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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.
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PengujianHipotesisuntukVarians PengujianHipotesis untukVarians SatuPopulasi DuaPopulasi Chi-Square test statistic F test statistic
SatuPopulasi PengujianHipotesisuntukVarians * SatuPopulasi H0: σ2 = σ02 HA: σ2≠σ02 Two tailed test H0: σ2σ02 HA: σ2<σ02 Lower tail test Chi-Square test statistic H0: σ2≤σ02 HA: σ2>σ02 Upper tail test
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
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.
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
Interval Estimation of 2 0.025 0.025 95% of the possible 2 values 2 0
NilaiKritis • Nilaikritis, ,dapatdilihatdaritabel chi-square 2 Upper tail test: H0: σ2≤σ02 HA: σ2>σ02 2 Do not reject H0 Reject H0 2
Lower Tail or Two Tailed Chi-square Tests Lower tail test: Two tail test: H0: σ2σ02 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
Contoh • Sebuahmeriamharusmemilikiketepatanmenembakdenganvariasi yang minimum. Spesifikasidaripabriksenjatamenyebutkanbahwastandardeviasidariketepatanmenembakmeriamjenistersebutmaksimumadalah 4 meter. Untukmengujihaltersebut, diambilsampelsebanyak 16 meriamdandiperolehhasils2 = 24 meter. Ujilahstandardeviasidarispesifikasitersebut! Gunakan = 0.05
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
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
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
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
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
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
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
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
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