Peter Theodorakakos

Sample Size Determination Peter Theodorakakos

A manufacturer of SMC Deck-Lid Panels has reduced the incidence of “pops” and “craters” on the panel surface during the painting process, at the Paint Booth of an Automotive Manufacturer. These defects occurred when catalyst trapped gases exited the formed panel surface due to the high cure temperature of the Paint Booth (425 F). Peter Theodorakakos

The defect reduction was the result of a process change at the SMC Deck-lid Panel manufacturing facility and involved raising the temperature of the press-mold from 385 F to 450 F, so the gasification was all done while the SMC blob was being pressed into the Panel Shape. Peter Theodorakakos

The initial action reduced the incidence of defects from 1/3 panels to 1/35 panels, an acceptable rate of defect observation for a Paint Booth, and certainly one that the Automotive Manufacturer is equipped to repair without disruption to the shipping schedule. Peter Theodorakako

Peter Theodorakakos

Furthermore, the supplier decided to paint the SMC Panel with a protective 2-K epoxy primer and seal the surface to further prevent the case for gasification from occurring, promising at least a 10% in defect rate reduction. Peter Theodorakakos

Peter Theodorakakos

The question is: What is the sample size, which will verify the stated improvement? If you dedicate 1 worker / shift, for 3 shifts per day, for 1 week, at 8 hours per day, each at the rate of $14/hr, the cost is $336 daily. The Automotive Manufacturer produces 45 cars per hour, or 1,080 daily. Peter Theodorakakos

Peter Theodorakakos

Statement of Work (S.O.W.)

Determine the sample size to verify that at least a 10% improvement has been achieved with a 5% error allowed for the test’s ability at “.9 Power Level” to detect the effect?

Estimate the amount which must be budgeted to account for the quality verification.

Peter Theodorakakos

Determination of Sample Size – “One Proportion Design Test”Peter Theodorakakos

Let’s set up test parameters to determine a minimum Sample Size of Deck-Lids Panels, which will verify a 10% improvement, with a 5% chance of error.

1. Baseline Proportion is 1/35 (p=.02857143), defective panels for regular primer

2. Comparison rate is 10% less, 1/39 (p=.02564103) defective panels with 2- K primer

3. “Power of Test” is set at .90 Peter Theodorakakos

We will use a “Sample Test of One Proportion” method. This method will specify the Sample Size required for verifying a 10% proportional improvement. The “Power of the Test” will assess the probability that the test will make the verification possible from the specified Sample Size.

Peter Theodorakakos

The required amount of Deck Lid Panels to tease out a 10% improvement at the Paint Booth facility is 26,452. That will require 25 days of production, which will account for 27,000 cars, a number higher than Minitab specifies. Peter

Theodorakakos

The budget must allocate funds for 25 days or $8,400. Peter Theodorakakos

Determination of Sample Size – “Full Factorial Design Test”Peter Theodorakakos

A Sintering manufacturer wishes to compare Ultimate Strength of Steel Test Shafts from three producers (A, B, and C), consolidated at 2 different pressures (A=30 tons/in^2 and B=60 tons/ in^2) and strengthened at 2 different temperatures (A=1700 F and B=1950 F).

The team wants to conduct a general full factorial experiment that includes runs at all possible combinations of supplier, pressure, and temperature. They want to find out what power they can achieve with 2, 3 or 4 replicates to detect a difference in main effect means of 5000, 7500, or 10000. (That will be psi for the Ultimate Strength readings)

Total MaximumReps Runs Power Difference 2 24 0.9 8119.85 3 36 0.9 6194.46 4 48 0.9 5248.73

What is shown is the for an acceptable error of 5%, the experiment must be run utilizing 4 replicates to achieve a 90% chance of detecting a difference of 5249 psi during the Ultimate Test, and perform 48 runs. If 3 replicates are used, you can expect the test to determine a 6194 psi difference due to the factor effect, and perform 36 runs. If 2 replicates are only used, a difference of over 8120 psi has a 90% chance of being determined sample tested, and perform 24 runs.

Determination of Sample Size – “Two-Level Factorial Design Test” Peter Theodorakakos

A MIM (Metal Injection Molder) is evaluating the porosity of the production output and contributes the outcome to 4 input factors. Mold Temperature, Mold Pressure, Binder Type and Binder Concentration. Being able to detect effects of a magnitude 5 or more is

deemed important, so how many replicates and how many tests must be run for an 80% chance to detect such difference in the sample obtained from the test? The quality department has indicated that the porosity standard deviation of 4.5 is a historical value and can be used as a reference point. Peter Theodorakakos

Choose a design which can be described as a ½ factorial, 4 Factors, 3 Center Points, and 8 runs. This will determine the “optimized settings” with a reasonable budget allocation and effort. Yet, how many replicates are required to determine the difference of porosity with a magnitude of 5, and how many runs? Peter Theodorakakos

Center TotalPoints Effect Reps Runs Power 3 5 1 11 0.157738 3 5 2 19 0.518929 3 5 3 27 0.730495 3 5 4 35 0.856508

Peter Theodorakakos

If we do not replicate our design (Reps = 1), we will only have a 16% chance of detecting effects that we have determined are important, and involves 11 runs. If we use 4 replicates of our 1/2 fraction design for a total 35 runs (32 corner points and 3 center points), we will have an 86% chance of finding important effects. Peter Theodorakakos

Peter TheodorakakosPeter Theodorakakos

Determination of Sample Size based on Power– “One Way ANOVA Test” Peter Theodorakakos

A MIM (Metal Injection Molder) is investing the effect of pressure on porosity, and whether or not 4 treatments (levels) affect the yield of a product using 5 observations per

treatment. You know that the mean of the control group should be around 7, and you would like to find significant differences of +3.5. Thus, the maximum difference you are considering is 3.5 units. Previous research suggests the population is 1.5. Peter Theodorakakos

One-way ANOVA

Alpha = 0.05 Assumed standard deviation = 1.5Factors: 1 Number of levels: 4

Maximum Sample TargetDifference Size Power Actual Power 3.5 7 0.9 0.936210

The sample size is for each level. Total runs are 28.

Therefore, the difference in porosity will require a 28 run experiment and the chance that 5 Pressure treatments (levels) will tease out a difference of 3.5 of the porosity scale, is 93%, as long as we assign 7 replicates per treatment. Peter Theodorakakos

Peter Theodorakakos

Determination of Sample Size based on Power– “2 - Variance Test” Peter Theodorakakos

The engineering manager of a MIM (Metal Injection Molder) outfit is reviewing the output of the 2 sintering ovens, for their effectiveness to remove binder and restore the

strength and ductility to the metal part “brown”. One of the ovens was just replaced, and the other is 15 years old. Operations manager is budgeting to replace the old oven but wants to know how the new versus the old oven performs in removing binder from the “green” part, before a replacement is ordered. Peter Theodorakakos

In preparation for the pilot study, the manager wants to know what the power of the 2 variances test is if they collect 50, 75, or 100 samples from each oven and are able to detect a standard deviation ratio of 0.75. Peter Theodorakakos

SampleRatio Size Power0.75 50 0.4552300.75 75 0.6241730.75 100 0.750227

The sample size is for each group. So you will make a run of 200 parts in total.

The recommendation is at least 100 parts must be run per oven to determine the .75 standard deviation ratio between the new and the old oven. Notice that the new oven is supposed to feature a smaller variance than the old one, thus the ratio will be below the value of 1, for (New St. Dev) / (Old St. Dev.). The chance of the experiment will determine such ratio of St. Deviations is only 75%. More samples tested will improve the Power of the Test. Peter Theodorakakos

Peter Theodorakakos

Determination of Sample Size based on Power– “1 - Variance Test” Peter Theodorakakos

The engineering manager of a MIM (Metal Injection Molder) outfit needs to control the porosity of the parts for a military application. He installs control sensors to reduce the variability at the Metal Injection Press and at the Sintering Oven. How many samples must be produced and measured, so that a 1 - Variance Test has Power of .9 and can detect a 20% improvement in reduced standard deviation compare to past performance.

Power and Sample Size

Test for One Standard Deviation

Sample TargetRatio Size Power Actual Power 0.8 90 0.9 0.901976

A 90 part sample must be produced and tested to determine a 20% reduction of Standard Deviation, provided that such improvement took place. The test power of .9 means that there is 90% chance that the collected sample will feature parts with reduced standard deviation (if it happened) by a level of 20%, or more, nut no less. Smaller difference can be found but the test power quickly diminishes. Peter

Theodorakakos

Peter TheodorakakosPeter Theodorakakos

Determination of Sample Size based on “1 – Sample Poison Rate” Peter Theodorakakos

The engineering manager of a MIM (Metal Injection Molder) outfit wishes to examine the number of porosity related defects on a Gas Turbine Impeller. He will conduct a 1-Sample Poison Rate Test to see if the flaws are significantly less that 17 per part (a number associated with size and complexity and material processed) before any

machining takes place. Since this is a critical component he wishes to specify .9 Test Power and thus determine the number of samples to establish a comparison rate of 15.

Power and Sample Size Peter Theodorakakos

Test for 1-Sample Poisson Rate

Testing rate = 17 (versus not = 17)Alpha = 0.05"Length" of observation = 1

Comparison Sample TargetRate Size Power Actual Power 15 43 0.9 0.903149

The manager will have to schedule a production run of 43 impellers and the test has a 90% chance of producing samples with a difference of 2 defects per part.

Peter Theodorakakos

Determination of Sample Size based on “2 – Sample Poison Rate” Peter Theodorakakos

The engineering manager of a MIM (Metal Injection Molder) outfit is interested in selecting a low defect rate Sintering Oven for a low cost application. Therefore, the post Sintering machining will be reduced and the profit margins will be safe. Historically, one

line’s defect rate is 17 defects per part. The other is not known. The manager wants to estimate the number of parts which must be budgeted to establish with .9 Power of Test, a difference of 3 defects from one line to the other. So, the number of defects he wishes to decipher is 14, or 20, since the Baseline Defect Rate is 17 per part.

Comparison Sample TargetRate Size Power Actual Power 14 37 0.9 0.906169 20 44 0.9 0.905161

The sample size is for each group.

As you can see, the sample must be 37 at least to detect a 14 Rate of Defects per part, and 44 at least to detect a 20 Rate of Defects per part. The chance of

either sample containing the 14, or 20 Rate of Defect is slightly over than 90%. Peter Theodorakakos

Peter Theodorakakos

Peter Theodorakakos

Determination of Sample Size based on “Paired - t” Peter Theodorakakos

The quality manager wants to prove the effectiveness of training operators of a MIM (Metal Injection Molder) outfit to spot defects on post Sintering process. He does so by

scheduling a pre and post test to a sample of participants. He wants a .8 and a .9 Power of Test to establish a Standard Deviation improvement of 6 defects. Based on historical data, the standard deviation of paired differences is 6. Peter Theodorakakos

For the Test Power of .8 he will require a group of 10 employees to review porosity on parts (before and after the training) and for the Test Power of .9 he will require a group of 13 employees (before and after the training). Peter Theodorakakos

This test specifies that there is a 90% chance that within a group of 13 employees will be able to find a 6 point standard deviation improvement in detecting defects due to training, if in fact such improvement tool place. Peter Theodorakakos

Peter Theodorakakos

Determination of Sample Size based on “1 - Sample t” Peter Theodorakakos

The drier of a Metal Injection Mold must maintain strict control on humidity level of the polymer/paraffin and metal powder mix in the hopper. The humidity cannot vary more

than .5% for a 1 gallon of mixed material. The tolerance is set so that the process standard deviation is .15%. Determine the sample size to measure for humidity with a confidence level of 99% for a Test Power of .9? Peter Theodorakakos

1-Sample t Test Peter Theodorakakos

Testing mean = null (versus not = null)Calculating power for mean = null + differenceAlpha = 0.01 Assumed standard deviation = 0.15

Sample TargetDifference Size Power Actual Power 0.5 5 0.9 0.946587

Peter Theodorakakos

Determination of Sample Size based on Margin of Error Peter Theodorakakos

A plating company wants to know the amount of Copper deposition onto the plated parts of Tank # 5 of their plating facility. Based on known performance the line has a range of

+/- .000025” deposition thickness. What sample size should be specified for study to within a 5% margin of error? Peter Theodorakakos

Let’s assume that the provided range is exactly a 3 sigma process. Therefore, the 1 sigma value is .000008”. Peter Theodorakakos

Parameter MeanDistribution NormalStandard deviation 8e-006 (estimate)Confidence level 95%Confidence interval Two-sided

Results

Margin Sampleof Error Size 0.05 2

 

Peter Theodorakakos