# Determining Sample Size needed for Operational Qualification (OQ)

M

#### mipelamo

Sorry if this is posted in the wrong forum. I am trying to determine the sample size needed for Operational Qualification (OQ). I have ran some parts in the past, thus I can calculate my current mean, std, and CpK etc. How do I figure out what the sample size should be to pass a particular tolerance? For example, if one of OQ output is length (1" ± 0.5") and currently my mean is 1.05 with stdv 0.005, how would I calculate the sample size? Any help would be appreciated. I have Excel and Minitab 15 at my disposal.

Thanks

Elsmar Forum Sponsor

#### Stijloor

Staff member
Super Moderator
A quick Bump!

Can someone help?

Thank you very much!

Stijloor.

#### mdurivage

##### Quite Involved in Discussions
I would try a tolerance interval. A tolerance interval will not give a sample size directly, but it will give an interval that one can state that you are 95% confident will contain 99% of the product produced. The calculation uses a sample size. So one may work backwords to get the required sample size.

*Normal distribution assumed
*Percent confidence and % contained can vary to the situation

Mark

Last edited:
M

#### mipelamo

Mark,

Thanks for the reply. I can calculate tolerance interval in Minitab, but do you know the equation?

Thanks

#### Bev D

##### Heretical Statistician
Staff member
Super Moderator
May I ask what you think the intent of OQ is?

OQ is defined as validating that a process can meet the output specification when the inputs are varied form their allowed minimum to their allowed maximum.

The use of a tolerance interval (intended to estimate the variation of the individual values) is not a bad way to iterate to a sufficient sample size. Common sample size formulas are based on estimations of the mean so they are not useful for OQ or PQ...the only caveat is that tolerance intervals rely heavily on the assumption of Normality of the data.

#### mdurivage

##### Quite Involved in Discussions
Here is a formula that I sometimes use to determine a sample size:

n= ln(1-C)/ln(R)

Where:
n= the required sample size
C = confidence level
R = reliability

For exaplme if you want to be 95% that you are 95% relaible you wood need 59 good (failure free parts).

Mark

Last edited:

#### Bev D

##### Heretical Statistician
Staff member
Super Moderator
can you provide the derivation or a reference?
...not questioning your formula, its just helpful to know where formulas come from so we can all understnad what they are appropriately used for...

#### mdurivage

##### Quite Involved in Discussions
can you provide the derivation or a reference?
...not questioning your formula, its just helpful to know where formulas come from so we can all understnad what they are appropriately used for...
Reliability Statistics by Robert Dovich page 41

M

#### mipelamo

Thanks for all the replies. I ended up just using the Power and Sample Size for One-Way ANOVA tool from Mini-Tab. I entered a standard deviation, Power value and maximum difference b/t the means, and then MiniTab calculated the sample size.