Could you please share the formula used to calculate the number of birds to weigh according to the accepted difference to the mean value?
Thank you.
Ricardo Hume
My name is Herman Fleuren from OPTICON AGRI SYSTEMS in Holland
See article about your question.
Variation in liveweight
Dear Sir,
If you take one bird at random and weigh it, how representative would it be of the average weight of all
the birds in the house?
The answer can only be given as a probability of the true average being within a given range. You will know
the terms “Standard Deviation (SD)” and “Mean” (which is the same as the average. You may or may not
know the term “Coefficient of Variation (CV)”, which is the SD as a percentage of the mean. In biology this
is usually 10%
So if the average weight of your population of birds is 1000 grams, the SD would be about 100 grams. You
will know that 95% of the individuals in a population is within 2 SDs of the mean. So if you weigh one bird
from a population that has a mean of 1000 grams, you will be 95% sure that the bird you weigh will be
1000 ± 200 grams or within the range of 800 grams to 1200 grams. Clearly that’s not accurate enough, so
you must weigh more birds and average them to get a more accurate estimate of the population mean.
How does the number that you weigh affect your estimate of the population mean? There is a simple
formula to calculate that.
The CV of a sample size of x compared to a sample size of 1 is 10% / vx (square root of x)
So if you weigh 100 birds, the CV of the average is 10%/10 = 1% and 1% of 1000 grams is 10 grams
So you would be 95% confident that the true mean of the population would be 1000 ± 20 grams or within
the range of 980 grams to 1020 grams.
At present you weigh 80 females, (the square root of that is about 9) and the CV is 10%/9 = 1.11% and
1.11% of 1000 grams is 11 grams. So you can be 95% confident that the true average of the pullets would
be 1000 ± 22 grams or within the range of 978 grams to 1022 grams. The same applies to the cockerels.
But if you consider the house average, you weigh 160 birds and the square root of that is about 12.65.
So the CV is 10%/12.65 = 0.79% and 0.79% of 1000 grams is about 8 grams. So you can be 95% confident
that the true average of all the birds would be 1000 ± 16 grams or within the range of 984 grams to 1016
grams.
If you had automatic bird weighers which would average 400 weighings a day on each of 4 scales, that
would give a CV of 10%/40 = 0.25% and 0.25%% of 1000 grams is 2.5 grams. So you would be 95%
confident that the true average of the all the birds would be 1000 ± 5 grams or within the range of 995
grams to 1005 grams! You would also get daily weighings and be able to automatically graph LWG daily.
That would enable you to see each day if GROWTH was on target! That’s quick enough to take quick
remedial action f things start to go wrong. Food for thought?!
Hello Ricardo,
We are manufacturing and improving live bird weighing systems since 1993. So we made a lot of improvements and special algorithms to obtain accurate data.
best regards
Herman
HI. I believe the P-value=0.3 is good for estimating sample size because if we consider P-value = 0.05 (like scientific experiments) we have weight large number of birds and this recommendation may not be practical. Please inform me your opinion.
Regards,
Nima.