Pengujian Hipotesis untuk Satu dan Dua Varians Populasi

Pengujian Hipotesis untukSatu dan Dua Varians PopulasiPengujian Hipotesis untuk VariansPengujian Hipotesisuntuk VariansSatu PopulasiDua PopulasiChi-Square test statisticF test statisticSatu PopulasiPengujian Hipotesis untuk VariansSatu PopulasiChi-Square test statisticH0: 2 = 02 HA: 2 02H0: 2 02 HA: 2 < 02H0: 2 02 HA: 2 > 02*Two tailed testLower tail testUpper tail testChi-Square Test StatisticPengujian Hipotesis untuk VariansSatu PopulasiChi-Square test statistic*Statistik Uji:

Dimana:2 = variabel standardized chi-squaren = jumlah sampels2 = varians sampel2 = varians yang dihipotesiskanChi-Square Distribution We can use the chi-square distribution to develop interval estimates and conduct hypothesis tests about a population variance. 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. The chi-square distribution is based on sampling from a normal population. The chi-square distribution is the sum of squared standardized normal random variables such as (z1)2+(z2)2+(z3)2 and so on.Examples of Sampling Distribution of (n - 1)s2/ 20With 2 degrees of freedom

With 5 degrees of freedomWith 10 degrees of freedomDistribusi chi-square tergantung dari derajat bebasnya: d.f. = n 1 95% of thepossible 2 values200.025

0.025

Interval Estimation of 2

Nilai KritisNilai kritis, , dapat dilihat dari tabel chi-squareDo not reject H0Reject H0222H0: 2 02 HA: 2 > 02Upper tail test:Lower Tail or Two Tailed Chi-square TestsH0: 2 = 02 HA: 2 02H0: 2 02 HA: 2 < 022/2Do not reject H0 Reject21-2Do not reject H0 Reject/221-/22/2 RejectLower tail test:Two tail test:ContohSebuah meriam harus memiliki ketepatan menembak dengan variasi yang minimum. Spesifikasi dari pabrik senjata menyebutkan bahwa standar deviasi dari ketepatan menembak meriam jenis tersebut maksimum adalah 4 meter. Untuk menguji hal tersebut, diambil sampel sebanyak 16 meriam dan diperoleh hasil s2 = 24 meter. Ujilah standar deviasi dari spesifikasi tersebut! Gunakan = 0.05Nilai kritis dari tabel chi-square :Do not reject H0Reject H0 = .05222= 24.9958= 24.9958 ( = 0.05 dan d.f. = 16 1 = 15)

Statistik Uji:Karena 22.5 < 24.9958,Tidak dapat menolak H0H0: 2 16 HA: 2 > 16Hipotesis:Pengujian Hipotesis untuk VariansDua PopulasiF test statistic*Dua PopulasiH0: 12 22 = 0HA: 12 22 0Two tailed testLower tail testUpper tail testH0: 12 22 0HA: 12 22 < 0H0: 12 22 0HA: 12 22 > 0Pengujian Hipotesis untuk VariansF test statistic*F Test untuk Perbedaan Dua Varians PopulasiDua Populasi

F test statistic : = Variance of Sample 1 n1 - 1 = numerator degrees of freedom n2 - 1 = denominator degrees of freedom = Variance of Sample 2

The F critical value is found from the F tableThe are two appropriate degrees of freedom: numerator and denominator

In the F table, numerator degrees of freedom determine the rowdenominator degrees of freedom determine the columnThe F Distributionwhere df1 = n1 1 ; df2 = n2 1

F 0 rejection regionNilai KritisF 0

rejection regionF F1- Reject H0Do not reject H0H0: 12 22 0HA: 12 22 < 0H0: 12 22 0HA: 12 22 > 0Do not reject H0 RejectNilai KritisF 0

rejection region for a two-tailed test isF1- /2H0: 12 22 = 0HA: 12 22 0Do notreject H0 Reject/2F/2 Reject/2F Test: An ExampleYou 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: NYSE NASDAQNumber 2125Mean3.272.53Std dev1.301.16

Is there a difference in the variances between the NYSE & NASDAQ at the = 0.1 level?

F Test: Example SolutionForm the hypothesis test:H0: 21 22 = 0 (there is no difference between variances)HA: 21 22 0 (there is a difference between variances)Find the F critical value for = 0.1:Numerator:df1 = n1 1 = 21 1 = 20Denominator: df2 = n2 1 = 25 1 = 24

F0.05, 20, 24 = 2.03

F0.95, 20, 24 = 0.48The test statistic is:0

/2 = 0.05F/2 =2.03Reject H0Do not reject H0H0: 12 22 = 0HA: 12 22 0F Test: Example SolutionF = 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(continued)Conclusion: There is no evidence of a difference in variances at = .05

F1-/2 =0.48/2 = 0.05Reject H0