BY

VEBYA LESTARI BAHRUN

105204003

ENGLISH DEPARTMENT

FACULTY OF LANGUAGES AND LITERATURE

STATE UNIVERSITY OF MAKASSAR

2012/1013

2010

A. INTRODUCTION

Construction a good language test is not so easy and simple as one may

imagine. It must fulfill certain criteria and not deviate from its aims. Language

testing is designed and performance of learners in the language they are learning.

This test is designed to pinpoint strengths and weakness in the learned

abilities of the students. The purpose of this test is for diagnosis and feedback.

Before making the test items, first of all, it is better to make test

specifications that provide the official statement about what the test tests and how

it tests it. Test specification is the part of test construction procedure and as the

blueprint to be followed by test/item writer.

The testees are the students of Junior High School in second class in SMP

Neg. I Soppeng Riaja. The test is conducted on Saturday, 5 June 2010. There are

20 testees.

This paper attempts to describe/interpret the best score. May this paper is

useful, even though it is still far from the perfection.

B. TEST SPESIFIFICATION

Table 1: Test Spesification

NO.Basic

CompetenceObjective/Indicators

DomainNumber

ItemsMethod of the Test

1 Listening

The students can listen and get the information that they have heard.

Cognitive 5Multiple choice

2 Speaking

The students can use the appropriate expression in communication.

Cognitive 5Multiple choice

3 Reading

The students can comprehend the text and get the information from the text.

Cognitive 5Multiple choice

4 Writing

The students can arrange the disorder words and sentences become a good arrangement.

Cognitive 5Multiple choice

5 Grammar

The students can use the suitable formula in constructing sentences

Cognitive 5Multiple choice

6 Vocabulary The students Cognitive 5 Multiple choice

can use their vocabulary in communication

C. TRY OUT

LEMBAR SOAL TRY- OUT

Mata Pelajaran : Bahasa Inggris

Kelas : XI (Sebelas)

Sekolah : SMAN 3 Takalar

PETUNJUK UMUM

1. Tulislah nama, kelas dan sekolah anda pada lembar jawaban yang

disediakan

2. Bacalah soal yang diberikan dengan teliti sebelum mengerjakannya

3. Pilihlah jawaban yang tepat dengan memberikan tanda (X) pada jawaban

benar a, b, c atau d di lembar jawaban yang disediakan.

4. Periksalah pekerjaan anda sebelum diserahkan kepada pengawas.

LISTENING SECTION

Listen to the dialogue, to answer the question number one until two)

1. Why does Vina look so gloomy?

a. She has an assignment from Mr.Clark

b. He want to meet Mr. Clark

c. She didn’t come to garden party last night

d. d. Mr. Clark is angry with her.

2. Eka suggest to vina to make an essay about her......

a. Student’s party c. Garden Party

b. Garden d. Kitchen Garden

Listen to the speaker to answer the question of number three to five)

My friend, Steve, is a diligent and smart student in his school. He gets up at

5.00 a.m. every morning. He always goes to school ...(3)..... than his friends. He goes

to school by public transport and he sometimes goes home by motorcycle with his

uncle. He’s never absent. After school he usually helps his parents. They have a ....

(4).... store beside their house. In the evening he studies his lesson and he does the

assignments given by his teacher. He seldom watches TV. He ....(5)...... ever goes to

bed late.

3. a. Lately c. Lastly

b. Early d. Earlier

4. a. Trousers c. Cloth

b. Shirt d. Skirt

5. a. Slowly c. Easily

b. Hardly slowly d. Fastly

SPEAKING SECTION

6. Shopkeeper: ............. to carry these books to your room, Sir?

Teacher : No, thanks, I can do it myself.

a. Do you want c. Can I help you

b. Shall I do d. Help me

7. Iwan  : I have to go to school, but look at the sky. It is very cloudy. I think it

is going

to rain.

Rani : ..........

a. You don’t take an umbrella.

b. You are going to take an umbrella.

c. I should take an umbrella.

d. You should take an umbrella.

6. Iwan : “My grandmother has passed away this morning.”

Anton : “………… I hope you are not sad.”

a. Oh, no! c. Poor you

b. How awful d. I’m sorry to hear that

9. Jane : What do you think about the food?

Madie : ................ . The food was so salty.

a. I love that taste

b. I like it

c. I am a little dissatisfied

d. I am satisfied with the food

10. Douhgter: Dad, May I go to the movie?

Dad : ............., You have to stay at home and finish your homework.

a. Yes, You can

b. Yes, Of course

c. No, you may not

d. Absolutely yes.

VOCABULARY SECTION

11. My uncle lives in town.

The synonym of the underline word is …

a. city c. village

b. country d. continent

12. Mother usually reminds us to put on our …... before going to bed.

a. pyjamas d. coat

b. suit e. sweater

13. A person whose age is between 13 – 19 years old is called a/an......

a. child d. teenager

b. adult e. grown up

14. In the market, we can find many sellers.

The antonym of the underline word is ….

a. shop c. market

b. buyer d. price

15. My father saves his money at the place where money is kept safely.

The underline phrase can be change by the word ….

a. hospital c. school

b. supermarket d. bank

GRAMMAR SECTION

16. Ardi and I … play tennis tomorrow.

a. am going to c. were going to

b. are going to d. is going to

17. When I … at his house, he still was sleeping.

a. arrive c. were arrived

b. arrived d. arrive

18. Alam : What is Lastri doing?

Bino : She ....... with Serly now.

a. swims c. was swimming

b. swim d. is swimming

19. I could not sleep last night … there was a loud noise in outside.

a. therefore c. although

b. before d. because

20. We ....... late for the movie last night. The movie ....... at 7, but we .......

a. was – started – will not

b. are – started – will not

c. were – started – didn’t arrive

d. weren’t – started – didn’t arrive

WRITING SECTION

21. (1) He is clever and diligent.

(2) One of his favorite games is playing hide and seek. Adi and his friends like

this game because it is a cheap one. They do not need any money to play

the game.

(3) When he has finished studying his lessons, he often plays some games

with his friends.

(4) Adi is a student.

The best paragraph is ….

a. 1-2-3-4 c. 4-3-2-1

b. 3-2-1-4 d. 4-1-3-2

22. to- come- will- she- wedding- not- the-party

1 2 3 4 5 6 7 8

The best arrangement of the word is ......

a. 7 – 8 – 3 – 6 – 1 – 5 – 4 – 2

b. 4 – 3 – 6 – 2 – 1 – 7 – 5 – 8

c. 7 – 5 – 8 – 6 – 3 – 2 – 1 – 4

d. 4 – 6 – 3 – 2 – 1 – 7 – 8 – 5

23. likes – the – very – game – much – he – playing

1 2 3 4 5 6 7

The correct sentence is ….

a. 6-7-2-4-1-3-5 c. 2-7-5-6-1-3-4

b. 2-4-1-6-7-3-5 d. 6-1-7-2-4-3-5

24. The following signs means that…

STAFF ONLY

a. The room is only for the staff

b. We should get in the room.

c. We can meet a director in this room

d. The staff cannot get in.

25. Choose the right order of the following sentences.

1. It takes place within three or more days.

2. The more important the person who dies, the more buffaloes they kill.

3. It is carried out in honour, of the dead person.

4. One of the Toraja traditions is the funeral.

5. It is done by slaughtering a water buffalo.

a. 2,5,4,3,1 c 4,3,1,5,2

b. 2,3,5,4,1 d. 4,1,5,3,2

READING SECTION

“Read this following text to answer questions numbers 26-30.

There are several animals that are usually looked after as pets. They are the birds,

cocks, cats, dogs, horses, etc. people like to keep them, not as breeding animals. Therefore

people do not need their meat or their eggs. People like pets because they can be nice and

funny.

A bird is one of the animals that are kept as a pet. People like birds because they can

sing beautifully. Birds can be expensive because of their song. The price of bird can be five

million rupiahs if it can be sing beautifully.

It is not difficult to get birds. We can buy them in the bird shop or pet shop. We can

choose whatever bird we want: pigeons, parrots, cecaks rowo, etc. we can also buy birds

cages in the bird shop.

26. The several animals that are people usually kept as a pet are …

a. birds c. butterfly

b. mouse d. wolf

27. Why do people like pets? Because …

a. they need their meet c. they can be nice and funny

b. they need their eggs d. they can be pleasant

28. The word they in the second sentence of the text refers to …

a. animals c. birds

b. people d. pets

29. How many price the bird which can be able to sing beautifully?

a. 2 million rupiahs c. 4 million rupiahs

b. 3 million rupiahs d. 5 million rupiahs

30. Where we can buy birds?

a. café c. restaurant

b. bird shop d. supermarket

D. ANALYSIS

1. Frequency Distribution

The following table contains the imaginary scores of a group of 23

students on a particular test consisting of 30 items. The table contains a

frequency distribution showing the number of students who obtained each

mark awarded; tallies, which are the strokes representing the number of

students obtaining the same scores; the frequency and the percentage of each

score.

TABLE 1: The Frequency Distribution of Scores

No. Raw Score

Final Score Tally Frequency Percentage

1 28 93,3 / 1 3.0

2 27 90 / 1 3

3 26 86,6 / 1 3.0

4 23 76,6 / 1 3,0

5 22 73,3 / 2 6.0

6 21 70 /////// 7 21

7 20 66,6 //// 6 18.0

8 19 63,3 /// 3 9.0

9 18 60 / 5 15

10 16 53,3 / 1 3.0

11 13 43.3 // 1 3.0

12 11 36,6 // 1 3.0

Total 30 30 100%

The distribution of the scores illustrated above can be presented in another way as in

the following frequency polygon:

FIGURE: The Distribution of the Scores

28 27 26 23 22 21 20 19 18 16 13 110

2

4

6

8

Students Score Frequency

2. Measures of Central Tendency

TABLE 2: The Frequency Distribution of

Scores

No X F Fx

1 7.3 1 7.3

2 6.7 1 6.7

3 6.6 1 6.6

4 6.0 2 12.0

5 5.7 4 22.8

6 5.3 5 26.5

7 5.0 1 5.0

8 4.7 1 4.7

9 4.3 2 8.6

10 4.0 2 8.0

Total 166 20 108.2

Mean:

X=∑ fx

N=

108 . 220

=5 . 41

3. Measures of Dispersion

TABLE 3: Standard Deviation

(d) = x – X

1. 7.3 – 5.42 = 1.89 8. 5.7 – 5.42 = 0.29 15.5.0 – 5.42 = -0.41

2. 6.7 – 5.42 = 1.29 9. 5.7 – 5.42 = 0.29 16.4.7 – 5.42 = -0.71

3. 6.6 – 5.42 = 1.19 10. 5.3 – 5.42 = -0.11 17.4.3– 5.42 = -1.11

4. 6.0 – 5.42 = 0.59 11. 5.3 – 5.42 = -0.11 18.4.3 – 5.42 = -1.11

5. 6.0 – 5.42 = 0.59 12. 5.3 – 5.42 = -0.11 19.4.0 – 5.42 = -1.41

No X d d2

1 7.3 1.89 3.5721

2 6.7 1.29 1.6641

3 6.6 1.19 1.4161

4 6.0 0.59 0.3481

5 6.0 0.59 0.3481

6 5.7 0.29 0.0841

7 5.7 0.29 0.0841

8 5.7 0.29 0.0841

9 5.7 0.29 0.0841

10 5.3 -0.11 0.0121

11 5.3 -0.11 0.0121

12 5.3 -0.11 0.0121

13 5.3 -0.11 0.0121

14 5.3 -0.11 0.0121

15 5.0 -0.41 0.1681

16 4.7 -0.71 0.5041

17 4.3 -1.11 1.2321

18 4.3 -1.11 1.2321

19 4.0 -1.41 1.9881

20 4.0 -1.41 1.9881

Total 108.2 0 14.858

6. 5.7 – 5.42 = 0.29 13. 5.3 – 5.42 = -0.11 20.4.0– 5.42 = -1.41

7. 5.7 – 5.42 = 0.29 14. 5.3 – 5.42 = -0.11

A. Range = 22 – 12

= 10

B. Standard Deviation

S . d=√∑ d 2

N

=√14 . 858

20

=√0 .7429

=0.86

4. Validity

r pbi=M p−M t

SD t √ pq

r pbi = Point bi-serial Correlation Coefficient, i.e. item validity coefficient

M p = Mean score of testees correctly answering the analyzed item

M t = Mean score of the total score

SDt = Standard deviation of the total score

p = Proportion of testees correctly answering the analyzed item

q = Proportion of testees incorrectly answering the analyzed item.

The following steps are recommended for calculation:

1) Determining the proportion of testes correctly answering the analyzed item:

p=∑ fx

N=

2020

=1

2) Calculating the mean score of total score:

Mt=32420

=16 . 2

3) Determining the proposition of testes correctly answering the analyzed item:

1−p=1−1=0

4) Calculating score of testes correctly answering the analyzed item:

Mp=¿22+20+19+18+18+17+17+17+17+16+16+ ¿16+16+16+15+14+13+13+12+12 ¿

20=32420

=16 , 2

5) Calculating the standard deviation of the total score SD=0,86

r pbi=16 .2−16 .2

0 ,86 √ 10

= 0

5. Reliability

The formula of Pearson product moment correlation is as follows:

r xy=N ∑ xy−(∑ x ) (∑ y )

√[ N∑ x2−(∑ x)2 ][ N∑ y2−(∑ y )2 ]r xy

= Pearson product moment correlation between variables x and y

N= number of students taking the test

∑ x= sum of variable x

∑ y= sum of variable y

∑ xy= sum of multiplication of variable x and variable y

∑ x2

= sum of square x

∑ y2= sum of square y

TABLE 3: Pearson Product Moment Correlation

No. X Y X2 Y2 XY

1 16 20 16 400 256 320

2 17 16 7 256 49 112

3 18 11 16 121 256 176

4 19 20 5 400 25 100

5 20 20 14 400 196 280

6 21 19 4 361 16 76

7 22 13 4 169 16 52

8 23 17 8 289 64 136

9 24 5 12 25 144 60

10 25 15 3 225 9 45

11 26 1 3 1 9 3

12 27 9 11 16 121 99

13 28 2 11 4 121 22

14 29 5 9 25 81 45

15 30 18 10 324 100 180

X= 191 Y= 133 X2= 3016 Y2= 1463 FX= 1706

r xy=34120−25403

√[ (60320−36481) ][ (29260−17689) ]

r xy=871716608 . 46377

=0 . 52

The result of this calculation is then analyzed using Spearman-Brown odd even model correlation to see the reliability of the test.

rtt=2r hh

1+rhh

rtt = Total test coefficient reliability (tt = total test)

rhh = Product moment Correlation Coefficient between the first half and the second half of the test (hh = half-half)

1 & 2 = constant numbers

rtt=2×0.521+0 . 52

=0 . 68

6. Item Difficulty and Item Discrimination

IF=UG+LGN

ID=UG−LGn

r xy=(20×1706 )−(191×133 )

√[ (20×3016) .(191)2 ] [(20×1463 ).(1332 ]

r xy=14117

√[ 23839 ]. [11571 ]

IF = index of facility;

ID = index of discrimination;

n = number of students in one group (½N);

UG = frequency of score by upper group (upper half); and

LG = frequency of score by lower group (lower half).

TABLE : The Indices of facility and

Discrimination

Item UG LG IF ID

1 10 10 1.00 0.00

2 8 8 0.80 0.00

3 4 7 0.55 -0.30

4 10 10 1.00 0.00

5 10 10 1.00 0.00

6 10 9 0.95 0.10

7 8 5 0.65 0.30

8 8 9 0.85 0.10

9 4 1 0.25 0.30

10 8 6 0.70 0.20

11 1 0 0.05 0.10

12 5 4 0.45 0.10

13 2 0 0.10 0.20

14 2 3 0.25 0.10

15 10 8 0.90 0.20

16 9 7 0.80 0.20

17 4 3 0.35 0.10

18 7 9 0.80 -0.20

19 3 2 0.25 0.10

20 6 6 0.60 0.00

21 3 1 0.20 0.20

22 3 1 0.20 0.20

23 6 2 0.40 0.40

24 7 5 0.60 0.20

25 3 0 0.15 0.30

26 2 1 0.15 0.10

27 8 3 0.55 0.50

28 5 6 0.55 -0.05

29 6 3 0.45 0.30

30 8 2 0.50 0.60

7. Full item analysis

1.

IF=UG+LGN

=10+1020

=1,0

ID=UG−LGn

=10−1010

=0.0

2.

IF=8+820

=0 .8

ID=0−010

=0.0

UG LG UG+LG

A* 10 10 20

B 0 0 0

C 0 0 0

D 0 0 0

Tot 10 10 20UG LG UG+LG

A 0 0 0

B 2 0 2

C* 8 8 16

D 0 2 2

Tot 10 10 20

3.

IF=4+720

=0 . 55

ID=4−710

=−0 . 30

4.

IF=10+1020

=1 .00

ID=10−1010

=0 .00

5

UG LG UG+LG

A* 4 7 11

B 0 0 0

C 2 1 3

D 4 2 6

Tot 10 10 20

UG LG UG+LG

A 0 0 0

B 0 0 0

C 0 0 0

D* 10 10 20

Tot 10 10 20

6.

IF=10+920

=0. 95

ID=10−910

=0 . 10

7.

8

IF=8+920

=0 .85

ID=8−910

=0. 10

UG LG UG+LG

A 0 0 0

B* 10 10 20

C 0 0 0

D 0 0 0

Tot 10 10 20

UG LG UG+LG

A 0 1 1

B* 10 9 19

C 0 0 0

D 0 0 0

Tot 10 10 20

UG LG UG+LG

A 0 1 1

B* 8 5 13

C 2 1 3

D 0 3 3

Tot 10 10 20

UG LG UG+LG

A 2 1 3

B* 8 9 17

C 0 0 0

D 0 0 0

Tot 10 10 20

IF=8+520

=0.65

ID=8−510

=0 .30

9.

IF=4+120

=0 .25

ID=4−110

=0 .30

10.

IF=10+820

=0,9

ID=10−810

=0,2

11

IF=1+020

=0 , . 0

ID=1−010

=0 .10

UG LG UG+LG

A 6 8 14

B 0 0 0

C 0 1 1

D* 4 1 5

Tot 10 10 20UG LG UG+LG

A* 8 7 15

B 2 2 4

C 0 1 1

D 0 0 0

Tot 10 10 20

UG LG UG+LG

A 0 1 1

B 10 5 15

C* 1 0 1

D 0 3 3

Tot 10 10 20

12.

IF=5+420

=0 . 45

ID=5−410

=0 .10

13.

IF=2+020

=0 .10

ID=2−010

=0 . 20

14.

IF=2+320

=0 .25

ID=2−310

=0 .10

UG LG UG+LG

A 0 0 0

B* 5 4 9

C 4 5 9

D 1 1 2

Tot 10 10 20

UG LG UG+LG

A 8 10 18

B* 2 0 2

C 0 0 0

D 0 0 0

Tot 10 10 20

UG LG UG+LG

A 6 2 8

B 1 3 4

C 1 2 3

D* 2 3 5

Tot 10 10 20

15.

IF=10+820

=0.90

ID=10−810

=0 .20

16.

IF=9+720

=0 .8

ID=9−710

=0.2

17.

IF=4+220

=0 .4

ID=4−210

=0 .2

UG LG UG+LG

A* 10 8 18

B 0 0 0

C 0 1 1

D 0 1 1

Tot 10 10 20

UG LG UG+LG

A 1 2 3

B 0 1 1

C* 9 7 16

D 0 0 0

Tot 10 10 20

UG LG UG+LG

A 0 3 3

B 4 4 8

C* 4 2 6

D 2 1 3

Tot 10 10 20

18.

IF=7+920

=0 .8

ID=7−910

=−0. 2

19.

IF=3+220

=0 .25

ID=3−210

=0 .1

20.

IF=6+620

=0 .6

ID=6−610

=0. 0

21.

UG LG UG+LG

A 1 1 2

B 2 0 2

C 0 0 0

D* 7 9 16

Tot 10 10 20

UG LG UG+LG

A 1 0 1

B 5 5 10

C* 3 2 5

D 1 3 4

Tot 10 10 20

UG LG UG+LG

A 3 1 4

B 1 2 3

C 0 1 1

D* 6 6 12

Tot 10 10 20

IF=3+120

=0 .2

ID=3−110

=0 .2

22.

IF=3+120

=0 .2

ID=3−110

=0 .2

23.

IF=6+220

=0 .4

ID=6−210

=0 .4

24.

UG LG UG+LG

A 7 4 11

B* 3 1 4

C 0 4 4

D 0 1 1

Tot 10 10 20

UG LG UG+LG

A* 3 1 4

B 2 5 7

C 3 3 6

D 2 1 3

Tot 10 10 20

UG LG UG+LG

A 1 7 8

B 0 0 0

C* 6 2 8

D 3 1 4

Tot 10 10 20

25.

IF=4+020

=0 . 2

ID=4−010

=0 . 4

26.

IF=2+120

=0 .15

ID=2−110

=0 .1

27.

UG LG UG+LG

A 0 0 0

B 0 5 5

C 3 0 3

D* 7 5 12

Tot 10 10 20

UG LG UG+LG

A 0 1 1

B 6 6 12

C 0 3 3

D* 4 0 4

Tot 10 10 20

UG LG UG+LG

A* 2 1 3

B 1 1 2

C 6 8 14

D 1 0 1

Tot 10 10 20

IF=7+520

=0.6

ID=7−510

=0 . 2

28.

IF=2+020

=0 .1

ID=2−010

=0 .2

29.

IF=7+320

=0.5

ID=7−310

=0 .4

30.

UG LG UG+LG

A 0 3 3

B 2 4 6

C 0 3 3

D* 8 0 8

Tot 10 10 20

UG LG UG+LG

A 3 2 5

B 0 0 0

C 5 8 13

D* 2 0 2

Tot 10 10 20

UG LG UG+LG

A 1 5 6

B* 7 3 10

C 2 2 4

D 0 0 0

Tot 10 10 20

IF=8+020

=0 . 4

ID=8−010

=0. 8

IF=0+220

=0 .1

ID=0−210

=−0 .2

E. CONCLUSION

The test was conducting in SMP Negeri 3 Watampone Kab. Bone. The testees

are the students in the second grade. The total of testees is 30 students.

From the test which held on Saturday, 9 Desember 2012, the writer can

conclude that the test made by tester seems difficult toward testees. We can

indicate such fact from the data of IF and ID. From the test we can see that the

UG LG UG+LG

A 2 4 6

B* 0 2 3

C 7 3 10

D 0 1 1

Tot 10 10 20

students are low in reading section. It can be showed in test validity where

number 11 just can be answered by one student.

The reliability is low since its number reached 0, 68. The method employed is

split half method that divides testees into two groups. The other weakness is about

the quality of distractors that are very low.

May in the next test everything will be better.