Statistika Farmasi

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613.123.15 Statistika Farmasi

Bab 4: Uji Hipotesis

Atina Ahdika, S.Si, M.Si

Statistika FMIPA Universitas Islam Indonesia

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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.

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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

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Pengujian Hipotesis

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

H0 benar H0 salah

Tidak menolak H0

√

Eror tipe II (β) Menolak H0 Eror tipe I (α)

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Pengujian 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 = µ0

H1: µ > µ0 atau H1 : µ < µ0

Uji hipotesis dua arah (two-tailed): H0: µ = µ0

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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

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Pengujian 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)

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Daerah penolakan uji hipotesis satu arah

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Pengujian Hipotesis

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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

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Pengujian Hipotesis

Uji Hipotesis Satu Rata-Rata

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Statistik Uji

i. Jika variansi (σ2_{) diketahui, n ≥ 30. Statistik ujinya:}

z0= ¯ x − µ0 σ √ n

ii. Jika variansi (σ2_{) tidak diketahui, n < 30. Statistik ujinya:}

t0= ¯ x − µ0 s √ n

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Pengujian Hipotesis

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Pengujian 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 = ¯x −µ0

σ/√n

5 _{Computation: ¯}_{x = 71.8 years, σ = 8.9 years, and hence}

z = 71.8−70

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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.

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Pengujian 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 = ¯x −µ0

s/√n with 11 degrees

of freedom

5 Computations: ¯x = 42 kilowatt hours, s = 11.9 kilowatt

hours, and n = 12. Hence t = 42 − 46

11.9/√12 = −1.16, 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.

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Kriteria Pengujian

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Pengujian Hipotesis

Statistik Uji

i. Jika variansi (σ2

1dan σ22) diketahui, n ≥ 30. Statistik ujinya:

z0= (¯x1− ¯x2) − d0 q σ2 1 n1 + σ2 2 n2 ii. Jika variansi (σ2

1dan σ22) tidak diketahui namun dianggap

sama, n < 30. Statistik ujinya: t0= (¯x1− ¯x2) − d0 sp q 1 n1 + 1 n2 dengan sp= q (n1−1)s12+(n2−1)s22 n1+n2−2 . Derajat bebas: ν = n1+ n2− 2.

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iii. Jika variansi (σ2₁ dan σ2₂) tidak diketahui namun dianggap berbeda, n < 30. Statistik ujinya:

t0 = (¯x1− ¯x2) − d0 q s2 1 n1 + s2 2 n2 Derajat bebas: ν = s2₁ n1+ s2₂ n2 s2₁ n1 !2 n1−1 + s2₂ n2 !2 n2−1 .

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Pengujian 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?

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Let µ1 and µ2 represent the population means of the abrasive wear

for material 1 and material 2, respectively.

1 H₀: µ₁− µ₂ = 2

2 _H₁_{: µ}₁− µ₂ _{> 2} 3 _{α = 0.05}

4 Critical region: t > 1.725, where t = (¯x1−¯x2)−d0

sp √ 1/n1+1/n2 with ν = 20 degrees of freedom 5 Computations: ¯ x1 = 85, s1= 4, n1 = 12 ¯ x2 = 81, s2= 5, n2 = 10

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Pengujian Hipotesis Hence, sp= r (11)(16) + (9)(25) 12 + 10 − 2 = 4.478 t = (85 − 81) − 2 4.478p1/12 + 1/10 = 1.04 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.

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Kriteria Pengujian

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Pengujian Hipotesis Statistik Uji t0 = ¯ d − d0 sd √ n dengan sd = s P d2₋(P d)2 n n − 1 dan n adalah jumlah pasangan data. Derajat bebas: ν = n − 1.

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1 H₀: µ₁ = µ₂ or µ_D = µ₁− µ₂= 0 2 H₁: µ₁ 6= µ₂ or µ_D = µ₁− µ₂6= 0 3 α = 0.05

4 _{Critical region: t < −2.145 and t > 2.145, where t =} d −d¯ 0

sD/

√ n

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 18.474/√15 = 2.06

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Pengujian 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.