Question

1. The length of industrial filters is a quality characteristic of interest. Thirty samples,
   each of size 5, are chosen from the process. The data yields an average length of
   110 mm, with the process standard deviation estimated to be 4 mm.
   (a) Find the warning limits for a control chart for the average length.
   (b) Find the 3sigma control limits. What is the probability of a type I error?
   (c) If the process mean shifts to 112 mm, what are the chances of detecting this shift
   by the third sample drawn after the shift?
   (d) What is the chance of detecting the shift for the first time on the second sample
   point drawn after the shift?
   (e) What is the ARL for a shift in the process mean to 112 mm? How many samples,
   on average, would it take to detect a change in the process mean to 116 mm?

Answer 1

Here X: “ Length of industrial filter” is a quality characteristic. Given that thirty samples each of size five gave the process average = 110 mm with process standard = 4 mm.

1. Warning limits are 2σ limits, therefore

UCL of warning limit =Process Average + 2 Process Standard Deviation

= 110+2*4= 118

LCL of warning limit = Process Average - 2 Process Standard Deviation

= 110-2*4= 102

1. 3 σ limits are

UCL of 3σ limit =Process Average + 3 Process Standard Deviation

= 110+3*4= 122

LCL of 3σ limit = Process Average - 3 Process Standard Deviation

= 110-3*4= 98

Probability of Type I error = P[ X > UCL] + P[X < LCL] = 0.0027

1. P=Probability that process shifts to 112 mm = P[X > 112]

P[Z > (112-110)/4] = P[Z > 0.5] =0.691462

Probability that shift in process average is detected on third sample =

Probability that shift is not detected on first, second but detected on third sample= (1-0.691462)²(0.691462) = 0.065824

1. Probability that shift in process average is detected on second sample for the first time =

      Probability that shift is not detected on first, but detected on second sample

       = (0.691462)(0.691462) = 0.213342

1. ARL for a shift in the process mean to 112 mm = 1/P where P is calculated in c)

Therefore ARL = 1/0.691462= 1.4462. It means shift in process will be detected on first or second sample only.