Statistika Farmasi

Uji Hipotesis

613.123.15 Statistika Farmasi Bab 4: Uji Hipotesis

Atina Ahdika, S.Si, M.Si

Statistika FMIPA Universitas Islam Indonesia


613.123.15 Statistika Farmasi

Uji Hipotesis

Definisi Pengujian Hipotesis

Definisi

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. Pengujian Hipotesis Suatu prosedur pengujian yang dilakukan dengan tujuan memutuskan apakah menerima atau menolak hipotesis mengenai parameter populasi.



Uji Hipotesis


Pengujian Hipotesis

Hipotesis Nol H0 Hipotesis yang diartikan sebagai tidak adanya perbedaan antara ukuran populasi dan ukuran sampel. Hipotesis Alternatif H1 Lawannya hipotesis nol, adanya perbedaan data populasi dengan sampel. Hipotesis alternatif ini biasanya merepresentasikan pertanyaan yang harus dijawab atau teori yang akan diuji



Uji Hipotesis


Pada pengujian hipotesis, terdapat empat kemungkinan keadaan yang menentukan apakah keputusan kita benar atau salah. Kemungkinan keadaan tersebut adalah sebagai berikut Tidak menolak H0 Menolak H0

H0 benar √ Eror tipe I (α)


H0 salah Eror tipe II (β) √


Uji Hipotesis


Langkah Pengujian Hipotesis



Uji Hipotesis


Formulasi Hipotesis Hipotesis nol H0 dirumuskan sebagai pernyataan yang akan diuji, hendaknya dibuat pernyataan untuk ditolak. Hipotesis alternatif H1 dirumuskan sebagai lawan/tandingan hipotesis nol. Jenis uji hipotesis: Uji hipotesis satu arah (one-tailed): H0 : µ = µ 0 H1 : µ > µ0 atau H1 : µ < µ0 Uji hipotesis dua arah (two-tailed): H0 : µ = µ 0 H1 : µ 6= µ0 Atina Ahdika, S.Si, M.Si


Uji Hipotesis


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.

H0 : µ = 1.5 H1 : µ > 1.5



Uji Hipotesis


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)



Uji Hipotesis


Daerah penolakan uji hipotesis satu arah



Uji Hipotesis


Daerah penolakan uji hipotesis dua arah



Uji Hipotesis


Kriteria Pengujian dan Statistik Uji

Bentuk keputusan menerima/menolak H0 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



Uji Hipotesis


Uji Hipotesis Satu Rata-Rata Kriteria Pengujian



Uji Hipotesis


Statistik Uji i. Jika variansi (σ 2 ) diketahui, n ≥ 30. Statistik ujinya: z0 =

x¯ − µ0 σ √ n

ii. Jika variansi (σ 2 ) tidak diketahui, n < 30. Statistik ujinya: t0 =


x¯ − µ0 √s

n


Uji Hipotesis


Tabel z



Uji Hipotesis


Tabel t



Uji Hipotesis


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

H0 : µ = 70 years

2

H1 : µ > 70 years

3

α = 0.05

4

Critical region: z > 1.645, where z =

5

Computation: x¯ = 71.8 years, σ = 8.9 years, and hence 71.8−70 √ z = 8.9/ = 2.02 100

x¯−µ √0 σ/ n

Desicion: Reject H0 and conclude that the mean life span today is greater than 70 years. Atina Ahdika, S.Si, M.Si


Uji Hipotesis


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.



Uji Hipotesis


1

H0 : µ = 46 kilowatt hours

2

H1 : µ < 46 kilowatt hours

3

α = 0.05

4

Critical region: t < −1.796, where t = of freedom

5

Computations: x¯ = 42 kilowatt hours, s = 11.9 kilowatt hours, and n = 12. Hence t=

42 − 46 √ = −1.16, 11.9/ 12

x¯−µ √0 s/ n

with 11 degrees

P = P(T < −1.16) ≈ 0.135

Desicion: Do not reject H0 and conclude that the average number of kilowatt hours used annually by home vacuum cleaners is not significantly less than 46.



Uji Hipotesis


Uji Hipotesis Dua Rata-Rata Kriteria Pengujian



Uji Hipotesis


Statistik Uji i. Jika variansi (σ12 dan σ22 ) diketahui, n ≥ 30. Statistik ujinya: z0 =

(¯ x1 − x¯2 ) − d0 q 2 σ1 σ22 n1 + n2

ii. Jika variansi (σ12 dan σ22 ) tidak diketahui namun dianggap sama, n < 30. Statistik ujinya: t0 =

(¯ x1 − x¯2 ) − d0 q sp n11 + n12

q (n1 −1)s12 +(n2 −1)s22 dengan sp = . n1 +n2 −2 Derajat bebas: ν = n1 + n2 − 2.




Uji Hipotesis

iii. Jika variansi (σ12 dan σ22 ) tidak diketahui namun dianggap berbeda, n < 30. Statistik ujinya: t0 =

(¯ x1 − x¯2 ) − d0 q 2 s1 s22 n1 + n2

s12 s2 + n2 n1 2 !2

Derajat bebas: ν =

s2 1 n1

n1 −1

+


s2 2 n2

!2

.

n2 −1


Uji Hipotesis


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?



Uji Hipotesis


Let µ1 and µ2 represent the population means of the abrasive wear for material 1 and material 2, respectively. 1

H0 : µ 1 − µ 2 = 2

2

H1 : µ 1 − µ 2 > 2

3

α = 0.05

4

Critical region: t > 1.725, where t =

(¯ x1 −¯ x2 )−d0 √ sp 1/n1 +1/n2

ν = 20 degrees of freedom 5

Computations: x¯1 = 85, s1 = 4, n1 = 12 x¯2 = 81, s2 = 5, n2 = 10



with

Uji Hipotesis


Hence, r

(11)(16) + (9)(25) = 4.478 12 + 10 − 2 (85 − 81) − 2 p t= = 1.04 4.478 1/12 + 1/10

sp =

P = P(T > 1.04) ≈ 0.16 Decision: Do not reject H0 . We are unable to conclude that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units.



Uji Hipotesis


Uji Hipotesis Data Berpasangan Kriteria Pengujian




Uji Hipotesis

Statistik Uji t0 =

d¯ − d0 sd √ n

dengan s sd =

P

d)2 n

P

d2 − ( n−1

dan n adalah jumlah pasangan data. Derajat bebas: ν = n − 1.



Uji Hipotesis




Uji Hipotesis




Uji Hipotesis


1

H0 : µ1 = µ2 or µD = µ1 − µ2 = 0

2

H1 : µ1 6= µ2 or µD = µ1 − µ2 6= 0

3

α = 0.05

4

Critical region: t < −2.145 and t > 2.145, where t = with ν = 14 degrees of freedom

5

Computations: The sample mean and standard deviation for the di are d¯ = 9.848 and sd = 18.474 Therefore, t=

9.848 − 0 √ = 2.06 18.474/ 15



¯ 0 d−d √ sD / n

Uji Hipotesis


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.



Statistika Farmasi

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