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Pengujian Hipotesis untuk Satu dan Dua Varians Populasi.

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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 Pengujian Hipotesis untuk Varians Satu PopulasiDua Populasi Chi-Square test statisticF test statistic

3 Satu Populasi Pengujian Hipotesis untuk Varians Satu Populasi Chi-Square test statistic H 0 : σ 2 = σ 0 2 H A : σ 2 ≠ σ 0 2 H 0 : σ 2  σ 0 2 H A : σ 2 < σ 0 2 H 0 : σ 2 ≤ σ 0 2 H A : σ 2 > σ 0 2 * Two tailed test Lower tail test Upper tail test

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

5 Chi-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) s 2 /  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. n The chi-square distribution is the sum of squared standardized normal random variables such as ( z 1 ) 2 +( z 2 ) 2 +( z 3 ) 2 and so on.

6 Examples of Sampling Distribution of ( n - 1) s 2 /  With 2 degrees of freedom of freedom With 2 degrees of freedom of freedom With 5 degrees of freedom of freedom With 5 degrees of freedom of freedom With 10 degrees of freedom of freedom With 10 degrees of freedom of freedom  Distribusi chi-square tergantung dari derajat bebasnya: d.f. = n – 1

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

8 Nilai Kritis  Nilai kritis,, dapat dilihat dari tabel chi-square Do not reject H 0 Reject H 0  22 22 22 H 0 : σ 2 ≤ σ 0 2 H A : σ 2 > σ 0 2 Upper tail test:

9 Lower Tail or Two Tailed Chi-square Tests H 0 : σ 2 = σ 0 2 H A : σ 2 ≠ σ 0 2 H 0 : σ 2  σ 0 2 H A : σ 2 < σ 0 2  2  /2 Do not reject H 0 Reject   2 1-  22 Do not reject H 0 Reject  /2  2 1-  /2 22  /2 Reject Lower tail test:Two tail test:

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 s 2 = 24 meter. Ujilah standar deviasi dari spesifikasi tersebut! Gunakan  = 0.05

11  Nilai kritis dari tabel chi-square : Do not reject H 0 Reject H 0  =.05 22 22 22 = = (  = 0.05 dan d.f. = 16 – 1 = 15) Statistik Uji: Karena 22.5 < , Tidak dapat menolak H 0 H 0 : σ 2 ≤ 16 H A : σ 2 > 16  Hipotesis:

12 Pengujian Hipotesis untuk Varians Dua Populasi F test statistic * Dua Populasi H 0 : σ 1 2 – σ 2 2 = 0 H A : σ 1 2 – σ 2 2 ≠ 0 Two tailed test Lower tail test Upper tail test H 0 : σ 1 2 – σ 2 2  0 H A : σ 1 2 – σ 2 2 < 0 H 0 : σ 1 2 – σ 2 2 ≤ 0 H A : σ 1 2 – σ 2 2 > 0

13 Pengujian Hipotesis untuk Varians F test statistic * F Test untuk Perbedaan Dua Varians Populasi Dua Populasi F test statistic : = Variance of Sample 1 n = numerator degrees of freedom n = denominator degrees of freedom = Variance of Sample 2

14  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 The F Distribution where df 1 = n 1 – 1 ; df 2 = n 2 – 1

15 F0  rejection region Nilai Kritis F0  rejection region  FF F 1-  Reject H 0 Do not reject H 0 H 0 : σ 1 2 – σ 2 2  0 H A : σ 1 2 – σ 2 2 < 0 H 0 : σ 1 2 – σ 2 2 ≤ 0 H A : σ 1 2 – σ 2 2 > 0 Do not reject H 0 Reject 

16 Nilai Kritis F0  rejection region for a two-tailed test is F 1-  /2 H 0 : σ 1 2 – σ 2 2 = 0 H A : σ 1 2 – σ 2 2 ≠ 0 Do not reject H 0 Reject  /2 F  /2 Reject  /2

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 2125 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: H 0 : σ 2 1 – σ 2 2 = 0 ( there is no difference between variances) H A : σ 2 1 – σ 2 2 ≠ 0 ( there is a difference between variances)  Find the F critical value for  = 0.1:  Numerator:  df 1 = n 1 – 1 = 21 – 1 = 20  Denominator:  df 2 = n 2 – 1 = 25 – 1 = 24 F 0.05, 20, 24 = 2.03 F 0.95, 20, 24 = 0.48

19  The test statistic is: 0  /2 = 0.05 F  /2 =2.03 Reject H 0 Do not reject H 0 H 0 : σ 1 2 – σ 2 2 = 0 H A : σ 1 2 – σ 2 2 ≠ 0 F Test: Example Solution  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 H 0 (continued)  Conclusion: There is no evidence of a difference in variances at  =.05 F 1- α /2 =0.48  /2 = 0.05 Reject H 0


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