assignment of english language testi
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TRANSCRIPT
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.
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