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APPENDIX C
SOLUTIONS TO PROBLEMS
C.1 (i) This is just a special case of what we covered in the text, with n = 4: E(
Y
) = µ and Var(
Y
) =
2/4.
C.2 (i) E(Wa) = a1E(Y1) + a2E(Y2) + + anE(Yn) = (a1 + a2 + + an)µ. Therefore, we must
have a1 + a2 + + an = 1 for unbiasedness.
C.3 (i) E(W1) = [(n 1)/n]E(
Y
) = [(n 1)/n]µ, and so Bias(W1) = [(n 1)/n]µ µ = µ/n.
Similarly, E(W2) = E(
Y
)/2 = µ/2, and so Bias(W2) = µ/2 µ = µ/2. The bias in W1 tends to
zero as n , while the bias in W2 is µ/2 for all n. This is an important difference.
C.4 (i) Using the hint, E(Z|X) = E(Y/X|X) = E(Y|X)/X =
X/X =
. It follows by Property CE.4,
the law of iterated expectations, that E(Z) = E(
) =
.
Y
Y
266
(ii) This follows from part (i) and the fact that the sample average is unbiased for the
population average: write
(iii) In general, the average of the ratios, Yi/Xi, is not the ratio of averages,
2/.W Y X=
(This
(iv) For the n = 17 observations given in the table which are, incidentally, the first 17
C.5 (i) While the expected value of the numerator of G is E(
Y
) =
, and the expected value of
the denominator is E(1
Y
) = 1
, the expected value of the ratio is not the ratio of the
C.6 (i) H0: µ = 0.
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(iv) The estimated reduction, about 33 ounces, does not seem large for an entire year’s
(v) The implicit assumption is that other factors that affect liquor consumption such as
income, or changes in price due to transportation costs, are constant over the two years.
C.7 (i) The average increase in wage is
d
d
= .24, or 24 cents. The sample standard deviation is
(iii) We have the mean and standard error from part (i): t = .24/.1164
2.062. The 5%
critical value for a one-tailed test with df = 14 is 1.761, while the 1% critical value is 2.624.
C.8 (i) For Mark Price,
y
= 188/429
.438.
C.9 (i) X is distributed as Binomial(200,.65), and so E(X) = 200(.65) = 130.
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(iv) The evidence is pretty strong against the dictator’s claim. If 65% of the voting
C.10 Since
y
y
)
.024. We can use the standard normal approximation for the