PROBLEM 2 (5 points) Samples of size n=5 are taken from a manufacturing process

Question

PROBLEM 2 (5 points) Samples of size n=5 are taken from a manufacturing process every hour. A CTQ characteristic is measured, X-bar and R are computed for each sample. After 25 samples have been analyzed, we have 25 25 i-1 R-9.00 i-1 =662.50 and The quality characteristic is normally distributed. (a) Find the control limits for the X-bar and R charts. (b) Assume that both charts exhibit control. If the specifications are 26.40 ± 0.50, estimate the number of defective parts produced (in PPMs). (c) If a special cause of variation made the process mean to shift to a value of 26.61, how many samples would take to the X-bar control chart to detect that change?

Explanation / Answer

Solution

Back-up Theory

Let xij represent the jth measurement in the ith sub-group, j = 1, 2,   , ni and j = 1, 2,   , k where ni = size of the ith sub-group and k = number of sub-groups.

Average of the ith sub-group, say xi(bar) = (sum over j of xij)/ni

Grand average, say x(double bar) = {sum over i of xi(bar)}/k

Sub-group Range, say Ri = Max measurement – min measurement of ith sub-group.

Average Range = (sum over i of Ri)/k

Control Limits for x(bar) = x(double bar) ± A2.R(bar) [values of A2 are available in Standard Tables of Control Charts Constants]

Control Limits for R = lower: D3.R(bar); upper: D4.R(bar) [values of D3 and D4 are available in Standard Tables of Control Charts Constants]

Standard Deviation, = R(bar)/d2 [values of d2are available in Standard Tables of Control Charts Constants]

Under the assumption that measurements follow Normal Distribution, given the specification for measurement as

s ± t, proportion of measurement not conforming to specification is given by

P[Z < {(s - t) - x(double bar)}/] + P[Z > {(s + t) - x(double bar)}/], where Z is the Standard Normal Variate whose probabilities can be read off from the Standard Normal Tables]

Now, to work out the solutions,

We have ni = 5 for i, k = 25, A2(5) = 0.577, d2(5) = 2.326, D3(5) = 0, D4(5) = 2.114

Part (a)

x(double bar) = 662.50/25 = 26.5; R(bar) = 9.00/25 = 0.36

Control Limits for x(bar) = 26.5 ± 0.577x0.36 = 26.5 ± 0.2077 = (26.2923, 26.7077) ANSWER

Control Limits for R = lower: 0; upper: 2.114x0.36 = 0.7610.

So, the limits are: (0, 0.761) ANSWER

Part (b)

= 0.36/2.326 = 0.1548

Given specification is: (25.90, 26.90), proportion of measurement not conforming to specification

= P[Z < {(- 0.6/0.1548)/ + P[Z >{(0.4/0.1548)}]

= P(Z < - 3.876) + P(X > 2.584) = 0.0001 + 0.0049 = 0.005 = 0.5% ANSWER

Part (c)

Given setting is at 26.40, replace x(double bar) by 26.61 in the above (b) part calculations.

Proportion of measurement not conforming to specification = P[Z < {(- 0.5/0.1548)/ + P[Z >{(0.3/0.1548)}]

= P(Z < - 3.230) + P(X > 1.937) = 0.0006 + 0.0244 = 0.025 = 2.5% ANSWER

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