Pengujian Hipotesis untuk Satu dan Dua Varians Populasi

Presentasi berjudul: "Pengujian Hipotesis untuk Satu dan Dua Varians Populasi"— Transcript presentasi:

1 Pengujian Hipotesis untuk Satu dan Dua Varians Populasi

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

3 * 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

4 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

5 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.

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 Distribusi chi-square tergantung dari derajat bebasnya: d.f. = n – 1

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

8 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

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 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

11 2 Nilai kritis dari tabel chi-square :
Hipotesis: H0: σ2 ≤ 16 HA: σ2 > 16 Nilai kritis dari tabel chi-square : 2 = ( = dan d.f. = 16 – 1 = 15) Statistik Uji: Karena 22.5 < , Tidak dapat menolak H0  = .05 2 Do not reject H0 Reject H0 2 =

12 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

13 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

14 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

15 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

16 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

17 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 Mean Std dev Is there a difference in the variances between the NYSE & NASDAQ at the  = 0.1 level?

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 F = is not greater than the critical F value of 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