Anava Regresi-simpangan Model Dan Galat Murni

UTS STATISTIK PROGRAM PASCA SARJANA PENGAJAR : TOTO WARSA, Ir., M.S.

PENYUSUN: Nama : Ade Setiawan NPM : 150220060003

PERTANYAAN DAN JAWABAN

Pertanyaan: a. Hitung nilai β0 dan β1 untuk model regresi Y = β0 + β1X + ε dan tentukanlah persamaan garis regresinya! b. Lakukan Uji statitik, Analisis Varians dan termasuk Simpangan Model dan Galat Murninya! c. Kerjakan point a dan b untuk data yang sudah di tranformasi logaritma

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Jawab: 1. Analisis data sebelum ditransformasikan: No 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47

X 20.0 20.0 20.5 20.5 20.5 20.5 21.0 21.0 21.5 21.5 21.5 21.5 21.5 21.5 21.5 21.5 21.5 21.5 21.5 21.5 21.5 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.5 22.5 22.5 22.5 22.5 22.5

Y 90 105 90 90 100 110 100 105 100 105 110 110 110 110 110 110 115 115 115 115 120 110 110 115 115 115 115 120 120 120 120 120 120 120 120 120 125 125 125 130 130 115 115 115 120 120 120

X2 400.00 400.00 420.25 420.25 420.25 420.25 441.00 441.00 462.25 462.25 462.25 462.25 462.25 462.25 462.25 462.25 462.25 462.25 462.25 462.25 462.25 484.00 484.00 484.00 484.00 484.00 484.00 484.00 484.00 484.00 484.00 484.00 484.00 484.00 484.00 484.00 484.00 484.00 484.00 484.00 484.00 506.25 506.25 506.25 506.25 506.25 506.25

Y2 8100 11025 8100 8100 10000 12100 10000 11025 10000 11025 12100 12100 12100 12100 12100 12100 13225 13225 13225 13225 14400 12100 12100 13225 13225 13225 13225 14400 14400 14400 14400 14400 14400 14400 14400 14400 15625 15625 15625 16900 16900 13225 13225 13225 14400 14400 14400

XY 1800.0 2100.0 1845.0 1845.0 2050.0 2255.0 2100.0 2205.0 2150.0 2257.5 2365.0 2365.0 2365.0 2365.0 2365.0 2365.0 2472.5 2472.5 2472.5 2472.5 2580.0 2420.0 2420.0 2530.0 2530.0 2530.0 2530.0 2640.0 2640.0 2640.0 2640.0 2640.0 2640.0 2640.0 2640.0 2640.0 2750.0 2750.0 2750.0 2860.0 2860.0 2587.5 2587.5 2587.5 2700.0 2700.0 2700.0

Ulangan

JKGM

2

112.50000

4

275.00000

2

12.50000

13

307.69231

20

573.75000

2

No 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Jumlah Rataan

X 22.5 22.5 22.5 22.5 22.5 22.5 22.5 22.5 22.5 22.5 22.5 22.5 22.5 22.5 22.5 22.5 22.5 23.0 23.0 23.0 23.0 23.0 23.0 23.0 23.0 23.0 23.5 23.5 23.5 23.5 24.0 24.0 25.0 1775 22.1875

X Y ΣX i ΣYi

Y 120 120 120 125 125 125 125 125 130 130 130 130 135 135 135 140 150 125 130 140 140 140 140 145 145 145 145 150 155 170 165 170 170 9910 123.875

= 22.1875 = 123.8750 = 1775 = 9910

X2 506.25 506.25 506.25 506.25 506.25 506.25 506.25 506.25 506.25 506.25 506.25 506.25 506.25 506.25 506.25 506.25 506.25 529.00 529.00 529.00 529.00 529.00 529.00 529.00 529.00 529.00 552.25 552.25 552.25 552.25 576.00 576.00 625.00 39443

Y2 14400 14400 14400 15625 15625 15625 15625 15625 16900 16900 16900 16900 18225 18225 18225 19600 22500 15625 16900 19600 19600 19600 19600 21025 21025 21025 21025 22500 24025 28900 27225 28900 28900 1250800

ΣX i 2 ΣYi ΣX i Y JKGM 2

XY 2700.0 2700.0 2700.0 2812.5 2812.5 2812.5 2812.5 2812.5 2925.0 2925.0 2925.0 2925.0 3037.5 3037.5 3037.5 3150.0 3375.0 2875.0 2990.0 3220.0 3220.0 3220.0 3220.0 3335.0 3335.0 3335.0 3407.5 3525.0 3642.5 3995.0 3960.0 4080.0 4250.0 220930

Ulangan

JKGM

23

1660.86957

9

388.88889

4

350.00000

2

12.50000 3693.7008

= 39443 = 1250800 = 220930 = 3693.7008

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a. Jika model regresinya Y = β0 + β1X + ε, tentukanlah persamaan garis regresinya! ΣX i ΣYi β 0 = Y − β1 X ΣX i Yi − = 123.8750 − 17.47664(22.1875) n β1 = = -263.8879 2 (ΣX i ) 2 ΣX i − n (1775)(9910) 220930 − 80 = (1775) 2 39443 − 80 = 17.47664

Sehingga persamaan regresinya: Y = -263.88785 + 17.47664 X Koefisien Korelasi:

ΣX i ΣYi n ryx = 2 ⎛ (ΣX i ) ⎞⎛ 2 (ΣYi ) 2 ⎞ ⎟ ⎟⎜ ΣYi − ⎜⎜ ΣX i 2 − n ⎟⎠⎜⎝ n ⎟⎠ ⎝ (1775)(9910) 220930 − 80 = 2 ⎛ (1775) ⎞⎛ (9910) 2 ⎜⎜ 39443 − ⎟⎟⎜⎜1250800 − 80 ⎠⎝ 80 ⎝ = 0.890182

Koefisien Determinasi: 2 R 2 = ryx = (0.890182) 2 = 0.792424

ΣX i Yi −

⎞ ⎟⎟ ⎠

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b. Uji Statistik (Analisis Varians termasuk Simpangan Model dan Galat Murni) (ΣΣi ) 2 ( 9910 ) 2 = FK = 80 n = 1227601.25

JKT = ΣYi − FK = 1250800-1227601.25 = 23198.75 2

ΣX i ΣYi ⎤ ⎡ JK Re gresi = β1 ⎢ ΣX i Yi − n ⎥⎦ ⎣ ( 1775 )( 9910 ) ⎤ ⎡ = 17.47664⎢220930⎥⎦ 80 ⎣ = 18383.235981 JK Re sidu = JKT-JK Re g = 23198.75 − 18383.24 = 4815.51 JKGM = 3693.700762 JKSM = JK Re sidu-JKGM = 4815.51 - 3693.700762 = 1121.813257 Tabel Analisis Varians Regresi: Sumber Ragam Regresi Residu SM GM Total

DB JK RJK 1 18383.235981 18383.24 78 4815.514019 61.73736 8 1121.813257 140.2267 70 3693.700762 52.76715 79 23198.750000

F-hit 297.77

F.05 3.96

F.01 6.97

2.66*

2.07

2.78

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2. Analisis data setelah ditransformasikan: No 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

X 1.30 1.30 1.31 1.31 1.31 1.31 1.32 1.32 1.33 1.33 1.33 1.33 1.33 1.33 1.33 1.33 1.33 1.33 1.33 1.33 1.33 1.34 1.34 1.34 1.34 1.34 1.34 1.34 1.34 1.34 1.34 1.34 1.34 1.34 1.34 1.34 1.34 1.34 1.34 1.34 1.34 1.35 1.35 1.35 1.35 1.35 1.35 1.35 1.35

Y 1.95 2.02 1.95 1.95 2.00 2.04 2.00 2.02 2.00 2.02 2.04 2.04 2.04 2.04 2.04 2.04 2.06 2.06 2.06 2.06 2.08 2.04 2.04 2.06 2.06 2.06 2.06 2.08 2.08 2.08 2.08 2.08 2.08 2.08 2.08 2.08 2.10 2.10 2.10 2.11 2.11 2.06 2.06 2.06 2.08 2.08 2.08 2.08 2.08

X2 1.69 1.69 1.72 1.72 1.72 1.72 1.74 1.74 1.77 1.77 1.77 1.77 1.77 1.77 1.77 1.77 1.77 1.77 1.77 1.77 1.77 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.82 1.82 1.82 1.82 1.82 1.82 1.82 1.82

Y2 3.8025 4.0804 3.8025 3.8025 4.0000 4.1616 4.0000 4.0804 4.0000 4.0804 4.1616 4.1616 4.1616 4.1616 4.1616 4.1616 4.2436 4.2436 4.2436 4.2436 4.3264 4.1616 4.1616 4.2436 4.2436 4.2436 4.2436 4.3264 4.3264 4.3264 4.3264 4.3264 4.3264 4.3264 4.3264 4.3264 4.4100 4.4100 4.4100 4.4521 4.4521 4.2436 4.2436 4.2436 4.3264 4.3264 4.3264 4.3264 4.3264

XY 2.5350 2.6260 2.5545 2.5545 2.6200 2.6724 2.6400 2.6664 2.6600 2.6866 2.7132 2.7132 2.7132 2.7132 2.7132 2.7132 2.7398 2.7398 2.7398 2.7398 2.7664 2.7336 2.7336 2.7604 2.7604 2.7604 2.7604 2.7872 2.7872 2.7872 2.7872 2.7872 2.7872 2.7872 2.7872 2.7872 2.8140 2.8140 2.8140 2.8274 2.8274 2.7810 2.7810 2.7810 2.8080 2.8080 2.8080 2.8080 2.8080

Ulangan

JKGM

2

0.00245

4

0.00570

2

0.00020

13

0.00492

20

0.00772

6

No 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

X 1.35 1.35 1.35 1.35 1.35 1.35 1.35 1.35 1.35 1.35 1.35 1.35 1.35 1.35 1.35 1.36 1.36 1.36 1.36 1.36 1.36 1.36 1.36 1.36 1.37 1.37 1.37 1.37 1.38 1.38 1.40 107.5

Y 2.08 2.10 2.10 2.10 2.10 2.10 2.11 2.11 2.11 2.11 2.13 2.13 2.13 2.15 2.18 2.10 2.11 2.15 2.15 2.15 2.15 2.16 2.16 2.16 2.16 2.18 2.19 2.23 2.22 2.23 2.23 167.12

X2 1.82 1.82 1.82 1.82 1.82 1.82 1.82 1.82 1.82 1.82 1.82 1.82 1.82 1.82 1.82 1.85 1.85 1.85 1.85 1.85 1.85 1.85 1.85 1.85 1.88 1.88 1.88 1.88 1.90 1.90 1.96 144.4772

1.3438

2.0890

1.805965

X Y ΣX i ΣYi

= 1.3438 = 2.0890 = 107.5 = 167.12

ΣX i 2 ΣYi ΣX i Y JKGM 2

Y2 4.3264 4.4100 4.4100 4.4100 4.4100 4.4100 4.4521 4.4521 4.4521 4.4521 4.5369 4.5369 4.5369 4.6225 4.7524 4.4100 4.4521 4.6225 4.6225 4.6225 4.6225 4.6656 4.6656 4.6656 4.6656 4.7524 4.7961 4.9729 4.9284 4.9729 4.9729 349.3934 4.367417 5

XY 2.8080 2.8350 2.8350 2.8350 2.8350 2.8350 2.8485 2.8485 2.8485 2.8485 2.8755 2.8755 2.8755 2.9025 2.9430 2.8560 2.8696 2.9240 2.9240 2.9240 2.9240 2.9376 2.9376 2.9376 2.9592 2.9866 3.0003 3.0551 3.0636 3.0774 3.1220 224.6412

Ulangan

JKGM

23

0.01918

9

0.00400

4

0.00260

2

0.00005 0.046825686

2.808015

= 144.4772 = 349.3934 = 224.6412 = 0.04682569

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a. Jika model regresinya Y = β0Xβ1ε, tentukanlah β0 dan β1 serta persamaan garis regresinya! Jika persamaan Y = β0Xβ1ε, di rubah jadi bentuk logaritma: Log Y = Log β0 + β1 Log X + Log ε atau Y* = β0* + β1X* + ε* ΣX i ΣYi n β1 = (ΣX i ) 2 2 ΣX i − n (107.5)(167.12) 224.6412 − 80 = (107.5) 2 144.4772 − 80 = 3.061267 ΣX i Yi −

β 0 * = Y − β1 X

= 2.0890 − 3.061267(1.3438) = -2.024577

Persamaan regresinya: Y* = -2.0246 + 3.0613 X Dari persamaan tersebut bisa didapatkan nilai β0 dan β1: β0 = 10β0 = 10(-2.024577362) = 0.0094498 ; dan β1 = 3.06127 Sehingga persamaan regresinya: Y = 0.00945 X 3.06127

Koefisien Korelasi:

ΣX i ΣYi n ryx = 2 ⎛ (ΣX i ) ⎞⎛ 2 (ΣYi ) 2 ⎞ ⎟⎜ ΣYi − ⎟ ⎜⎜ ΣX i 2 − n ⎟⎠⎜⎝ n ⎟⎠ ⎝ (107.5)(167.12) 224.6412 − 80 = 2 ⎛ (107.5) ⎞⎛ (167.12) 2 ⎜⎜144.4772 − ⎟⎟⎜⎜ 349.3934 − 80 ⎠⎝ 80 ⎝ = 0.898096

Koefisien Determinasi: 2 R 2 = ryx = (0.890182) 2 = 0.792424

ΣX i Yi −

⎞ ⎟⎟ ⎠

8

b. Uji Statistik (Analisis Varians termasuk Simpangan Model dan Galat Murni) (ΣYi ) 2 ( 167.12 ) 2 FK = = n 80 = 349.113680 JKT = ΣYi − FK = 349.393400 -349.113680 = 0.279720 2

ΣX i ΣYi ⎤ ⎡ JK Re gresi = β1 ⎢ ΣX i Yi − n ⎥⎦ ⎣ ( 107.5 )( 167.12 ) ⎤ ⎡ = 3.061267 ⎢224.6412⎥⎦ 80 ⎣ = 0.225615 JK Re sidu = JKT-JK Re g = 0.279720 − 0.225615 = 0.054105 JKGM = 0.046826 JKSM = JK Re sidu-JKGM = 0.054105 - 0.046826 = 0.007279

Tabel Analisis Varians Regresi: Sumber Ragam Regresi Residu SM GM Total

DB 1 78 8 70 79

JK 0.225615 0.054105 0.007279 0.046826 0.279720

RJK F-hit F.05 F.01 0.225615 325.26 3.96 6.97 0.000694 2.07 2.78 0.000910 1.36tn 0.000669

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