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Suppose that the effects of copper on a second species (say, species B) of fish show the variance of ln(LC50) measurements to be .8. If the population means of ln(LC50) for the two species are equal, find the probability that, with random samples of ten measurements from each species, the sample mean for species A exceeds the sample mean for species B by at least 1 unit.

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Answer:

Hello the reference related to your question is missing attached below is the reference

answer : p ( x - y ≥ 1 ) = 0.0019

Step-by-step explanation:

Assuming 'X' and 'Y' are independent random samples

X : mean value = [tex]U_{1}[/tex] , variance = 0.4 /10 = 0.04

Y : mean value = [tex]U_{2}[/tex] , variance = 0.8/10 = 0.08

since the values are independent

V [ X - Y ] = V [ X ] + V [Y ] = 0.04 + 0.08 = 0.12

Determine the probability that the sample mean for species A exceeds the sample mean for species B by at least 1 unit

P ( x - y ≥ 1 ) = 0.0019

attached below is the detailed solution

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