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: SAMPLE 1 2 3 NUMBER OF DEFECTIVE ITEMS IN THE SAMPLE 2 0 3 1 O 3 1 n 15 15 15 15 15 15 15 15 15 15 5 6 7 0 0 10 a. Determine the PSUCI and LCL for a p-chart of 95 percent confidenco (1.96 standard deviations) (Leave no colla blank . be certain to enter "9" wherever required. Round your answers to 3 decimal places) P S UCL LCL

Answer 1

n 15 Day No of defects 1 2 2 0 3 3 4 1 5 0 p value Calculation p value (2/15) 0.133 (0/15) 0.000 (3/15) 0.200 (1/15) 0.067 (0/15) 0.000 (3/15) 0.200 (1/15) 0.067 (0/15) 0.000 (0/15) 0.000 (0/15) 0.000 0.067 6 3 7 1 8 0 9 0 10 0 p_bar Average value

p_bar*(1-p_bar) = 0.067*(1-0.067) 0.06222 p Chart 0.00415 0.220 (p_bar*(1-p_bar}}/n =0.06222/15 Std Deviation=SQRT{p_bar/n) = SQRT(0.00415) - - - - - 1 1 - - 0.064 - - - - 1 1 1 1 0.170 As the control limits must include 95% confidence, we need to develop control charts using 2 sigma limits 0.120 - - - - - - - - 1 - 1 p value 0.070 p value 0.195 UCL = p_bar + 2*std deviation = 0.067 + (2*0.064) LCL = p_bar - 2*std deviation = 0.067 - (2*0.064) -0.062 0.020 1 2 3 4 5 6 7 8 9 10 UCL CL LCL -0.030 1 - 1 - 1 - - - - - - 1 We can see that all p values do not lie within the control limits, hence the process is not in control. -0.080 Answers p_bar = 0.067 Std Dev = 0.064 UCL = 0.195 LCL= -0.062