Operations Management question (Process control): A new coin machine printing nickels is being tested by the Mint. Ten samples, with each using 10 nickles, are weighed. Each nickle should weigh 5...
Operations Management question (Process control):
A new coin machine printing nickels is being tested by the Mint. Ten samples, with each using 10 nickles, are weighed. Each nickle should weigh 5 g and each sample should weigh 50g. The data collected is shown in the chart below:
- Determine the upper and lower control limits and the overall mean for the x bar chart and the R chart
- Draw the charts and plot the values of the sample means and range.
- Does the data indicate that the mint should invest in this new machine? Explain
I will answer this question using the R Software. I will also include the R Code used.
Determine the upper and lower control limits and the overall mean for the x bar chart and the R chart.
Note that the question says that "Each nickle should weigh 5 g and each sample should weigh 50 g". Looking at the attached data table, we observe that the column titled sample mean actually contains total weights, rather than sample means of each sample of 10 nickels. As such we have to compute a new column of sample means (by dividing each sample total by sample size). We enter the data in R as follows:
> sample_mean <- c(49.998,50.001,50.025,50.121,50.012,49.953,49.999,50.002,50.007,49.099)
> sample_range<- c(0.017,0.011,0.025,0.005,0.001,0.004,0.007,0.022,0.001,0.014)
> sample_mean <- sample_mean/10
 4.9998 5.0001 5.0025 5.0121 5.0012 4.9953 4.9999 5.0002 5.0007 4.9099
> xbarbar <- mean(sample_mean)
> rbar <- mean(sample_range)
We can see that the X-bar-bar = 4.99217 and the R-bar = 0.0107.
Other values are calculated as follows:
UCL (R-chart) = R-Bar x D4 = 0.0107 x 1.777 = 0.0190139
LCL (R-chart) = R-bar x D3 = 0.0023861
UCL (X-bar) = X-bar-bar + (A2 x R-bar) = 4.995466
LCL (X-bar) = X-bar-bar - (A2 x R-bar) = 4.988874
The values D3, D4 and A2 are found in the control chart constants table.
Draw the charts and plot the values of the sample means and range.
The images are attached below. They are produced in R using the following code:
> plot(1:10,sample_mean,ylim = c(4.8,5.1),xlim = c(1,10),type = 'b', xlab = "sample no.", ylab = "mean", main = "X-Bar Chart")
> abline(h = xbarbar)
> abline(h = 4.995466)
> abline(h = 4.988874)
> plot(1:10,sample_range,xlim = c(1,10),type = 'b', xlab = "sample no.", ylab = "range", main = "R Chart")
Does the data indicate that the mint should invest in this new machine? Explain.
We first look at the R-chart. We observe that there are 2 points below the LCL and 2 points above the UCL. Therefore, the R-chart is out of control and the limits on the X-bar chart are meaningless. As such the mint should not invest in the machine.
1) The xbar and R chart are implemented together. The center lines are respectively the average item weight (calculated from all samples) and the average range over the n (=10) samples.
Therefore the center lines are respectively
xbar = Sum(xi)/mn where m=10 is the number of samples, n=10 is the sample size and xi are the sample total weights
xbar = 4.99217
Rbar = Sum(Ri)/m = 0.0107
Now, the control limits for the two charts are respectively given by
xbar +/- A2*Rbar
where A2, D3 and D4 are sample-size-specific anti-biasing constants available from statistical tables. For n=10 we find from the tables that we should set
A2 = 0.308, D3 = 0.223, D4 = 1.777
so that the control limits for the xbar chart and R chart respectively are
(4.99217+/- 0.308*0.0107), (0.223*0.0107, 1.777*0.0107) =
(4.988874,4.995466), (0.0023861,0.0190139) =
(4.989,4.995), (0.0024,0.0190) approx. With the center lines at
xbar = 4.992, Rbar = 0.0107.
2) See attached figure for drawing of the charts
3) The R chart indicates that the process is not 'in control' since a number of the sample range values fall outside of the control limits. This indicates that the variance of the samples is not constant. On this evidence alone, regardless of what the xbar chart indicates, the mint should not invest in the machine as it does not perform predictably to specification. It is noticeable that the final sample mean value is very low, and hence weighs down the overall mean estimate to somewhat below the target of 5g, but this is not of immediate interest since analysis does not proceed beyond inspection of the R chart, which indicates poor performance of the machine.