Question

The following are quality control data for a manufacturing process at Kensport Chemical Company. The data show the temperature in degrees centigrade at five points in time during a manufacturing cycle. Sample x R 1 95.72 1.0 2 95.24 0.9 3 95.18 0.9 4 95.48 0.4 5 95.46 0.5 6 95.32 1.1 7 95.40 1.0 8 95.44 0.3 9 95.08 0.2 9 10 10 95.50 0.6 11 95.80 0.6 12 95.22 0.2 13 95.60 1.3 14 95.22 0.5 15 95.04 0.8 16 95.72 1.1 17 94.82 0.6 18 95.46 0.5 19 95.60 0.4 20 95.74 0.6 The company is interested in using control charts to monitor the temperature of its manufacturing process. Compute the upper and lower control limits for the R chart. (Round your answers to three decimal places.) UCL = ICI =

Answer 1

To find the control limits of R chart, we first need to find $\bar R$ (R-bar) which is the mean of all Ranges.

R 1+0.9 +0.9 + 0.4+ 0.5 + 1.1+1+0.3 +0.2 + 0.6 + 0.6 + 0.2 +1.3+0.5 + 0.8 +1.1 +0.6 + 0.5 + 0.4 +0.6 20
$\bar R = \frac{13.5}{20} = 0.675$

For the R chart the control limits are given as below

LCL = D₃*R-bar

UCL = D₄*R-bar

For sample size of 5, we have D₃ = 0 and D₄ = 2.115. Hence the control limits for range control chart are

LCL = 0*0.675 = 0

UCL = 2.115*0.675 = 1.428

The R-chart is plotted as below

R-chart 2 1.75 1.5 1.25 Sample Range 1 0.75 w 0.5 0.25 0 0 5 10 15 20 Sample Number

Thus we can see that the correct graph is as per option c.

To find the control limits of x-bar chart we need to find the mean of all the means(x-double bar) along with the R-bar

$\bar {\bar x} = \frac{95.72+95.24+95.18+95.48+95.46+95.32+95.4+95.44+95.08+95.5+95.8+95.22+95.6+95.22+95.04+95.72+94.82+95.46+95.6+95.74 }{20}$$\bar {\bar x} = \frac{1908.04}{20} = 95.402$

Now for the x-bar chart the control limits are given as per below formula

Control Limits = (x-double bar) $\small \pm$ A₂*(R-bar)

From the control chart tables for the sample size of 5, A₂ = 0.577

Hence Control Limits for x-bar chart are

UCL = 95.402 + 0.577*0.675 = 95.791

LCL = 95.402 - 0.577*0.675 = 95.013