Suppose the amount for Calcium Y to be recovered is uniformly distributed betwee

Question

Suppose the amount for Calcium Y to be recovered is uniformly distributed between 4 and 7 mg. The amount of calcium recovered by one method is the random variable W1=0.2281+0.9948y+E1, where the error term E1 has mean 0 and variance 0.0427 and is independent of Y. A second method has random variable W2= -0.0748+1.0024Y+E2, where the error term E2 has mean 0 and variance 0.0159 and is independent of Y. The better technique should have a mean as close as possible to the mean of Y(which is 5.5) and a variance as small as possible. Compare the two methods on the basis of mean and variance.

Explanation / Answer

mean of W1= 5.6995 and mean of W2=5.4384. difference from mean is 0.1995 & -0.0616 respectively. Variance of W1 = (0.9948)^2 V(Y) + V(E1) = 0.7849 Variance of W2 = (1.0024)^2 V(Y)+V(E2) = 0.7695 Thus we notice that for W2, distance from mean is lower and variance is also lower Thus 2nd method is the better method.

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