Question

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

Answer:

Given,

Average length = 110 mm

Standard deviation() = 4 mm

Now to give the warning limits for a control chart for the average length

Upper control limit(3) = Average length + 2*standard deviation

substitute known values

= 110 + 2*4

= 110 + 8

UCL = 118

Lower limit = 110 - 2*4

= 110 - 8

LCL = 102

b)

To give the 3 sigma control limits & the probability of a type I error

UCL = mean + 3*

= 110 + 3*4

= 110 + 12

UCL = 122

LCL = mean - 3*

= 110 - 3*4

= 110 - 12

LCL = 98

Now to give the required probability

P(Type 1 error) = P(X > UCL) + P(X < LCL)

= P(X > 122) + P(X < 98)

= P(z > (112 - 110)/4 ) + P(z < (98 - 110)/4)

= P(z > 0.5) + P(z < - 3)

= 0.3085375 +  0.0013499 [since from z table]

= 0.3098874

P(Type 1 error) = 0.3099

c)

To give the probability of the chances of detecting this shift by the third sample drawn after the shift

P(X > 112) = P(z > (112 - 110)/4)

= P(z > 2/4)

= P(z > 0.5)

= 0.3085375 [since from z table]

P(X > 112) = 0.3085

P(Shift in average length is detected on the third sample) = (1 - 0.3085)^2 * 0.3085

= 0.4782*0.3085

Required probability = 0.1475

d)

To give the chance of detecting the shift for the first time on the second sample point drawn after the shift

i.e.,

= 0.3085*0.3085

= 0.0952

e)

ARL = 1 / P(X > 112)

= 1 / 0.3085

= 3.2415

Here we can say that the shift in process will be detected on third sample