7-1 (a) True
(b) False. m is a population parameter and Xbar is the sample mean, an unbiased estimator of m .
(c) False. A quadrupling of the sample size doubles the accuracy of Xbar in estimating m
(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)/200=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, but lets try. 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) (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) m= =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.