Sample Size Calculation for Attribute Data
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 This topic has 13 replies, 5 voices, and was last updated 16 years, 3 months ago by M Jay.

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August 11, 2006 at 7:14 pm #25339
Hi everyone!
I am looking for some help on calculating sample size for attribute related data. I am used to doing this for continous data….
HELP!
Thanks.0August 12, 2006 at 6:49 pm #61042Attributes Data: We use the formula
n=[2* Z(a/2)/ w]2 *p*q Where:
n = sample size
Z(a/2) = probability value associated with the desired confidence level
w = desired precession of the estimate
p = population proportion (use 0.5 if unknown)
q = 1p
Example: How large of a sample is required to estimate the proportion of charts having nonuse abbreviations on them with a sample precision of ± 5% at a 95% confidence level?
· First, find the value for Z(a/2) for 95% from the Z table
Z(a/2) = 1.96
· Next get w. Since we have ± 5%, the range is 5 to +5 or a 10% spread. So, w = 10% or 0.10
So, in order for the sample estimate to be within ± 5% of the actual population value with 95% confidence, 385 charts would need to be sampled randomly from the population0August 12, 2006 at 6:49 pm #61043Attributes Data: We use the formula
n=[2* Z(a/2)/ w]2 *p*q Where:
n = sample size
Z(a/2) = probability value associated with the desired confidence level
w = desired precession of the estimate
p = population proportion (use 0.5 if unknown)
q = 1p
Example: How large of a sample is required to estimate the proportion of charts having nonuse abbreviations on them with a sample precision of ± 5% at a 95% confidence level?
· First, find the value for Z(a/2) for 95% from the Z table
Z(a/2) = 1.96
· Next get w. Since we have ± 5%, the range is 5 to +5 or a 10% spread. So, w = 10% or 0.10
So, in order for the sample estimate to be within ± 5% of the actual population value with 95% confidence, 385 charts would need to be sampled randomly from the population0August 12, 2006 at 6:54 pm #61044Sorry about that – somethign went a little screwy when I was typing in the message.
Note that the p value is a porportionate value between 0 and 1. The default given a normal, unweighted event, is to have p=.5. This makes q=1.5 or .5.
0August 13, 2006 at 6:58 pm #61047
madtexanParticipant@madtexan Include @madtexan in your post and this person will
be notified via email.Dear Doug,
I read your question and Brit’s reply. I am concerned about both. You
state that the data is attribute but then say it is continuous data. Or
at least that is the way that I read it. Brit’s reply was for binomial
distributions. Can you be more specific about the data you have
collected. For sample size collection, general formulas exist for both
continous and binomial data, if the sample sizes are large enough and
follow a normal or continous distribution. Of course, maybe they
don’t. So, I would like some more information about your data before
I can help. You can also contact me through
http://www.bryantsstatisticalconsulting.com0August 14, 2006 at 1:05 pm #61049Texan – I think if you read his post again it was for attribute data. He said he had only done this for continuous data in the past.
0August 15, 2006 at 1:37 pm #61050That is correct Brit! Thanks for your entry…it helped a lot!
A little background, coming from manufacturing, I dealt a lot with continuous (or variable depending on what you want to call it) data…not a lot with attribute…now working in healthcare…almost everything seems to be attribute!
Thanks again to all who answered!
Doug0August 15, 2006 at 2:06 pm #61051Good morning Brit!
One quick question for you…in your example, you use +/5% for sample precision – would this be the same thing as reliability? What I mean is, if I want to have a 95% confidence level and 90% reliability that there would be 0 defects…then Z = 1.96, w = 0.10? Is this the same thing?
And by the way, thanks again for your help.
Doug0August 15, 2006 at 4:09 pm #61052No – reliability in the way you mention below is really your probabiltiy of not making a defect. This can be calculated using the binomial probability function for attribute yes/no type data or the poisson probability for rate type data. For example, out of 50 samples, what is the probabilty of me finding 5 or less defects.
The confidennce interval and margin of error reads like this if you used a 95% conf interval and +/5% margin of error:
I am 95% confident that the result I get from my sample will be within +/5% of the true population result.
Note: the smaller the margin or the larger the confidence = larger sample size.0August 15, 2006 at 10:00 pm #61053Doug:
I have worked in mfg, svc and gov’t and have landed in healthcare as well. Yes – I would say that 80% of my data is attribute with the most being time or rates with an upper limit only.0August 16, 2006 at 1:17 pm #61056Brit,
Thanks! That makes sense! I appreciate you taking the time to straighten me out!
Doug0August 28, 2006 at 3:49 pm #61070
Eugene LanningParticipant@EugeneLanning Include @EugeneLanning in your post and this person will
be notified via email.Perhaps it would be insighful/helpful to get a copy of “Sampling Procedures and Tables for Inspection by Attributes”, ANSI/ASQ Z1.42003
0August 28, 2006 at 8:03 pm #61071Doug if you gate a chance please contact me, i would like to know what kind of attibute project you have.
Thank you
MJS0August 28, 2006 at 8:08 pm #61072Sample Size relating to what test ie; Gage R&R (30 Samples), Chi Square (?) or Proportion Test. What kind of test are you trying to run? And a P Value of 0.05 will give you a 95% CI…… Not .5 – Watch your decimal point.
MJS0 
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