Question

A process that is considered to be in control measures an ingredient in ounces. Below are the last 10 samples (each of size n=5) taken. The population process standard deviation, ? is 1.36.
A process that is considered to be in control meas



If z = 3, what are the control limits for the mean chart? What are the control limits for the range chart? Is the process in control?

Answer 1

Note: The 10 samples referred to in the question are in the table contained in the file attached.

Answer:

Upper Control Limit for mean chart = 11.62

Lower Control Limit for mean chart = 7.98

Upper Control Limit for range chart = 8.88

Lower Control Limit for range chart = 0

The process is in control

Step-by-step explanation:

Step 1: Calculate the mean $\bar{X}$ and the range R for each of the 10 samples.

The $\bar{X}$ and R of each sample are in each column of the table in the second file attached.

e.g. For sample 1,

$\bar{X} = (11+9+9+10+12)/5 = 10.2\\R = highest no. - smallest no. = 12-9 = 3$

Step 2: Calculate $\bar{R}$ and $\bar{\bar{X}}$

$\bar{R} = (3+3+6+2+3+5+4+5+6+5)/10 = 4.2$$\bar{\bar{X}} = (10.2+9.4+9.8+9.6+10.6+10.6+9+9.4+9.4+10)/10 = 9.80$Step 3: Calculate Standard Deviation

If z = 3, $\sigma_{x} = 1.36$, n = 5

Standard deviation of $\bar{X}$, $\sigma_{\bar{x}} = \frac{\sigma_{x} }{\sqrt{n} }$

$\sigma_{\bar{x}} = \frac{1.36 }{\sqrt{5} }$ = 0.6082

Step 4: Calculate the Control limits for the mean chart:

Upper Control Limit(UCL)

$UCL = \bar{\bar{X}} + z* \sigma_{\bar{x}}\\UCL = 9.8+(3*0.6082) = 11.62$

Lower Control Limit (LCL)

$LCL = \bar{\bar{X}} + z* \sigma_{\bar{x}}\\LCL = 9.8-(3*0.6082) = 7.98$

All the $\bar{X}$ for the 10 samples fall within the LCL and UCL for the mean chart

Step 5: Calculate the Control limits for the range chart:

Upper control limit (UCL)

$UCL = \bar{R} * D_{4}$

$D_4 = 2.114$ for n = 5 (from the control chart constant table)

UCL = 2.114 * 4.2 = 8.88

Lower control limit (LCL)

$LCL = \bar{R} * D_3$

$D_3 = 0$ for n = 5 (from the control chart constant table)

LCL = 4.2*0 = 0

All the  for the 10 samples fall within the LCL and UCL for the range chart

Since the average for the 10 samples fall within the LCL and UCL for both the mean and range charts, the process is in control.