LAMPIRAN
Lampiran 1. Proses Pendugaan Galat pada RBSL dengan Satu Data Hilang 𝜇𝜇̂ 𝑦𝑦 = 𝑌𝑌�… =
𝜇𝜇̂ 𝑥𝑥 = 𝑋𝑋�… =
𝛾𝛾� =
549 = 34,3125 16
531 = 33,1875 16
𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥𝑥𝑥 153,25 = = 1,17124688 177 𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥
𝑎𝑎�𝑖𝑖 = (𝑌𝑌�𝑖𝑖.. − 𝑌𝑌�… ) − 𝛾𝛾�(𝑋𝑋�𝑖𝑖.. − 𝑋𝑋�… ) 𝑎𝑎�1 = �
125 124 − 34,3125� − 1,17124688 � − 33,1875� = −0,50039746 4 4
𝑎𝑎�2 = �
135 129 − 34,3125� − 1,17124688 � − 33,1875� = 0,53554395 4 4
𝑎𝑎�4 = �
115 111 − 34,3125� − 1,17124688 � − 33,1875� = 0,80615489 4 4
𝑎𝑎�3 = �
174 167 − 34,3125� − 1,17124688 � − 33,1875� = −0,84130139 4 4
𝛽𝛽̂𝑗𝑗 = �𝑌𝑌�.𝑗𝑗 . − 𝑌𝑌�… � − 𝛾𝛾��𝑋𝑋�.𝑗𝑗 . − 𝑋𝑋�… � 𝛽𝛽̂1 = �
145 132 − 34,3125� − 1,17124688 � − 33,1875� = 2,15710879 4 4
130 132 𝛽𝛽̂2 = � − 34,3125� − 1,17124688 � − 33,1875� = −1,59289121 4 4 133 129 − 34,3125� − 1,17124688 � − 33,1875� = 0,03554395 𝛽𝛽̂3 = � 4 4
𝛽𝛽̂4 = �
141 138 − 34,3125� − 1,17124688 � − 33,1875� = −0,59976153 4 4
𝜏𝜏̂ 𝑘𝑘 = (𝑌𝑌�..𝑘𝑘 − 𝑌𝑌�… ) − 𝛾𝛾�(𝑋𝑋�..𝑘𝑘 − 𝑋𝑋�… ) 𝜏𝜏̂𝐴𝐴 = �
145 139 − 34,3125� − 1,17124688 � − 33,1875� = 0,10742675 4 4
𝜏𝜏̂ 𝐵𝐵 = �
132 125 − 34,3125� − 1,17124688 � − 33,1875� = 0,95679082 4 4
133 134 𝜏𝜏̂ 𝐶𝐶 = � − 34,3125� − 1,17124688 � − 33,1875� = −1,42851465 4 4
Universitas Sumatera Utara
139 133 𝜏𝜏̂ 𝐷𝐷 = � − 34,3125� − 1,17124688 � − 33,1875� = 0,36429707 4 4
𝜀𝜀̂𝑖𝑖𝑖𝑖𝑖𝑖 = 𝑌𝑌𝑖𝑖𝑖𝑖𝑖𝑖 − 𝑌𝑌�𝑖𝑖𝑖𝑖𝑖𝑖 = 𝑌𝑌𝑖𝑖𝑖𝑖𝑖𝑖 − 𝑌𝑌�… − 𝑎𝑎�𝑖𝑖 − 𝛽𝛽̂𝑗𝑗 − 𝜏𝜏̂ 𝑘𝑘 − 𝛾𝛾��𝑋𝑋𝑖𝑖𝑖𝑖𝑖𝑖 − 𝑋𝑋�… �
𝜀𝜀̂11𝐴𝐴 = 27 − 34,3125 − (−0,50039746) − 2,15710879 − 0,10742675 −1,17124688(24 − 33,1875) = 1,68419260
𝜀𝜀̂12𝐵𝐵 = 25 − 34,3125 − (−0,50039746) − (−1,59289121) − 0,95679082 −1,17124688(26 − 33,1875) = 0,24233477
𝜀𝜀̂13𝐶𝐶 = 40 − 34,3125 − (−0,50039746) − 0,03554395 − (−1,42851465) −1,17124688(38 − 33,1875) = 1,94424256
𝜀𝜀̂14𝐷𝐷 = 33 − 34,3125 − (−0,50039746) − (−0,59976153) − 0,36429707 −1,17124688(36 − 33,1875) = −3,87076993
𝜀𝜀̂21𝐵𝐵 = 35 − 34,3125 − 0,53554395 − 2,15710879 − 0,95679082 −1,17124688(32 − 33,1875) = −1,57108789
𝜀𝜀̂22𝐶𝐶 = 22 − 34,3125 − 0,53554395 − (−1,59289121) − (−1,42851465) −1,17124688(26 − 33,1875) = −1,40830116
𝜀𝜀̂23𝐷𝐷 = 34 − 34,3125 − 0,53554395 − 0,035543953 − 0,36429707 −1,17124688(31 − 33,1875) = 1,31421758
𝜀𝜀̂24𝐴𝐴 = 44 − 34,3125 − 0,53554395 − (−0,59976153) − 0,10742675 −1,17124688(40 − 33,1875) = 1,66517147
𝜀𝜀̂31𝐶𝐶 = 53 − 34,3125 − (−0,84130139) − 2,15710879 − (−1,42851465) −1,17124688(50 − 33,1875) = −0,89138088
𝜀𝜀̂32𝐷𝐷 = 42 − 34,3125 − (−0,84130139) − (−1,59289121) − 0,36429707 −1,17124688(40 − 33,1875) = 1,77827618
𝜀𝜀̂33𝐴𝐴 = 33 − 34,3125 − (−0,84130139) − 0,03554395 − 0,10742675 −1,17124688(35 − 33,1875) = −2,73705428
𝜀𝜀̂34𝐵𝐵 = 46 − 34,3125 − (−0,84130139) − (−0,59976153) − 0,95679082 −1,17124688(42 − 33,1875) = 1,85015898
𝜀𝜀̂41𝐷𝐷 = 30 − 34,3125 − 0,80615489 − 2,15710879 − 0,36429707 −1,17124688(26 − 33,1875) = 0,77827618
Universitas Sumatera Utara
𝜀𝜀̂42𝐴𝐴 = 41 − 34,3125 − 0,80615489 − (−1,59289121) − 0,10742675 −1,17124688(40 − 33,1875) = −0,61230979
𝜀𝜀̂43𝐵𝐵 = 26 − 34,3125 − 0,80615489 − 0,03554395 − 0,95679082 −1,17124688(25 − 33,1875) = −0,52140586
𝜀𝜀̂44𝐶𝐶 = 18 − 34,3125 − 0,80615489 − (−0,59976153) − (−1,42851465) −1,17124688(20 − 33,1875) = 0,35543947
Universitas Sumatera Utara
Lampiran 2. Langkah-langkah Analisis Kovarian pada RBSL dengan Satu Data Hilang Diketahui: 𝑟𝑟
� 𝑥𝑥𝑖𝑖𝑖𝑖𝑖𝑖 = (24 + 26 + ⋯ + 20) = 531 𝑖𝑖𝑖𝑖𝑖𝑖 𝑟𝑟
� 𝑥𝑥𝑖𝑖𝑖𝑖𝑖𝑖 2 = (242 + 262 + ⋯ + 202 ) = 18643 𝑖𝑖𝑖𝑖𝑖𝑖 𝑟𝑟
� 𝑦𝑦𝑖𝑖𝑖𝑖𝑖𝑖 = (27 + 25 + ⋯ + 18) = 549 𝑖𝑖𝑖𝑖𝑖𝑖 𝑟𝑟
� 𝑦𝑦𝑖𝑖𝑖𝑖𝑖𝑖 2 = [272 + 252 + ⋯ + 182 ] = 20203 𝑖𝑖𝑖𝑖𝑖𝑖
𝑥𝑥…2 (531)2 = = 17622,5625 𝑟𝑟 2 16 2 2 𝑦𝑦… (549) = = 18837,5625 𝑟𝑟 2 16 𝑥𝑥… 𝑦𝑦… (531)(549) = = 18219,9375 16 𝑟𝑟 2 � � 𝑥𝑥𝑖𝑖𝑖𝑖𝑖𝑖 𝑦𝑦𝑖𝑖𝑖𝑖𝑖𝑖 = [(24)(27) + (26)(25) + ⋯ + (20)(18)] = 19359
Menghitung Jumlah Kuadrat Total (JKT) dan Jumlah Hasil Kali Total (JHKT) untuk variabel X dan Y 𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥 = 18643 − 17622,5625 = 1020,4375 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 = 20203 − 18837,5625 = 1365,4375 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥𝑥𝑥 = 19359 − 18219,9375 = 1139,0625
Menghitung Jumlah Kuadrat Baris (JKB) dan Jumlah Hasil Kali Baris (JHKB) untuk variabel X dan Y (124)2 + (129)2 + (167)2 + (111)2 − 17622,5625 = 434,1875 4 2 2 2 2 (125) + (135) + (174) + (115) 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 = − 18837,5625 = 500,1875 4
𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥 =
𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥𝑥𝑥 =
(124)(125) + (129)(135) + (167)(174) + (111)(115) 4 −18219,9375 = 464,5625
Menghitung Jumlah Kuadrat Kolom (JKK) dan Jumlah Hasil Kali Kolom (JHKK) untuk variabel X dan Y (132)2 + (132)2 + (129)2 + (138)2 − 17622,5625 = 10,6875 4 2 2 2 2 (145) + (130) + (133) + (141) − 18837,5625 = 36,1875 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 = 4 (132)(145) + (132)(130) + (129)(133) + (138)(141) 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥𝑥𝑥 = 4 −18219,9375 = 8,8125 𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥 =
Universitas Sumatera Utara
Menghitung Jumlah Kuadrat Perlakuan (JKP) dan Jumlah Hasil Kali Perlakuan (JHKP) untuk variabel X dan Y (139)2 + (125)2 + (134)2 + (133)2 − 17622,5625 = 25,1875 4 (145)2 + (132)2 + (133)2 + (139)2 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 = − 18837,5625 = 27,1875 4 (139)(145) + (125)(132) + (134)(133) + (133)(139) 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥𝑥𝑥 = 4 −18219,9375 = 21,0625 𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥 =
Menghitung Jumlah Kuadrat Galat (JKG) dan Jumlah Hasil Kali Galat (JHKG) untuk variabel X dan Y 𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥 = 1020,4375 − 434,1875 − 10,6875 − 25,1875 = 550,375 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 = 1365,4375 − 500,1875 − 36,1875 − 27,1875 = 801,875 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥𝑥𝑥 = 1139,0625 − 464,5625 − 8,8125 − 21,0625 = 644,625
Menghitung Jumlah Kuadrat Terkoreksi
(644,625)2 = 46,85998183 550,375 (1109,1875)2 = 52,47508411 𝐽𝐽𝐽𝐽(𝐵𝐵 + 𝐺𝐺) terkoreksi = 1302,0625 − 984,5625 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 terkoreksi = 52,47508411 − 46,85998183 = 5,61510228 (653,4375)2 𝐽𝐽𝐽𝐽(𝐾𝐾 + 𝐺𝐺) terkoreksi = 838,0625 − = 77,04110505 561,0625 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 terkoreksi = 77,04110505 − 46,85998183 = 30,18112322 (665,6875)2 𝐽𝐽𝐽𝐽(𝑃𝑃 + 𝐺𝐺) terkoreksi = 829,0625 − = 59,13769139 575,5625 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 terkoreksi = 59,13769139 − 46,85998183 = 12,27770956 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 terkoreksi = 801,875 −
Menghitung db terkoreksi untuk galat baris, kolom, dan perlakuan db galat terkoreksi = (r – 1)(r – 2) -1= 5 db baris terkoreksi = r – 1 = 3 db kolom terkoreksi = r – 1 = 3 db perlakuan terkoreksi = r – 1 = 3
Menghitung Kuadrat Tengah (KT) 𝐾𝐾𝐾𝐾𝐾𝐾 terkoreksi =
46,85998183 = 9,37199637 5
𝐾𝐾𝐾𝐾𝐾𝐾 terkoreksi =
30,18112322 = 10,06037441 3
𝐾𝐾𝐾𝐾𝐾𝐾 terkoreksi =
𝐾𝐾𝐾𝐾𝐾𝐾 terkoreksi =
5,61510228 = 1,87170076 3
12,27770956 = 4,09256985 3
Universitas Sumatera Utara
Lampiran 3. Proses Pendugaan Galat pada RBSL dengan Dua Data Hilang 𝜇𝜇̂ 𝑦𝑦 = 𝑌𝑌�… =
𝜇𝜇̂ 𝑥𝑥 = 𝑋𝑋�… = 𝛾𝛾� =
548 = 34,25 16
531 = 33,1875 16
𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥𝑥𝑥 631,375 = = 1,14717238 550,375 𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥
𝑎𝑎�𝑖𝑖 = (𝑌𝑌�𝑖𝑖.. − 𝑌𝑌�… ) − 𝛾𝛾�(𝑋𝑋�𝑖𝑖.. − 𝑋𝑋�… ) 𝑎𝑎�1 = �
125 124 − 34,25� − 1,14717238 � − 33,1875� = −0,49056041 4 4
𝑎𝑎�2 = �
131,5 129 − 34,25� − 1,14717238 � − 33,1875� = −0,29952589 4 4
𝑎𝑎�4 = �
113,5 111 − 34,25� − 1,14717238 � − 33,1875� = 0,36274983 4 4
𝑎𝑎�3 = �
178 167 − 34,25� − 1,14717238 � − 33,1875� = 0,42733648 4 4
𝛽𝛽̂𝑗𝑗 = �𝑌𝑌�.𝑗𝑗 . − 𝑌𝑌�… � − 𝛾𝛾��𝑋𝑋�.𝑗𝑗 . − 𝑋𝑋�… � 𝛽𝛽̂1 = �
141,5 132 − 34,25� − 1,14717238 � − 33,1875� = 1,34009482 4 4
134 132 𝛽𝛽̂2 = � − 34,25� − 1,14717238 � − 33,1875� = −0,53490518 4 4
131,5 129 𝛽𝛽̂3 = � − 34,25� − 1,14717238 � − 33,1875� = −0,29952589 4 4
𝛽𝛽̂4 = �
141 138 − 34,25� − 1,14717238 � − 33,1875� = −0,50566375 4 4
𝜏𝜏̂ 𝑘𝑘 = (𝑌𝑌�..𝑘𝑘 − 𝑌𝑌�… ) − 𝛾𝛾�(𝑋𝑋�..𝑘𝑘 − 𝑋𝑋�… ) 𝜏𝜏̂𝐴𝐴 = �
145 139 − 34,25� − 1,14717238 � − 33,1875� = 0,20754315 4 4
𝜏𝜏̂ 𝐵𝐵 = �
127 125 − 34,25� − 1,14717238 � − 33,1875� = −0,27735351 4 4
133 134 𝜏𝜏̂ 𝐶𝐶 = � − 34,25� − 1,14717238 � − 33,1875� = −1,35849137 4 4
143 133 − 34,25� − 1,14717238 � − 33,1875� = 1,42830173 𝜏𝜏̂ 𝐷𝐷 = � 4 4
Universitas Sumatera Utara
𝜀𝜀̂𝑖𝑖𝑖𝑖𝑖𝑖 = 𝑌𝑌𝑖𝑖𝑖𝑖𝑖𝑖 − 𝑌𝑌�𝑖𝑖𝑖𝑖𝑖𝑖 = 𝑌𝑌𝑖𝑖𝑖𝑖𝑖𝑖 − 𝑌𝑌�… − 𝑎𝑎�𝑖𝑖 − 𝛽𝛽̂𝑗𝑗 − 𝜏𝜏̂ 𝑘𝑘 − 𝛾𝛾��𝑋𝑋𝑖𝑖𝑖𝑖𝑖𝑖 − 𝑋𝑋�… �
𝜀𝜀̂11𝐴𝐴 = 27 − 34,25 − (−0,49056041) − 1,34009482 − 0,20754315 −1,14717238(24 − 33,1875) = 2,23256870
𝜀𝜀̂12𝐵𝐵 = 25 − 34,25 − (−0,49056041) − (−0,53490518) − (−0,27735351) −1,14717238(26 − 33,1875) = 0,29812060
𝜀𝜀̂13𝐶𝐶 = 40 − 34,25 − (−0,49056041) − (−0,29952589) − (−1,35849137) −1,14717238(38 − 33,1875) = 2,37781058
𝜀𝜀̂14𝐷𝐷 = 33 − 34,25 − (−0,49056041) − (−0,50566375) − 1,42830173 −1,14717238(36 − 33,1875) = −4,90849989
𝜀𝜀̂21𝐵𝐵 = 31,5 − 34,25 − (−0,29952589) − −1,34009482 − (−0,27735351) −1,14717238(32 − 33,1875) = −2,15094822
𝜀𝜀̂22𝐶𝐶 = 22 − 34,25 − (−0,29952589) − (−0,53490518) − (−1,35849137) −1,14717238(26 − 33,1875) = −1,81177606
𝜀𝜀̂23𝐷𝐷 = 34 − 34,25 − (−0,29952589) − (−0,29952589) − 1,42830173 −1,14717238(31 − 33,1875) = 1,43018964
𝜀𝜀̂24𝐴𝐴 = 44 − 34,25 − (−0,29952589) − (−0,50566375) − 0,20754315 −1,14717238(40 − 33,1875) = 2,53253464
𝜀𝜀̂31𝐶𝐶 = 53 − 34,25 − 0,42733648 − −1,34009482 − (−1,35849137) −1,14717238(50 − 33,1875) = −0,94577561
𝜀𝜀̂32𝐷𝐷 = 46 − 34,25 − 0,42733648 − (−0,53490518) − 1,42830173 −1,14717238(40 − 33,1875) = 2,61415512
𝜀𝜀̂33𝐴𝐴 = 33 − 34,25 − 0,42733648 − (−0,29952589) − 0,20754315 −1,14717238(35 − 33,1875) = −3,66460368
𝜀𝜀̂34𝐵𝐵 = 46 − 34,25 − 0,42733648 − (−0,50566375) − (−0,27735351) −1,14717238(42 − 33,1875) = 1,99622417
𝜀𝜀̂41𝐷𝐷 = 30 − 34,25 − 0,36274983 − −1,34009482 − 1,42830173 −1,14717238(26 − 33,1875) = 0,86415512
𝜀𝜀̂42𝐴𝐴 = 41 − 34,25 − 0,36274983 − (−0,53490518) − 0,20754315 −1,14717238(40 − 33,1875) = −1,10049966
Universitas Sumatera Utara
𝜀𝜀̂43𝐵𝐵 = 24,5 − 34,25 − 0,36274983 − (−0,29952589) − (−0,27735351) −1,14717238(25 − 33,1875) = −0,14339655
𝜀𝜀̂44𝐶𝐶 = 18 − 34,25 − 0,36274983 − (−0,50566375) − (−1,35849137) −1,14717238(20 − 33,1875) = 0,37974109
Universitas Sumatera Utara
Lampiran 4. Langkah-langkah Analisis Kovarian pada RBSL dengan Dua Data Hilang Diketahui: 𝑟𝑟
� 𝑥𝑥𝑖𝑖𝑖𝑖𝑖𝑖 = (24 + 26 + ⋯ + 20) = 531 𝑖𝑖𝑖𝑖𝑖𝑖 𝑟𝑟
� 𝑥𝑥𝑖𝑖𝑖𝑖𝑖𝑖 2 = (242 + 262 + ⋯ + 202 ) = 18643 𝑖𝑖𝑖𝑖𝑖𝑖 𝑟𝑟
� 𝑦𝑦𝑖𝑖𝑖𝑖𝑖𝑖 = (27 + 25 + ⋯ + 18) = 548 𝑖𝑖𝑖𝑖𝑖𝑖 𝑟𝑟
� 𝑦𝑦𝑖𝑖𝑖𝑖𝑖𝑖 2 = [272 + 252 + ⋯ + 182 ] = 20246,5 𝑖𝑖𝑖𝑖𝑖𝑖
𝑥𝑥…2 (531)2 = = 17622,5625 𝑟𝑟 2 16 2 2 𝑦𝑦… (548) = = 18769 𝑟𝑟 2 16 𝑥𝑥… 𝑦𝑦… (531)(548) = = 18186,75 16 𝑟𝑟 2 � � 𝑥𝑥𝑖𝑖𝑖𝑖𝑖𝑖 𝑦𝑦𝑖𝑖𝑖𝑖𝑖𝑖 = [(24)(27) + (26)(25) + ⋯ + (20)(18)] = 19369,5
Menghitung Jumlah Kuadrat Total (JKT) dan Jumlah Hasil Kali Total (JHKT) untuk variabel X dan Y 𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥 = 18643 − 17622,5625 = 1020,4375 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 = 20246,5 − 18769 = 1477,5 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥𝑥𝑥 = 19369,5 − 18186,75 = 1182,75
Menghitung Jumlah Kuadrat Baris (JKB) dan Jumlah Hasil Kali Baris (JHKB) untuk variabel X dan Y (124)2 + (129)2 + (167)2 + (111)2 − 17622,5625 = 434,1875 4 2 2 2 (125) + (131,5) + (178) + (113,5)2 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 = − 18769 = 601,88 4
𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥 =
𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥𝑥𝑥 =
(124)(125) + (129)(131,5) + (167)(178) + (111)(113,5) 4 −18186,75 = 510,25
Menghitung Jumlah Kuadrat Kolom (JKK) dan Jumlah Hasil Kali Kolom (JHKK) untuk variabel X dan Y (132)2 + (132)2 + (129)2 + (138)2 − 17622,5625 = 10,6875 4 2 2 2 (141,5) + (134) + (131,5) + (141)2 − 18769 = 18,88 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 = 4 (132)(141,5) + (132)(134) + (129)(131,5) + (138)(141) 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥𝑥𝑥 = 4 −18186,75 = 10,125 𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥 =
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Menghitung Jumlah Kuadrat Perlakuan (JKP) dan Jumlah Hasil Kali Perlakuan (JHKP) untuk variabel X dan Y (139)2 + (125)2 + (134)2 + (133)2 − 17622,5625 = 25,1875 4 (145)2 + (127)2 + (133)2 + (143)2 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 = − 18769 = 54,0 4 (139)(145) + (125)(127) + (134)(133) + (133)(143) 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥𝑥𝑥 = 4 −18186,75 = 31,0 𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥 =
Menghitung Jumlah Kuadrat Galat (JKG) dan Jumlah Hasil Kali Galat (JHKG) untuk variabel X dan Y 𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥 = 1020,4375 − 434,1875 − 10,6875 − 25,1875 = 550,375 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 = 1477,5 − 601,88 − 18,88 − 54,0 = 802,75 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝑥𝑥𝑥𝑥 = 1182,75 − 510,25 − 10,125 − 31,0 = 631,375
Menghitung Jumlah Kuadrat Terkoreksi
(631,375)2 = 78,45403702 550,375 (1141,625)2 = 80,88707802 𝐽𝐽𝐽𝐽(𝐵𝐵 + 𝐺𝐺) terkoreksi = 1404,63 − 984,5625 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 terkoreksi = 80,88707802 − 78,45403702 = 2,428041 (641,5)2 𝐽𝐽𝐽𝐽(𝐾𝐾 + 𝐺𝐺) terkoreksi = 821,63 − = 88,16046675 561,0625 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 terkoreksi = 88,16046675 − 78,45403702 = 9,70142973 (662,375)2 𝐽𝐽𝐽𝐽(𝑃𝑃 + 𝐺𝐺) terkoreksi = 856,75 − = 94,46850907 575,5625 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 terkoreksi = 94,46850907 − 78,45403702 = 16,01447205 𝐽𝐽𝐽𝐽𝐽𝐽𝑦𝑦 terkoreksi = 802,75 −
Menghitung db terkoreksi untuk galat baris, kolom, dan perlakuan db galat terkoreksi = (r – 1)(r – 2) -2= 4 db baris terkoreksi = r – 1 = 3 db kolom terkoreksi = r – 1 = 3 db perlakuan terkoreksi = r – 1 = 3
Menghitung Kuadrat Tengah (KT) 𝐾𝐾𝐾𝐾𝐾𝐾 terkoreksi =
78,45403702 = 19,61350926 4
𝐾𝐾𝐾𝐾𝐾𝐾 terkoreksi =
9,70142973 = 3,23380991 3
𝐾𝐾𝐾𝐾𝐾𝐾 terkoreksi =
𝐾𝐾𝐾𝐾𝐾𝐾 terkoreksi =
2,428041 = 0,809347 3
16,01447205 = 5,33815735 3
Universitas Sumatera Utara
Lampiran 5. Daftar Nilai Kritik Sebaran F pada Taraf Kritis 5%
Universitas Sumatera Utara
