Data Analysis-Take Home Final Second Part2

Data Analysis-Take Home Final Second Part2 - 1 8-21 A Hi st...

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1.) 8-21 A) 3.5 3.0 2.5 2.0 1.5 1.0 0.5 9 8 7 6 5 4 3 2 1 0 Time (min) Frequency Mean 2.090 StDev 0.7590 N 30 Normal  Histogram of Wally Burgers' Drive-Thru Service Time (min) The histogram seems to be almost normal but seem to be more skewed right B.) 28 25 22 19 16 13 10 7 4 1 4 3 2 1 0 Obser vation Individual Value _ X= 2.090 UCL= 4.191 LCL= -0.011 28 25 22 19 16 13 10 7 4 1 2.4 1.8 1.2 0.6 0.0 Moving Range __ MR= 0.79 UCL= 2.581 LCL= 0 4 I -MR Chart of Wally Burgers' Drive-Thru Service Time The data seems to be in staticical control
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C.) standard deviation is .70035 process mean is 2.090 D.) E.) X=.85 The new mean is .85 min. in order for the UTL no to exceed 3.4 customers per million. F.) X=.4244 new ơ is .4244 so ơ 2 = .18015. ơ 2 =
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2.) 9-3 A 28 25 22 19 16 13 10 7 4 1 10.010 10.005 10.000 O bse r v a t ion I n d i v u a l   V e _ X= 10.00321 UCL= 10.00876 LCL= 9.99767 28 25 22 19 16 13 10 7 4 1 0.0060 0.0045 0.0030 0.0015 0.0000 O bse r v a t ion M o g R __ MR= 0.002085 UCL= 0.006812 LCL= 0 9 -3 a
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Both charts are within statistical control according to the above Minitab charts. B ) σ measurement = (MR/d 2 ) 2 >> ( 0.002085 / 1.128) 2 = 0.0000034166 Since reproducibility error = 0 because there is only a single technician taking the measurements σ measurement = σ repeatabiliyt = 0.000342% C.) (1 / SNR) 2 = σ observed / σ measurement σ observed = (R/d 2 ) 2 >> (0.03865 / 2.059) 2 = 0.0003524 σ observed / σ measurement = (0.01877 / 0.001848) = 10.157 Because SNR > 10, the measurement error is negligible with respect to the size of the process variation. D ) C p = (ET / NT) >> (0.2 / 0.11208) = 1.784 ET = USL – LSL >> 10.1 - 9.9 = 0.2 NT = 6σ process >> (6*0.01868) = 0.11208 C pk = {C pl , C pu } >> {2.225, 1.344} C pl = (LSL - μ) / 3σ process >> (9.9 - 10.02471) / (3*0.01868) = 2.225 C pk = (USL - μ) / 3σ process >> (10.1 - 10.02471) / (3*0.01868) = 1.344 C pk > 1.33, so the process is capable and there are less than 63 defective parts per million e) z = (USL - x) / σ x —> (10.1 - 10.02471) / 0.01877 = 0.000032% z = (LSL - x) / σ x —> (9.9 - 10.02471) / 0.01877 ≤ 0.000001%
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f) Upper limit defects(same Z) = 30.3593 defects per million Lower limit defects( same Z) = < 0.0002 defects per million g) C pm = ET / 6σ target —> 0.2 / (6*0.031) = 1.074 σ target = √((σ process ) 2 + (μ - Target) 2 ) —> √((0.01868) 2 + (10.02471 - 10) 2 ) = 0.031 P pk = {P pl , P pu } = {2.215, 1.337} P pl = (LSL - μ) / 3σ observed >> (9.9 - 10.0247) / (3*0.01877) = 2.215 P pu = (USL - μ) / 3σ observed > > (10.1 - 10.0247) / (3*0.01877) = 1.337 The process is capable. 9-26) a) 28 25 22 19 16 13 10 7 4 1 256 254 252 250 Sample Sample Mean _ _ X= 253.668 UCL= 255.216 LCL= 252.121 28 25 22 19 16 13 10 7 4 1 6.0 4.5 3.0 1.5 0.0 Sample Range _ R= 3.202 UCL= 6.417 LCL= 0 1 1 1 1 Xbar-R Chart of C1, . .., C6 Because the variance between plate to plate is high, some of the plates become out of control.
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b) The R-bar chart captures the variability within the plates. The X-bar chart captures the variability between plate to plate. X-bar = total average points/30 = 253.668 σ= R-bar/d 2 = 3.202/2.534 = 1.2636 The average shows the average of all averages of all 30 plates within each locations. The standard deviation shows how far the variance is off of the average. I am very confident in the average. Since the data is out of control I am not as confident in the estimate of standard deviation and how well it shows the variation.
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c) 28 25 22 19 16 13 10 7 4 1 256.8 255.6 254.4 253.2 252.0 Sample Sample Mean _ _ X= 253.917 UCL= 256.580 LCL= 251.255 28 25 22 19 16 13 10 7 4 1 8 6 4 2 0
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Data Analysis-Take Home Final Second Part2 - 1 8-21 A Hi st...

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