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Nama

: Ratna Puspitasari

NPM

: 20187279089

Kelas

: 2B

Dosen

: Dr. Mamiek Suendarti, M.Pd



Bab 2 Latihan Hal. 12

2-15. The diameter of a ball bearing was measured by 12 inspectors, each using two different kinds of calipers. The results were. Inspectors

Caliper 1

Caliper 2

1

0.265

0.264

2

0.265

0.265

3

0.266

0.264

4

0.267

0.266

5

0.267

0.267

6

0.265

0.268

7

0.267

0.264

8

0.267

0.265

9

0.265

0.265

10

0.268

0.267

11

0.268

0.268

12

0.265

0.269

(a) Is there a significant difference between the means of the population of measurements from which the two samples were selected? Use 𝛼 = 0.05 (b) Find the P-value for the test in part (a). (c) Construct a 95 percent confidence interval on the difference in mean diameter measurements for the two types of calipers.

Jawab: Input SPSS :

No.

A

Y

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2

.265 .265 .266 .267 .267 .265 .267 .267 .265 .268 .268 .265 .264 .265 .264 .266 .267 .268 .264 .265 .265 .267 .268 .269

Keterangan : A : Cliper Y : Diameter of ball bearing

Output SPSS: Univariate Analysis of Variance Between-Subjects Factors Value Label

N

1

Caliper 1

12

2

Caliper 2

12

Caliper

Tests of Between-Subjects Effects Dependent Variable: The diameter of a ball bearing Source

Type III Sum of

Df

Mean Square

F

Sig.

Squares Corrected Model

3.750E-007a

1

3.750E-007

.164

.689

Intercept

1.700

1

1.700

744164.940

.000

A

3.750E-007

1

3.750E-007

.164

.689

Error

5.025E-005

22

2.284E-006

Total

1.700

24

5.062E-005

23

Corrected Total

a. R Squared = .007 (Adjusted R Squared = -.038)

Analisis Data Hipotesis Statistik H0 : μ A = μ B H0 : μ A ≠ μ B Hipotesis Penelitian H0 : Terdapat pengaruh yang tidak signifikan Caliper terhadap Diameter of Ball bearing H1 : Terdapat pengaruh yang signifikan Caliper terhadap Diameter of Ball bearing

Hasil Statistik Nilai F0 = 0,164 dan Sig = 0,689 > 0,05. Dari hasil data di atas dapat disimpulkan bahwa terdapat pengaruh yang tidak signifikan Caliper terhadap Diameter of Ball bearing.

2-19. In semiconductor manufacturing wet chemical etching is often used to remove silicon from the backs of wafers prior to metalization. The etch rate is an important characteristic of this process. Two different etching solutions are being evaluated. Eight randomly selected wafers have been etched in each solution and the observed etch rates are shown below. Solution 1

Solution 2

9.9

10.2

9.4

10.0

10.0

10.7

10.3

10.5

10.6

10.6

10.3

10.2

9.3

10.4

9.8

10.3

(a) Do the data indicate that the claim that both solutions have the same mean etch rate is valid? Use 𝛼 = 0.05 and assume equal variances. (b) Find a 95 percent confidence interval on the difference in mean etch rates. (c) Use normal probability plots to investigate the adequacy of the assumptions of normality and equal variances. Jawab: Input SPSS :

A

No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Y 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2

9.9 9.4 10.0 10.3 10.6 10.3 9.3 9.8 10.2 10.0 10.7 10.5 10.6 10.2 10.4 10.3

Keterangan : A : Solution Y : Semiconductor manufacturing

Output SPSS : Between-Subjects Factors Value Label

N

1

Solution 1

8

2

Solution 2

8

Solution

Tests of Between-Subjects Effects Dependent Variable: Semiconductor Source

Type III Sum of Squares

Df

Mean Square

F

Sig.

,681a

1

,681

5,297

,037

1650,391

1

1650,391

12845,292

,000

,681

1

,681

5,297

,037

Error

1,799

14

,128

Total

1652,870

16

2,479

15

Corrected Model Intercept A

Corrected Total

a. R Squared = ,275 (Adjusted R Squared = ,223)

Analisis Data Hipotesis Statistik H0 : μ A = μ B H1 : μA ≠ μB

Hipotesis Penelitian H0

: Terdapat pengaruh yang tidak signifikan Solution terhadap Semiconductor

H1

: Terdapat pengaruh yang signifikan Solution terhadap Semiconductor

Hasil Statistik Nilai F0 = 5,297 dan Sig = 0,037 < 0,05. Dari hasil data di atas dapat disimpulkan bahwa terdapat pengaruh yang signifikan Solution terhadap Semiconductor.

 Bab 3 Latihan Hal. 20 3-13 Four chemists are asked to determine the percentage of methyl alcohol in a certain chemical compound. Each chemist makes three determinations, and the results are the following. Chemists

Percentage of Methyl Alcohol

1

84.99

84.04

84.38

2

85.15

85.13

84.88

3

84.72

84.48

85.16

4

84.20

84.10

84.55

(a) Do chemists differ significantly ? use 𝛼 = 0.05 (b) Analyze the residuals from this experiment (c) If chemist 2 is a employee, construct a meaningful set of orthogonal contracts that might have been useful at the start of the experimen Jawab : Input SPSS : No 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

A 1 1 1 2 2 2 3 3 3 4 4 4

Y 84.99 84.04 84.38 85.15 85.13 84.88 84.72 84.48 85.16 84.20 84.10 84.55

Output SPSS: Between-Subjects Factors N 1

3

2

3

3

3

4

3

Chemical

Keterangan :

A : Compound Y : Precentage of methyl alcohol

Tests of Between-Subjects Effects Dependent Variable: Precentage of methyl alcohol Source

Type III Sum of

Df

Mean Square

F

Sig.

Squares 1.045a

3

.348

3.246

.081

85984.084

1

85984.084

801529.565

.000

1.045

3

.348

3.246

.081

Error

.858

8

.107

Total

85985.987

12

1.903

11

Corrected Model Intercept A

Corrected Total

a. R Squared = .549 (Adjusted R Squared = .380)

Analisis data Hipotesis Statistik 𝐻0 : 𝜇𝐴 = 𝜇𝐵 = 𝜇𝐶 = 𝜇𝐷 𝐻1 : Bukan 𝐻0 Hipotesis Penelitian H0 : Terdapat pengaruh yang tidak signifikan dari methyl alcohol terhadap chemical compound. H1 : Terdapat pengaruh yang signifikan dari methyl alcohol terhadap chemical compound. Hasil statistik : Nilai F0 = 3.246 dan sig. = 0.081. dikatakan signifikan apabila nilai Sig. < 0,05 Dari hasil data di atas dapat disimpulkan bahwa terdapat pengaruh yang tidak signifikan dari methyl alcohol terhadap chemical compound.

3. 15 Four catalysts that may affect the concentration of one component in a three component liquid mixture are being investigated. The following concentrations are obtained. Catalyst 1 58.2 57.2 58.4 55.8 54.9

2 56.3 54.5 57.0 55.3

3 50.1 54.2 55.4

4 52.5 49.9 50.0 51.7

(a) Do the four catalysts have the same effect on the concentration ? (b) Analyze the residuals from this experiment (c) Construct a 99 percent confidence interval estimate of the mean response for catalyst 1. Jawab : Input SPSS : No 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

A

Y

1 1 1 1 1 2 2 2 2 3 3 3 4 4 4 4

58.2 57.2 58.4 55.8 54.9 56.3 54.5 57.0 55.3 50.1 54.2 55.4 52.9 49.9 50.0 51.7

Keterangan :

A : Catalysts Y : Concentration of one component in a three- component liquid

Output SPSS : Between-Subjects Factors N 1

5

2

4

3

3

4

4

Catalyst

Tests of Between-Subjects Effects Dependent Variable: Concentration Source

Type III Sum of

df

Mean Square

F

Sig.

Squares 85.676a

3

28.559

9.916

.001

45584.001

1

45584.001

15827.015

.000

A

85.676

3

28.559

9.916

.001

Error

34.562

12

2.880

Total

47622.440

16

Corrected Model Intercept

Corrected Total

120.237

15

a. R Squared = .713 (Adjusted R Squared = .641)

Analisis data Hipotesis Statistik 𝐻0 : 𝜇𝐴 = 𝜇𝐵 = 𝜇𝐶 = 𝜇𝐷 𝐻1 : Bukan 𝐻0 Hipotesis Penelitian H0 : Terdapat pengaruh yang tidak signifikan dari catalyst terhadap concentration of one component in

a three- component liquid. H1 : Terdapat pengaruh yang signifikan dari catalyst terhadap concentration of one component in a

three- component liquid. Hasil statistik : Nilai F0 = 9, 9169 dan sig. = 0,001 dikatakan signifikan apabila nilai Sig. < 0,05 Dari hasil data di atas dapat disimpulkan bahwa terdapat pengaruh yang signifikan dari catalyst terhadap

concentration of one component in a three- component liquid.

 BAB 4 latihan hal 52 4.13 An industrial engineer is conducting an experiment on eye focus time. He is interested in the effect of the distance of the object from the eye on the focus time. Four different distances are of interest. He has five subjects available for the experiment. Because there may be difference among individuals, he decides to conduct the experiment in a randomized block design. The data obtained follow, analyze the data from this experiment (use 𝛼 = 0,05) and draw appropriate conclusions. Distance (Ft)

subject 1

2

3

4

5

4

10

6

6

6

6

6

7

6

6

1

6

8

5

3

3

2

5

10

6

4

4

2

3

Jawab Input SPSS :

No 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. : 12. 13. 14. 15. 16. 17. 18. 19. 20.

A

B

Y

4 4 4 4 4 6 6 6 6 6 8 8 8 8 8 10 10 10 10 10

1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

10 6 6 6 6 7 6 6 1 6 5 3 3 2 5 6 4 4 2 3

A : Distance B : Subject Y : Eye Focus Time

Output SPSS : Univariate Analysis of Variance Between-Subjects Factors N 4

5

6

5

8

5

10

5

1

4

2

4

3

4

4

4

5

4

Distance

Subject

Tests of Between-Subjects Effects Dependent Variable: Eye Focus Time Source

Type III Sum of

df

Mean Square

F

Sig.

Squares Corrected Model

69.250a

7

9.893

7.759

.001

Intercept

470.450

1

470.450

368.980

.000

A

32.950

3

10.983

8.614

.003

B

36.300

4

9.075

7.118

.004

Error

15.300

12

1.275

Total

555.000

20

84.550

19

Corrected Total

a. R Squared = .819 (Adjusted R Squared = .713)

Analisis Data Hipotesis Statistik Pertama 𝐻0 : 𝜇01 = 𝜇02 𝐻1 : 𝜇01 ≠ 𝜇02 Kriteria Pengujian Hipotesis Tolak 𝐻0 dan terima 𝐻1 : jika nilai sig. < 0.05 Tolak 𝐻0 dan tolak 𝐻1 : jika nilai sig. > 0.05

Kedua 𝐻0 : 𝜇10 = 𝜇20 𝐻1 : 𝜇10 ≠ 𝜇20 Kriteria Pengujian Hipotesis Tolak 𝐻0 dan terima 𝐻1 : jika nilai sig. < 0.05 Tolak 𝐻0 dan tolak 𝐻1 : jika nilai sig. > 0.05 Hipotesis Penelitian Pertama 𝐻0 : terdapat pengaruh yang tidak signifikan distance terhadap eye focus time 𝐻1 : terdapat pengaruh yang signifikan distance terhadap eye focus time Kedua 𝐻0 : terdapat pengaruh yang tidak signifikan Subject terhadap eye focus time 𝐻1 : terdapat pengaruh yang signifikan Subject terhadap eye focus time Hasil Statistik 1. Nilai 𝐹0 = 8.614 dan sig. = 0.003. dari hasil data di atas dapat disimpulkan bahwa terdapat pengaruh yang tidak signifikan distance terhadap eye focus time 2. Nilai 𝐹0 = 7.118 dan sig. = 0.004. dari hasil data di atas dapat disimpulkan bahwa terdapat pengaruh yang signifikan subject terhadap eye focus time. 4.23 Suppose that in Problem 4-15 the engineer suspect that the workplace used by the four operators may represent an additional source of variation . a fourth factor, workplace (𝛼, 𝛽, 𝛾, 𝛿) may be introduced and another experiment conducted, yielding the GraecoLatin square that follower. Analyze the data from this experiment (use 𝛼 = 0.05) and drew conclusions Order of Asembly 1 2 3 4 Jawab: A : Order of Assembly B : Operator C : Latin Square D : Graeco Y : Workplaces used

Operator 1 𝐶𝛽 =11 𝐵𝛼 = 8 𝐴𝛿 = 9 𝐷𝛾 = 9

2 𝐵𝛾 = 10 𝐶𝛽 = 12 𝐷𝛼 = 11 𝐴𝛽 = 8

3 𝐷𝛿 = 14 𝐴𝛾 = 10 𝐵𝛽 = 7 𝐶𝛼 = 18

4 𝐴𝛼 = 8 𝐷𝛽 = 12 𝐶𝛾 = 15 𝐵𝛽 = 6

Input SPSS: No 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

A

B

C

D

Y

1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4

1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4

3 2 4 1 2 3 1 4 1 4 2 3 4 1 3 2

2 3 4 1 1 4 3 2 4 1 2 3 3 2 1 4

11 10 14 8 8 12 10 12 9 11 7 15 9 8 18 6

Output SPSS : Between-Subjects Factors Value Label

N

1

4

2

4

3

4

4

4

1

4

2

4

3

4

4

4

Order of Assembly

Operator

Latin Square

1

A

4

2

B

4

3

C

4

4

D

4

1

4

2

4

3

4

4

4

Graeco

Tests of Between-Subjects Effects Dependent Variable: Workplace used Source

Type III Sum of

df

Mean Square

F

Sig.

Squares Corrected Model

122.500a

12

10.208

1.114

.529

Intercept

1764.000

1

1764.000

192.436

.001

A

.500

3

.167

.018

.996

B

19.000

3

6.333

.691

.616

C

95.500

3

31.833

3.473

.167

D

7.500

3

2.500

.273

.843

Error

27.500

3

9.167

Total

1914.000

16

150.000

15

Corrected Total

a. R Squared = .817 (Adjusted R Squared = .083)

Analisis Data Hipotesis Statistik Pertama 𝐻0 : 𝜇01 = 𝜇02 𝐻0 : 𝜇01 ≠ 𝜇02 Kriteria Pengujian Hipotesis Tolak 𝐻0 dan terima 𝐻1 : jika nilai sig. < 0.05 Tolak 𝐻0 dan tolak 𝐻1 : jika nilai sig. > 0.05 Kedua 𝐻0 : 𝜇10 = 𝜇20 𝐻1 : 𝜇10 ≠ 𝜇20 Kriteria Pengujian Hipotesis Tolak 𝐻0 dan terima 𝐻1 : jika nilai sig. < 0.05 Tolak 𝐻0 dan tolak 𝐻1 : jika nilai sig. > 0.05 Ketiga 𝐻0 : 𝜇20 = 𝜇30 𝐻1 : 𝜇20 ≠ 𝜇30 Kriteria Pengujian Hipotesis Tolak 𝐻0 dan terima 𝐻1 : jika nilai sig. < 0.05 Tolak 𝐻0 dan tolak 𝐻1 : jika nilai sig. > 0.05 Keempat

𝐻0 : 𝜇30 = 𝜇40 𝐻1 : 𝜇30 ≠ 𝜇40 Kriteria Pengujian Hipotesis Tolak 𝐻0 dan terima 𝐻1 : jika nilai sig. < 0.05 Tolak 𝐻0 dan tolak 𝐻1 : jika nilai sig. > 0.05 Hipotesis Penelitian Pertama 𝐻0 : terdapat pengaruh yang tidak signifikan Order of Assembly terhadap Workplace used 𝐻1 : terdapat pengaruh yang signifikan Order of Assembly terhadap Workplace used Kedua 𝐻0 : terdapat pengaruh yang tidak signifikan Operator terhadap Workplace used 𝐻1 : terdapat pengaruh yang signifikan Operator terhadap Workplace used Ketiga 𝐻0 : terdapat pengaruh yang tidak signifikan Latin Square terhadap workplace used 𝐻1 : terdapat pengaruh yang signifikan Latin Square terhadap workplaced used Keempat 𝐻0 : terdapat pengaruh yang tidak signifikan Graeco terhadap workplaced used 𝐻1 : terdapat pengaruh yang signifikan Graeco terhadap workplaced used Hasil Statistik 1. Nilai 𝐹0 = 0.018 dan sig. = 0.996. dari hasil data di atas dapat disimpulkan bahwa terdapat pengaruh yang tidak signifikan Order of Assembly terhadap Workplace used 2. Nilai 𝐹0 = 0.691 dan sig. = 0.616 dari hasil data di atas dapat disimpulkan bahwa terdapat pengaruh yang tidak signifikan Operator terhadap Workplace used 3. Nilai 𝐹0 = 3 473 dan sig. = 0.167 dari hasil data di atas dapat disimpulkan bahwa terdapat pengaruh yang tidak signifikan Latin Square terhadap workplace used 4. Nilai 𝐹0 = 0. 273dan sig. = 0.843 dari hasil data di atas dapat disimpulkan bahwa terdapat pengaruh yang yang tidak signifikan Graeco terhadap workplaced used