Pengujian Hipotesis untuk Satu dan Dua Varians Populasi

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

* Satu Populasi Pengujian Hipotesis untuk Varians 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

Chi-Square Test Statistic Pengujian Hipotesis untuk Varians Statistik Uji: Satu Populasi * Chi-Square test statistic Dimana: 2 = variabel standardized chi-square n = jumlah sampel s2 = varians sampel σ2 = varians yang dihipotesiskan

Chi-Square Distribution The chi-square distribution is 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 Distribusi chi-square tergantung dari derajat bebasnya: d.f. = n – 1

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

Nilai Kritis 2 Nilai kritis, , dapat dilihat dari tabel chi-square 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: 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

Contoh Sebuah 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  = 0.05

2 Nilai kritis dari tabel chi-square : Hipotesis: H0: σ2 ≤ 16 HA: σ2 > 16 Nilai kritis dari tabel chi-square : 2 = 24.9958 ( = 0.05 dan d.f. = 16 – 1 = 15) Statistik Uji: Karena 22.5 < 24.9958, Tidak dapat menolak H0  = .05 2 Do not reject H0 Reject H0 2 = 24.9958

Pengujian Hipotesis untuk Varians Dua Populasi Pengujian Hipotesis untuk Varians * Dua Populasi 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 untuk Perbedaan Dua Varians Populasi Pengujian Hipotesis untuk Varians F test statistic : Dua Populasi * 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 value is 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

Nilai Kritis H0: σ12 – σ22  0 HA: σ12 – σ22 < 0 H0: σ12 – σ22 ≤ 0   F F F Do not reject H0 Reject H0 Reject Do not reject H0 F1-  rejection region rejection region

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

F Test: An Example 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: NYSE NASDAQ Number 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: 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 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