Question

The samples below give the readings for volume of a precious liquid in ml in an industrial chemical manufacturing plant.Determine if the process of manufacturing is in control.

sample	1	2	3	4	5
1	4.03	4.02	4.13	4.03	4.04
2	4.07	4.03	4.30	4.02	4.05
3	4.09	4.07	4.07	4.00	4.05
4	4.08	4.50	4.08	4.06	4.05
5	4.06	4.07	4.05	4.10	4.05
6	4.06	4.01	3.23	4.02	4.04
7	4.08	4.01	4	4.24	4.05
8	4.04	3	4.09	4.07	4.1
9	4.07	4.01	4.07	4.13	3.30
10	4.07	4.02	4.06	4.06	4.02

Answer 1

Solution:

Mean and Range are calculated as:

Mean = Sum of observations / Total number of observations

Range = Maximum observation — Minimum observation

The below table summarizes the mean and range of each sample:

From the above table,

X-bar = 4.019

R-bar = 0.403

From the three-sigma control limits chart for constants,

for n = 5 (Since there are 5 observations in each sample)

A2 = 0.58

D3 = 0

D4 = 2.11

The LCL and UCL for X-chart is calculated as below:

UCLx = X-bar + (A2 x R-bar)

LCLx = X-bar - (A2 x R-bar)

Putting the values in the above formula;

UCLx = 4.019 + (0.58 x 0.403)

UCLx = 4.253

LCLx = 4.019 - (0.58 x 0.403)

LCLx = 3.785

The LCL and UCL for R-chart is calculated as below:

UCLr = D4 x R-bar

LCLr = D3 x R-bar

Putting the values in the above formula;

UCLr = (2.11 x 0.403)

UCLr = 0.850

LCLr = (0 x 0.403)

LCLr = 0

The mean (X-bar) of all the samples falls within the control limits for X-chart (4.253 and 3.785). However, for R-chart with control limits as 0 and 0.850, the Range for sample 8 (R = 1.1) falls outside control limits. Therefore, the process is not in statistical control.