Dian Abdul Fatah M. Fikri Firdaus M. Reza Pahlefi Nice Rahma Zika Kelompok 9 Dian Abdul Fatah M. Fikri Firdaus M. Reza Pahlefi Nice Rahma Zika

Jawaban no.1 Db = n1 – 1 Db = n2-1 = 5 – 1 Db = 7-1 = 4 Db = 6 X1 = 𝟔𝟑,𝟓 𝟓 = 12,7 X2 = 𝟗𝟏,𝟐 𝟕 = 13,1 S1 = Σ𝑿𝟏 𝟐 − (Σ𝑿𝟏) 𝟐 /𝒏 𝒏−𝟏 = (𝟏𝟐,𝟔 𝟐 , 𝟏𝟑,𝟒 𝟐 , 𝟏𝟏,𝟗 𝟐 , 𝟏𝟐,𝟔 𝟐 , 𝟏𝟑,𝟎 𝟐 ) − (𝟔𝟑,𝟓) 𝟐 𝟓 𝟓−𝟏 = 𝟖𝟎𝟕,𝟔𝟗 − 𝟒𝟎𝟑𝟐,𝟐𝟓 𝟓 𝟒 - 𝟖𝟎𝟕,𝟔𝟗−𝟖𝟎𝟔,𝟒𝟓 𝟒 = 𝟏,𝟐𝟒 𝟒 = 𝟎,𝟑𝟏 = 0,557 S2 = Σ𝑿𝟏 𝟐 − (Σ𝑿𝟏 𝟐 /𝒏) 𝒏−𝟏 = (𝟏𝟑,𝟏 𝟐 , 𝟏𝟑,𝟒 𝟐 , 𝟏𝟐,𝟖 𝟐 , 𝟏𝟑,𝟓 𝟐 , 𝟏𝟑,𝟑 𝟐 ,𝟏𝟐,𝟕 𝟐 , 𝟏𝟐,𝟒 𝟐 ) − (𝟗𝟏,𝟐) 𝟐 𝟕 𝟕−𝟏 = 𝟏𝟏𝟖𝟗,𝟐 − 𝟖𝟑𝟏𝟕,𝟒𝟒 𝟕 𝟔 - 𝟏𝟏𝟖𝟗,𝟐−𝟏𝟏𝟖𝟖,𝟐𝟏 𝟔 = 𝟎,𝟗𝟗 𝟔 = 𝟎,𝟏𝟔𝟓 = 0,406

Lanjutan jawaban no.1 S 2 𝑝= 𝑑𝑏1 2 .𝑠1 2 + 𝑑𝑏2 2 .𝑠2 2 𝑑𝑏1+𝑑𝑏2 = 4 (0,557) 2 + 6 (0,406) 2 4+6 = 4 0,310 +6 (0,165) 10 = 1,24+0,99 10 = 2,23 10 = 0,223 Thit = x̄1 −x̄2 −(μ1 −μ2) S 2 𝑝 ( 1 𝑛1 + 1 𝑛2 ) = 12,7 −13,1 − 0 0,223 ( 1 5 + 1 7 ) = −0,4 0,223 𝑥 0,34 = −0,4 0,1 = −0,4 0,32 = -1,25 maka Keputusan = Thit < Ttabel  Terima H0 Keputusan = Kandungan protein pada suatu varietas gandum di daerah 1 sama dengan daerah 2

Jawaban no.2 S 2 𝑝 = 𝒅𝒃𝟏 . S1 2 + 𝒅𝒃𝟐 . S2 2 𝒅𝒃𝟏+𝒅𝒃𝟐 = 𝟗 . (3000) 2 + 𝟏𝟏 . (2500) 2 𝟗+𝟏𝟏 = 𝟗 𝒙 𝟗.𝟎𝟎𝟎.𝟎𝟎𝟎 +𝟏𝟏 𝒙 𝟔.𝟐𝟓𝟎.𝟎𝟎𝟎 𝟐𝟎 = 𝟖𝟏.𝟎𝟎𝟎.𝟎𝟎𝟎 + 𝟔𝟖.𝟕𝟓𝟎.𝟎𝟎𝟎 𝟐𝟎 = 𝟏𝟒𝟗.𝟕𝟓𝟎.𝟎𝟎𝟎 𝟐𝟎 = 7.487.500 Tbit = x̄𝟏 −x̄𝟐 −(μ𝟏 −μ𝟐) S 𝟐 𝒑 ( 𝟏 𝒏𝟏 + 𝟏 𝒏𝟐 ) = 𝟏𝟐.𝟎𝟎𝟎 −𝟏𝟒.𝟓𝟎𝟎 − 𝟎 𝟕.𝟒𝟖𝟕.𝟓𝟎𝟎 ( 𝟏 𝟏𝟎 + 𝟏 𝟏𝟐 ) = −𝟐.𝟓𝟎𝟎 𝟕.𝟒𝟖𝟕.𝟓𝟎𝟎 𝒙 𝟎,𝟏𝟖 = −𝟐.𝟓𝟎𝟎 𝟎,𝟏 = -2,15 Keputusan : Thit > Ttab  Tolak H0 signifikan. Maka, lokasi pedagang kaki lima di Jln.Cihideung Tasikmalaya dan di Jln.Hz. Mustofa ada pengaruhnya terhadap besarnya pendapatan

jawaban no.3 S1 2 = 𝑃1 − 𝑄1 𝒏𝟏 S2 2 = 𝑃2 − 𝑄2 𝒏𝟐 S ( P1 – P2 ) = S1 2 + S2 2 = 𝟎,𝟑𝟐 − 𝟎,𝟔𝟖 𝟒𝟗𝟔 = 𝟎,𝟐𝟓 − 𝟎,𝟕𝟓 𝟓𝟖𝟗 = 𝟎,𝟎𝟎𝟎𝟒𝟑𝟗+𝟎,𝟎𝟎𝟎 = 0,00439 = 0,00318 = 𝟕,𝟓𝟕 𝒙 10 −4 = 0,028 Thit = 𝑷𝟏 −𝑷𝟐 𝑺 (𝑷𝟏 −𝑷𝟐) db = n1 + n2 – 2 = 𝟎,𝟑𝟐 − 𝟎,𝟐𝟓 𝟎,𝟎𝟐𝟖 = 496 + 589 – 2 = 1083 = 2,5 α = 5% = 0,05 Keputusan : Thit > Ttab  Tolak H0, signifikan. Maka derajat pelunasan pajak di kedua daerah tersebut ada perbedan.

jawaban no.4 S ( pA – pB ) = 𝑺𝒂 𝟐 + 𝑺𝒃 𝟐 db = nA + nB - 2 = 𝟐 𝟐 + 𝟑 𝟐 = 100 + 50 - 2 = 𝟒+𝟗 = 148 = 𝟏𝟑 α = 1% = 0,01 = 3,6 Ttab = 2,576 Interval Keyakinan ( pA – pB ) – t α 𝟐 . S ( pA – pB ) < p̂A – p̂B < ( pA – pB ) + t α 𝟐 . S ( pA – pB ) ( 12 – 11 ) – 2,576 . 3,6 < p̂A – p̂B < ( 12 – 11 ) + 2,576 . 3,6 1- 9,2736 < p̂A – p̂B <1+9,2736 - 8,2736 < p̂A – p̂B < 10,2736 Jadi, selisih p̂A – p̂B ada di anara – 8,2736 dan 10,2736 dengan taraf keyakinan 99%