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Published data suggests that the mean standard lactation yield, 305 days, for Holstei
About 8000kg and the sd of milk yield is 1425kg. Suppose you want to compare the mean milk yields of Holsteins cowin in the two largest counties in Wisconsin. How many animals need to be sampled from each counties if I require a 85% chance of detecting a difference in the mean milk yield of 250kg at the 5% of significance?
Please help, because I don't understand how to do. Don't even have formula for it |
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