,7 0,3 0,7 0,3 0,4 0,7

G

s

G

s

G

13 13 13 13 10 13

0,65 0,65 0,65 0,65 0,5 0,65

M M M M M M

Criterion of Difficulty Index

0,00 - 0,30 : Difficult 0,30 - 0, 70 : Moderate 0,70 - 1,00 : Easy

(6)

Number of

Students 1 2 3

I 4 4 4

2

4

4 4

3 4 4 4

4 4 4 4

5 4 4 4

6 4 2 2

7 4 4 4

8 4 4 3

9 4 4 3

10 4 4 4

I I 2 4 4

12 4 4 4

13

2

4

2

14 4 4 3

15 2 2 4

16 4 2 2

17 4 4 4

18 4 4 3

19 4 4 3

20 4 4 2

Vi

0,53 0,53 0,66

Evi

44,35

Vt

=

135,905

r

=

0,725 (high )

Notes:

X

=

total amount

Vi

=

sum of variance

Evi

=

sum of total variance

Vt =total sum of variance

r

=

reliability

4 4 3 2 3 4 2 4 4 4 4 4 2 4 4

2

4 I

4

4 1 1,22

The Try-Out Results

"TOEFL"

Number of Items

5 6 7 8 9 10

6 3 6 3 9 9

5 3 6 3 7 5

3 3 3 6 5 10

6 6 3 6 5 10

6 6 6 6

5

9

3 I 3 3 4 7

5

6 6 6 10 5

6 5 4 5 8 9 6 3 5 5 5 10 6 3 4

5

8 9 3 3 6 6

5

6 5 6 6 5 8

3 6 6 6

5

6 6 4 5 8 • 10

3 3

3

6 9 10

6 6 6 3 9 5

3 2 I 3 3 4

6 6

4

5

8 10

6 6 3 3 9 7

1

2

3

3 4 4

2,61 3,01 2,35 1,80 4,57 5,41

X

11 12 13 14

5 5 10 9 81

10 9 10 10 83

5

10 5 69

5 5

5

5 71

10 10 5 5 84

5

10 9 8 63

10 5 5 10 84

10 9 5 10 86

10 10 10 5 84

10 10 10 8 89

10 9

5

4 70

5

10 9 10 84

5

63

10 5 10 10 89

8 10 9 8 79

9 10 5 9 80

9 5 5

5

53

10 10 10 10 94

9 10 10 7 85

5

9 5 8 55

(7)

The Calculation of Discrimination Power and Difficulty Index

Number Or

Students 1 2 3 4

18 1 1 1 1

14 I 1 I I

10 I 1 1 I

8 l 1 1 I

19 I 1 I I

12 1 1 1 0

9 I I I I

7 1 I 1 I

5 1 1 1 1

2 1 1 I 1

Correct

Answer ( U) 10 10 10 9

1 I I I I

16 I 0 0 I

15 0 0 I 0

4 I I I 1

II 0 I I 1

3 I I 1 0

13 0 1 0 1

6 1 0 0 0

20 1 I 0 0

17 I 1 1 0

Correct

Answer (L) 7 7 6 5

DP 0,3 0,3 0,4 0,4

(U-L)

IN

Interpretation

s

Correctly (C) 17 17 16 14

Dl= C /Total 0,85 0,85 0,85 0,7

Interpretation

E

Criterion of Discrimination Power

0,00 · 0,20 : Poor 0,20 • 0,40 : Satisfactory 0,40 • 0,70 :

Good

0, 70 - 1,00 : Excellent

5 1 I I 1 I 1 I I 1 1 10 I I 0 I 0 0 0 0 0 0 3 0,7 E l3 0,65 M

Number Of Items

6 7 8

1 l 1

1 I 1

0 I 1

l I I

1 0 0

I l 1

0 I I

1 l l

I I 1

0 I 0

7 9 8

0 l 0

I l 0

0 0 I

1 0 I

0 1 I

0 0 1

1 l I

0 0 0

3 4 5

0,4 0,5 0,3

s

G

s

10 l3 13

0,5 0,65 0,65

M

9 10 11 12 13 14

l 1 1 1 1 1

1 I I 0 1 1

1 1 1 1 1 1

I 1 I 1 0 I

1 I 1 1 1 1

0 l 0 1 1 1

0 1 I I 1 0

1 0 1 0 0 1

0 1 I 1 0 0

I 0 1 1 1 I

7 8 9 8 7 8

1 I 0 0 I I

I 0 I 1 0 I

I I I I I I

0 . I 0 0 0 0

0 0 1 1 0 0

0 1 0 0 1 0

0 0 0 0 0 0

0 1 0 1 I 1

0 0 0 I 0 1

0 0 1 0 0 0

3 5 4 5 4 5

0,4 0,3 0,5 0,3 0,3 0,3

s

G

s

10 13 13 13 II 13

0,5 0,65 0,65 0,65 0,55 0,65

M M M M M M

Criterion of Difficulty Index

0,00 - 0,30 : Difficult 0,30 • 0, 70 : Moderate 0,70 • 1,00 : Easy

(8)

Number of

Students I 2 3 4

I 4 4 4 4

2 2 2 2 2

3 2 4 4 4

4 4 4 4 4

5 4 4 4 4

6 2 4 2 4

7 4 4 4 4

8 4 4 4 4

9 4 4 2 2

10 4 4 2 2

11 4 4 2 2

12 4 4 4 4

13 2 2 4 4

14 4 4 4 4

15 4 4 4 2

16 4 4 2 2

17 4 4 4 4

18 4 4 4 4

19 4 4 4 4

20 4 2 4 2

Vi 0,67 0,53 0,88 0,95

Evi 42,14

Vt = 117,69

r = 0,691 (high )

Notes:

X

= total amount

Vi = sum of variance Evi = sum of total variance Vt =total sum of variance r

= reliability

The Try-Out Results

"ADVERTISING"

Number of Items X

5 6 7 8 9 10 II 12 13 14

3 3 3 3 5 9 5

5

9 5 66

3 3 6 3 9 5 5 6 6 5 59

6 6 6 6 5 5 5 10

5

5 73

3 3 3 3 10 10 5 5 8 5 71

3 6 3 6 10 10 5 10 10 10 89

6 3 3 6 5 5 9 9 5 8 71

6 6 3 3 10 5 10 10 8 8 85 3 6 6 6 5 10 10 8 10 10 90

6 6 3 6

5

9 9 5 5 5 71

3 6 6 3 5

5

5 5 9 9 68

3 3 3 3 8

5

6 5

5

58

3 6 6 6 10 10 10 10 7 7 91

6 3 3 3 10 8 5 5 5 8 68

6 6 5 3 10 - 10 8 10 10 7 91

6 6 2 6 10 10 10

5

9 83

3 6 6 6 5 5 5 9 9 5 71

6 6 6 6 10 10 10 10 5 5 90

6 6 6 6 10 10 5 5 10 5 85

6 3 6 3 10 10 5 10 9 5 83

6 3 6 6 9 9 10 9 8 5 83

(9)

The Calculation of Discrimination Power and DifficultY Index Number Of

Students I 2 _, 4

14 I I I I

12 I I I I

8 I I I I

17 I I I I

5 I I I 1

18 1 1 1 1

7 I I I I

19 I I I 1

15 I 1 I 0

20 1 0 1 0

Correct

Answer ( U) 10 9 10 8

2 0 I I I

4 l I 1 1

9 1 I 0 0

6 0 I 0 1

16 1 1 0 0

10 I I 0 ()

13 0 0 I 1

I 1 I I I

2 0 0 0 0

11 1 I 0 0

Correct

Answer (L) 6 6 4 5 DP 0,4 0,3 0,6 0,3 (U-L) IN

Interpretation

s

G

s

Correctly (C) 16 15 14 13 Dl= C I Total 0,8 0,75 0,7 0,65 Interpretation E E E M

Criterion of Discrimination Power 0,00 - 0,20 : Poor

0,20 • 0,40 Satisfactorv 0,40 -

o,

70 : Good . 0,70 - 1,00 : Excellent

5 I 0 () I 0 1 I I I 1 7 I 0 I I 0 0 I 0 0 0 4 0.3

s

II 0,55 M

Number Of Items

6 7 8

I I 0 I I I

I I I

I I 1 I 0 I

l 1 1

I 0 0

0 I 0

I 0 I

0 1 I

8 7 7

l I l

0 0 0

I 0 I

0 0 I

1 I I

I I 0

0 0 0

0 1 0

0 0 0

4 4 4 0,4 0,3 0,3

s

12 II II 0,6 0,55 0,55

M M M

9 10 II 12 13 14

I I I I I I

0 I I 1 I I

I I I I () ₀

I I 0 I I I

I 1 0 0 I 0

I 0 I I I 1

I I 0 I I 0

I I I 0 0 I

I I I I 1 0

9 9 7 8 8 6

0 0 0 I 0 0

I 1 0 0 I 0

0 I I 0 0 0

0 • 0 I 1 0 I

0 0 0 I 1 0

0 0 0 0 1 I

I 1 0 0 0 I

0 I 0 0 I 0

1 0 0 1 1 0

1 0 I 0 0 0

4 4 3 4 5 3 0,5 0.5 0,4 0.4 0.3 0,3

s

13 13 10 12 13 9

0,65 0,65 0.5 0,6 0,65 0,45

M M M M M M

Criterion of Difficulty Index 0,00 - 0,30 : Difficult 0,30 - 0.70 : Moderate 0, 70 - I ,00 : Easy

(10)

THE EXPERIMENT RESULTS BASED ON THE READI:\'G PASSAGES

I. THE RESULT OF THE FIRST TREATI\IEI'IT ( TOEFL )

GROUP C (Traditional Reading)

Number of Students I 2 3 4 5 6 7 8 9 10 II 12 13 14 15 16 17 18 19 20

EXc = 1490

Xc

=

74,50 I 4 4 3 2 4 4 4 4 3 4 4 4 2 3 3 4 3 4 2 3

Sc

=

10,0969

Notes:

2 3

3 2

2 3

4 4

4 3

4 4

3

2

2 4

4 4

3 4 2 3

4 3

4 2

4 4

4 3

4 4

EXc

= total scores of group c

Xc

=

mean of group c

4 5

4 4

4 6

4

5

4 4

4 6

4 5

4 4

3

5

4 6

3 4

4 3

4 4

3 4

4 4

4 6

4 5

4 4

Sc

=

standard deviation of group c

Number of Items

6 7 8 9 10

5 4 3 5 6

3 5 5 7 5

5 5 6 8 6 5 3 5 7 5

6 5 6 8 6

4 6 6 9 8

6 5 6 7 9

6 4

5

8 7

3 3 5 7 5

4 4 6 7 -1

6 6 6 9 7

6 6 6 9 8

5 4 4 7 - 8

5 4 5 8 5

4 5 4 6 7 2 3 5 7 5 5 4 6 5 6

4 5 6 8 8

3 5 4 6 7

6 3 5 7 6

II 12 13 14

-1 7 3 6

7 8 7 4

-1 6 8 7

8 7 6 7

7 5 7 8

10 8 9 8

8 9 7 9

6 8 9 6

5 9 7 6

5

7 7 8

10 9 9 10

6 8 6 8

6 8 5 7

7 7 8 6

4 5 7 8

6 6 3 4

8 9 7 9

7 9 5 6

3 6 2 6

8 9 8 7

(11)

K TOEFL ..

GROUP D (Critical Reading)

Number of Students I

2

3 4 5 6 7 8 9

10

11

12

13

14

15

16

17

18

19

20

EXd

=

1640 Xd

=

82

I

4

3 3 4 4

3

4 4 2 4 2 3 4 4 2 4 4 4 4 3

Sd

=

10,2649

Notes:

2

3 4

4 4 4

4 3 4

4 3 3

2

3 4

4 4 4

3

4 4

4 3 4

3

3 4

3 2 3

3

2

4

2 3 3

4 4

4

4 4 4

4

4 4

3 4

2

4 4 4 4 4 4

EXd = total scores of group d Xd

=

mean of group d

5 6 5 6 5 6 6 6 5 4 6 3 5 6 6 5 5 6 4 6 5

Sd

=

standard deviation of group d

Number of Items

6 7 8 9

10

6 5 6

10

8

6 5 5 9 7

4 3 6 8 6

6 4 6 7

8

6 6 6

10

6 6 6 9

7

5 6 6 9 9

5 6 5 8 9

6 6 5 7

7

6 5 6 8

8

2

4 2

7

4

3

4

8 6

3 5 5 7

8

6 4 6

8

9

.

6 5 6 9

7

5 6 6

9

6 6 6

10

7

5 6 5 8 6

6 6 6

8

9

6 5 6 9 7

II

12

13

14

9

10

9

10

95

8

7 6

80

7 8 6 8 75 8

8

7

6 78

8

9

8 9

94

8 10 9 8

90

7

10

8 10 92

6

9

7

10

85

5 8 8 6 75

10

7

5 I)

80

6

7

4 6 58

5

4

6 6 63

6

7

5 6

70

8 7 7 8 85 6

9

7

9

83

6

10

8 9 89

8

10

7 9

91

8

9

8

80

(12)

HO :

~to

=

~Lc

;

there is no difference between the mean groups.

HA : 1-!o

>

/.lc ; the mean score of group D is greater than the one of group C.

T-test ; where df= nc

+

nd- 2

=

20

+

20-2

=

38

t ( 0,05)

=

1,686

IX

XC

=

74,50

n

n.

IX

2 -

(IX

i

n (n-1)

IX

XD = =

82,00

n

n.

IX

2 -

(IX

i

n (n-1)

=

10,0969

=

10,2649

Xo -Xc

to=

=

2)3295

(no-1) 5o

2

+

(nc-1 )oc

2

no+ nc-

2

[

~D

+

~J

-1,686

0

1,686

So HO is rejected since to= 2,3295

>

Tt

=

+

1,686

(13)

II. THE RESULT OF THE SECOND TREATMENT ( ADVERTISING )

GROUP C (Traditional Reading)

Number of Students 1 2 3 4 5 6 7 8 9 10 II 12 13 14 15 16 17 18 19 20

EXc

=

1506

Xc

=

75,3

I 4 4 3 2 4 3 4 3 4 3 3 4 4 4 4 2 3 4 4 4

Sc

=

11,7388

Notes:

2 3 4

4 2 4

3 4 4

4 3 4

4 4 3

4 3 4

2 4 3

3 4 2

2 3 3

4 4 4 3 4 3 2 4 2

4 4 4

4 3 4

4 4 4

3 4 3

4 4 4

EXc

=

total scores of group c

Xc

=

mean of group c

5

2

5

6

5

6 4 6

5

6 5 4

5

6 6 5 4 5 6 5 4

Sc

=

Number of Items

6 7 8 9 10 11

6 3

2

3 4 2

6 4

5

6 9 8

4 5 4 7 6 7

6 6 6 10 10 8

5 6 3 10 7 8

6 3 4 5 4 5

6 4

5

8 10 9 3 6 5 6 4 8

4 4 3 4 7 5

6 6 6 10 10 7

5 3

5

7 6 4

3

4 3 5 8 6

6 3 6 9 7 8

2 6 6 10 10 8

4 5 4 8 • 3 2

5 6 3 6 5 3

3 4 3 2 6 4 6 3 5 7 8 6

4 4 6 8 9 6

6 6 3 8 9 7

12 13 14

6 7 5 54

9 8 7 80

7 8 9 77

9 10 7 90

10 9 8 87

7 6 4 60

8 6 4 79

9 7

5

70

9 8 8 70

10 10 10 95

7 6 3 63

7 7 5 65

9 8 8 86

10 8 8 90

8 5 6 65

9 8 7 70

8 7 5 60

9 8 6 80

10 8 9 85

(14)

"ADVERTISING"

GROUP D (Critical Reading)

Number of Students I

l 4

2 3

3 4

4 3

5 4

6 4

7 4

8 2

9 4

10 4

II 4

12 4

13 4

14 4

15 2

16 4

17 3

18 4

19 3

20 4

EXd

=

1672

Xd

=

83,6

Sd

=

8,9466

Notes:

2 3 4

4 4 3 4 3 4 4 4 4 4 3 4 4 4 4 4 4 2 4 4 4 2 3 3 3 4 4 4 4 4

2 3 3

4 3 4 3 4 4 4 4 4 3 4 2 4 3 3 4 3 4 4 4 4 3 4 3 4 4 4

EXd

=

total scores of group d

Xd

=

mean of group d

5 6 6 6 6 6

5

6 6

5

6 4 5 4 6 5 6 5 6 4 6

Sd

=

Number of Items 6 7 8 9 10

5 6 6 10 8 6

5

6 4 8 6 6 6 7 10 5 4 3 6 8 6

5

6 10 9 6 4 6 9 10 4 6 5 9 10 6 2

5

7 6 6 6 4 9 8 6 6 6 10 9 6 4 5 4 7 2

5

6 9 8 5 6 6 10 10 6 6 6 9 -8 4 2 3 5 6 5 5 6 8 6

6 4 5 6 7

4 6 5 9 7

6

5

6 8 7 3 6 6 10 9

II 12 13 14

10 8 9 7 90 9 7 7 8 80 9 8 7 7 88

7 9 8 8 78

(15)

HO :

llD

=

)lc ;

there is no difference between the mean groups.

HA :

llD

>

)lc ;

the mean score of group D is greater than the one of group

C.

+ nd- 2

=

20 + 20-2

= 38

t (

0,05 )

=

1,686

rx

XC

= =

75,30

n

8c

==~

n .

L:X

2 - {

L:X

)

2

n(n-1)

rx

XD=

=

83,60

n

8o

=~

n . L:X

2 - (

L:X

}

2

n (n-1)

=11,7388

=

8,9466

Xo -Xc

to=

~

(no- 1) 8o

2

+ (nc-1)oc

2

no+

llc-

2

-

2,5149

[

~D

+

~J

-1,686

0

1,686

So HO is rejected since to=

2,5149

>

Tt = + 1,686

(16)

III. THE RESULT OF THE POST TEST ( NATIVE OR NON-NATIVE TEACHERS )

GROUP C (Traditional Reading)

Number of

Students 1

1 4

2 4

3 4

4 4

5 4

6 4

7 4

8 3

9 4

10 4

11 4

12 3

13 2

14 3

15 2

16 4

17 2

18 4

19 4

20 4

EXc

=

1397

Xc

=

69,85

Sc

=

13,2438

Notes:

2 3

4 4

2 4

4 4

2 4

4 4

4 3

4 4

3 4 3 4

4 3

4

3

2 4

4 2

3 4 4 4

4 4

EXc

=

total scores of group c

Xc

=

mean of group c

4

5

4 6

3

5

3 6

3 3

4 6

2 4

4 6

4 5

4 6

4 4

1 5

4

5

3 2

2 4

4 3

3 6 4 4

4 5

4

6

4 4

Sc

=

Number of Items

6 7 8 9 10

6 6 4 8 7

4 4

5

8 9

3 3 4 8 6

4 5 3 4 6

3 6 2 8 10

4 2 5

5

6

6 6 6 7 7

5 6 5 8 4

5 6 5 lO 8

5 4 6 9 8

6 3 4 5 3

4 5 5 8 6

1 2 5 2 4

3 4 5 5 - 7

1 5 3 3 1

4 5 6 8 8 4 4 4 6 3 3 5 6 7 4

6 6 6 8 8

2 4 3 5 7

11 12 13 14

8 7

5

4

7 7 4 3

9 6 4 6

5

6 2 3

7 6 8 8

3 6 6 7

8 10 8 10

5 6 7 5

6 7 10 9

8 9 4 5

4 6 7 3

4 7 9 6

5

8 4 5

3 5 6 7

2 4 5 6

6 7 8 9

7 6 6 3

5 7

5

7

8 10 10 10

4 7 9 9

(17)

"NATIVE OR NON-NATIVE TEACHERS ..

GROUP D (Critical Reading)

Number of Students I 2 3 4 5 (_)

7

8 9 10

11

12

13 14

15

I(,

17

18

19

20

EXd

=

1608

Xd

= 80,4

I

4

4 3 3 4 4 4 I 4 4 4

4

4 4 4 4 3 4 3 4

Sd

=

12,3006

Notes

2 3 4

4 4 4

_, ₃ ₄

4 4 3

2

4

2

4 4 4

4 4

4

4 4 4 2 3 2

4

_, ₄ ₂

2 4 3

4

4 4 4

4

4 4 4 4

4 2 4

4

4 3

4 4 4

4 3 I

4 4 4

EXd

=

total scores of group d

Xd

=

mean of group d

5 6 4 5 4 6 6 6 3 6 5 5 6 6 6 5 6 5 6 2 6

Sd

=

Number of Items

6

7

g 9

10

5 6 6 Ill 8

3 5 4 6 5

5 6 4 5 8

2 4 3 4 3

3 () 6

10

[(J

6 6 5 7

10

6 6 6 8

7

5 3 6 (, 4

5 6 6 10 8

4 6 2

7

7 5

4

2 6 8

(, 6 6 10

10

6 6 6

10

4 6 6 9 9

6 6 6 8

7

3 6 5 9 10

4

5 4 8 8

6 6 6

10

5 5 3

2

6

5 6 6 8 6

II 9 8 6 6 8

7

10

3 9 5 4 IU <) 9

10

8

7

9 3

7

12

13

14

10

9 95 8 7 6 70

9 7 8 77

8 7 8 60

6 8 10 89

6

7

8 84

7

10 8

90

7

6 8 59

6 6 8 86 8 6

7

70

7 8 6 68

9 10

7

96

7 (, 8 90

6 8 9 88

7

6 8 85

6 9 6 82

9 5 9

78

7

9 10

95

8 6

7

58

(18)

HO :

J.lD

=

).lc ;

there is no difference between the mean groups.

HA :

J.ln

>

).lc ;

the mean score of group D is greater than the one of f,>roup C.

+

nd- 2

= 20

+

20-2

=

38

t ( 0,05)

=

1,686

LX

XC=

=

69,85

n

8c

=~

n.2:X

2 -

{ 2:X

22

n (n-1)

LX

XD=

=

80,40

n

8o

=~

n.2:X

2 -

{ 2:X

22

n (n-1)

=

13,2438

=

12,3006

Xo- Xc

to=

-

2,6103

(no- 1 )

on

2

+ (

nc- 1 ) 8c

2

no+

nc-

2 [

~D

+

~J

-1,686

0

1,686

So HO is rejected since to= 2,6103 > Tt =

+

1,686

(19)

THE EXPERIMENT RESULTS BASED ON THE TYPE OF QUESTIOl'iS USED IN THE TREATMENT

I. THE FIRST TREATMENT ( TOEFL)

Number of

KNOWLEDGE QUESTION

(In Reading Passage TOEFL)

Group C

Group D

Students

( Traditional Reading )

( Critical Reading )

1

2

total

I

2

total

X

I

4

3

7

4 4

8

2

4

2

6

3

4

7

3

4

7

3

4

7

4

2

4

6

4

2

6

5

4 4

8

4 4

8

6

4 4

8

3

4

7

4 4

8

4 4

8

4 4

8

4

3

7

9

3

6

2

4

6

10

4

2

6

4

3

7

II

4

8

2

3

5

12

4

8

3

6

13

2

3

5

4

2

6

14

3

2

5

4

-

4

8

15

3

4

7

2

4

6

16

4 4

8

4 4

8

17

3

4

7

4

8

18

4

8

4

3

7

19

2

4

6

4

8

20

3

4

7

3

4

7

HO : llo

=

).lc ;

there is no difference

between

the mean groups.

HA : llo

>

).lc ;

the mean score of group

D

is greater than

the

one of group

C.

T-test : where

df= nc

+

nd-

2

=

20

+

20-2

=

38

t (

0,05)

=

1,686

:EXc

=

139

Xc

=

6,95

Oc

=

1,0501

to=

0,1603

:EXo

=

140

Xo

=

7,00

Oo

=0,9177

(20)

Number of

COMPREHENSIO~ QUESTION

Group C

Group D

Students

( Traditional Reading)

(Critical Reading )

3

4

total

3

4

total

X

I

2

4

6

4

8

2

3

4

7 3

4

7

3

4

8

3 3

6

4

3

4

7

3

4

7

5

4

8

4

8

6

4

8

4

8

7

4

8

4

8

4

8

4

8

9

2

4

6

3

4

7

10

4

3

7

3

4

7

II

4

8

2

3

5

12

4

8

2

4

6

13

4

3

7

3

6

14

3

4

7

4

8

15

3

4

7

4

8

16

2

3

5

4

8

17

4

8

4

8

18

4

8

4

-

2

6

19

3

4

7

4

8

20

4

8

4

8

HO :

).to

=

1-tc ;

HA :

).to

>

1-tc ;

C.

T-test :where

df= nc

+

nd-

2

=

20

+

20-2

=38

t (

0,05)

=

1,686

LXc

=

146

Xc

=

7,30

8c

=

0,8645

to=

-0,1724

LXo= 145

Xo

=

7,25

8o

=

0,9665

(21)

Number of

APPLICATION QUESTION

Group C

Group

D

Students

(Traditional Reading)

(Critical Reading )

5

6

total

5

6

total

X

1

4

5

9

6

12

2

6

3

9

5

6

11

3 5

5

10

6

4

10

4

5

9

5

6

II

5

6

12

6

12

6

4

10

6

12

7

6

12

6

5

II

8

5

6

11

5

10

9

4

3

7

4

6

10

5

4

9

6

12

11

6

12

3

2

5

12

6

12

5

4

9

13

4

5

9

6

3

9

14

3

5

8

6

12

15

4

8

5

6

11

16

4

2

6

5

10

17

4

5

9

6

12

18

6

4

10

4

-

5

9

19

5

3

8

6

12

20

4

6

10

5

6

11

HO : l!o =

J..lc ;

HA :

l!o

> J..lc ;

D

is greater than the one of group

C.

T-test :where df=

nc

+

nd- 2

= 20

+

20-2

=38

t (

0,05) = 1,686

L.Xc = 190

Xc

=

9,50

8c

= 1, 7014

to= 1,9520

L.Xo = 211

Xn=10,55

8

0

= 1,7006

>

Tt = 1,686

(22)

ANALYSIS QUESTION

Number of

Group C

Group D

Students

7

8

total

7

8

total

X

1

4

3

7

5

6

11

2

5

10

5 5

10

3

5

6

II 3

6

9

4

3 5

8

4

6

10

5

6

11

6

12

6

12

6

12

7

5

6

11

6

12

8

4

5

9

6

5

11

9

3

5

8

6

5 II

10

4

6

10

5

6

11

6

12

4

2

6

12

6

12

3

4

7

!3

4

8

5

10

14

4

5

9

4

6

10

15

5

4

9

5

6

11

16

3

5

8

6

12

17

4

6

10

6

12

18

5

6

11

6

-

5

II

19

5

4

9

6

12

20

3

5

8

5

6

ll

HO :

I-to

=

1-tc ;

HA :

I-to

>

1-lc ; the mean score of group D is greater than the one of group C.

T-test :where df= nc

+

nd-

2

=

20

+

20-2

=38

t (

0,05)

=

1,686

LXc

=

193

Xc

=

9,65

oc

=

1,5652

to= 1,7766

LX

o

=

211

Xo =10,55

oo

=

1,6376

So HO is rejected since to=

1,7766

>

Tt

=

1,686

(23)

Number of

SYNTHESIS QUESTION

..

Group C

Group D

Students

(Traditional R.eading)

( Critical

Readi.n~

)

9

10

11'

total

9

10

11

total

X

I

5

6

4

15

10

8

9

27

2

7

5

7

19

9

7

8

24

3

8

6

4

18

8

6

7

21

4

7

5

8

20

7

8

23

5

8

6

7

21

10

8

28

6

9

8

10

27

9

7

8

24

7

9

8

24

9

7

25

8

7

6

21

8

9

6

23

9

7

5

17

7

5

19

10

7

4

5

16

8

10

26

II

9

7

10

26

7

6

20

12

9

8

6

23

8

6

5

19

13

7

8

6

21

7

8

6

21

14

8

5

7

20

8

9

8

25

15

6

7

4

17

9

7

6

22

16

7

5

6

18

9

6

24

17

5

6

8

19

10

7

8

25

18

8

7

23

8

6

8

22

19

6

7

3

16

8

9

10

27

20

7

6

8

21

9

7

24

HO :

Jlo

=

J..1c ;

HA :

Jlo

>

J..1c ;

the mean score of group D is greater than the one of group C.

T-test :where

df=

nc

+ nd-

2

= 20

+

20-2

=

38

t ( 0,05 ) = 1,686

LXc

= 402

Xc = 20,1

8c

= 3,3071

to= 3,5586

LX

0

=469

Xo =23,45

8

0

= 2,6052

>

Tt = 1,686

(24)

Number of

EVALUATION QUESTION

..

Group C

Group D

Students

(Traditional R.eading)

( Critical

Readi.n~

)

12

13

14.

total

12

13

14

total

X

I

7

3

6

16

10

9

10

29

2

8

7

4

19

8

7

6

21

3

6

8

7

21

8

6

8

22

4

7

6

7

20

8

7

6

21

5

7

8

20

9

8

9

26

6

8

9

8

25

10

9

8

27

7

9

7

9

25

10

8

10

28

8

9

6

23

9

7

10

26

9

7

6

22

8

6

22

10

7

8

22

7

5

9

21

11

9

10

28

7

4

6

17

12

8

6

8

22

4

6

16

13

8

5

7

20

7

5

6

18

14

7

8

6

21

7

8

22

15

5

7

8

20

9

7

9

25

16

6

3

4

13

10

8

9

27

17

9

7

9

25

10

7.

9

26

18

9

5

6

20

9

8

25

19

6

2

6

14

10

8

10

28

20

9

8

7

24

8

7

6

21

HO : llo

=

11c

;

HA : llo

>

11c

;

D

C.

+

nd-

2

=

20

+

20-2

=

38

t (

0,05)

=

1,686

!:Xc = 420

Xc = 21,00

oc

=

3,6992

to= 2,0155

!:Xo= 468

X

₀

=23,40

8

₀

=

3,8306

So

HO

is rejected since to= 2,0155

>

Tt

=

1,686

(25)

II. THE SECOND TREATME:vf ( ADVERTISEMENT) KNOWLEDGE QUESTION

(In Reading Passage ADVERTISING)

Number of

Group C

Group D

Students

(Traditional Reading)

I

2

total

I

2

total

X

I

4

8

4

8

2

4

3

7

3

4

7

3 3

4

7

4

8

4

2

4

6

3

4

7

5

4

8

4

8

6

3

2

5

4

8

7

4

3

7

4

8

3

6

2

4

9

4

2

6

4

3

7

10

~

4

7

4

8

.)

II

3

6

4

2

6

12

4

2

6

4

8

13

4

8

4

3

7

14

4

8

4

8

15

4

8

2

3

5

16

2

4

6

4

8

17

3 3

6

3

-

4

7

18

4

8

4

8

19

4

8 3 3

6

20

4

8

4

8

HO :

!lo

=

11c ;

HA :

!lo

> llc ;

D

C.

+

nd- 2

=

20

+

20-2

=

38

t ( 0,05) = 1,686

LXc

=

139

Xc = 6,95

oc

= 0,9987

to= 0,7334

LX

0

= 144

Xo =7,20

Oo

= 1,1517

(26)

COMPREHENSION QUESTION

Number of

Group C

Group D

Students

3

4

total

3

4

total

X

I

2

4

6

4

3 7

2

4

8

3

4

7

3 3

4

7

4

8

4

3

7

3

4

7

5

3

4

7

4

8

6

4

3

7

4

2

6

7

4

2

6

4

8

4

2

6

3 3

6

9

3

6

4

8

10

4

8

4

8

II

4

3

7

3 3

6

12

4

2

6

3

4

7

13

4

8

4

8

14

4

8

4

8

15

3

4

7

4

2

6

16

4

8

3

6

17

4

3

7

3

4

7

18

4

8

4

-

4

8

19

4

8

4

3

7

20

4

8

4

8

HO :

).lo = ).lc ;

HA :

).lo

>

).lc ;

the mean score of group Dis greater than the one of group

C.

T-test

:where

df=

nc

+

nd- 2

=

20

+

20-2

=

38

t (

0,05)

=

1,686

IXc

=

143

Xc

=

7,15

eSc

=

0,8127

to=

0,1921

IX

0

=

144

Xo

=

7,20

8o

= 0,8335

So HO is accepted since Tt

= -

1,686

<

to

=

0,1921

<

Tt

=

+

1,686

(27)

APPLICATION QUESTION

(In Reading

Pas~·age

ADVERTISING)

Number of

Group C

Group D

Students

5

6

total

5

6

total

X X

1

2

6

8

6

5

11

2

3

6

9

6

12

3

6

4

10

6

12

4

5

6

II

6

5

I 1

5

6

5

I 1

6

12

6

4

6

10

5

6

I I

7

6

12

6

4

10

8

5

3

8

6

12

9

6

4

10

5

6

I 1

10

5

6

II

6

12

1 I

4

5

9

4

6

10

12

5

3

8

5

2

7

13

6

12

4

5

9

14

6

2

8

6

12

15

5

4

9

5

4

9

16

4

5

9

6

5

11

17

5

3

8

5

-

6

11

18

6

12

6

4

10

19

5

4

9

4

6

10

20

4

6

10

6

3

9

HO :

J.lo

=

Jlc ;

HA : Po

>

Jlc ;

C.

+

nd- 2

= 20

+

20-2

= 38

t (

0,05) = 1,686

l:Xc

= 194

Xc

= 9,70

8c

=1,4179

to= 2,0534

2:X

0

= 212

Xo

=10,60

8

0

=

1,3534

>

Tt = 1,686

(28)

Al"'AL YSIS QUESTION

Number of

Group C

Group D

Students

7

8

total

7

8

total

X

1

3

2

5

6

12

2

4

5

9

5

6

11

3

5

4

9

6

12

4

6

12

4

3

7

5

6

3

9

5

6

11

6

3

4

7

4

6

10

7

4

5

9

6

5

11

8

6

5

11

2

5

7

9

4

3

7

6

4

10

6

12

6

12

11

3

5

8

4

5

9

12

4

3

7

5

6

11

13

3

6

9

5

6

12

14

6

12

6

12

15

5

4

9

2

3

5

16

6

3

9

5

6

11

17

4

3

7

4

5

9

18

3

5

8

6

5

II

19

4

6

10

5

6

11

20

6

3

9

6

12

HO : llo

=

).lc ;

HA : llo

>

).lc ;

T-test :where

df= nc

+

nd- 2

=

20+ 20-2

=

38

t

(

0,05)

=

1,686

EXc

=

178

Xc

=

8,90

8c

=

1,8610

to=

2,3065

EXo

=

206

Xo

=

10,30

8o

=

1,9762

>

Tt

=

1,686

(29)

SY!I<THESIS QUESTION

(In Reading Passage ADVERT/SL'VG)

..

Number of Group C Group D

Students (Traditional R~ading) ( Critical Readi_n~ ) 9 10

11'

total 9 10 11 total

X

1 3 4 2 9 10 8 10 28

2 6 9 8 ?' _-~ ₄ ₈ ₉ ₂₁

3 7 6 7 20 7 10 9 26

4 10 10 8 28 6 8 7 21

5 10 7 8 25 10 9 8 27

6 5 4 5 14 9 10 8 27

7 8 10 9 27 9 10 8 27

8 6 4 8 18 7 6 5 18

9 4 7 5 16 9 8 7 24

10 10 10 7 27 10 9 8 27

11 7 6 4 17 4 7 6 17

12 5 8 6 19 9 8 8 25

13 9 7 8 24 10 10 8 28

14 10 10 8 28 9 8 7 24

15 8 .) ' 2 13 5 6 4 15

16 6

5

3 14 8 6 7 21

17 2 6 4 12 6 7 7 20

18 7 8 6 21 9 7 8 24

19 8 9 6 23 8 7 6 21

20 8 9 7 24 10 9 10 29

HO :

I!D

=

llc ;

HA : l!o

> llc ;

T-test :where

df= nc

+

nd-

2

= 20+ 20-2

= 38

t ( 0,05) = 1,686

L:Xc = 402

Xc = 20,10

<lc

=5,7938

to= 2,1517

L:X0 = 470

Xo =23,50

<3

0 = 4,0458

So HO is rejected smce to = 2,1517

>

Tt = 1,686

(30)

EVALUATION QUESTION

(In Reading Passage AD VERT/SING)

..

Number of

Group C

Group D

Students

( Traditional R.eading )

12

13

14 .

total

12

13

14

total

X

I

6

7

5

18

8

9

7

24

2

9

8

7

24

7

8

22

3

7

8

9

24

8

7

22

4

9

10

7

26

9

8

25

5

10

9

8

27

10

9

10

29

6

7

6

4

17

10

8

28

7

8

6

4

18

9

8

9

26

8

9

7

5

21

8

7

23

9

8

25

9

10

9

28

10

30

8

9

6

23

II

7

6

3

16

10

30

12

7

5

19

9

7

6

22

13

9

8 8

25

10

30

14

10

8

28

9

8

25

15

8

5

6

19

7

5

6

18

16

9

8

7

24

10

30

17

8

7

5

20

8

9

7

24

18

9

8

6

23

7

6

7

20

19

10

8

9

27

9

6

8

23

20

8

7

6

21

9

7

8

24

HO : Jlo = 11c ;

HA : Jlo

>

11c ;

+

nd- 2

=

20

+

20-2

=

38

t (

0,05 )

=

1,686

:!:Xc

=

450

Xc

=

22,50

8c

=

3,9001

to= 1,9772

!:Xo

=

496

Xo

=

24,80

8o

=

3.4428

1,9772

> Tt = 1,686

(31)

THE EXPERIMEJioT RESULTS BASED ON THE TYPE OF QUESTIONS USED IN THE POST TEST

KNOWLEDGE QUESTION

(In Reading Passage NATIVE OR NON-NATIVE TEACHERS)

Number of

Group C

Group D

Students

1

2

total

1

2

total

X

1

4

8

4

8

2

4

8

4

3 7

3

4

8

3

4

7

4

2

6

3

2

5

4

8

4

8

6

4

2

6

4

8

7

4

8

4 4

8

3

4 7

I

2

3

9

4

4 8

4

8

10

4

3 7 4

3

7

11

4

3 7

4

2

6

12

3

4

7

4

8

13

2

4

6

4

8

14

3

2

5

4 4

8

15

2

4

8

16

4

8

4 4

8

17

2

3

5

3

4

7

18

4

8

4

8

19

4 4

8

3

4

7

20

4 4

8

4

8

HO :

!lo

= !lc ;

HA :

!lo

>

!lc ;

the mean score of group Dis greater than the one of group C.

+

nd- 2

= 20

+

20-2

=

38

t (

0,05) = 1,686

l:Xc=l40

Xc

=

7,00

Oc

= 1,2566

to= 0,6202

l:X

0

= 145

X0

=

7,25

8

0

=

1,2927

So HO is accepted since Tt

= -

1,686

<

to= 0,6202

<

Tt =

+

1,686

(32)

COMPREHE~SIO~

QUESTION

Number of

Group C

Group D

Students

3

4

total

3

4

total

X

1

4

8

4

8

2

4

3

7

3

4

7

3

4

3

7

4

3

7

4

3

7

4

2

6

5

4

8

4

8

6

4

2

6

4

8

7

4

8

4

8

3

4

7

3

2

5

9

4

8

4

8

10

4

8

4

2

6

11

4

I

5

4

3

7

12

3

4

7

4

8

13

3

6

4

8

14

4

2

6

4

8

15

4

8

4

8

16

2

3

5

2

4

6

17

4

8

4

3

7

18

4

8

4

8

19

4

8

3

1

4

20

4

8

4

8

HO :

J..lo

= l-Ie ;

HA :

J..lo

>

l-Ie ;

D

C.

+

nd- 2

= 20

+

20-2

=

38

t ( 0,05) = 1,686

:EXc = 143

Xc = 7,15

Oc

=

1,0400

to=O

:EXo=l43

Xo =7,15

8

0

=I, 1821

So

HO

is accepted since Tt

= -

1,686

<

to= 0

<

Tt =

+

1,686

(33)

APPLICATION QUESTION

Number of

Group C

Group D

Students

5 6

total

5 6

total

X X

1 6 6 12 6 5

11

2

5 4 9 4

3

7

3 6 3 9

5

10

4 3 4 7 4 2 6

5 6 3 9 6 3 9

6 4 4

8

6 6 12

7 6 6 12 6 6 12

8

5 5 10 3 5

8

9 6 5 1 1 6 5 11

10 4 5 9 5 4 9

11 5 6 11 5 5 10

12 5 4 9

6

6 12

l3 2 1 3 6 6 12

14 4 3 7 6 4 10

15 3 1 4 5 6 11

16 6 4 10

6

3 9

17 4 4

8

5

4 9

18 5 3 8

6

12

19

6

12 2

5

7

20 4 2

6

5

11

HO :

).ln

=

!1c

;

HA :

).lo

>

!1c

;

+

nd- 2

= 20

+

20-2

= 38

t ( 0,05) = 1,686

~Xc=174

Xc = 8,7 Oc = 2,4730

to=

1,7339

~Xo = 198 Xo

=

9,90 Oo = 1.8610

So HO is rejected since to

=

I, 7339

>

Tt

=

I ,686

(34)

ANALYSIS QUESTION

Number of

Group C

Group D

Students

7

8

total

7

8

total

X

I

6

4

10

6

12

2

4

5

9

5

4

9

3

4

7

6

4

10

4

5 3

8

4

3

7

5

6

2

8

6

12

6

2

5

7

6

5 II

7

6

12

6

12

8

6

5 II 3

6

9

6

5

II

6

12

10

4

6

10

6

2

8

II

3

4

7

4

2

6

12

5

10

6

12

l3

2

5

7

6

12

14

4

5

9

6

12

15

5

3

8

6

12

16

5

6

II

6

5

II

17

4 4

8

5

4

9

18

5

6

II

6

12

19

6

12

5

3

8

20

4

3

7

6

12

HO :

).to

=

J.lc

;

HA :

).to

>

J.lc

;

+ nd-

2

= 20

+

20-2

=

38

t ( 0,05) = 1,686

LXc

=

!83

Xc

=

9,15

oc

= 1,7852

to= 2,0945

LXo = 208

Xo =10,40

Oo

= 1,9841

>

Tt

=

1,686

(35)

SYNTHESIS QUESTION

..

Number of

Group C

Group D

Students

( Traditional R_eading ) ( Critical

Readi_n~

)

9 10

II .

total

9 10

II

total

X

I

8 7 8 23 10 8 9 27

2 8 9 7 24 6 5 8 19

3 8 6 9 ? '

_

_,

₅ ₈ ₆ ₁₅

4 4 6 5 15 4 3 6 12

5 8 10 7 25 10 10 8 28

6 5 6 ' .) 14 7 10 7 27

7 7 7 8 22 8 7 10 30

8 8 4 5 17 6 4 3 9

9 10 8 6 24 10 8 9 27

10 9 8 8 25 7 7 5 19

II

5 3 4 12 6 8 4 12

12

8

6 4 18 10 10 10 30

13 2 4 5

II

10 10 9 29

14 5 7 3 15 9 9 9 27

15 3

I

2 6

8

7 10 25

16 8 8 6 22 9 10

8

27

17 6

3

7

16 8 8 7 ?'

_

_,

18 7 4 5 16 10 10 9 29

19 8 8 8 24 2 6 3 7

20 5 7 4 16 8 6 7 21

HO : Jlo

=

1..1c ;

HA :

J.lo

>

1..1c ;

+

nd-

2

=

20

+

20-2

= 38

t ( 0,05)

=

1,686 IXc = 368 Xc

=

18,40

oc

= 5,4328

to=

1,8166

IXo

=

443 Xo = 22,15 Oo

=

7,4641

So

HO

is rejected smce to

= I ,8166

>

Tt

=

1,686

(36)

EVALUATION QUESTION

..

Number of

Group C

Group D

Students

( Traditional R_eading )

( Critical

Readi_n~

)

12

13

14.

total

12

13

14

total

X

I

7

5

4

16

10

9

29

2

7

4

3

14

8

7

6

21

3

6

4

6

16

9

7

8

24

4

6

2

3 II

8

7

8

23

5

6

8

22

6

8

10

24

6

7

19

6

7

8

21

7

10

8

10

28

7

10

8

25

8

6

7

5

18

7

6

8

21

9

7

10

9

26

6

8

20

10

9

4

5

18

8

6

7

21

II

6

7

3

16

7

8

6

21

12

7

9

6

22

9

10

7

26

13

8

4

5

17

7

6

8

21

14

5

6

7

18

6

8

9

23

15

4

5

6

15

7

6

8

21

16

7

8

9

24

6

9

6

21

17

6

3

15

9

5

9

23

18

7

5

7

19

7

9

10

26

19

10

30

8

6

7

21

20

7

9

25

10

9

28

HO :

!lD

= 11c ; there is no difference between the mean groups.

HA :

llo

>

11c ;

+

nd- 2

= 20

+

20-2

=

38

t ( 0,05) = 1,686

:EXc = 389

Xc = 19,45

Oc

= 5,0321

to= 3,0041

:EXo = 460

Xo = 23,00

Oo

=2,6157

3,0041

>

Tt = 1,686

(37)

APPENDIX ( 2 )

(38)

Native and Non-native

Teacher: A l\latter to

Think Over

ANDREA M. DE A. MATTOS

E\t·r ~itwt· I ~tarkd lt•arhin~ F.nglis.h as a fort•iJ,!n lmt~Wl.J.!t' in my c•ntmtry~

I

han~

lu't'll l'lliH't•ntrtl 'w't·ith tlw rnlP of tlu-

non-proft•.;..:innalnnli\o't- lt"ncltt•r of F.nglis.h as a

fun·i~n lnn~un~P (EFL). In nrazil a

non-Eu~ti~h "'P''akin~ t'ntmtry. this has l.Jeen a

11111~1 .. t•riou~ rnath•r for many English !'Whool .... pnrtit·ttlarl~· prhatP language in-!'1-titult• ... itwP hiring ~uod tf!achers

di-n•dl~· t•ontrihutt·~ to tht• .. un·ival of the

in-!-Oiitution.