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Uji Hipotesis Dua Populasi

Uji Hipotesis Dua Populasi. Uji Hipotesis Rata-rata Dua Populasi. Dua Populasi , Sampel Independent. Lower tail test: H 0 : μ 1  μ 2 H A : μ 1 < μ 2 atau , H 0 : μ 1 – μ 2  0 H A : μ 1 – μ 2 < 0. Upper tail test: H 0 : μ 1 ≤ μ 2 H A : μ 1 > μ 2 atau ,

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Uji Hipotesis Dua Populasi

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  1. UjiHipotesisDuaPopulasi

  2. UjiHipotesisRata-rata DuaPopulasi DuaPopulasi, Sampel Independent Lower tail test: H0: μ1μ2 HA: μ1 < μ2 atau, H0: μ1 – μ2 0 HA: μ1 – μ2< 0 Upper tail test: H0: μ1≤μ2 HA: μ1>μ2 atau, H0: μ1 – μ2≤ 0 HA: μ1 – μ2> 0 Two-tailed test: H0: μ1 = μ2 HA: μ1≠μ2 atau, H0: μ1 – μ2= 0 HA: μ1 – μ2≠ 0

  3. UjiHipotesisuntukμ1 – μ2 Rata-rata populasi, sampel independent Gunakanstatistikujiz σ1 and σ2diketahui σ1 and σ2tdkdiketahui, n  30 Gunakansuntukmengestimasiσ , perkirakandenganstatistikujiz σ1 and σ2tdkdiketahui, n < 30 Gunakansuntukmengestimasiσ , gunakanstatistikujit

  4. σ1 and σ2Diketahui Rata-rata populasi, sampel independent Statistikujiuntuk μ1 – μ2adalah: * σ1 and σ2diketahui σ1 and σ2tdkdiketahui, n  30 σ1 and σ2tdkdiketahui, n < 30

  5. σ1 and σ2TidakDiketahui, SampelBesar Rata-rata populasi, sampel independent Statistikujiuntuk μ1 – μ2adalah: σ1 and σ2diketahui * σ1 and σ2tdkdiketahui, n  30 σ1 and σ2tdkdiketahui, n < 30

  6. σ1 and σ2TidakDiketahui, Sampel Kecil Asumsi: 12=22 Statistikujiuntukμ1 – μ2adalah: Rata-rata populasi, sampel independent σ1 and σ2diketahui σ1 and σ2tdkdiketahui, n  30 Where t/2has (n1 + n2 – 2) d.f., and * σ1 and σ2tdkdiketahui, n < 30

  7. σ1 and σ2TidakDiketahui, Sampel Kecil Asumsi: 1222 Statistikujiuntukμ1 – μ2adalah: Rata-rata populasi, sampel independent σ1 and σ2diketahui σ1 and σ2tdkdiketahui, n  30 Denganderajatbebas: * σ1 and σ2tdkdiketahui, n < 30

  8. UjiHipotesisuntukμ1 – μ2 DuaPopulasi, Sampel Independent Lower tail test: H0: μ1 – μ2 0 HA: μ1 – μ2< 0 Upper tail test: H0: μ1 – μ2≤ 0 HA: μ1 – μ2> 0 Two-tailed test: H0: μ1 – μ2= 0 HA: μ1 – μ2≠ 0 a a a/2 a/2 -za za -za/2 za/2 TolakH0jikaz < -za TolakH0jikaz > za TolakH0jikaz < -za/2 atauz > za/2

  9. Untukmelihatapakahterdapatperbedaandalampembayarandevidenantarasaham yang tercatatdalam IHSG danIndeks LQ? Andamengambilsampelsecara random sebagaiberikut: IHSGLQJumlahsampel10 12 Rata2 Sampel 3.27 2.53 Std dev Sampel 1.30 1.16 Denganmengasumsikankedua varianssama, apakahterdapat perbedaan rata2 dlmpembayaran deviden( = 0.05)? Contoh

  10. PenghitunganStatistikUji

  11. H0: μ1 - μ2 = 0 i.e. (μ1 = μ2) HA: μ1 - μ2≠ 0 i.e. (μ1 ≠μ2)  = 0.05 df = 10 + 12 - 2 = 20 Critical Values: t = ± 2.086 Test Statistic: Solution Reject H0 Reject H0 .025 .025 t 0 -2.086 2.086 1.5073 Decision: Conclusion: Do not Reject H0 at a = 0.05 There is no evidence of a difference in means.

  12. UjiHipotesisuntuk Data Berpasangan Statistikujiuntukd : Data Berpasangan Derajatbebasuntukt/2= n - 1

  13. UjiHipotesisuntuk Data Berpasangan (continued) Data Berpasangan Lower tail test: H0: μd 0 HA: μd < 0 Upper tail test: H0: μd≤ 0 HA: μd> 0 Two-tailed test: H0: μd = 0 HA: μd≠ 0 a a a/2 a/2 -ta ta -ta/2 ta/2 TolakH0jikat < -ta TolakH0jikat > ta TolakH0jikat < -ta/2 ataut > ta/2 Derajatbebasuntuk t/2 = n - 1

  14. Contoh • Suatuperusahaantelahmengirimkaryawannyanyamelakukanpelatihan “customer service”. Apakahpelatihantsbefektif? Untukitudiambilsampel random sbb:  Banyaknyakomplain: (2) - (1) KaryawanSebelum(1)Setelah(2)Difference,di A64- 2 B206 -14 C32 - 1 D00 0 E40- 4 -21 di d = n = -4.2

  15. Solution • Apakahpelatihanmemberikanpebedaan rata-rata jumlahkomplainkonsumen ( = 0,01)? Tolak Tolak H0:μd = 0 HA:μd 0 /2 /2  = .01 d = - 4.2 - 4.604 4.604 - 1.66 Critical Value = ± 4.604 d.f. = n - 1 = 4 Decision: Do not reject H0 (t stat is not in the reject region) Test Statistic: Conclusion:There is not a significant change in the number of complaints.

  16. UjiHipotesisuntukDuaProporsiPopulasi ProporsiPopulasi Lower tail test: H0: p1p2 HA: p1 < p2 i.e., H0: p1 – p2 0 HA: p1 – p2< 0 Upper tail test: H0: p1≤ p2 HA: p1> p2 i.e., H0: p1 – p2≤ 0 HA: p1 – p2> 0 Two-tailed test: H0: p1 = p2 HA: p1≠ p2 i.e., H0: p1 – p2= 0 HA: p1 – p2≠ 0

  17. DuaProporsiPopulasi Statistikujiuntuk p1 – p2: ProporsiPopulasi

  18. DuaProporsiPopulasi Dimana :

  19. UjiHipotesisuntukDuaProporsiPopulasi ProporsiPopulasi Lower tail test: H0: p1 – p2 0 HA: p1 – p2< 0 Upper tail test: H0: p1 – p2≤ 0 HA: p1 – p2> 0 Two-tailed test: H0: p1 – p2= 0 HA: p1 – p2≠ 0 a a a/2 a/2 -za za -za/2 za/2 TolakH0jikaz < -za TolakH0jikaz > za TolakH0jikaz < -za/2 atauz > za/2

  20. Contoh Apakahadaperbedaan yang signifikanantaraproporsilaki-laki & proporsiperempuan yang akanmenyatakanYauntuksuatupertanyaanA? • Dalamsuaturandom sample, 36 dari72 laki-lakidan31 dari50 perempuanmenyatakanakanmengatakanYa • Ujidengantingkatkesalahan 0.05

  21. Solution H0: p1 – p2= 0 HA: p1 – p2≠0 • Proporsisampel: • Laki-laki: p1= 36/72 = .50 • Perempuan: p2= 31/50 = .62 • Pendugaproporsigabungan

  22. Solution (continued) Reject H0 Reject H0 Nilaistatistikujiuntukp1 – p2: .025 .025 -1.96 1.96 -1.31 Keputusan: TidakmenolakH0 Kesimpulan:Tidakterdapatcukupbuktiuntukmengatakanbahwaterdapatperbedaanproporsilaki-lakidanperempuanuntukmengatakanYapadapertanyaantersebut. Critical Values = ±1.96 For  = .05

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