1. CHAPTER IV RESEARCH FINDINGS AND DISCUSSION This chapter
presents the data that was collected during the experimental
research. First analysis focuses on the homogeneity of the sample;
the second analysis focuses on the validity, reliability, index
difficulty, and discriminating power of instruments. And the third
analysis represents the result of pre-test and post-test that was
done both in experimental and control group. A. First Analysis The
first analysis was homogeneity test of the sample. That was
previous summative score of students XI IPS 4 as experimental group
and students of XI IPS I as control group. The analysis was meant
to get homogeneous class of XI IPS 4 and IPS 1. In this study, the
homogeneity of the test was measured by comparing the obtained
score (Fscore) with Ftable. Thus, if the obtained score (Fscore)
was lower than Ftable or equal, it could be said that Ho was
accepted. It meant those the classes were homogeneous. The analysis
of homogeneity test could be seen in table I. Table .I. Test of
Homogeneity Experimental G Control G 2244 2236 37 38 60.65 58.84
29.790 43.164 5.46 6.57 Variants (s2) Standart deviation (s) n X
Variant Sources Sum By knowing the mean and the variance, the
researcher was able to test the similarity of the two variants with
the homogeneity test from students previous score between IPS 4 and
IPS 1. The computation of the rest of homogeneity as follows: F =
Biggest Variance Smallest Variance 43.164 29.790 F = F =
1.448938879 On a 5 % with df1 numerator =37, df2 =38, it was found
Ftable = 1.721142152. Because Fscore < Ftable / 1.448938879
rtable Calculation: Below is the example of the item validity of
number 1 3. 15 22 10 30 18 6016 121 100 100 29 T-02 T-03 T-06 T-07
T-08 72 42 24 27 24 72 81 30 24 56 39 54 42 36 45 18 196 144 81 144
324 225 81 4 100 144 196 169 324 196 144 225 32416 9 4 9 4 16 4 1 9
4 16 9 9 9 9 9 14 12 9 12 18 9 11 10 10 9 T-15 T-32 T-12 T-13 10 12
14 13 18 14T-28 T-35 T-36 T-37 T-33 T-04 31 25 26 27 28 T-11 T-16
T-24 T-26 T-27 2 2 1 3 2 21 22 23 24 30 4 3 2 3 2 4 4 19 20 3 2 4 3
3 3 3 3 13 14 15 16 17 18 8 30 16 51 27 39 56 12 15 18 9 9 16 9 4
100 196 81 9 4 No Kode X 9 Y X2 Y2 XY 51 2 T-20 3 15 9 1 T-09 3 16
9 256 3 17 9 289 9 289 225 45 48 51 9 225 3 T-25 4 16 16 256 3 174
T-29 6 T-10 3 15 289 45 5 T-05 3 17 9 289 51 64 80 64 51 7 T-14 8
T-21 3 17 9 289 9 T-30 169 4 20 16 400 17 13 326 7044 1482 T-17
T-22 4 3 3 9 14 36 37 T-33 10 11 32 33 34 35 12 2 12 T-34 3 T-01
T-18 T-19 3 2 10 T-31 1 8 106 496 144 24 1 64 8 4 Where N = 37 X2 =
326 X = 106 Y2 = 7044 Y = 496 XY = 1482 37 37 2 37 2 = = 1482 106
0.650 326 rxy 496 496 106 7044 Because of rxy > rtable, so item
no 1 is valid. 2. Reliability of Try Out Test After validity items
had been done, the next analysis was to test the reliability of
instrument. It was done to find out whether a test had higher
critical score and gave the stability or consistency of the test
scores or not. From the computation of reliability of the try out
instruments using advertisements video, it was obtained 0,743, for
5 % with N = 37. It was obtained 0,725. It could be concluded that
the instruments that were used in his research was reliable. The
complete analysis and the computation as follow : 4. The
computation of reliability using advertisement video Formula :
Criteria : Criteria : The Try out is reliable if > rtable
Calculation: = + + = + + = + . . .+ 0.92490.6201 1.0751 0.7973 + .
. .+ 4.4474 Variance 2 S2 = 7044 10.970 37 36 496 = Index
Reliability 5 1 5 10.970 4.447 10.970 = 0.743 = The result shows
that 0.743 is more than 0.725, it meant the items of instrument
were valid. 3. Discriminating Power of Try Out Test The
discriminating power of the five items analysis of speaking was
satisfied. It showed that all speaking items had strong
discrimination. The complete analysis and the sample of computation
as follow. The Computation of Discriminating Power 5. Formula:
Where: MH : The average upper class ML :The average lower class x1
2 : The number of individual quadrate deviation upper class x2 2 :
The number of individual quadrate deviation lower class Ni : 27% x
N, with N= the number of students Criteria: The result of tscore is
compared with ttable ,df= (N1-1) + (N2-1) and alpha =5%. If tscore
> ttable, so the discriminating power of the item is
significant. Calculation: Below is the example of the computation
of discriminating power on item number 1. Ni = 27% x 45=12 = = = t
= + 10 10 1 1.60 8.40 1.80 8.401.60 2.603.20 The result obtained
Discriminating power is = 1.80 Because ttable = 1.725 and tscore
> ttable, so the item number 1 is significant. Because the
result tscore > ttable. So the item number 1 is satisfactory. 4.
Difficulty Level Of Try Out Test 6. From the computation of
difficulty level of the five items analysis of speaking, it was
found that the difficulty level is easy. So it could be concluded
that the final total items analysis for the instruments were
categorized satisfactory. The sample of computation is as follow.
The Computation of Difficulty Index Formula : DL = N G x 100%
Where: DL: Difficulty level G : The number of students who fail in
answer N : The number of students Criteria: < < < <
< < 0.27 Easy 0.27 DL 0.72 medium 0.00 Interval P Criteria DL
1.00 Difficult DL 0.72 Calculation : Below is the example of the
computation of difficulty level on item number 1. = % DL = 37 10 x
100% 27.03 The result obtained DL = 27.03% Because of the result is
between 0.27-0.72, so the item number 1 is medium. C. Third
Analysis 7. The second analysis represents the result of pre-test
and post-test that was done both in experimental and control group.
This analysis will answer the research question Is advertisement
video effective to teach speaking of advertisement? We can conclude
advertisement video is effective when the result of post-test of
the experimental class (using advertisement video) and control
class (using conventional media) has significant differences or the
assumption that those classes is equal is not fulfilled. Before the
researcher tested the hypothesis that had been mentioned in the
chapter two, the researcher analyzed and tested hypothesis
prerequisites which contained of normality test and homogeneity
test. Second analysis dealt with normality test, homogeneity test,
and t-test (test difference two variants) in pre-test and
post-test. 1. Analysis of Pre-test The experimental group (XI IPS
4) was given a pre test on Mei 21, 2012 and control group (XI IPS
1) was given a pre-test on Mei 22, 2012. They were asked to
practiced became advertiser to advertise certain product. a. Test
of Normality Test of normality was used to find out whether data of
control and experimental group which had been collected from the
research come from normal distribution normal or not. The result
computation of Chi- quadrate ( X2 score ) then was compared with
table of Chi-quadrate ( X2 table ) by using 5 % alpha of
significance. If X2 score < X2 table meant that the data spread
of research result distributed normally. Based on the research
result of XI IPS 1 student in the control group before they were
taught speaking of advertisement text without advertisement video,
they reached the maximum score 72 and minimum score 48. The
stretches of score were 24. So there were 6 classes with length
with classes 4. From the computation of frequency distribution, it
was found ( f i Xi ) = 2309, and ( fi Xi 2 ) = 141990.9. So the
average score (X )was 72,25 and the standard deviation (S) was
6.75587. After counting the average sore and standard deviation,
table of observation frequency was needed to measure Chi-quadrate (
X2 score ). Table 4. 1 Table of the Observation Frequency of
Control Group 8. 47.50 -1.96 -0.475 48 52 0.0725 2.7544 4 0.5633
52.00 -1.30 -0.403 53 56 0.1667 6.3352 7 0.0698 56.50 -0.63 -0.236
57 61 0.2220 8.4363 9 0.0377 61.00 0.04 0.014 62 65 0.2444 9.2876 8
0.1785 65.50 0.70 0.258 66 70 0.1558 5.9215 6 0.0010 70.00 1.37
0.414 71 74 0.0648 2.4611 4 0.9622 74.50 2.03 0.479 hitung = 1.8125
large area Ei OiClass Class limit Zi P(Zi) Based on the
Chi-quadrate table X2 table for 5% alpha significance with 6-3= 3,
it was found X2 table = 7.81. Because of X2 score < X2 table, so
the initial data of experimental group distributed normally. While
from the result of XI IPS 4 students in experimental group, before
they were taught speaking of advertisement by using advertisement
video, was found that the maximum score was 76 and minimal score
was 52. The stretches of score were 24. So, there were 6 classes
(fi xi ) = 2434. 25, and ( f i xi 2 ) = 161853.8. So, the average
score (X) was 65.7905 and the standard deviation ( S ) was 6.87829.
After counting the average score and standard deviation, table of
observation frequency was needed to measure Chi-quadrate ( X2 score
). Table IV.2 Table of the Observation Frequency of Experimental
Group 51.50 -2.08 -0.481 52 56 0.0584 2.1622 3 0.3246 56.00 -1.42
-0.423 57 60 0.1436 5.3127 7 0.5359 60.50 -0.77 -0.279 61 65 0.2334
8.6340 6 0.8036 65.00 -0.11 -0.046 66 69 0.1594 5.8981 9 1.6313
69.50 0.54 0.205 70 74 0.1785 6.6049 7 0.0236 74.00 1.19 0.384 75
78 0.0840 3.1085 5 1.1510 78.50 1.85 0.468 X = 4.4700 Ld OiP(Zi)
EiZiLcClass Based on the Chi-quadrate table ( X2 table ) for 5%
alpha of significance with df 6-3=3, it was found X2 table = 7.81.
Because of X2 score < X2 table, so the initial data of
experimental group distributed normally. b. Test of Homogeneity 9.
Test of Homogeneity was done to know whether sample in the research
come from the population that had same variance or not. In this
study, the homogeneity of the test was measured by comparing the
obtained score ( Fscore ) with Ftable . Thus, if the obtained score
( Fscore ) was lower than the Ftable or equal, it could be said
that the Ho was accepted. It meant that the variance was
homogenous. The analysis of homogeneity test could be seen in table
IV.3. Table IV.3. Test of Homogeneity (Pre-Test) Eksperiment G
Control G 2244 2236 37 38 60.65 58.84 29.790 43.164 5.46 6.57
Varians (s 2 ) Standard deviation (s) n X Variant Sources Sum By
knowing the mean and the variance, the writer was able to test the
similarity of the two variants in the pre-test between experimental
and control group. The computation of the test of homogeneity as
follows: F = Biggest Variance Smallest Variance 43.164 29.790 F = F
= 1.448938879 On a 5% with df 1 = 37, df 2 = 38, it was found
Ftable = 1.721142152. Because of Fscore Ftable, so it concluded
that both experimental and control group had no differences. The
result showed both groups had similar variants (homogenous). c.
Test difference two variants in pre-test between experiment and
control group After counting standard deviation and variance, it
could be concluded that both group have no differences in the test
of similarity 10. between two variances in pre-test score. So, to
differentiate whether the students result of speaking of
advertisement in experimental and control group were significant or
not, the writer used t-test to test the hypothesis that had been
mentioned in the chapter two. The writer used formula: 21 21 11 nn
s xx t + = Where: 2 )1()1( 21 2 22 2 11 + + = nn snsn s Based on
table IV.3, first the writer had to find out S by using the formula
above: (37-1) . 29.790 + (38-1) . 43.164 S 2 = 36.568 S = 6.047 =S
2 37 + 38 -2 = After S was found, the next step was to measure
t-test: t = - 1 1 37 38 60.649 58.842 6.047 + 1.807 1.397 = = 1.293
After getting t-test result, then it would be consulted to the
critical score of ttable to check whether the difference is
significant or not. For a = 5% with df = 37+ 38-2= 73, it was found
ttable (0.975)(80)= 1.99. Because of tscore < ttable, so it
could be concluded that were no significance of difference between
the experimental and control group. It meant that both experimental
and control group had same condition before getting treatments. 2.
Analysis of Post-Test The experimental group was given post-test on
Mei 28, 2012 and control group was given a pos test Mei 29, 2012.
Post test was conducted after all treatments were done.
Advertisements video was used as medium in the teaching speaking of
11. advertisement text to students in experimental group. While for
students in control group, they were given treatments without
advertisements video. Post-test was aimed to measure students
ability after they got treatments. They were asked to advertise
certain products. a. Test of Normality Test of normality was used
to find out whether data of control and experimental group, which
had been collected after they got treatments, come from normal
distribution normal or not. The formula, that was used, was
Chi-Quadrate. The result computation of Chi-Quadrate ( X2 score ),
then was compared with table of Chi-Quadrate ( X2 score ) then was
compared with table of Chi-quadrate ( X2 table ) by using 5% alpha
of significance. If X2 score < X2 table meant that the data
spread of research result distributed normally. Based on the
research result of XI IPS 1 students in the control group after
they got usual treatments in the teaching speaking of
advertisements they reached the maximum score 80 and minimum score
60. The stretches score 20. So there were 6 classes with length of
4 classes. From the computation of frequency distribution it was
found fi xi = 2709 and ( fi xi 2 ) = 194457.5. So the average score
( X ) was 71.2895 and the standard deviation (S) was 6.00521. It
meant that there was an improvement of students score after they
got treatments. After counting the average score and standard
deviation, table of observation frequency was needed to measure Chi
quadrate ( X2 score ). Table IV.4 Table of the Observation
Frequency of Control Group 59.50 -1.96 -0.475 60 63 0.0725 2.7544 4
0.5633 63.50 -1.30 -0.403 64 67 0.1667 6.3352 7 0.0698 67.50 -0.63
-0.236 68 71 0.2220 8.4363 9 0.0377 71.50 0.04 0.014 72 75 0.2444
9.2876 8 0.1785 75.50 0.70 0.258 76 79 0.1558 5.9215 6 0.0010 79.50
1.37 0.414 80 83 0.0648 2.4611 4 0.9622 83.50 2.03 0.479 hitung =
1.8125 Luas Daerah Ei OiClass Lc Zi P(Zi) 12. Based on the
Chi-quadrate table ( X2 table ) for 5 % alpha of significance with
df 6-3 = 3, it was found X2 table = 7.81. Because of X2 score <
X2 table, so the data of experimental group after getting
treatments distributed normally. While from the result of XI IPS 4
students in experimental group, after they were taught using
advertisement video, was found that the maximum score was 92, and
minimal score was 64. The stretches of score were 28. So, there
were 6 classes with length of classes 5. From the computation of
frequency distribution, it was found ( fiXi ) = 2939, and ( fiXi 2
) = 235237, so the average score (X) was 79. 3784 and the standard
deviation ( S ) was 7.64254. By seeing the average score of
students in experimental group, it could be concluded that there an
improvement of students score after they got treatments by using
advertisement video. After counting the average score and standard
deviation, table of observation frequency was needed to measure
Chi-quadrate ( X2 score ). Table IV. 5 Table of the Observation
Frequency of Experimental Group 63.50 -2.08 -0.481 64 68 0.0584
2.1622 3 0.3246 68.50 -1.42 -0.423 69 73 0.1436 5.3127 7 0.5359
73.50 -0.77 -0.279 74 78 0.2334 8.6340 6 0.8036 78.50 -0.11 -0.046
79 83 0.1594 5.8981 9 1.6313 83.50 0.54 0.205 84 88 0.1785 6.6049 7
0.0236 88.50 1.19 0.384 89 93 0.0840 3.1085 5 1.1510 93.50 1.85
0.468 X = 4.4700 P(Zi) Ei Luas Daerah OiZiLcClass Based on the
Chi-quadrate table (X2 table) for 5 % alpha of significance with df
= 6-3 = 3, it was found X2 table 7.81. Because of X2 score < X2
table, so the data of experimental group after getting treatments
distributed normally. b. Test of Homogeneity 13. The writer
determined the mean and variance of the students score either in
experimental or control group. By knowing the mean and variance,
the writer was able to test the similarity of the two variance in
the post-test between experimental and control group. Table. IV.6
Test of Homogeneity (Post-test) Eksperimen Kontrol 2860 2700 37 38
77.30 71.05 53.381 32.376 7.31 5.69 Varians (s 2 ) Standart
deviation (S) n X Varians Sources Jumlah The computation of the
test of homogeneity as follows: F = Biggest Variance Smallest
Variance 32.376 53.381 F = F = 0.60649486 On a 5 % with df 1 = 37,
df 2 = 38,it was found Ftable = 1.721142152. Because of Fscore
Ftable, so it could be concluded that both experimental and control
group had no differences. The result showed both groups had similar
variance (homogenous). c. Test of difference two variants in
post-test between experiment and control group After counting
standard deviation and variance, it could be concluded that both
group have no differences in the test of similarity between two
variances in the post-test score. So, to differentiate if the
students result of speaking of advertisement text in experimental
and control group after getting treatments were significant or not,
the writer used t-test to test the hypothesis that had been
mentioned in chapter two. To see the difference between the
experimental and control group, the writer used formula : 14. 21 21
11 nn s xx t + = Where: 2 )1()1( 21 2 22 2 11 + + = nn snsn s Based
on the table IV.6, first the writer had to find out S by using the
formula above: (37-1) . 53.381 + (38-1) . 32.376 S 2 = 42.735 S =
6.537 =S 2 37 + 38 -2 = After S was found, the next step was to
measure t-test: - 1 1 37 38 6.245 1.510 t = 4.136 t = = 77.297
71.053 6.537 + t = After getting t-test result, then it would be
consulted to the critical score of ttable to check whether the
difference is significant or not. For a = 5% with df = 37+38-2=73,
it was found ttable (0.05)(95)= 1.67. Because of tsore >
ttable,, so it could be concluded that there was significance of
difference between the experimental and control group. It meant
that experimental group was better than control group after getting
treatments. Since the obtained t-score was higher than the critical
score on the table, the difference was statistically significance.
Therefore, based on the computation there was a significance
difference between the teaching of speaking of advertisement using
advertisement video and teaching speaking of advertisement without
advertisement video for the eleventh grade students of SMA Negeri 1
Pegandon Kendal. Teaching speaking of advertisement using
advertisement video to be more effective than teaching speaking of
advertisement without using advertisement video. It can be seen
from the result of the test where the students taught speaking 15.
of advertisement by using advertisement video got higher scores
than the students taught speaking of advertisement without
advertisement video. D. Discussions The data were obtained from the
students achievement scores of the test of speaking of
advertisement. They were pre-test and post-test scores from the
experimental and control group. The average score for experimental
group was 60.65(pre-test) and 77.3 (post-test). The average score
for control group was 58.84(pre-test) and 71.65(post-test). The
following was the simple tables of pre and post-test students
average score and students average score of each speaking
components. Table IV. 7 The Pre-test and Pos-test Students Average
Scores of the Experimental and Control Group No Group The Average
Percentage of Pre-test The Average Percentage of Post-test 1
Experimental 60.65 77.3 2 Control 58.84 71.65 Table IV. 8 The
Pre-test and Post-test StudentsAverage Scores of the Experimental
and Control No Component of Speaking Group The Average score of
Pre-test The Average score of Post-test 1 Pronunciation
Experimental 2.9 3.8 Control 2.9 3.3 2 Grammar Experimental 2.8 3.8
Control 3 3.6 3 Vocabulary Experimental 3.1 3.8 Control 2.9 3.6 4
Fluency Experimental 3 3.8 Control 2.8 3.3 5 Comprehension
Experimental 3.3 4.2 Control 3.1 3.9 16. a. Students Condition in
Control Group In this study, source of data that become as control
group was class IPS 1. In the control group, there was not a new
treatment in a teaching learning process. They were given a usual
treatment. They were taught speaking of advertisement in
conventional media. By listen teacher explanation and use the
realia media and pictures. Teacher only give little example in
speaking how to advertise certain product. Students could not enjoy
in practicing their skill in speaking because they only saw
teachers way in advertise certain product. It could not effective
to teach students in speaking of advertisement because the teacher
seldom practice became good advertiser. It was proven with the
control groups average in the post test (71.65) which was lower
than experimental group (77.3). b. Students condition in
experimental group 1. Analysis students speaking before treatment
(pre-test) In the pre-test, students ability in speaking of
advertisement was low. Pre- test was conducted before treatment.
From the result of pre-test, it was known was students faced many
difficulties in speaking of advertisement. Sentences which were
used by students to convoy the idea, were influenced by Indonesian.
Moreover they dont know what should they say when they want to
convey their meaning. Students ability was in low level when they
had to arrange words to be a good sentence that comprehensible by
considering main function. It meant that the idea was not clearly
stated and the sentences were not well-organized to support to
transformation of meaning. Students word voice (Pronunciation and
fluency) was also far from being perfect. Not only the way they
convey their idea was not clear but also there were many
difficulties in grammar and vocabulary; therefore, students ability
of speaking of advertisement text was hard to be understood. To
minimize the number of students mistakes of their speaking, the
researcher collected students speaking in writing form after they
do advertise certain product, gave correction and returned the
paper to them. From the correction of their mistakes, students were
supposed to learn more and improve and improve their ability in
speaking of advertisement. 2. Analysis students speaking after
treatment (post-test) 17. Based on the analysis of students
ability, it was found that students ability after getting treatment
was improved. In the treatment, students were doing speaking of
advertisement that was in line with the function of some
expressions they learn. The vocabulary choice, sentences
arrangement, and the way they produce the word were good and
relevance to the topic or (their meaning) so the meaning were easy
to be understood. Their speaking was still comprehensible however;
there were some mistakes in grammar and pronunciation. The finding
that shows students ability in namely the increasing of students
average score. There were still some mistakes that students had
made like grammar and pronunciation. But it was very human. So, it
could be concluded that the implementation of using advertisement
video as media in the teaching speaking of advertisement was
effective. It was proven with students average score in
experimental group was higher than control group. By considering
the students final score after getting treatment, the teaching of
speaking of advertisement using advertisement video as media was
better than without advertisement video. Based on t-test analysis
was done, it was found that t-score (4.136) was higher than t-table
using 5 % alpha of significance (1.67). Since tscore > ttable ,
it proved that there was significant difference between the
improvement of students achievement that was given a new treatment
(using advertisement video) and the improvement of students
achievement that was given a usual treatment. c. The advantages and
disadvantages of using advertisement video in teaching of speaking
of advertisement 1) The advantages of using advertisement video in
teaching of speaking of advertisement After conducting the
research, there were some advantages of using advertisement video
in teaching speaking of advertisement: a. Advertisement video gave
students the real situations of a chronological action. It helped
students to express their imagination to be transferred in to the
reality. The use advertisement video was actually meant to help
them expressing their ideas easily. b. Students boredom i learning
speaking could be avoided. The treatment gave students different
nuances of teaching and learning process so they were interested in
following the lesson. Video advertisement that gives students
chance to notice how to be good advertiser and how use the
expression appropriately according the situation, so they will
motivated to 18. imitate practiced to be good advertiser by using
clear pronunciation and appropriate expression. It also could
measure how far their speaking skill is. 2) The disadvantages of
using advertisement video in teaching of speaking of advertisement
The disadvantages were described below: a. It spent a lot of time,
because the students skill was too low, they cant directly speaking
in front of class to advertise certain product after teacher
distributed certain product to be advertised. They need time to
prepare their performance. b. It was not easy enough to manage the
class because, sometime the students will be very hysteric when
they see their friends practicing in front of them. Their voice can
disturb another class. E. Limitation of research The writer
realized that there were some hindrances and barriers in doing this
research. The hindrances and barriers occurred was not caused by
inability of the researcher but caused by limitation of the
research like time, fund, and equipment of research. 19. .