Question

A technician in a quick oil change shop had the following service times (in minutes) for 20 randomly selected cars:

Sample 1	Sample 2	Sample 3	Sample 4
3.4	3.5	3.4	3.6
3.1	3.4	3.5	3.5
3.1	3.3	3.3	3.7
3.2	3.6	3.3	3.4
3.2	3.2	3.5	3.8

1. Determine the mean of each sample.
2. Estimate the mean of the process.
3. Estimate the standard deviation of the sample means (assuming that the estimate of the std. deviation of the process is 0.192).
4. What would Three Sigma control limits for the sample mean be?
5. Using the control limits of 3.14 and 3.86, are any sample means beyond these control limits? If so, which one(s)?
6. Construct the control charts for sample mean and sample range using the Alternate method. Are any sample means and ranges beyond the control limits? If so, which one(s)?

Answer 1

a)

Mean of each sample

Formula

b)

Mean of the process = x-bar-bar = 3.4 as computed in above screenshot

c)

Standard deviation of sample means = process standard deviation/sqrt(sample size) = 0.192/sqrt(5) = 0.08586501

d)

3-sigma control limits:

UCL = x-bar-bar + 3*Standard deviation of sample means = 3.4+3*0.08586501 = 3.65759503

LCL = x-bar-bar - 3*Standard deviation of sample means = 3.4-3*0.08586501 = 3.14240497

e)

No, using the control limits of 3.14 and 3.86, there are no sample means beyond these control limits

f)

Using alternate method, control limit for mean chart,

UCL = x-bar-bar + A2*R-bar = 3.4+0.577*0.44 [ for sample size = 5, A2 = 0.577] = 3.65388

LCL = x-bar-bar - A2*R-bar = 3.4-0.577*0.44 = 3.14612

Using alternate method, control limit for range chart,

UCL =D4*R-bar =2.114*0.44 [ for sample size = 5, D4 = 2.114] = 0.93016

LCL = D3*R-bar = 0 [ for sample size = 5, D3 = 0]

No, sample means and ranges are beyond the control limits