Please show all work and steps for each question. Also may support answers with

Question

Please show all work and steps for each question. Also may support answers with Minitab where it applies.
2) A process being controlled with a fraction nonconforming control chart. The process average has been shown to be 0.07. The procedure states to take daily sample of size of 500 items. a) Calculate the upper and lower three-sigma control limits b) Calculate the upper and lower 0.05 control limits. ) If the Process average suddenly shifts to 0.10, what is the probability that the process shift is detected? d) What is the ARL for p-0.1? e What is the smallest sample size for the lower control limit to be positive?

Explanation / Answer

(a) process average p0 = 0.07

sample size = 500

upper three - sigma control limit = p0 + 3 * sqrt [p0 * (1-p0)/N] = 0.07 + 3 * sqrt (0.07 * 0.93/500) = 0.07 + 3 * 0.01141 = 0.1042

Lower three sigma control limit = p0 - 3 * sqrt [p0 * (1-p0)/N] = 0.07 - 3 * sqrt (0.07 * 0.93/500) = 0.07 - 3 * 0.01141 = 0.0358

(b) Upper 0.05 control limit = p0 + 1.96 * sqrt [p0 * (1-p0)/N] = 0.07 + 1.96 * sqrt (0.07 * 0.93/500) = 0.07 + 1.96 * 0.01141 = 0.0924

Lower 0.05 control limit = p0 - 1.96 * sqrt [p0 * (1-p0)/N] = 0.07 - 1.96 * sqrt (0.07 * 0.93/500) = 0.07 - 1.96 * 0.01141 = 0.0476

(c) Here proces mean is shifted to 0.10, then probaility that the process shift is detected.

Pr(p > 0.10 ; 0.07 ; 0.01141) = 1 - Pr(p < 0.10; 0.07 ; 0.01141)

Z = (0.10 - 0.07)/ 0.01141 = 2.63

Pr(p > 0.10 ; 0.07 ; 0.01141) = 1 - Pr(p < 0.10; 0.07 ; 0.01141) = 1- Pr(Z < 2.63) = 1 - 0.9957 = 0.0043

(d) ARL for p = 0.1 is

ARL = 1/ 0.0043 = 234.23 or 234

(e) Here lets say smalles sample size is n

so to be positive,

LCM = p0 - 3 * sqrt [p0 * (1-p0)/N] > 0

0.07 - 3 *sqrt [0.07 * 0.93/n] > 0

0.072 > 9 * (0.07 * 0.93/n)

n > 9 * (0.07 * 0.93)/0.072

n > 119.57

or n = 120

so smallest sample size mus be 120 for lower control limit to be positive.

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