HW 2 (SPC) (2)

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California State University, Northridge *

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Statistics

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Feb 20, 2024

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HW #2. Solve the following two problem sets. Show your work. Worth 20 points in total. Due 3/10 (Thurs) . I. Suppose that you have collected the following output data from a process. Observations Sample Number 1 2 3 4 1 11.3 10.2 10.6 10.3 2 9.3 10.6 9.7 10.4 3 9.8 10.1 11 10.7 1. Determine CL, UCL, LCL for an R -chart. Observations Range Sample Number 1 2 3 4 1 11.3 10.2 10.6 10.3 1.10 2 9.3 10.6 9.7 10.4 1.3 3 9.8 10.1 11 10.7 1.2 R average = 1.2 CL=R-bar = 1.2 UCL= D4*R-bar UCL= (2.115)(1.2)= 2.538 LCL = D3*R-bar = (0)(1.2) = 0 2. Determine CL, UCL, LCL for an -chart . Observations x-bar Sample Number 1 2 3 4 1 11.3 10.2 10.6 10.3 10.6 2 9.3 10.6 9.7 10.4 10.0 3 9.8 10.1 11 10.7 10.4 The average of x-bar= 10.333
CL= X-double bar = 10.333 UCL= X-double bar + A2*R-bar = 10.333+ (0.729)(1.2) = 11.208 LCL= X-double bar -A2*R-bar = 10.333- (0.729)(1.2) = 9.458 3. Suppose the specifications for the process output are 10 ± 1 . If the standard deviation, σ , of the process output is estimated to be 0.215 and the firm is seeking four sigma qualities, is the process capable of producing at the target quality level? NV= 10 USL = 10+1 = 11 LSL = 10-1 = 9 Xdouble bar = 10.333 Sigma= 0.215 Cpk = min[double x bar-LSL/3sigma, USL-double X bar/3sigma] [ 10.333-9/3(0.215), 11-10.333/3(0.215) ] = [ 2.06, 1.034] Since the minimum Cpk, which is 1.034, less than 1.33, the process does not have the 4-sigma quality. II. In order to establish a p -chart with 3-sigma control limits, you have collected the following 10 samples of size 300. Sample Defects Sample Defects 1 15 6 15 2 22 7 21 3 17 8 15 4 42 9 16 5 16 10 18 1. Determine CL, UCL, and LCL for the p -chart.
Sample n Defects P 1 300 15 0.05 2 300 22 0.07 3 300 17 0.06 4 300 42 0.14 5 300 16 0.053 6 300 15 0.05 7 300 21 0.07 8 300 15 0.05 9 300 16 0.053 10 300 18 0.06 To find P-bar, p-bar = number of defects / Total number of observations 197/ 3000 = 0.0657 Next, we have to caculate the sandard diviation of sample propotion: Sp = pbar ( 1 pbar )/ n = 0.0657 ( 1 0.0657 )/ 300 =0.0143 To determine the CL and Control limits: CL = Pbar = 0.0657 UCL = Pbar+3Sp = 0.0657+3(0.0143) = 0.1086 LCL= Pbar-3Sp = 0.0657 – 3(0.0143) = 0.0228 2. Is the process in statistical control? Explain. The process is not in statistical control because the defect rate for sample 4 is 0.14 which is outside control limit.
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