Question

QUESTION #10:

Consider a process by which coils are manufactured. Samples of size n = 5 are randomly selected from the process, and the resistance values (in ohms) of the coils are measured for 25 subgroups (samples). The data values are given in the following table.

Sample	x1	x2	x3	x4	x5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25	20 19 25 20 19 22 18 20 21 21 20 22 19 20 20 21 20 20 20 22 23 21 21 20 19	22 18 18 21 24 20 20 18 20 19 20 21 22 21 24 20 18 24 19 21 22 18 24 22 20	21 22 20 22 23 18 19 23 24 20 23 20 19 22 24 24 18 22 23 21 22 18 24 21 21	23 20 17 21 22 18 18 20 23 20 22 22 18 21 23 20 20 23 20 24 20 17 23 21 21	22 20 22 21 20 19 20 21 22 20 20 23 19 22 23 21 20 23 19 22 22 19 23 20 22

1. Use these data to determine the control limits for the x and R control charts for this coil resistance process.
2. Use Excel and the X-Y (scatter) chart tool to plot the control charts for the 25 samples. Is this process in statistical control? (1 mark)
3. Suppose that the assignable cause(s) responsible for any out-of-control points found in part (b) has been identified and adjustments made to the process to correct its performance. Calculate the revised center line and control limits if necessary. Are there any other points plotting outside the control limits? If so, assume that, on further investigation, no special causes could be identified for this sample(s). So, the revised limits will then be used to monitor future observations. (1 mark)
4. Estimate the (unknown) process average μ and (unknown) process standard deviation σ. (1 mark)
5. Suppose that the specifications on coil resistance are 21 ± 3 (in ohms). Use the results of part (d) to determine the proportion of the output that is nonconforming (i.e. does not meet specifications), assuming that the quality characteristic (coil resistanc is normally distributed.
6. If the daily production rate is 10,000 coils and if coils with a resistance less than the LSL (lower specification limit) cannot be used for the desired purpose, what is the loss to the manufacturer if the unit cost of scrap is 50 cents? (1 mark).

Answer 1

(a)

Sample size (n) = 5, using the standard 3-sigma tables for X-bar/ R chart, we get A2 = 0.577, D3 = 0, D4 = 2.114

X-bar chart:

UCLx = X_double-bar + A2 * R-bar = 20.84 + 0.577 * 3.48 = 22.85
CLx = X_double-bar = 20.84
LCLx = X_double-bar - A2 * R-bar = 20.84 - 0.577 * 3.48 = 18.83

Range Chart:

UCLr = D4 * R-bar = 2.114 * 3.48 = 7.357
CLr = R-bar = 3.48
LCLr = D3 * R-bar = 0.000 * 3.48 = 0.000

(b)

The process is out of control for sample 22 and 23 in the mean chart and sample 3 in the range chart.

(c)

Remove sample 3, 22, and 23 and recalculated the mean and range chart UCL and LCL as follows:

X-bar chart:

UCLx = X_double-bar + A2 * R-bar = 20.864 + 0.577 * 3.273 = 22.75
CLx = X_double-bar = 20.86
LCLx = X_double-bar - A2 * R-bar = 20.864 - 0.577 * 3.273 = 18.98

Range Chart:

UCLr = D4 * R-bar = 2.114 * 3.273 = 6.92
CLr = R-bar = 3.27
LCLr = D3 * R-bar = 0.000 * 3.273 = 0.000

The revised chart contains just one point (sample-14) outside the limit in the mean chart marginally.

(d)

Process mean (μ) is estimated by the X_double-bar i.e. 20.86
Process stdev (σ) = R-bar / d2

The coefficient d2 is to be noted from the standard table and will be equal to 2.326 for n=5.

So,

σ = 3.27 / 2.326 = 1.41

(e)

USL = 21 + 3 = 24
LSL = 21 - 3 = 18

μ = 20.86
σ = 1.41

Prob{LSL < Output < USL} = Prob(Output < USL} - Prob{Output < LSL}

= NORM.DIST(24, 20.86, 1.41, 1) - NORM.DIST(18, 20.86, 1.41, 1)

= 0.966

So, Proportion non-conforming = 1 - Prob{LSL < Output < USL} = 1 - 0.966 = 0.034

(f)

Prob{Output < LSL} = NORM.DIST(18, 20.86, 1.41, 1) = 0.021

So, expected number of loss = 0.021 * 10,000 = 210 units

So, cost of scrap per day = 210 * $0.50 = $105