7-1 (a) True
(b) False. mu is a population parameter and Xbar is the sample mean, an unbiased estimator of mu .
(c) False. A quadrupling of the sample size doubles the accuracy of Xbar in estimating mu
(d) True.
7-2 (a) Both Xbar and U are unbiased.
(b) Var(Xbar)=s2/2, Var(U)=(1/9)*s2 +(4/9)*s2 =(5/9)*s2, The relative efficiency of Xbar to U is Var(U)/Var(Xbar)=1.11, hence Xbar is 11% more efficient.
7-7 (a) Var(P*)=Var((1/2)*P1+(1/2)*P2)= (1/4)*Var(P1) +(1/4)*Var(P2)=(1/4)*(0.00105+0.0001824)=0.0003081
(b) Var(P2)=(.24*.76)/1000=0.0001824. Both P2 and P* are unbiased for estimating the population proportion p, hence the relative efficiency of P2 to P* is the ratio of their variances: Var(P*)/Var(P2)= 0.0003081/0.000182=1.693, so P2 is 69.3% more efficient.
(c) Var(P)=(0.25*0.75)/1200=0.00015625. As you know P is unbiased for estimating p, so relative efficiency is a ratio of variances: Var(P*)/Var(P)=0.000308/0.000156=1.974%, so 97.4% more efficient
(d) True, as part (a) showed.
7-8 (a) From the left to the right: A, C, B
(b) B is biased, its mean pointed to the left; B and C have minimum variance (they look to be the same); C has the smallest MSE and so is most efficient. Which of A or B is least efficient (this will also answer the question 'which has largest MSE?')? It is hard to tell by looking. A good answer would be that the MSE of both are approximately equal. If you HAD to pick one with lower MSE, however, you might try something like this: If we assume, as it looks to be, that Bias(A)=0, then MSE(A)=Var(A). Now assume that SD(A) is approximately equal to Bias(B), because the spread of the shots in A appear to have a standard deviation equal to the distance that the clump of shots in C is from the bullseye. (I'm just eye-balling it), then MSE(B)=Var(B)+Bias(B)2 = Var(B)+Var(A). Since B is biased then MSE(B) must be larger than MSE(A). So it looks like B has highest MSE and is least efficient.
7-10 (a) mu= =0*.40+1*.24+2*.20+3*.12+4*.04=1.16
(b) Yes. Survey ii gets a small, representative sample.
(c) Survey ii's MSE equals its variance (there is no bias). The variance in the population reached by survey ii is s2=1.4144. The variance of a sample mean, sample size = 25, from this populion is s2/25 = 0.056576. The mean of the population reached by survey i is 0.61, its variance is 0.8379. With 200 replies, the variance of a sample mean from this population is 0.8379/200 = 0.0041895. This estimator is biased due to the non-response problem, the bias is (0.61-1.16)=-0.55. Its MSE is 0.0041895+(-0.55)2=0.3066895. The MSE from the smaller, no non-response bias, survey is the lower of the two.