Statistic Process Control Week 3 Ananda Sabil Hussein, SE, MCom.

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1 Statistic Process Control Week 3 Ananda Sabil Hussein, SE, MCom

2 Latar Belakang Pertengahan tahun 80 an pangsa pasar pager Motorola di rebut oleh produk-produk Jepang seperti halnya NEC, TOSHIBA dan Hitachi. Motorola melakukan perubahan radikal dengan memperbaiki mutu, pengembangan produk dan penurunan biaya yang berbasis statistik

3 Statistical Process Control Teknik statistik yang secara luas digunakan untuk memastikan bahwa proses yang sedang berjalan telah memenuhi standar.

4 Start Produce Good Provide Service Stop Process Yes No Assign. Causes? Take Sample Inspect Sample Find Out Why Create Control Chart

5 Variasi Alami dan Khusus Variasi alami adalah sumber-sumber variasi dalam proses yang secara statistik berada dalam batas kendali Variasi Khusus/dapat dihilangkan yaitu variasi yang muncul disebabkan karena peralatan yang tidak sesuai, karyawan yang lelah atau kurang terlatih serta bahan baku baru.

6 Diagram Pengendalian

7 17 = UCL 15 = LCL 16 = Mean Sample number |||||||||||| 123456789101112

8 Konsep Rata-rata dan Jarak Rata-rata

9 Menentukan Batas Diagram Rata-rata Batas Kendali Atas (UCL) = Batas Kendali Bawah (LCL) = = rata-rata dari sampel = = Standar deviasi = 2 (95.5%) 3(99.7%) = Standar deviasi rata-rata sampel

10 Cara Lain Batas Kendali Atas = Batas Kendali Bawah Dimana : = rentangan rata-rata sampel = Nilai batas kendali = rata-rata dari sampel rata-rata

11 Batas Bagan Rentangan

12 Bagan Rata-rata(a) These sampling distributions result in the charts below (Sampling mean is shifting upward but range is consistent) R-chart (R-chart does not detect change in mean) UCL LCL x-chart (x-chart detects shift in central tendency) UCL LCL

13 R-chart (R-chart detects increase in dispersion) UCLLCL Figure S6.5 (b) These sampling distributions result in the charts below (Sampling mean is constant but dispersion is increasing) x-chart (x-chart does not detect the increase in dispersion) UCLLCL Bagan Jarak

14 Bagan Kendali Atribut Mengukur persentase penolakan dalam sebuah sampel, bagan-p Menghitung jumlah penolakan, bagan-c

15  For variables that are categorical  Good/bad, yes/no, acceptable/unacceptable  Measurement is typically counting defectives  Charts may measure  Percent defective (p-chart)  Number of defects (c-chart) Control Charts for Attributes

16 Control Limits for p-Charts Population will be a binomial distribution, but applying the Central Limit Theorem allows us to assume a normal distribution for the sample statistics UCL p = p + z  p ^ LCL p = p - z  p ^ wherep=mean fraction defective in the sample z=number of standard deviations  p =standard deviation of the sampling distribution n=sample size ^ p(1 - p) n p =p =p =p = ^

17 Contoh Soal JamRata2JamRata2JamRata2 117.1516.5916.3 218.8616.41016.5 314.5715.21114.2 414.8816.41217.3

18 Ditanyakan : Batas kendali proses 9 boks yang mencakup 99.7% Jawab : UCLx = = 16 + 3 LCLx == 16 - 3

19 Setting Control Limits Process average x = 16.01 ounces Average range R =.25 Sample size n = 5

20 Setting Control Limits UCL x = x + A 2 R = 16.01 + (.577)(.25) = 16.01 +.144 = 16.154 ounces Process average x = 16.01 ounces Average range R =.25 Sample size n = 5 From Table S6.1

21 Setting Control Limits UCL x = x + A 2 R = 16.01 + (.577)(.25) = 16.01 +.144 = 16.154 ounces LCL x = x - A 2 R = 16.01 -.144 = 15.866 ounces Process average x = 16.01 ounces Average range R =.25 Sample size n = 5 UCL = 16.154 Mean = 16.01 LCL = 15.866

22 Contoh Soal SampleNumberFractionSampleNumberFraction Numberof ErrorsDefectiveNumberof ErrorsDefective 16.06116.06 25.05121.01 30.00138.08 41.01147.07 54.04155.05 62.02164.04 75.051711.11 83.03183.03 93.03190.00 102.02204.04 Total = 80 (.04)(1 -.04) 100  p = =.02 ^ p = =.04 80(100)(20)

23 .11.11 –.10.10 –.09.09 –.08.08 –.07.07 –.06.06 –.05.05 –.04.04 –.03.03 –.02.02 –.01.01 –.00.00 – Sample number Fraction defective |||||||||| 2468101214161820 p-Chart for Data Entry UCL p = p + z  p =.04 + 3(.02) =.10 ^ LCL p = p - z  p =.04 - 3(.02) = 0 ^ UCL p = 0.10 LCL p = 0.00 p = 0.04

24 .11.11 –.10.10 –.09.09 –.08.08 –.07.07 –.06.06 –.05.05 –.04.04 –.03.03 –.02.02 –.01.01 –.00.00 – Sample number Fraction defective |||||||||| 2468101214161820 UCL p = p + z  p =.04 + 3(.02) =.10 ^ LCL p = p - z  p =.04 - 3(.02) = 0 ^ UCL p = 0.10 LCL p = 0.00 p = 0.04 p-Chart for Data Entry Possible assignable causes present

25 Control Limits for c-Charts Population will be a Poisson distribution, but applying the Central Limit Theorem allows us to assume a normal distribution for the sample statistics wherec=mean number defective in the sample UCL c = c + 3 c LCL c = c - 3 c

26 c-Chart for Cab Company c = 54 complaints/9 days = 6 complaints/day |1|1 |2|2 |3|3 |4|4 |5|5 |6|6 |7|7 |8|8 |9|9 Day Number defective 14 14 – 12 12 – 10 10 – 8 8 – 6 6 – 4 – 2 – 0 0 – UCL c = c + 3 c = 6 + 3 6 = 13.35 LCL c = c - 3 c = 3 - 3 6 = 0 UCL c = 13.35 LCL c = 0 c = 6