Alpha Beta Error when using Finite Correction Factor
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 This topic has 10 replies, 4 voices, and was last updated 11 years, 7 months ago by Draper.

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March 10, 2010 at 2:52 am #53342
I calculated a certain sample size based on a selected level of alpha and beta. Since the population turned out to be smaller than my calculated sample size, I needed to use the finite population correction factor to reduce the original sample size. Question is what happens to the alpha and beta error in the new computed and revised sample size? Does it stay the same as the original values or change since the sample size is now smaller? Thanks.
0March 10, 2010 at 4:48 am #189686
SeverinoParticipant@Jsev607 Include @Jsev607 in your post and this person will
be notified via email.Just a stupid question: If you calculated a sample size and were okay with that level of sampling and then found out that your sample size was larger than your population size why move to inferential statistics at all? Why not just use the population?
0March 10, 2010 at 8:47 am #189691Hi Darth,
the Finite Poiulation Correction will reduce the standard error of the mean by a given factor (SQRT((Nn)/N1)). So IMHO you could calculate the alpha and beta errors by simply introducing the new standard error value into the respective formulas.BTW the Correction has an effect as if you increased the sample size.
Regards
Sandor0March 10, 2010 at 11:53 am #189692Because the revised sample size is still smaller than the population and since sampling costs money it is still cost effective to sample rather than use the entire population.
For example, using my desired alpha (.05), beta (.90), precision and an estimate of variation let’s say I calculate a sample size of 350. Unfortunately, the population is 250. I either can sample the whole 250 as your suggest or apply the pfcf. By applying the correction factor my revised sample size, let’s say, reduces to 160. Since sampling costs money, I choose to sample the 160. So, now the question remains does the alpha and beta, .05/.90 still apply to my statistical statement or does my confidence and power change?
0March 10, 2010 at 7:51 pm #189695Hi,
I just did a quick example with Minitab and it seems that all the values remain the same.The alpha and the precision are chosen values, so presumably you shall calculate the new sample size so, that they will stay the unchanged. The only question is the beta value, or better, the power of the test.
I worked through the example and could check that if i calculated the new sample size using the finite population correction and recalculated the power using the corrected standard deviation (as in my post above) then the power remains the same.
0March 11, 2010 at 3:27 am #189701Sandor,
Thanks for taking a shot at this. Using your approach and logic I came to the conclusion that the alpha remains the same but Power increases. The guy I am working with this week is trying to develop a model that will work this issue out. The logic seems to be that the std. error reduces by applying the fpc factor and since the sample size is a larger proportion of the population the confidence intervals shrink. A reducing confidence interval occurs when the s.d. goes down and/or the n goes up. This relates to Power so conceptually we would expect the Power to go up. Our problem at this point is that the Power goes up really high. Hopefully the Model that is being developed will make some sense. If it works, I will let you and be glad to share with you.0March 11, 2010 at 4:17 am #189702
SeverinoParticipant@Jsev607 Include @Jsev607 in your post and this person will
be notified via email.Darth wrote:
Because the revised sample size is still smaller than the population and since sampling costs money it is still cost effective to sample rather than use the entire population.
Most assuredely. I was just pointing out the obvious in case it was not considered. There are often times where the work is cheaply bought, but the cost of incorrect inference is intolerable. That’s why airlines don’t sample their preflight checks. In any event, the thread was helpful to me if for nothing else than the correction factor. Cheers.
0March 11, 2010 at 7:13 am #189703Hi Darth,
“The logic seems to be that the std. error reduces by applying the fpc factor and since the sample size is a larger proportion of the population the confidence intervals shrink.”
I think the trick is that if you apply the sample size formula with the new, reduced stdev you will get a sample size that will keep the confidence interval, so, there will be no shrinkage and consequently tha power should stay the same.
The formula I used to calculate the new sample size was:
D=1,96*s*SQRT((Nn)/(n*(N1))) where D is the halflength of the confidence interval, s the assumed standard deviation of the population, n the sample size and N the population size. If you set the D to be the same as before you can calculate n from this formula.
P.S the formula is almost the same as the “standard” one
D=1,96*s/SQRT(n) except that s is multiplied by the correction factor SQRT((Nn)/N1))
Thanks for letting me know your results, I am really interested.
0March 16, 2010 at 4:23 am #189766
DraperParticipant@DonDraper Include @DonDraper in your post and this person will
be notified via email.Darth,
Did I just read you humble yourself?
I’ve never read you post such a topic which you couldnt figure out.
I am glad you are turning a new leaf.
Don
0March 16, 2010 at 1:18 pm #189770Don,
Thanks for the kind words. If you check the archives, you will see that this is not the first time that I have asked for help. The problem is that the questions I do ask is usually quite technical in nature so it is rare that I get a complete and adequate answer from the Forum. Folks sometimes give it a good try but I don’t believe I have had a great epiphany. Even Minitab was stumped on this one. I and a friend are figuring this out and coming up with a solution that will allow a plug and chug to get the answer. Now, if you have a suggestion how to answer the question, please feel free to offer it.0March 17, 2010 at 2:43 am #189789
DraperParticipant@DonDraper Include @DonDraper in your post and this person will
be notified via email.Darth,
My apologies. It was a poor attempt at humor.
I enjoy your techincal expertise and advice, 90% of the time. I have to admit I dont have anything close to a suggestion for you. I would like to know the outcome/answer once you find out….
Don
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