ELT:Try Out for First Grade of SMA Negeri 1 Pangkajene

ENGLISH LANGUAGE TESTINGTRY OUT FOR THE FIRST GRADE

STUDENTSOF SMA NEGERI 2 PANGKAJENE

BY:

MUTIAH NUR ADZRA

1252042006

ENGLISH EDUCATION D

FACULTY OF LANGUAGE AND LITERATURE

STATE UNIVERSITY OFMAKASSAR

2014TABLE OF CONTENTS

TABLE OF CONTENTS.........................................i

CHAPTER I INTRODUCTION....................................1

A. Background........................................1

B. The Constructed Tests.............................1

C. Answer Keys.......................................1

D. Scoring Rubric....................................1

CHAPTER II TEST SPECIFICATION.............................1

A. The Constructed Tests.............................1

B. Answer Keys.......................................1

C. Scoring Rubric....................................1

CHAPTER III ANALYSIS......................................1

A. Data Presentation.................................1

B. Measures of Central Tendency......................1

C. Standard Deviation................................1

D. Reliability.......................................1.

E. Validity..........................................1

F. Difficulty Level..................................1

G. Discrimination Power..............................1

H. Distractor Power..................................1

CHAPTER IV CONCLUSION.....................................1

APPENDIX..................................................1

A. Surat Izin........................................1

B. Samples of Students’ Answer Sheets................1

C. Curriculum/Syllabus...............................1

D. Surat Keterangan Telah Melakukan Ujicoba Tes......1

E. Curriculum Vitae..................................1

CHAPTER I

INTRODUCTION

A. Background

Language testing is designed to measure the actual

competence and performance of the learners in the

language they are learning. The actual question is what

is to be included in the language test. This is often

difficult because a mastery of language skill is

assessed rather that\n areas of knowledge or content.

The construction of a language test is even more

difficult if the coverage of the syllabus is not well

defined.

All forms of language testing are one of

measurement. Tests of language abilities may be

inaccurate or unreliable in the sense that repeated

measures may give different results. These measures may

also be invalid in the sense that the other abilities

are mixed in our test of reading comprehension on

closes examination may turn out be mixed in test, to be

useful, must provide us with reliable and valid

measurement for a variety of purpose.

B. Coverage

The test should be covered with four skills in

language and two elements of language. The form of the

test is multiple choices which have ten items in

vocabulary, ten items in grammar, and ten items in

reading.. Multiple choice makes easy to analyze the

test in order to know the difficulty level, index of

discrimination, and full item analysis of the test.

C. Place of Tryout

The test was given toward 21 students of the 10th

grade of SMA NEGERI 2 PANGKAJENE. 24th of May 2014. I

choose this school because it was high school that one

of my friends used to be a student. The result of the

test will be discussed in the next chapter.

CHAPTER II

TEST SPESIFICATION

(Blueprint)

A. Test Specification

Subject : English

Semester/Class : II / X

School : SMA Negeri 2 Pangkajene

Test Techniques : Multiple Choice

Test Type : Written Test

Test Description :

Table.1 Test Specification

No.

BasicCompeten

ce

Objective/Indicators Domain Techniqu

e Items

1Vocabula

ry

The studentscan usetheirvocabularyincommunication.

C1. C2Multiplechoice 10

2

Grammar

The studentscan use thesuitableformula inconstructingsentences.

C1. C2Multiplechoice 10

3 Reading

The studentscancomprehendthe text andget theinformationfrom thetext.

C1.C2Multiplechoice 10

B. The Constructed Test

Subject : EnglishUnit of Education : Senior High SchoolClass/Semester : X/IIDuration : 50 minutes

GENERAL INSTRUCTION1. Write your name, class, and school on the place

provided2. Check and read the questions carefully before you

work3. Use time effectively and efficiently4. Please check your answer before submitted to the

supervisory

NAME : CLASS :SCHOOL :Answer the following questions by circle the letter of the correct option !Vocabulary Section1. Mr. Hadi is a columnist he often sends his article

to the newspaper.

The synonym of the underlined word is ….a. letter c. music sectionb. advertisement d. writing material

2. …… is someone who writes for a newspaper.a. Editorb. Journalistc. Novelist d. Reporter

3. There are many kinds of interesting …. on thetelevision.a. channelsb. announcersc. programmesd. tone

4. We need……… to make a call at the public telephone.a. coins b. money c. keys d. telephone book .

Fill in the missing words to the following passage .When Hari Raya Idul Fitri comes, people are going

to the mosque doing pray. After that, they will (5) …..for forgiveness from their elders. This is on importantcustom of the muslims.5. a. say

b. askc. telld.talk

The following text is for questions 6 to 7Text 1

Koes Plus was well known as pop music group in theseventies until the eighties. At that time most Indonesia people were ____(6) about their songs because

they were nice and simple. Every radio station broadcasted these songs and always put them in the top of pop music. Their music shows were always full of _______ (7).6. a. fanatical

b. dissatisfiedc. crazyd. disappointed

7. a. followersb. spectatorsc. participantsd. speakers


A tiger once caught a fox while hunting for food.The fox was very bold. “I am the king of the forest,”he said. But the tiger grew ____ (8) and said that hewould eat the fox at once. “If you don’t believe me,come for a walk with me, “answered the fox quitecalmly. “You’ll soon see whether all the other animalsare afraid of me or not. “Tiger agreed to go with thefox. _____ (9) all the animals saw them coming, ____(10) ran away as fast as they could. The tiger neverfound out that animals were really frightened of himand the fox.8. a. righten

b.quite friendlyc. very angryd.very strong

9. a. whenb.soc.becaused.before

10. a. he

b. shec. itd.they

Grammar Section11. Please be quite. I ... to concentrate.

a. triedb. am tryingc. was tryingd. am to try

12. All these years, Ina’s family ... in proverty.a. livedb. has been livingc. had been livingd. was living

13. Fatimah : Grand ma, Can I help you to bring thisflowers?Grand ma : …...... my grand child, That’s very kind of you.a. All rightb. No,thanksc. Thank youd. Fine

14. Rahmat : Do you feel like coming to my house next Sunday?Hasan : …….. .Thanks for inviting me.a. Sorryb. I am displeasedc. I’d love tod. I am happy

15. Mr. Dhani composed a song for the new singer.A song … by him for the new singer.

a. is composedb. was composedc. had been composed

d. were composed16. Father said to me, “Do you close the windows at

night?”The indirect form is: Father asked me …. At nighta. if I closed the windows b. whether you closed the windowsc. that I closed the windowsd. when I closed the windows

17. Ana : “They say you will be pointed as the best student this year”Rida : “Are you sure?”

Ana : “Congratulation for your success, Rida” Rida : “Thank you, Ana”

The underlined sentence above is about the expression of … a. congratulation b. agreement

c. sympathy d. hope

18. Yusuf : Do you feel like coming to my birthday party next weekend?Septi : I’d love to but I’m afraid I

can’t. I have to take my parents to my brother’s house in Jakarta.

What does Septi mean with her statement?a. She object to the invitation b. She wants to come to the partyc. She approves of the invitationd. She declines the invitation

19. Mother : Have you washed the dishes?Rini : Sorry, I can’t hear you Mom?

Mother : I asked you ....... the dishes.a. have you washed

b. had you washedc. if you had washedd. whether you can wash

Reading Section20. Amri : Can you come to my house for English

discussion tonight?Hadi : That’s sound good. I’d love to

The underlined sentence expresses … a. agreement

b. Abilityc. possibilityd. invitation


AnnouncementMany customers have received emails claiming to be

from Buwana Bank. These emails were not genuine. These emails ask you to reply, giving your :

Account number Password Credit card number Date of birth Address

Buwana Bank does not send emails asking for this information. If you ever received an email asking for this information, please do not reply.

Send the email immediately to [email protected]. If you have given anyone your details, call our helpline immediately on 0885 2000 117.

Be warned. Criminals are trying to access your accounts21. What is the announcement about?

a. It is a caution for customers.

mailto:[email protected]

b. It is about a new bank product.c. It is an offer from Buwana Bank.d. It is a claim about Buwana Bank.

22. Where should you address the emails asking for your Account number?

a. Buwana Bankb. Criminal courtc. 0885 2000 117d. [email protected]

23. “Be warned. Criminals are trying to access your accounts.” (The last sentence)The above sentence has the same meaning as ...

a. Attention! Criminals have place in our bank.b. Be careful! Do not let criminals know your

account numberc. Do not target. Open your new bank account at

our bank nowd. Remember! Our bank is safe. Criminals cannot

get into your accountThe following text is for questions 24 to 28Text 4

Jakarta: The spreading of Lapindo hot mudflow postexplosion of Pertamina’s gas pipe forced the residentof Kedungbendo and Renokenongo villages to evacuate.

The mudflow also flooded half of the TangggulanginAnggun Sejahtera housing complex in Sidoarjo.Mudflow of more than half a meter in depth flooded66,448 houses in this complex.

“The residents were evacuated to Pasar Baru, whichwas previously a shelter for 2,605 families whosehouses were flooded by the mudflows”, said SyaifulIllah, Deputy Regent of Sidoarjo, yesterday.

From Tempo’s monitoring, the refugees camecontinuously by trucks, public cars and motorcycles.

mailto:[email protected]

They straightway occupied the market, kiosks and stallsby spreading out mats and pillows for sleeping.

Government officials appeared to be busy preparingmass kitchen and supplying rice, instant noodles,cooking oil and other needs. “We took the food from theSidoarjo Social Service. There hasn’t been any aidprovided by Lapindo”, said Syaiful illah.

In the mean time, the National Team for MudflowRelief in Sidoarjo has not been able to block mudflowsin the gas pipe explosion.24. The type of text above is a/ an …

a. procedure c. recounte. narrative

b. news item d. descriptive25. The text as a whole reports …

a. Tanggulangin Anggun Sejahtera housingb. The victims of hot mudflow c. The damage of hot mudflow d. The evacuation of Kedungbendo and renokenongo’s

resident26. Which statement is TRUE according to the text?

a. Pasar Baru is a place where the resident evacuatedb. Mudflow flooded Tanggulangin Anggun Sejahtera c. The refugees came continuously by motorcycles.d. Mudflow can be blocked by Lapindo’s team

27. They straightway occupied the market, kiosks andstalls (paragraph 4). The word “They” refers to …..a. housesb. public carsc. marketsd. the refugees

28. What is the purpose of the text above?a. To inform readers about the resident evacuation b. To describe about hot mudflow

c. To amuse readers about the story of Lapindod. To explain about hot mudflow

The following text is for questions 29 to 30Text 5How to make a Ballon Powered Rocket

You will need a ballon, sticky tape, sensors, string, aplastic drinking straw.

1. Thread the string carefully through the drinking straw.

2. The one end of the string to an object (tree, doorhandle, post, and so on)

3. The other end of the string to something ten meters away making sure that the string is tight.

4. Cut two pieces of sticky tape.5. Gently blow a little air into the ballon.6. Hold the end of the ballon tightly so the air does

not escape.7. Tape the ballon firmly to the straw.8. Blow more air into the ballon and again hold the

end tighly.9. Quickly release the end of the ballon and watch it

travel along the string.

29. What is the genre of the text?a. Procedureb. Reportc. Expositiond. Narrative

30. What is the communicative purpose of the text?a. To tell the story of a ballon b. To retell the eventc. To explain the steps how to make a ballon power

rocketd. To inform the reader how to do something

C. Answer KeyI. VOCABULARY III. READING

1.D 6. C 21. A26. A

2.B 7. B 22. D27. D

3.C 8. C 23. B28. A

4.A 9. A 24. B29. A

5.B 10. D 25. D30. C

II. GRAMMAR11. B 16. A12. B 17. A13. C 18. D14. C 19. C 15. B 20. D

D. Scoring Rubric

In scoring rubrics, the items measure as follows:

1. The correct answer will be scored 1

2. The error answer will be scored 0

3. No answer will be scored 0

CHAPTER III

RESULT OF ANALYSIS

Interpreting Test Score

List of Students’ Score

List of the score below is the result of language

testing of the first grade of SMA Negeri 2 Pangkajene.

Table 2: List of the Students’ Score

No. Name Score

1 Amalia Fidzah Adhari 27

2 Humairah 22

3 Ainun Nur Zakina 22

4 Alifia Fitrah Ridhayani 21

5 Asriana 20

6 Citra Handayani 20

7 Baso Amirullah 18

8 St. Nurkaya Basri 16

9 Ade Irna Novita Sari 16

10 Ameliyah Nur Rasyidah 15

11 Devi Tirta Sari 15

12 Abdul Rahman 15

13 St. Ayunka Ainayyah 14

14 Khairunnisa. R 13

15 Wulandari 13

16 Ulfadillah 12

17 Sukmawati 12

18 Abdul Rahman 11

19 Widya Salsabila Amri 11

20 Fadel Muhadi 10

21 Irnawati 3

ANALYSIS

A. Frequency DistributionThe following table contains the imaginary

scores of a group of 21 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 3.Frequency Distribution of Scores

No. Raw Score Final Tally Frequency Percentage

Score (%)1. 27 90 / 1 4.82. 22 73 // 2 9.53. 21 70 / 1 4.84. 20 67 // 2 9.55. 18 60 / 1 4.86. 16 53 // 2 9.57. 15 50 /// 3 14.28. 14 47 / 1 4.89. 13 43 // 2 9.510. 12 40 // 2 9.511. 11 37 // 2 9.512. 10 33 / 1 4.813. 3 10 / 1 4.8 Total 21 21 100

The distribution of the scores illustrated above

can be presented in another way as in the following

frequency polygon:

FIGURE 1: The Distribution of the Scores

Students’ Score Frequency

3 10 11 12 13 14 15 16 18 20 21 22 270

0.5

1

1.5

2

2.5

3

3.5

Tally

a. Mean

The mean score is the arithmetical average.

Table 4: The Frequency Distribution of Scores

No. X F Fx

1 9.0 1 9.0

2 7.3 2 14.6

3 7.0 1 7.0

4 6.7 2 13.4

5 6.0 1 6.0

6 5.3 2 10.6

7 5.0 3 15.0

8 4.7 1 4.7

9 4.3 2 8.6

10 4.0 2 8.0

11 3.7 2 7.4

12 3.3 1 3.3

13 1.0 1 1.0

Tot

al67.3 21 fx = ∑ 108.6

Mean:

X=∑fxN

=108.621

=5.17

B. Measures of Dispersiona. Range

One simple way of measuring the spread of marks

is based on the difference between the highest and

the lowest scores. Thus, if the highest score on a

30 items is 9.0 and the lowest is 1.0, the range

from 9.0 to 1.0 is 8.0; (9.0 – 1.0 = 8.0).

b. Standard Deviation

It shows how all the scores are spread out

and thus gives a fuller description of the test

scores

than the

range.

Table 5. Standard

Deviation

No. X d (x-mean) d2

1 9.0 3.83 14.6689

2 7.3 2.13 4.5369

3 7.3 2.13 4.5369

4 7.0 1.83 3.3489

5 6.7 1.53 2.3409

6 6.7 1.53 2.3409

7 6.0 0.83 0.6889

8 5.3 0.13 0.0169

9 5.3 0.13 0.0169

10 5.0 -0.17 -0.0289

11 5.0 -0.17 -0.0289

12 5.0 -0.17 -0.0289

13 4.7 -0.47 -0.2209

14 4.3 -0.87 -0.7569

15 4.3 -0.87 -0.7569

16 4.0 -1.17 -1.3689

17 4.0 -1.17 -1.3689

18 3.7 -1.47 -2.1609

19 3.7 -1.47 -2.1609

20 3.3 -1.87 -3.4969

21 1.0 -4.17 -17.3889

Total 108.6 0.00 Σd2 =

38.280

Standard deviation:x = students final score

d=x−x=9.0−5.17=3.83d2=3.832

¿14.6689

∑d2=38.280

s.d=√∑d2N

s.d=√38.28021s.d=√1.8228s.d=1.35

C. Validity & Reliabilitya. Validity

In terms of the test validity, we can show

the tests to the colleagues for face validity,

compare the course objective and the test items

for content validity, check whether the students

respond in the way they are expected in doing the

test for response validity, and calculate the

point bi-serial correlation for item validity

using the following formula:

rpbi=Mp−MtSDt √pq

(Sudjiono, 2003:185)

rpbi = Point bi-serial Correlation Coefficient,

i.e. item validity coefficient

Mp = Mean scores of testees correctly answering

the analyzed item

Mt = Mean scores of the total scores

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.

Item 1

q=1 – p

= 1 – 0.52

= 0.48

Mt= Raw x

Jumlah siswa

= 326

21

= 15.52

Mp= 27+22+21+20+16+15+14+13+12+11+1111

= 18211

= 16.54

p=∑ xN

=1121

=0.52

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 16.54−15.52

1.35×√0.520.48

= 1.021.35×1.04

= 0.78

Item 2

q=1 – p

= 1 – 0.80

= 0.20

Mt= 15.52

Mp=27+22+22+21+20+20+18+16+16+15+15+15+13+13+12+11+3

17

= 27917

= 16.41

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 16.41−15.52

1.35×√0.800.20

= 0.891.35×2

p=∑ xN

=1721

=0.80

= 1.31

Item 3

q=1 – p

= 1 – 0.61

= 0.39

Mt= 15.52

Mp= 27+22+22+21+20+20+18+16+15+15+13+13+1013

= 23213

= 17.84

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 17.84−15.52

1.35×√0.610.39

= 2.321.35×1.25

= 2.14

Item 4

q=1 – p

= 1 – 0.80

= 0.20

p=∑ xN

=1321

=0.61

p=∑ xN

=1721

=0.80

Mt= 15.52

Mp=27+22+22+21+20+20+18+16+16+15+15+15+14+13+12+11+11

17

= 28817

= 16.94

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 16.94−15.52

1.35×√0.800.20

= 1.421.35×2

= 2.10

Item 5

q=1 – p

= 1 – 0.47

= 0.53

Mt= 15.52

Mp= 27+22+22+16+16+15+15+15+13+1110

= 17210

= 17.2

p=∑ xN

=1021

=0.47

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 17.2−15.52

1.35×√0.530.47

= 1.681.35×1.06

= 1.31

Item 6

q=1 – p

= 1 – 0.04

= 0.96

Mt= 15.52

Mp= 111= 11

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 11−15.52

1.35×√0.040.47

= 1.681.35×0.29

= 0.36

Item 7

p=∑ xN

=121

=0.04

p=∑ xN

=021

=0.00

q=1 – p

= 1 – 0.00

= 1

Mt= 15.52

Mp= 00= 0

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 0−15.52

1.35×√01

= −15.521.35

×0

= 0

Item 8

q=1 – p

= 1 – 0.38

= 0.62

Mt= 15.52

Mp= 27+22+18+15+15+13+13+118

= 1348

p=∑ xN

=821

=0.38

= 16.75

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 16.75−15.52

1.35×√0.380.62

= 1.531.35×0.78

= 0.88

Item 9

q=1 – p

= 1 – 0.33

= 0.67

Mt= 15.52

Mp= 27+22+22+21+20+20+167

= 1487

= 21.14

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 21.14−15.52

1.35×√0.330.67

= 5.621.35×0.70

p=∑ xN

=721

=0.33

= 2.91

Item 10

q=1 – p

= 1 – 0.71

= 0.29

Mt= 15.52

Mp=27+22+22+21+20+20+18+16+16+15+15+14+13+11+10

15

= 26015

= 17.33

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 17.33−15.52

1.35×√0.710.29

= 2.111.35×1.56

= 2.43

Item 11

q=1 – p

= 1 – 0.52

p=∑ xN

=1521

=0.71

p=∑ xN

=1121

=0.52

= 0.48

Mt= 15.52

Mp= 27+22+22+20+18+16+15+15+13+11+1011

= 18911

= 17.18

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 17.18−15.52

1.35×√0.520.48

= 1.661.35×1.04

= 1.27

Item 12

q=1 – p

= 1 – 0.57

= 0.43

Mt= 15.52

Mp= 27+22+18+15+15+14+13+13+12+12+11+1012

= 18212

= 15.16

p=∑ xN

=1221

=0.57

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 15.16−15.52

1.35×√0.570.43

= −0.361.35

×1.15

= -0.30

Item 13

q=1 – p

= 1 – 0.42

= 0.58

Mt= 15.52

Mp= 27+22+21+20+18+15+14+12+129

= 1619

= 17.88

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 17.88−15.52

1.35×√0.420.58

= 2.361.35×0.85

= 1.48

p=∑ xN

=921

=0.42

Item 14

q=1 – p

= 1 – 0.57

= 0.43

Mt= 15.52

Mp= 27+22+22+21+20+20+18+15+15+14+13+1012

= 21712

= 18.08

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 18.08−15.52

1.35×√0.570.43

= 2.561.35×1.15

= 2.18

Item 15

q=1 – p

= 1 – 0.42

= 0.58

Mt= 15.52

p=∑ xN

=1221

=0.57

p=∑ xN

=921

=0.42

Mp= 27+22+21+20+20+16+15+12+119

= 1649

= 18.22

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 18.22−15.52

1.35×√0.420.58

= 2.71.35

×0.85

= 1.7

Item 16

q=1 – p

= 1 – 0.23

= 0.77

Mt= 15.52

Mp= 27+22+20+20+155

= 1045

= 20.8

s.d = 1.35

rpbi = Mp−MtSDt

×√pq

p=∑ xN

=521

=0.23

= 20.8−15.521.35

×√0.230.77

= 5.281.35×0.54

= 2.11

Item 17

q=1 – p

= 1 – 0.66

= 0.34

Mt= 15.52

Mp= 22+22+21+20+20+18+16+16+15+14+13+13+12+1114

= 23314

= 16.64

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 16.64−15.52

1.35×√0.660.34

= 1.121.35×1.39

= 1.15

p=∑ xN

=1421

=0.66

Item 18

q=1 – p

= 1 – 0.52

= 0.48

Mt= 15.52

Mp= 27+22+22+21+20+18+15+15+15+13+1211

= 20011

= 18.18

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 18.18−15.52

1.35×√0.520.48

= 2.661.35×1.04

= 2.04

Item 19

q=1 – p

= 1 – 0.23

= 0.77

Mt= 15.52

p=∑ xN

=1121

=0.52

p=∑ xN

=521

=0.23

Mp= 27+20+13+13+125

= 855

= 17

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 17−15.52

1.35×√0.230.77

= 1.481.35×0.54

= 0.59

Item 20

q=1 – p

= 1 – 0.66

= 0.34

Mt= 15.52

Mp= 27+22+22+21+20+20+18+16+16+15+15+14+13+1114

= 25014

= 17.85

s.d = 1.35

rpbi = Mp−MtSDt

×√pq

p=∑ xN

=1421

=0.66

= 17.85−15.521.35

×√0.660.34

= 2.331.35×1.39

= 2.39

Item 21

q=1 – p

= 1 – 0.52

= 0.48

Mt= 15.52

Mp= 27+22+22+21+20+16+15+15+13+12+1111

= 19411

= 17.63

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 17.63−15.52

1.35×√0.520.48

= 2.111.35×1.04

= 1.62

Item 22

p=∑ xN

=1121

=0.52

p=∑ xN

=1521

=0.71

q=1 – p

= 1 – 0.71

= 0.29

Mt= 15.52

Mp=27+22+22+21+20+20+18+16+16+15+15+14+12+11+10

15

= 25915

= 17.26

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 17.26−15.52

1.35×√0.710.29

= 1.741.35×1.56

= 2.01

Item 23

q=1 – p

= 1 – 0.57

= 0.43

Mt= 15.52

Mp= 27+22+22+21+20+20+18+16+16+15+15+1412

p=∑ xN

=1221

=0.57

= 22612

= 18.83

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 18.83−15.52

1.35×√0.570.43

= 3.311.35×1.15

= 2.81

Item 24

q=1 – p

= 1 – 0.33

= 0.67

Mt= 15.52

Mp= 27+22+21+20+16+15+127

= 1337

= 19

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 19−15.52

1.35×√0.330.67

p=∑ xN

=721

=0.33

= 3.481.35×0.70

= 1.80

Item 25

q=1 – p

= 1 – 0.57

= 0.43

Mt= 15.52

Mp= 27+21+20+16+15+15+14+13+12+11+11+312

= 17812

= 14.83

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 14.83−15.52

1.35×√0.570.43

= −0.691.35

×1.15

= -0.58

Item 26

q=1 – p

= 1 – 0.38

p=∑ xN

=1221

=0.57

p=∑ xN

=821

=0.38

= 0.62

Mt= 15.52

Mp= 27+22+20+20+18+12+12+118

= 1428

= 17.75

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 17.75−15.52

1.35×√0.380.62

= 2.231.35×0.78

= 1.28

Item 27

q=1 – p

= 1 – 0.52

= 0.48

Mt= 15.52

Mp= 27+22+22+21+20+18+16+16+15+12+1011

= 19911

= 18.09

p=∑ xN

=1121

=0.52

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 18.09−15.52

1.35×√0.520.48

= 2.571.35×1.04

= 1.97

Item 28

q=1 – p

= 1 – 0.57

= 0.43

Mt= 15.52

Mp= 27+22+22+21+20+20+16+16+14+12+12+1112

= 21312

= 17.75

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 17.75−15.52

1.35×√0.570.43

= 2.231.35×1.15

= 1.89

p=∑ xN

=1221

=0.57

Item 29

q=1 – p

= 1 – 0.90

= 0.10

Mt= 15.52

Mp=27+22+22+21+20+20+18+16+16+15+15+15+14+13+13+12+12+11+10

19

= 31219

= 16.42

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 16.42−15.52

1.35×√0.900.10

= 0.91.35

×3

= 2.01

Item 30

q=1 – p

= 1 – 0.85

= 0.15

p=∑ xN

=1921

=0.90

p=∑ xN

=1821

=0.85

Mt= 15.52

Mp=27+22+22+21+20+20+18+16+16+15+14+13+13+12+12+11+10+3

18

= 25518

= 14.16

s.d = 1.35

rpbi = Mp−MtSDt

×√pq= 14.16−15.52

1.35×√0.850.15

= −1.361.35

×2.38

= -2.39

Table 6. Test Validity

Valid: 0.33-1No. Mp Mt SD P Q rpbi Quality1 16.54 15.52 1.35 0.52 0.48 0.78 VALID2 16.41 15.52 1.35 0.80 0.20 1.31 INVALID3 17.84 15.52 1.35 0.61 0.39 2.14 INVALID4 16.94 15.52 1.35 0.80 0.20 2.10 INVALID5 17.2 15.52 1.35 0.47 0.53 1.31 INVALID6 11 15.52 1.35 0.04 0.96 0.36 VALID7 0 15.52 1.35 0 1 0 INVALID8 16.75 15.52 1.35 0.38 0.62 0.88 VALID9 21.14 15.52 1.35 0.33 0.67 2.91 INVALID10 17.33 15.52 1.35 0.71 0.29 2.43 INVALID11 17.18 15.52 1.35 0.52 0.48 1.27 INVALID12 15.16 15.52 1.35 0.57 0.43 -0.30 INVALID

13 17.88 15.52 1.35 0.42 0.58 1.48 INVALID14 18.08 15.52 1.35 0.57 0.43 2.18 INVALID15 18.22 15.52 1.35 0.42 0.58 1.7 INVALID16 20.18 15.52 1.35 0.23 0.77 2.11 INVALID17 16.64 15.52 1.35 0.66 0.34 1.15 INVALID18 18.18 15.52 1.35 0.52 0.48 2.04 INVALID19 17 15.52 1.35 0.23 0.77 0.59 VALID20 17.85 15.52 1.35 0.66 0.34 2.39 INVALID21 17.63 15.52 1.35 0.52 0.48 1.62 INVALID22 17.26 15.52 1.35 0.71 0.29 2.01 INVALID23 18.83 15.52 1.35 0.57 0.43 2.81 INVALID24 19 15.52 1.35 0.33 0.67 1.80 INVALID25 14.83 15.52 1.35 0.57 0.43 -0.58 INVALID26 17.75 15.52 1.35 0.38 0.62 1.28 INVALID27 18.09 15.52 1.35 0.52 0.48 1.97 INVALID28 17.75 15.52 1.35 0.57 0.43 1.89 INVALID29 14.16 15.52 1.35 0.90 0.10 2.01 INVALID30 16.42 15.52 1.35 0.85 0.15 -2.39 INVALID

b. Reliability

Reliability has to do with accuracy of

measurements. This kind of accuracy is reflected

in the obtaining of similar results when

measurement is repeated on the different occasion

or different instrument or by different person. In

terms of test reliability, we can use single test

single triad method with split half reliability;

applying Person Product Moment correlation and Spearman-

Brown odd even model correlation. This calculation may

be processed through SPSS program, based on the

level of significant of 5. The formula Person

Product Moment correlation is as follows:

rxy=N∑xy−(∑x) (∑y )

√{N∑x2−(∑x )2}{N∑y2−(∑y )2}

(Surapranata,

2004:58)

rxy = Person product moment correlation between

variable x and y

N = Number of students taking 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 7.Pearson Product Moment Correlation (Split

Half)

No. x y x2 y2 xy1 9 6 81 36 54

2 6 8 36 64 48

3 10 12 100 144 120

4 11 11 121 121 121

5 11 10 121 100 110

6 8 8 64 64 64

7 7 6 49 36 42

8 5 10 25 100 50

9 3 9 9 81 27

10 7 8 49 64 56

11 5 6 25 36 30

12 6 7 36 49 42

13 8 4 64 16 32

14 7 11 49 121 77

15 4 6 16 36 24

16 10 10 100 100 100

17 5 6 25 36 30

18 1 2 1 4 2

19 13 14 169 196 182

20 9 7 81 49 63

21 9 11 81 121 99

X = Y X2= Y2 XY=¿

The calculation above used split half reliability

where the scores are divided into two parts, odd and

even. The score of each part are correlated with other

half.

rxy=N∑xy−(∑ x) (∑y )

√[N∑ x2−(∑ x)2 ][N∑y2−(∑y )2 ]

=2345√[3626 ]. [3449 ]

=2345√12506074

=23453536

rxy=0.66

The result of this calculation 0.66 is quite high

which shows that there is a correlation between the two

sets of scores. The result of this calculation is then

analyzed using Spearman-Brown odd even model

correlation to see the reliability of the test.

rtt=2rhh

1+rhh

(Sudjiono,

2003:206)

rtt = total test coefficient reliability (tt = total

test)

rxy=(21×1373)−(154×172)

√[ (21×1302 )−(154 )2 ][ (21×1573 )−(172)2 ]

rhh = Product moment Correlation Coefficient between

first half and the second half of the test (hh =

half-half)

1∧2 = constant number

rtt=2rhh

1+rhh

= 2×0.661+0.66= 1.321.66

rtt=0.79

E. Items Analysis

a. Item DifficultyThe index of difficulty:

0,00 – 0,30 : difficult

0,31 – 0,70 : normal

0,71 – 1,00 : very easy

If the number of item is on level too difficult or too easy, it is not good to

include them on the test.

IF=UG+LGN

IF is the index of facility,UG represents the number of

correct answers by upper group, LG is by the lower

group, and N the number of students taking the test.

Here are the indexes of facility of the tested test.

1.IF=5+520

=0.5

2.IF=10+620

=0.8

3.IF=9+420

=0.65

4.IF=10+620

=0.8

5.IF=6+320

=0.45

6.IF=0+120

=0,05

7.IF=0+020

=0

8.IF=4+420

=0.4

9.IF=7+020

=0.35

10.IF=9+520

=0.7

11.IF=7+420

=0.55

12.IF=4+720

=0.55

13.IF=5+420

=0.45

14.IF=7+420

=0.55

15.IF=6+320

=0.45

16.IF=4+020

=0.20

17.IF=8+520

=0.65

18.IF=7+320

=0.50

19.IF=2+320

=0.25

20.IF=9+420

=0.65

21.IF=7+420

=0.55

22.IF=10+420

=0.70

23.IF=10+220

=0.60

24.IF=6+120

=0.35

25.IF=5+620

=0.55

26.IF=5+320

=0.4027.IF=8+320

=0.55

28.IF=8+420

=0.60

29.IF=10+820

=0.90

30.IF=9+820

=0.85

b. Item DiscriminationThe index of discrimination:

High discrimination index : +1,00 - 0,50

Low discrimination : 0,40 - 0,00

Negative discrimination : -0,10 - 1,00

ID = index of discrimination;

N = number of students in one group (1/2

N);

ID =UG−LG12N

UG = frequency of score by upper group

(upper half);

LG = frequency of score by lower group

(lower half).

1.ID=5−510

=0

2.ID=10−610

=0.40

3.ID=9−410

=0.50

4.ID=10−610

=0.40

5.ID=6−310

=0.3

6.ID=0−110

=−0.1

7.ID=0−010

=0

8.ID=4−410

=0

9.ID=7−010

=0.7

10.ID=9−510

=0.4

11.ID=7−410

=0.3

12.ID=4−710

=−0.3

13.ID=5−410

=0.1

14.ID=7−410

=0.3

15.ID=6−310

=0.3

16.ID=4−010

=0.4

17.ID=8−510

=0.3

18.ID=7−310

=0.4

19.ID=2−310

=−0.1

20.ID=9−410

=0.5

21.ID=7−410

=0.3

22.ID=10−410

=0.623.ID=10−210

=0.8

24.ID=6−110

=0.5

25.ID=5−610

=−0.1

26.ID=5−310

=0.2

27.ID=8−310

=0.5

28.ID=8−410

=0.4

29.ID=10−810

=0.2

30.ID=9−810.

=0.1

Table 8. The Indicates of Facility and Discrimination

Item UG LG UG+LG UG-LG IF ID REMARK1 5 5 10 0 0.5 0 IMPROPER2 10 6 16 4 0.8 0.40 IMPROPER3 9 4 13 5 0.65 0.5 PROPER4 10 6 16 4 0.8 0.4 IMPROPER5 6 3 9 3 0.45 0.3 PROPER6 0 1 1 -1 0.05 -0.1 IMPROPER7 0 0 0 0 0 0 IMPROPER8 4 4 8 0 0.4 0 IMPROPER9 7 0 7 7 0.35 0.7 PROPER10 9 5 14 4 0.70 0.4 PROPER11 7 4 11 3 0.55 0.3 PROPER12 4 7 11 -3 0.55 -0.3 IMPROPER13 5 4 9 1 0.45 0.1 IMPROPER14 7 4 11 3 0.55 0.3 PROPER15 6 3 9 3 0.45 0.3 PROPER16 4 0 4 4 0.20 0.4 IMPROPER17 8 5 13 3 0.65 0.3 PROPER18 7 3 10 4 0.50 0.4 PROPER19 2 3 5 -1 0.25 -0.1 IMPROPER20 9 4 13 5 0.65 0.5 PROPER

21 7 4 11 3 0.55 0.3 PROPER22 10 4 14 6 0.70 0.6 PROPER23 10 2 12 8 0.60 0.8 PROPER24 6 1 7 5 0.35 0.5 PROPER25 5 6 11 -1 0.55 0.5 PROPER26 8 3 11 5 0.40 0.2 IMPROPER27 8 3 11 5 0.55 0.5 PROPER28 8 4 12 4 0.60 0.4 PROPER29 10 8 18 2 0.90 0.2 IMPROPER30 9 8 17 1 0.85 0.1 IMPROPER

On the result listed in table above, only 17 items

could be safely used in future without rewritten.

Number 3, 5, 9, 10, 14, 15, 17, 18, 20, 21, 22, 23, 24,

25, 27, and 28 are appropriate to be put in the future

test as these items are good to students to answer.

Besides, the items 1, 2, 4, 6, 7, 8, 12, 13, 16, 19,

26, 29 and 30 are also improper since these items are

too difficult or too easy for students to answer.

c. Full item analysis

Item 1

IF=5+520

=0.5

ID=5−510

=0

UG LG UG +LGA 1 0 1B 4 4 8C 0 1 1D* 5 5 100 0 0 0TOTAL 10 10 20

Item 2

IF=10+620

=0.8

ID=10−610

=0.40

Item 3

IF=9+420

=0.65

ID=9−410

=0.50

Item 4

IF=10+620

=0.8

ID=10−610

=0.40

UG LG UG +LGA* 10 6 16B 0 2 2C 0 0 0D 0 2 20 0 0 0Total 10 10 20

UG LG UG +LGA 1 6 7B 0 0 0C* 9 4 13D 0 0 00 0 0 0TOTAL 10 10 20

UG LG UG +LGA 0 0 0B* 10 6 16C 0 2 2D 0 2 20 0 0 0TOTAL 10 10 20

Item 5

IF=6+320

=0.45

ID=6−310

=0.3

Item 6

IF=0+120

=0,05

ID=0−110

=−0.1

Item 7

IF=0+020

=0

ID=0−010

=0

Item 8




IF=4+420

=0.4

ID=4−410

=0

Item 9

IF=7+020

=0.35

ID=7−010

=0.7

Item 10

IF=9+520

=0.7

ID=9−510

=0.4

Item 11

UG LG UG +LGA* 7 0 7B 0 4 4C 2 1 3D 1 5 60 0 0 0TOTAL 10 10 20



IF=7+420

=0.55

ID=7−410

=0.3

Item 12

IF=4+720

=0.55

ID=4−710

=−0.3

Item 13

IF=5+420

=0.45

ID=5−410

=0.1




Item 14

IF=7+420

=0.55

ID=7−410

=0.3

Item 15

IF=6+320

=0.45

ID=6−310

=0.3

Item 16

IF=4+020

=0.20

ID=4−010

=0.4

Item 17

IF=8+520

=0.65UG LG UG +LG

A* 8 5 13B 1 0 1C 1 3 4D 0 2 20 0 0 0TOTAL 10 10 20




ID=8−510

=0.3

Item 18

IF=7+320

=0.50

ID=7−310

=0.4

Item 19

IF=2+320

=0.23

ID=2−310

=−0.1

Item 20

IF=9+420

=0.65

UG LG UG +LGA 4 3 7B 2 2 4C* 2 3 5D 2 2 4E 0 0 0TOTAL 10 10 20


UG LG UG +LGA 0 2 2B 0 0 0C 1 4 5D* 9 4 13E 0 0 0TOTAL 10 10 20

ID=9−410

=0.5

Item 21

IF=7+420

=0.55

ID=7−410

=0.3

Item 22

IF=10+420

=0.70

ID=10−410

=0.6

Item 23

IF=10+220

=0.60

ID=10−210

=0.8

Item 24




IF=6+120

=0.35

ID=6−110

=0.5

Item 25

IF=5+620

=0.55

ID=5−610

=−0.1

Item 26

IF=5+320

=0.40

ID=5−310

=0.2




Item 27

IF=8+320

=0.55

ID=8−310

=0.5

Item 28

IF=8+420

=0.60

ID=8−410

=0.4

Item 29

IF=10+820

=0.90

ID=10−810

=0.2

Item 30




IF=9+820

=0.85

ID=9−810.

=0.1


CONCLUSION

Based on the test analysis, which have been done,

the writer can conclude that there are just ten

questions which are valid. The other questions are

invalid. Some of them do not have effective choice, too

difficult, too easy and discriminated in the wrong way.

In addition, there are several questions which are

different with the material that have been taught by

the teacher.

The analysis shows that the mean score is 4.95,

standard deviation is 0.81. There are 22 improper

questions and there are 8 questions that are proper.

The score of reliability is 0.97, which means that the

evaluated test is reliable enough.

From the analysis, we can say that in arranging

the test, tester must make the questions based on the

blue print and the questions must be appropriate with

the material that has been taught by the teacher.

CURICULUM VITAE

PERSONAL IDENTITY

1. Name : Mutiah Nur Adzra

2. ID : 1252042006

3. Place & Date of Birth : Sengkang, on February 25th,

1996

4. Occupation : Student of State University of

Makassar

5. Status : Single

6. Address : Jl. A.P. Pettarani V No.4a Makassar

7. Mobile Phone : 08991529652

8. Email : [email protected]

EDUCATION BACKGROUND

(2000 – 2002) TK Pertiwi Sengkang

(2002 – 2008) SD Negeri 2 Maddukkelleng

(2008 – 2010) SMP Negeri 6 Sengkang

(2010 – 2012) SMA Negeri 3 Sengkang

(2012 – …) State University of Makassar, English

Department

ACTIVITIES BACKGROUND

1. Being a member of LDK Fosdik Al-‘Umdah UNM

70

71

ELT:Try Out for First Grade of SMA Negeri 1 Pangkajene

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