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 0.8 4 95.42 0.4 5 95.46 0.5 6 95.32 1.1 7 95.40 0.8 8 95.44 0.3 9 95.08 0.2 10 95.50 0.6 11 95.80 0.6 12 95.22 0.2 13 95.60 1.3 14 95.22 0.4 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

Answer 1

Answer:

See explaination

Step-by-step explanation:

Refer to attached file for table used in solving mean.

The mean of range is

\bar{R}=\frac{13.3}{20}=0.665

The mean of all six means:

\bar{\bar{x}}=\frac{1907.96}{20}=95.398

(a)

Here sungroup size is 5:

Range chart:

From constant table we have

D_{4}=2.114

So upper control limit:

UCL_{R}=D_{4}\cdot \bar{R}=2.114\cdot 0.665=1.40581

Lower control limit:

LCL_{R}=0.0000

Central limit: \bar{R}=0.665

Since all the range points are with in control limits so this chart shows that process is under control.

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X-bar chart:

From constant table we have

A_{2}=0.577

So upper control limit:

UCL_{\bar{x}}=\bar{\bar{x}}+A_{2}\cdot \bar{R}=95.398+0.577\cdot 0.665=95.78

Lower control limit:

LCL_{\bar{x}}=\bar{\bar{x}}-A_{2}\cdot \bar{R}=95.398-0.577\cdot 0.665=95.01

Central limit: \bar{\bar{x}}=95.398

Sample number 94.82 is not in teh limits of x-bar chart so it seems that process is not in control