CHAPTER IV. DATA ANALYSIS AND FINDINGS
B. Data Analysis
B.1 The Significant effect of applying three-steps interview technique on students’ achievement
9 RRN 83.33 66.6 14 15 15 15 14 73 74.33
10 DP 100 33.3 16 15 15 14 13 73 68.78
11 DY 83.33 50 16 15 16 16 16 79 70.78
12 ARS 33.33 50 14 14 15 14 15 72 51.78
13 RAZ 33.33 50 15 16 16 15 17 79 54.11
14 TNP 33.33 33.3 15 15 16 16 15 77 47.89
15 QP 33.33 50 17 15 15 15 16 78 53.78
16 SK 50 50 15 16 16 17 16 80 60
17 MYA 33.33 33.3 16 15 16 15 14 76 47.56
18 GS 66.67 33.3 16 15 15 15 14 75 58.33
19 EJL 66.67 66.6 16 16 16 16 17 81 71.44
20 IS 33.33 33.3 15 15 16 16 16 78 48.22
21 MI 50 50 15 15 14 15 14 73 57.67
22 AK 50 66.6 15 16 15 15 15 76 64.22
23 BR 66.67 50 15 15 15 16 16 77 64.56
24 AA 50 66.6 16 16 16 17 17 82 66.22
25 MR 66.67 33.3 15 15 16 16 16 78 59.33
26 IA 33.33 66.6 15 16 16 16 15 78 59.33
27 IP 33.33 33.3 15 15 15 16 15 76 47.56
28 WO 33.33 66.6 15 16 15 16 16 78 59.33
29 NT 33.33 66.6 15 15 16 16 16 78 59.33
30 RD 66.67 50 15 15 15 16 16 77 64.56
Mean 53.33 ∑1780
Table 4.5
The Score Pre-test and Post-test in Experimental Group
NO Students’
Initial Name
Score
Pre-Test Post-Test
(X1) (X1)2 (X2) (X2)2 X2-X1
1 AW 61.55556 3789.09 82.5 6815.4 21
2 PR 49.44444 2444.75 82.2 6760.5 32.7778
3 RA 43.55556 1897.09 88.7 7881.5 45.2222
4 AP 28.22222 796.494 66.5 4429.6 38.3333
5 NAL 28.22222 796.494 78.3 6136.1 50.1111
6 MSR 51.11111 2612.35 91.1 8301.2 40
7 NAH 45.22222 2045.05 82.8 6870.6 37.6667
8 RAR 67.77778 4593.83 90.7 8240.6 23
9 SA 55.33333 3061.78 88.1 7763.6 32.7778
10 AL 44.55556 1985.2 83.2 6925.9 38.6667
11 NSAN 27.55556 759.309 67.5 4563.8 40
12 RA 27.55556 759.309 67.2 4518.8 39.6667
13 DY 23.33333 544.444 84.2 7093.4 60.8889
14 DN 52.44444 2750.42 80.6 6507.1 28.2222
15 MS 55.66667 3098.78 84.8 7206.1 29.2222
16 SV 63.88889 4081.79 81.3 6615.1 17.4444
17 VH 46.88889 2198.57 79.3 6293.8 32.4444
18 DP 46.88889 2198.57 79.6 6346.8 32.7778
19 HM 35.44444 1256.31 84.5 7149.6 49.1111
20 SA 68.11111 4639.12 79.3 6293.8 11.2222
21 DAL 57 3249 73.4 5394.1 16.4444
22 AN 42 1764 79 6241 37
23 RS 72.66667 5280.44 84.2 7093.4 11.5556
24 SNS 57.33333 3287.11 74.1 5492.5 16.7778
25 WL 44.88889 2015.01 83.5 6981.5 38.6667
26 WLD 48.77778 2379.27 72.1 5200 23.3333
27 AL 48.44444 2346.86 66.8 4474.1 18.4444
28 SS 37.33333 1393.78 89.7 8060 52.4444
29 IC 32.11111 1031.12 83.2 6925.9 51.1111
30 RA 26.22222 687.605 78.6 6188.4 52.4444
∑1389.556 ∑69742.9 ∑2408.3 ∑194764 ∑1018.78
Mean 46.31852 80.27778
Table 4.6
The Score Pre-test and Post-test in Control Group
NO Students’
Initial Name
Score Pre-Test
(X1)2 Post-Test
(X2) (X2)2 X2-X1
(X1)
1 KN 39.6 1573.4 47.2 2229.9 7.55
2 SH 39.3 1547.1 57.6 3325.4 18.3
3 PCR 32.7 1074.3 45.8 2105.7 13.1
4 IM 38 1444 42.3 1792.1 4.3
5 DM 32.7 1074.3 53.7 2892.0 21
6 NF 44.5 1985.1 69.4 4822.5 24.8
7 HN 54 2916 85.4 7300.7 31.4
8 NM 66.1 4370.6 69.1 4776.3 3
9 RRN 49.4 2444.7 74.3 5525.4 24.8
10 DP 49.1 2411.9 68.7 4730.3 19.6
11 DY 32.7 1074.3 70.7 5009.4 38
12 ARS 27.2 741.0 51.7 2680.9 24.5
13 RAZ 38 1444 54.1 2928.0 16.1
14 TNP 39.3 1547.1 47.8 2293.3 8.5
15 QP 50.4 2544.6 53.7 2892 3.3
16 SK 32.7 1074.38 60 3600 27.2
17 MYA 41 1681 47.5 2261.5 6.5
18 GS 50.4 2544.6 58.3 3402.7 7.8
19 EJL 45.2 2045.0 71.4 5104.3 26.2
20 IS 48.7 2379.2 48.2 2325.3 -0.55
21 MI 45.2 2045.04 57.6 3325.4 12.44
22 AK 45.2 2045.0 64.2 4124.4 19
23 BR 61.8 3830.2 64.5 4167.4 2.6
24 AA 57.3 3287.1 66.2 4385.3 8.88
25 MR 61.8 3830.2 59.3 3520.4 -2.5
26 IRA 37.6 1418.7 59.3 3520.4 21.6
27 IP 48.7 2379.2 47.5 2261.5 -1.22
28 WO 43.5 1897.0 59.3 3520.4 15.7
29 NT 60.2 3626.7 59.3 3520.4 -0.88
30 RD 38 1444 64.5 4167.4 26.5
∑1351.55 ∑63720.
9 ∑1780 ∑10851
2 ∑428.44
Mean 45.05 59.33
Based on the table above or table 4.6 showed that the mean of the Pre-test in control group was 45.05 and the mean of the Post-test was 59.33. And based on the table 4.5 showed that the mean of the Pre-test in experimental group was 46.31 and the mean of the Post-test was 80.27.
After seeing both of table (4.5 and 4.6) showed that the mean score of Post-test in experimental group was 80.27 and the mean score of the control group was 59.33. The data showed that be mean score of student in experimental group who were taught by applying Three- Steps Interview Technique was greater than the mean score of students in control group who were taught by discussion method. After getting the students’ score in Pre-test and Post-test of both classes, it can be known that there was a difference of students’ ability after receiving the treatment.
B.2. The Students’ Achievement in Speaking by Using Three-Steps Interview Technique The determination was calculated as follows;
1. Normality Test
Normality test used to determine if a data set is well-modeled by a normal distribution and to compete how likely it is fpt random variable underlying the data to be normally distribution.
a. Normality Test of X Variable
The normality test of variable x used Liliefors test:
1. Listing the students’ score from the lowest to the highest 2. The score made to Z1,Z2,Z3,…………Zn by using formula:
ZI = 𝑋−𝑋̃
𝑆
3. The table of Zi could be seen from the table of normal curve.
4. F(Zi) = 𝑓𝑘
𝑛 = 1
30 = 0.033 Standard Deviation of X Variable S = √𝑛.∑ 2−(∑ 𝑋)𝑋 2
𝑛(𝑛−1)
= √30.194764−(2408.33)2 30(29)
=√5842920−5800053 870
= √42867
870
= √49.2
= 7.01
Table 4.7
Normality Test of X variable
NO Xi F F kum Zi F(Zi) S(Zi) F(Zi)-S(Zi)
1 66.5556 1 1 -1.955 0.02529 0.03333 -0.008
2 66.8889 1 2 -1.908 0.02823 0.06667 -0.0384
3 67.2222 1 3 -1.86 0.03144 0.1 -0.0686
4 67.5556 1 4 -1.813 0.03495 0.13333 -0.0984 5 72.1111 1 5 -1.164 0.12231 0.16667 -0.0444
6 73.4444 1 6 -0.974 0.16514 0.2 -0.0349
7 74.1111 1 7 -0.879 0.18982 0.23333 -0.0435 8 78.3333 1 8 -0.277 0.39088 0.26667 0.12421
9 78.6667 1 9 -0.23 0.40923 0.3 0.10923
10 79 1 10 -0.182 0.42777 0.33333 0.09444
11 79.3333 1 11 -0.135 0.44648 0.36667 0.07982
12 79.3333 1 12 -0.135 0.44648 0.4 0.04648
13 79.6667 1 13 -0.087 0.46531 0.43333 0.03198 14 80.6667 1 14 0.0554 0.52209 0.46667 0.05543
15 81.3333 1 15 0.1504 0.55977 0.5 0.05977
16 82.2222 1 16 0.277 0.60912 0.53333 0.07579 17 82.5556 1 17 0.3245 0.62723 0.56667 0.06056
18 82.8889 1 18 0.372 0.64506 0.6 0.04506
19 83.2222 1 19 0.4195 0.66257 0.63333 0.02924 20 83.2222 1 20 0.4195 0.66257 0.66667 -0.0041
21 83.5556 1 21 0.467 0.67974 0.7 -0.0203
22 84.2222 1 22 0.562 0.71293 0.73333 -0.0204 23 84.2222 1 23 0.562 0.71293 0.76667 -0.0537
24 84.5556 1 24 0.6095 0.72889 0.8 -0.0711
25 84.8889 1 25 0.6569 0.74439 0.83333 -0.0889 26 88.1111 1 26 1.116 0.86779 0.86667 0.00113
27 88.7778 1 27 1.211 0.88705 0.9 -0.0129
28 89.7778 1 28 1.3535 0.91205 0.93333 -0.0213 29 90.7778 1 29 1.4959 0.93266 0.96667 -0.034
30 91.1111 1 30 1.5434 0.93864 1 -0.0614
Based on the data on the table 4.7 Lhitung was 0.124 and the Lilifors test in significance α
= 0.05 with n = 30 Ltable was 0.161. So the Lhitung < Ltable (0.124< 0.161). So it can be concluded that the data was normally distributed.
b. Normality Test of Y Variable
The normality test of variable x used Liliefors test:
1. Listing the students’ score from the lowest to the highest 2. The score made to Z1,Z2,Z3,…………Zn by using formula:
ZI = 𝑋−𝑋̃
𝑆
3. The table of Zi could be seen from the table of normal curve.
4. F(Zi) = 𝑓𝑘
𝑛 = 1
30 = 0.033
Standard Deviation of Y Variable S = √𝑛.∑ 2−(∑ 𝑋)𝑋 2
𝑛(𝑛−1)
= √30.108512−(1780)2 30(29)
=√3255360−3168400 870
= √86960
870
= √99.9
= 9.88
Table 4.8
Normality Test of Y variable
NO Xi F F kum Zi F(Zi) S(Zi) F(Zi)-
S(Zi) 1 27.2222 1 1 -1.8046 0.03557 0.03333 0.00223 2 32.7778 1 2 -1.2423 0.10706 0.06667 0.04039
3 32.7778 1 3 -1.2423 0.10706 0.1 0.00706 4 32.7778 1 4 -1.2423 0.10706 0.13333 -0.0263 5 32.7778 1 5 -1.2423 0.10706 0.16667 -0.0596
6 37.6667 1 6 -0.7475 0.22739 0.2 0.02739
7 38 1 7 -0.7137 0.23769 0.23333 0.00436
8 38 1 8 -0.7137 0.23769 0.26667 -0.029
9 38 1 9 -0.7137 0.23769 0.3 -0.0623
10 39.3333 1 10 -0.5788 0.28136 0.33333 -0.052 11 39.3333 1 11 -0.5788 0.28136 0.36667 -0.0853
12 39.6667 1 12 -0.5451 0.29286 0.4 -0.1071
13 41 1 13 -0.4101 0.34086 0.43333 -0.0925
14 43.5556 1 14 -0.1514 0.43981 0.46667 -0.0269
15 44.5556 1 15 -0.0502 0.47997 0.5 -0.02
16 45.2222 1 16 0.01724 0.50688 0.53333 -0.0265 17 45.2222 1 17 0.01724 0.50688 0.56667 -0.0598
18 45.2222 1 18 0.01724 0.50688 0.6 -0.0931
19 48.7778 1 19 0.37712 0.64696 0.63333 0.01362 20 48.7778 1 20 0.37712 0.64696 0.66667 -0.0197
21 49.1111 1 21 0.41086 0.65941 0.7 -0.0406
22 49.4444 1 22 0.44459 0.67169 0.73333 -0.0616 23 50.4444 1 23 0.54581 0.7074 0.76667 -0.0593
24 50.4444 1 24 0.54581 0.7074 0.8 -0.0926
25 54 1 25 0.90568 0.81745 0.83333 -0.0159
26 57.3333 1 26 1.24306 0.89308 0.86667 0.02641
27 60.2222 1 27 1.53546 0.93766 0.9 0.03766
28 61.8889 1 28 1.70415 0.95582 0.93333 0.02249 29 61.8889 1 29 1.70415 0.95582 0.96667 -0.0108
30 66.1111 1 30 2.1315 0.98348 1 -0.0165
Based on the data on the table 4.8 Lobserved was 0.040 and the Lilifors test in significance α
= 0.05 with n = 30 Ltable was 0.161. So the Lobserved < Ltable (0.040 < 0.161). So it can be concluded that the data was normally distributed.
c. Homogeneity Test
Homogeneity test performed to determine whether the variances of data equal from two distribution group.
The data of variable X and variable Y:
a. Variable X 𝑥̃ = 80.27 𝑆𝑥2 = 49.2 N = 30 b. variable Y 𝑥̃ = 59.33 𝑆𝑥2 = 99.9 N = 30
F = THE HIGHHEST VARIANCE THE LOWEST VARIANCE
= 99.9
49.2
= 2.03
The value of Ftable with the significance α = 0.05 with n = 30 was 2.215. and the Fhitung
was 2.03. So the Fhitung < Ftabel or ( 2.03 < 2.215). So it can be conclude that the data was homogeny.
C. Testing Hypothesis
After calculating the data, the result was showed the rules of statistics normality and homogeneity was fulfilled so the next is testing hypothesis.
Table 4.9
The Calculation Table
NO X Y Xi(x-𝑥̃) Yi(y-y ̃) Xi2 Yi2 XiYi
1 82.5 47.2 2.3 -12 5.29 144 -27.6
2 82.2 57.6 2 -1.6 4 2.56 -3.2
3 88.7 45.8 8.5 -13.4 72.25 179.56 -113.9
4 66.5 42.3 -13.7 -16.9 187.69 285.61 231.53
5 78.3 53.7 -1.9 -5.5 3.61 30.25 10.45
6 91.1 69.4 10.9 10.2 118.81 104.04 111.18
7 82.8 85.4 2.6 26.2 6.76 686.44 68.12
8 90.7 69.1 10.5 9.9 110.25 98.01 103.95
9 88.1 74.3 7.9 15.1 62.41 228.01 119.29
10 83.2 68.7 3 9.5 9 90.25 28.5
11 67.5 70.7 -12.7 11.5 161.29 132.25 -146.1
12 67.2 51.7 -13 -7.5 169 56.25 97.5
13 84.2 54.1 4 -5.1 16 26.01 -20.4
14 80.6 47.8 0.4 -11.4 0.16 129.96 -4.56
15 84.8 53.7 4.6 -5.5 21.16 30.25 -25.3
16 81.3 60 1.1 0.8 1.21 0.64 0.88
17 79.3 47.5 -0.9 -11.7 0.81 136.89 10.53
18 79.6 58.3 -0.6 -0.9 0.36 0.81 0.54
19 84.5 71.4 4.3 12.2 18.49 148.84 52.46
20 79.3 48.2 -0.9 -11 0.81 121 9.9
21 73.4 57.6 -6.8 -1.6 46.24 2.56 10.88
22 79 64.2 -1.2 5 1.44 25 -6
23 84.2 64.5 4 5.3 16 28.09 21.2
24 74.1 66.2 -6.1 7 37.21 49 -42.7
25 83.5 59.3 3.3 0.1 10.89 0.01 0.33
26 72.1 59.3 -8.1 0.1 65.61 0.01 -0.81
27 66.8 47.5 -13.4 -11.7 179.56 136.89 156.78
28 89.7 59.3 9.5 0.1 90.25 0.01 0.95
29 83.2 59.3 3 0.1 9 0.01 0.3
30 78.6 64.5 -1.6 5.3 2.56 28.09 -8.48
Total 80.2 59.2 1 2.6 1428.12 2901.3 636.27
The table 4.9 above, calculating table that explained formula for Post-test in Experimental and Post-test in control group was implemented to find t-critical value both groups as the basic to the hypothesis of the research.
The following formula of t-test was implemented to find out the t-observed value both groups as the basis to test hypothesis of this research.
a. Coefficient r
Rxy = 𝑛 ∑ 𝑋𝑖𝑦𝑖−(∑ 𝑥𝑖)(∑ 𝑦𝑖)
{𝑛 ∑ 𝑥12−(𝑥𝑖)2}{𝑛 ∑ 𝑦12−(𝑦𝑖)2}
= 30(636.27−(1)(2.6))
√30(1428.12−1)(30(2901.3−(2.6)2))
= 19088.1−2.6
√(42843.6−1)(87039−6.79)
= 19085.5
√3728687445.4
= 19085.5
61062.9
= 0.312
b. Examining the Statistical Hypothesis
Ha : P # O There is significance effect of applying three-steps interview technique towards the students’ speaking achievement.
Ho : P # O There is no a significance effect of applying three-steps interview technique towards the students’ speaking achievement.
t = 𝑥̅̅̅̅̅̅̅̅̅1−𝑥2
√𝑆12 𝑁1+𝑆2
2
𝑁2−2𝑅(𝑆1
√𝑁1)(𝑆2
√𝑁2)
= 80.27−59.33
√49.230+99.930−2(0,31(7.01
√30)(9.88
√30)
= 20.94
√1.64+3.33−(0.62)(1.27)(1.80)
= 20.94
√3.56
= 20.94
1.88
= 11.13
After measuring the data above by using t-test formula. It showed that t-observed value was 11.13. After seeking the table of the distribution of t-observed as the basis of accounting in certain degree of freedom (df). The calculation showed that:
Df = N1+ N2-2 = 30+30-2 = 58
In the line of 58 showed that ttable was 1.29. to>ttable which was 11.13>1.29 the fact hypothesis Ha was accepted and Ho was rejected.
D. The Findings
1. The result of the t-test showed that the tobserved was higher than ttable (11.13>1.29). it means that whole three-steps interview technique gave a significant effect on students’ speaking achievement.
2. The student were taught by applying Three-Steps Interview Technique got higher score that those taught by discussion method. The mean score of post-test in experimental group was 80.27 and the mean score of control group was 59.33 it can be known that the students’
achievement who taught by applying three-steps interview technique got higher score than the students who taught by using discussions method .
CHAPTER V
CONCLUSIONS ANDSUGGESTIONS
A. Conclusions
Based on the data analysis, there are some conclusions that can be described as follow:
3. The result of the t-test showed that the tobserved was higher than ttable (11.13>1.29). it means that whole three-steps interview technique gave a significant effect on students’ speaking achievement.
4. The student were taught by applying Three-Steps Interview Technique got higher score that those taught by discussion method. The mean score of post-test in experimental group was 80.27 and the mean score of control group was 59.33 it can be known that the students’
achievement who taught by applying three-steps interview technique got higher score than the students who taught by using discussions method .
B. Suggestions
in relation on the conclusions above, suggestions are put forward as follows:
1. To the English teacher
a. They should consider that the applying of three-steps interview technique towards the students’ speaking achievement can be enjoyable and fun way in teaching speaking especially in Expressing of Agreeing and Disagreeing Opinion. They should try to make variation in teaching speaking until the students comprehend and more confidence, so they can practice with other.