Statistika Farmasi

Transkripsi

1 Bab 4: Uji Hipotesis Statistika FMIPA Universitas Islam Indonesia

2 Uji Hipotesis Hipotesis Suatu pernyataan tentang besarnya nilai parameter populasi yang akan diuji. Pernyataan tersebut masih lemah kebenarannya dan perlu dibuktikan. Dengan kata lain, hipotesis adalah dugaan yang sifatnya masih sementara. Suatu prosedur pengujian yang dilakukan dengan tujuan memutuskan apakah menerima atau menolak hipotesis mengenai parameter populasi.

3 Uji Hipotesis Hipotesis Nol H 0 Hipotesis yang diartikan sebagai tidak adanya perbedaan antara ukuran populasi dan ukuran sampel. Hipotesis Alternatif H 1 Lawannya hipotesis nol, adanya perbedaan data populasi dengan sampel. Hipotesis alternatif ini biasanya merepresentasikan pertanyaan yang harus dijawab atau teori yang akan diuji

4 Pada pengujian hipotesis, terdapat empat kemungkinan keadaan yang menentukan apakah keputusan kita benar atau salah. Kemungkinan keadaan tersebut adalah sebagai berikut Tidak menolak H 0 Menolak H 0 H 0 benar Eror tipe I (α) H 0 salah Eror tipe II (β)

5 Langkah

6 Formulasi Hipotesis Uji Hipotesis Hipotesis nol H 0 dirumuskan sebagai pernyataan yang akan diuji, hendaknya dibuat pernyataan untuk ditolak. Hipotesis alternatif H 1 dirumuskan sebagai lawan/tandingan hipotesis nol. Jenis uji hipotesis: Uji hipotesis satu arah (one-tailed): H 0 : µ = µ 0 H 1 : µ > µ 0 atau H 1 : µ < µ 0 Uji hipotesis dua arah (two-tailed): H 0 : µ = µ 0 H 1 : µ µ 0

7 A manufacturer of a certain brand of rice cereal claims that the average saturated fat content does not exceed 1.5 grams per serving. State the null and alternative hypotheses to be used in testing this claim. H 0 : µ = 1.5 H 1 : µ > 1.5

8 Taraf Nyata (Significant Level) Taraf nyata adalah besarnya toleransi dalam menerima kesalahan hasil hipotesis terhadap nilai parameter populasinya. Taraf nyata (significant level) disimbolkan dengan α Tingkat kepercayaan (confident level) disimbolkan dengan 1 α Pemilihan taraf nyata tergantung pada bidang penelitian masing-masing. Biasanya di bidang sosial menggunakan taraf nyata 5%10%, di bidang eksakta menggunakan 1%2%. Besarnya kesalahan disebut sebagai daerah kritis pengujian (daerah penolakan)

9 Daerah penolakan uji hipotesis satu arah

10 Daerah penolakan uji hipotesis dua arah

11 Kriteria Pengujian dan Statistik Uji Bentuk keputusan menerima/menolak H 0 Ada banyak jenis pengujian, dalam materi ini yang akan dipelajari adalah: a. Uji hipotesis satu rata-rata b. Uji hipotesis dua rata-rata c. Uji hipotesis data berpasangan

12 Uji Hipotesis Satu Rata-Rata Kriteria Pengujian

13 Statistik Uji i. Jika variansi (σ 2 ) diketahui, n 30. Statistik ujinya: z 0 = x µ 0 σ n ii. Jika variansi (σ 2 ) tidak diketahui, n < 30. Statistik ujinya: t 0 = x µ 0 s n

14 Tabel z Uji Hipotesis

15 Tabel t Uji Hipotesis

16 A random sample of 100 recorded death in the United States during the past year showed an average life span of 71.8 years. Assuming a population standard deviation of 8.9 years, does this seem to indicate that the mean life span today is greater than 70 years? Use a 0.05 level of significance. Solution: 1 H 0 : µ = 70 years 2 H 1 : µ > 70 years 3 α = Critical region: z > 1.645, where z = x µ 0 σ/ n 5 Computation: x = 71.8 years, σ = 8.9 years, and hence z = / 100 = 2.02 Desicion: Reject H 0 and conclude that the mean life span today is greater than 70 years.

17 The Edison Electric Institute has published figures on the number of kilowatt hours used annually by various home appliances. It is claimed that a vacuum cleaner uses an average of 46 kilowatt hours per year. If a random sample of 12 homes included in a planned study indicates that vacuum cleaners use an average of 42 kilowatt hours per year with a standard deviation of 11.9 kilowatt hours, does this suggest at the 0.05 level of significance that vacuum cleaners use, on average, less than 46 kilowatt hours annually? Assume the population of kilowatt hours to be normal.

18 1 H 0 : µ = 46 kilowatt hours 2 H 1 : µ < 46 kilowatt hours 3 α = Critical region: t < 1.796, where t = x µ 0 of freedom s/ n with 11 degrees 5 Computations: x = 42 kilowatt hours, s = 11.9 kilowatt hours, and n = 12. Hence t = 11.9/ = 1.16, P = P(T < 1.16) Desicion: Do not reject H 0 and conclude that the average number of kilowatt hours used annually by home vacuum cleaners is not significantly less than 46.

19 Uji Hipotesis Dua Rata-Rata Kriteria Pengujian

20 Statistik Uji i. Jika variansi (σ1 2 dan σ2 2 ) diketahui, n 30. Statistik ujinya: z 0 = ( x 1 x 2 ) d 0 σ 2 1 n 1 + σ2 2 n 2 ii. Jika variansi (σ1 2 dan σ2 2 ) tidak diketahui namun dianggap sama, n < 30. Statistik ujinya: t 0 = ( x 1 x 2 ) d 0 1 s p (n dengan s p = 1 1)s1 2+(n2 1)s2 2 n 1+n 2 2. Derajat bebas: ν = n 1 + n 2 2. n n 2

21 iii. Jika variansi (σ1 2 dan σ2 2 ) tidak diketahui namun dianggap berbeda, n < 30. Statistik ujinya: Derajat bebas: ν = t 0 = ( x 1 x 2 ) d 0 s 2 1 n 1 + s2 2 n 2 ( s 1 2 ) + s2 2 n 1 n2 ( ) s n 1 ( ) s n 2 n n 2 1.

22 An experiment was performed to campare the abrasive wear of two different laminated materias. Twelve pieces of material 1 were tested by exposing each piece to a machine measuring wear. Ten pieces of material 2 were similarly tested. In each case, the depth of wear was observed. The samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4, while the samples of material 2 gave an average of 81 with a sample standard deviation of 5. Can we conclude at the 0.05 level of significance that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units? Assume the populations to be approximately normal with equal variances?

23 Let µ 1 and µ 2 represent the population means of the abrasive wear for material 1 and material 2, respectively. 1 H 0 : µ 1 µ 2 = 2 2 H 1 : µ 1 µ 2 > 2 3 α = Critical region: t > 1.725, where t = ( x 1 x 2 ) d 0 s p 1/n1 +1/n 2 ν = 20 degrees of freedom 5 Computations: x 1 = 85, s 1 = 4, n 1 = 12 x 2 = 81, s 2 = 5, n 2 = 10 with

24 Hence, (11)(16) + (9)(25) s p = = (85 81) 2 t = /12 + 1/10 = 1.04 P = P(T > 1.04) 0.16 Decision: Do not reject H 0. We are unable to conclude that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units.

25 Uji Hipotesis Data Berpasangan Kriteria Pengujian

26 Statistik Uji t 0 = d d 0 s d n dengan d 2 ( d) 2 s d = n 1 dan n adalah jumlah pasangan data. Derajat bebas: ν = n 1. n

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29 1 H 0 : µ 1 = µ 2 or µ D = µ 1 µ 2 = 0 2 H 1 : µ 1 µ 2 or µ D = µ 1 µ α = Critical region: t < and t > 2.145, where t = d d 0 s D / n with ν = 14 degrees of freedom 5 Computations: The sample mean and standard deviation for the d i are d = and s d = Therefore, t = / 15 = 2.06

30 Though the t-statistics is not significant at the 0.05 level, P = P( T > 2.06) 0.06 As a result, there is some evidence that there is a difference in mean circulating levels of androgen.