Question

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2. A process has a X-bar control chart with UCL-17.4. LCL-12.79 using a 3-sigma limit. The subgroup sample size is 3. (a) Estimate the standard deviation of the process, sigma hat. (6 pts) (b) Suppose that the process mean shifts to 13. Find the probability that this shift will be detected on the next sample. (8 pts) (c) Calculate the average run length (ARL) before detecting the shift. (6 pts)

Answer 1

ANSWER:

0 Given details, UCL 174 LCL = 12-11 / Sample size in): 3 (subgroup) a. from the control chart table (2) for n=3, A2=1.023 , dz -1.693 "ULL = x +A₂R ) 17.4 = +1.023 R LCL = x - A₂ Ř ) 12.79 M - 1.023 Ř - ② After solving these two equation and 2.253 X = 15.095 2.253 -1.331 now 1,693 d2 ol-33) 1 b. given û'-13'; where M': shifted mean. ...P{ no detection there is a shift] - P[LCL <7 <UCLU M= 13) » B =P(12.79<# < 17:41M'=13] 12.79-133-13 âlho alun 17.4-13 2 2 =P ô lin

4.4 2 -0.21 22 L -PL 1-331) 1-331). -P[-0.273<225.73) $15.73) – iⓇ (-0.273) -0.99 - (0.5-P[0<220-273]] (due to symmetric nature of normal Curve) 0.99-20.5 -0.1064] = 0.5964 B = 0.5964 Now : P[detection at next sample) - B(1-6) = (0.5964) (-0.5964) = 0.241 C. ARL I-B 1-0.5964 2.48 That means on average we can say after second point we will get a point outside the control 3rd limits,