1 / 19

Pengujian Hipotesis untuk Satu dan Dua Varians Populasi

Pengujian Hipotesis untuk Satu dan Dua Varians Populasi. Pengujian Hipotesis untuk Varians. Pengujian Hipotesis untuk Varians. Satu Populasi. Dua Populasi. Chi-Square test statistic. F test statistic. Satu Populasi. Pengujian Hipotesis untuk Varians. *. Satu Populasi.

gray-boone
Download Presentation

Pengujian Hipotesis untuk Satu dan Dua Varians Populasi

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Pengujian Hipotesis untukSatu dan Dua Varians Populasi

  2. PengujianHipotesisuntukVarians PengujianHipotesis untukVarians SatuPopulasi DuaPopulasi Chi-Square test statistic F test statistic

  3. SatuPopulasi PengujianHipotesisuntukVarians * SatuPopulasi H0: σ2 = σ02 HA: σ2≠σ02 Two tailed test H0: σ2σ02 HA: σ2<σ02 Lower tail test Chi-Square test statistic H0: σ2≤σ02 HA: σ2>σ02 Upper tail test

  4. Chi-Square Test Statistic PengujianHipotesisuntukVarians StatistikUji: SatuPopulasi * Chi-Square test statistic Dimana: 2 = variabel standardized chi-square n = jumlahsampel s2 = varianssampel σ2 = varians yang dihipotesiskan

  5. Chi-Square Distribution • The chi-square distributionis the sum of squared standardized normal random variables such as (z1)2+(z2)2+(z3)2 and so on. • The chi-square distribution is based on sampling • from a normal population. • The sampling distribution of (n - 1)s2/ 2 has a chi- square distribution whenever a simple random sample of size n is selected from a normal population. • We can use the chi-square distribution to develop • interval estimates and conduct hypothesis tests • about a population variance.

  6. Examples of Sampling Distribution of (n - 1)s2/ 2 With 2 degrees of freedom With 5 degrees of freedom With 10 degrees of freedom 0 • Distribusichi-squaretergantungdariderajatbebasnya: d.f. = n – 1

  7. Interval Estimation of 2 0.025 0.025 95% of the possible 2 values 2 0

  8. NilaiKritis • Nilaikritis, ,dapatdilihatdaritabel chi-square 2 Upper tail test: H0: σ2≤σ02 HA: σ2>σ02  2 Do not reject H0 Reject H0 2

  9. Lower Tail or Two Tailed Chi-square Tests Lower tail test: Two tail test: H0: σ2σ02 HA: σ2<σ02 H0: σ2 = σ02 HA: σ2≠σ02  /2 /2 2 2 Reject Do not reject H0 Do not reject H0 Reject Reject 21- 21-/2 2/2

  10. Contoh • Sebuahmeriamharusmemilikiketepatanmenembakdenganvariasi yang minimum. Spesifikasidaripabriksenjatamenyebutkanbahwastandardeviasidariketepatanmenembakmeriamjenistersebutmaksimumadalah 4 meter. Untukmengujihaltersebut, diambilsampelsebanyak 16 meriamdandiperolehhasils2 = 24 meter. Ujilahstandardeviasidarispesifikasitersebut! Gunakan = 0.05

  11. Hipotesis: H0: σ2≤16 HA: σ2>16 • Nilaikritisdaritabel chi-square : 2 = 24.9958 ( = 0.05 dand.f. = 16 – 1 = 15) StatistikUji: Karena 22.5 < 24.9958, Tidakdapatmenolak H0  = .05 2 Do not reject H0 Reject H0 2 = 24.9958

  12. DuaPopulasi PengujianHipotesisuntukVarians * DuaPopulasi H0: σ12 – σ22 = 0 HA: σ12 – σ22 ≠ 0 Two tailed test H0: σ12 – σ22 0 HA: σ12 – σ22 < 0 Lower tail test F test statistic H0: σ12 – σ22≤ 0 HA: σ12 – σ22 > 0 Upper tail test

  13. F Test untukPerbedaanDuaVariansPopulasi PengujianHipotesisuntukVarians F test statistic : DuaPopulasi * F test statistic = Variance of Sample 1 n1 - 1 = numerator degrees of freedom = Variance of Sample 2 n2 - 1 = denominator degrees of freedom

  14. The F Distribution • The F critical valueis found from the F table • The are two appropriate degrees of freedom: numerator and denominator • In the F table, • numerator degrees of freedom determine the row • denominator degrees of freedom determine the column where df1 = n1 – 1 ; df2 = n2 – 1

  15. NilaiKritis H0: σ12 – σ22 0 HA: σ12 – σ22 < 0 H0: σ12 – σ22≤ 0 HA: σ12 – σ22 > 0   0 F 0 F F Reject Do not reject H0 Do not reject H0 Reject H0 F1- • rejection region • rejection region

  16. NilaiKritis H0: σ12 – σ22 = 0 HA: σ12 – σ22 ≠ 0 /2 /2 0 F F/2 Do not reject H0 Reject Reject F1-/2 • rejection region for a two-tailed test is

  17. You are a financial analyst for a brokerage firm. You want to compare dividend yields between stocks listed on the NYSE & NASDAQ. You collect the following data: NYSENASDAQNumber 21 25 Mean 3.27 2.53 Std dev 1.30 1.16 Is there a difference in the variances between the NYSE & NASDAQ at the =0.1 level? F Test: An Example

  18. F Test: Example Solution • Form the hypothesis test: H0: σ21 – σ22 = 0 (there is no difference between variances) HA: σ21 – σ22 ≠ 0 (there is a difference between variances) • Find the F critical value for  = 0.1: • Numerator: • df1 = n1 – 1 = 21 – 1 = 20 • Denominator: • df2 = n2 – 1 = 25 – 1 = 24 F0.05, 20, 24 = 2.03 F0.95, 20, 24 = 0.48

  19. F Test: Example Solution (continued) • The test statistic is: H0: σ12 – σ22 = 0 HA: σ12 – σ22 ≠ 0 /2 = 0.05 /2 = 0.05 0 • F = 1.256 is not greater than the critical F value of 2.327 or not less than the critical F value of 0.48, so we do not reject H0 Reject H0 Do not reject H0 Reject H0 F1-α/2=0.48 F/2=2.03 • Conclusion: There is no evidence of a difference in variances at  = .05

More Related