Quality Control Charts

Industrial Revolution led to Assembly Lines

Many workers are now responsible for a finished product rather than one person.

How do we detect and correct mistakes before the finished product is made?

The process used to check is called Quality Control.

Uses small samples and a control chart with constants to detect deviations from the norm.

Control Chart uses the following:a. Meansb. Mediansc. Rangesd. Standard Deviations

Quality Control

Understanding the quality control process means understanding the variations that exist when a product is manufactured.

For example, a company wishes to make rope that has a breaking strength of 50 pounds. There WILL always be small variation in each rope. Therefore, the manufacturer will set limits – a maximum and minimum acceptable value.

These limits are called the upper and lower control limits.

Control Limits

Once a graph is made to analyze, use the following rule to decide if a process is “in control” or “out of control.”

a) The graph exceeds the upper and/or lower

control limits OUT OF CONTROLb) The graph is within the upper and/or

lower control limits IN CONTROL

In Control??????

Example – OUT OF CONTROL

Example – IN CONTROL

Select several samples of the product and test their means.

Compute a variation measure. Use the range for an - chart.

Computer the upper and lower control limits and the grand mean. Represent these on a graph with lines

Control Chart Construction

Plot the means of the samples

Analyze it to see if it is in control or out of control.

If out of control, find the problem and fix it.

Do the process again.

Control Chart Construction….

There are 2 types of control charts:

◦ Variable Charts – used to analyze measurements such as lengths, weights, breaking strength, etc.

◦ Attribute Charts – used to analyze a manufacturing process that can be classified as acceptable or defective.

Types of Control Charts

We are going to construct a type of variable chart called an - bar chart.

Find the mean and range of each sample.

Find the grand mean, , and the mean of the ranges, .

Find the upper and lower control limits using the formulas and

where is found from the table of constants.

Draw the chart using the grand mean as the center line and the calculated upper & lower limits.

Plot the mean for each sample and analyze. Decide if it is in control or out of control.

- bar chart construction

= + = -

A manufacturer of rope selects six samples of five ropes each and tests the breaking strength of each. (See next slide)

Construct an - bar chart for the data.

Analyze the results.

Example 1

Sample

Breaking Strength (in pounds)

1 46 47 45 46 47

2 50 51 52 53 49

3 48 51 50 50 49

4 52 50 49 50 51

5 51 47 46 48 47

6 49 51 50 51 52

Now find the mean and range for all 6 samples.

= 46.2= 51 = 49.6 = 50.4 = 47.8 = 50.6

Sample

Breaking Strength (in pounds)

1 46 47 45 46 47

2 50 51 52 53 49

3 48 51 50 50 49

4 52 50 49 50 51

5 51 47 46 48 47

6 49 51 50 51 52

= 2= 4 = 3 = 3 = 5 = 3

= 46.2= 51 = 49.6 = 50.4 = 47.8 = 50.6

= 2= 4 = 3 = 3 = 5 = 3

Next, find the Grand Mean, (the mean of the sample means), and the mean of the Ranges.

= = 49.3

=

Now find the Upper and Lower Control Limits using the following formulas.

= +

= -

is found from the table of constants

How do you find ? You must use “n” to find the value for .

“n” is the number of items in each sample. In our example n = 5.

We’ll use the value of 0.577 for .

Now find the Upper and Lower Control Limits using the following formulas.

= + = -

= 49.3+(0.577) (3.33)

= 51.22

= 49.3 – (0.577) (3.33)

= 47.38

= 49.3

Now, put these on a graph and plot the mean of all of the samples to see if it is in/out of control

= 46.2= 51 = 49.6 = 50.4 = 47.8 = 50.6

Example 2 A pressure valve is designed to release at

200 psi. Ten sample of six valves each were selected and tested. The mean and range for each sample are shown in the table. Construct and analyze a quality control chart.

Sample 1 2 3 4 5 6 7 8 9 10200.2 198.9 199.3 201.1 200.8 202.6 201.3 203.7 205.6 206.1

R 3.1 0.8 2.2 1.0 3.6 2.3 1.4 2.2 2.3 2.4

Sample 1 2 3 4 5 6 7 8 9 10200.2 198.9 199.3 201.1 200.8 202.6 201.3 203.7 205.6 206.1

R 3.1 0.8 2.2 1.0 3.6 2.3 1.4 2.2 2.3 2.4 = = 201.96

= =

= 0.483 (at n = 6)

= + =201.96+(0.483)(2.13) = 202.99

= - =200.93