Statistika Farmasi

Uji Hipotesis Latihan

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


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.





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





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) √




Langkah 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 = 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




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





Taraf Nyata (Significance 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)





Daerah penolakan uji hipotesis satu arah





Daerah penolakan uji hipotesis dua arah





Kriteria Pengujian dan Statistik Uji

Bentuk keputusan menerima/menolak H0 Ada banyak jenis pengujian, dalam materi ini yang akan dipelajari adalah: a. b. c. d. e.

Uji Uji Uji Uji Uji

hipotesis hipotesis hipotesis hipotesis hipotesis

satu rata-rata dua rata-rata data berpasangan satu variansi dua variansi populasi





Uji Hipotesis Satu Rata-Rata Kriteria Pengujian





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




Tabel z





Tabel t





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




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.





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 Dua Rata-Rata Kriteria Pengujian





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.





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




An experiment was performed to compare 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?





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



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 Data Berpasangan Kriteria Pengujian





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.





Table below shows the results of a bioavailability study comparing a new formulation (A) to a marketed form (B) with regard to the area under the blood-level curve. The average difference is 18.5 and the standard deviation of the differences is 13. Test at the 0.05 level of significance that there is difference of the bioavailability between A and B. Animal 1 2 3 4 5 6

A 136 168 160 94 200 174


B 166 184 193 105 198 197

di 30 16 33 11 -2 23




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 < −2.571 or t > 2.571, where x1 −¯ x2 )−d0 √ with ν = 5 degrees of freedom t = (¯ sp

5

1/n1 +1/n2

Computations: the sample mean and standard deviation for the di are d¯ = 18.5 dan sd = 13 then t=

d¯ − d0 18.5 − 0 √ = √ = 3.48 sd / n 13/ 6

Decision: Reject H0 which mean that there is difference at the bioavailibility of formulation A and B. Atina Ahdika, S.Si, M.Si




Uji Hipotesis Satu Variansi Kriteria Pengujian





Statistik Uji χ2 =

(n − 1)s 2 σ02

di mana n adalah ukuran sampel, s 2 adalah variansi sampel, dan σ02 adalah nilai σ 2 yang diberikan oleh H0 . Jika H0 benar, χ2 adalah nilai dari distribusi chi-squared dengan derajat bebas ν = n − 1.





Tabel Distribusi Chi-Squared





A manufacturer of car batteries claims that the life of company’s batteries is approximately normally distributed with a standard deviation equal to 0.9 year. If a random sample of 10 of these batteries has a standard deviation of 1.3 years, do you think that σ > 0.9 year? Use a 0.05 level of significance.




1

H0 : σ 2 = 0.81

2

H1 : σ 2 > 0.81

3

α = 0.05

4


Critical region: H0 is rejected when χ2 > 16.919, where 2 χ2 = (n−1)s with ν = 9 degrees of freedom. σ2 0

5

Computations: s 2 = 1.44, n = 10, and χ2 =

(9)(1.69) = 18.78 0.81

Decision: Reject H0 which mean that there is evidence that σ > 0.9.





Uji Hipotesis Dua Variansi Kriteria Pengujian

Catatan: f1−α(ν1 ,ν2 ) =

1 fα(ν2 ,ν1 )





Statistik Uji f =

s12 s22

di mana s12 dan s22 variansi dari kedua sampel. Jika dua populasi tersebut menghampiri distribusi normal dan H0 s2 benar, maka rasio f = s12 adalah sebuah nilai dari distribusi F 2 dengan derajat bebas ν1 = n1 − 1 dan ν2 = n2 − 1.





In testing for the difference in the abrasive wear of the two materials in example before, we assumed that the two unknown population variances were equal. Were we justified in making this assumption? Use a 0.10 level of significance.





Let σ12 and σ22 be the population variances for the abrasive wear of material 1 and 2, repectively. 1

H0 : σ12 = σ22

2

H1 : σ12 6= σ22

3

α = 0.10

4

Critical region: fα/2(ν1 ,ν2 ) = f0.05(11,9) = 3.11 and f1−0.05(11,9) = f 1 = 0.34. 0.05(9,11) Therefore, H0 is rejected when f < 0.34 or f > 3.11.

5

Computations: s12 = 16, s22 = 25, and hence f =

16 25

= 0.64

Decision: Do not reject H0 . Conclude that there is insufficient evidence that variances differ.




Latihan

Latihan

1. According to a dietary study, high sodium intake may be related to ulcers, stomach cancer, and migrain headaches. The human requirement for salt is only 220 milligrams per day, which is surpassed in most single servings of ready-to-eat cereals. If a random sample of 20 similar servings of a certain cereal has a mean sodium content of 244 milligrams and a standard deviation of 24.5 milligrams, does this suggest at the 0.05 level of significance that the average sodium content for a single serving of such cereal is greater than 220 milligrams?




Latihan

2. According to Chemical Engineering, an important property of fiber is its water absorbency. The average percent absorbency of 25 randomly selected pieces of cotton fiber was found to be 20 with a standard deviation of 1.5. A random sample of 25 pieces of acetate yielded an average percent of 12 with a standard deviation of 1.25. Is there strong evidence that the population mean percent absorbency is significantly higher for cotton fiber than for acetate? Assume that the population variances in percent absorbency for the two fibers are the same. Use a significance level of 0.05.




Latihan

3. In a study conducted by the Department of Human Nutrition and Foods at Virginia Tech, the following data were recorded on sorbic acid residuals, in parts per million, in ham immediately after dipping in a sorbate solution and after 60 days of storage.

Is there sufficient evidence, at the 0.05 level of significance, to say that the length of storage influences sorbic acid residual concentrations? Atina Ahdika, S.Si, M.Si



Latihan

4. Aflotoxins produced by mold on peanut crops in Virginia must be monitored. A sample of 64 batches of peanuts reveals levels of 24.17 ppm, on average, with a variance of 4.25 ppm. Test the hypothesis that σ 2 = 4.2 ppm against the alternative that σ 2 6= 4.2 ppm.




Latihan

5. A study is conducted to compare the lengths of time required by men and women to assemble a certain product. Past experience indicates that the distribution of times for both men and women approximately normal but the variance of the times for women is less than that for men. A random sample of times for 11 men and 14 women produced the following data

Test the hypothesis that σ12 = σ22 against the alternative that σ12 > σ22 .



Statistika Farmasi

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