Ten samples of 15 parts each were taken from an ongoing process to establish a p

Question

Ten samples of 15 parts each were taken from an ongoing process to establish a p-chart for control. The samples and the number of defectives in each are shown in the following table NUMBER OF DEFECTIVE ITEMS IN THE SAMPLE SAMPLE n 15 15 15 15 15 2 9 10 15 a. Determine the Sp, UCL and LCL for a p-chart of 95 percent confidence (1.96 standard deviations). (Leave no cells blank-be certain to enter "O" wherever required. Round your answers to 3 decimal places.) UCL LCL b. What comments can you make about the process? Process is out of statistical control Process is in statistical control

Explanation / Answer

a) p-bar = Total number of defect/(Total number of sample) = = (2+2+0+2+1+3+2+1+3+2)/(15*10) = 0.120

Sp = sqrt(p-bar*(1-p-bar)/n) = sqrt(0.12*(1-0.12)/15) = 0.084

UCL = p-bar + 1.96*Sp = 0.120 + 1.96*0.084 = 0.285

LCL = p-bar - 1.96*Sp = 0.120 - 1.96*0.084 = -0.045. As it cannot be negative, hence 0

b) UCL is 0.285*15 = 4.275

As all defectives are within UCL & LCL. So, process is under control

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