Homework #13 Solutions: Stat 430, Spring 2011
Chapter 15 problems:
10. There are n ranks to be assigned. Thus, T+ + T = sum of all ranks =
n(n+1)/2
12. a. Define di to be the difference between the math score and the art score for the ith stud
Homework #2 Solutions: Stat 430, Spring 2011
36. An unbiased estimator of is
standard error of is
, and since
.
38. To find an unbiased estimator of
estimator of
, an unbiased estimator of the
, note that E(Y) =
. Further,
so Y is an unbiased
so
unbiased
Homework #3 Solutions: Stat 430, Spring 2011
80. The 95% CI, based on a tdistribution with 21 1 = 20 degrees of freedom, is
26.6 2.086
= 26.6 3.37 or (23.23, 29.97).
82. a. The 90% CI for the mean verbal SAT score for urban high school seniors is
505 1.72
Homework #4 Solutions: Stat 430, Spring 2011
4. We have that V( ) = (n + 1)2V(Y(n) =
. Similarly, V(
Thus, the ratio of these variances is as given.
6. Both estimators are unbiased and V(
7. The estimator
the mean) and V(
is unbiased so MSE(
) = 2.
Homework #5 Solutions: Stat 430, Spring 2011
24. a. From Chapter 6,
b. Note that
probability.
is chisquare with with n degrees of freedom.
and
. Thus, as n , Wn
26. a. We have that
in
.
If > ,
and
. Thus,
If < ,
,
= 1.
. So,
.
b. The result follo
Homework #6 Solutions: Stat 430, Spring 2011
42. The likelihood function is L(p) =
. By Theorem 9.4,
sufficient for p with
is
and h(y) = 1.
50. As in Ex. 9.49, we will define the uniform distribution on the interval (1, 2) as
.
The likelihood func
Homework Set #7 Solutions
Stat 430 Spring 2011
1. # 9.80 (a) The MLE is = Y .
(b) E ( = , V () = /n.
(c) Since is unbiased and has a variance that goes to zero with increasing
n, it is consistent.
(d) By the invariance property, the MLE for P (Y = 0) is e
Homework #8 Solutions: Stat 430, Spring 2011
20. H0: 64, Ha: < 64. Using the large sample test for a mean, z = 1.77, and w/ = .01, z.
01 = 2.326. So, H0 is not rejected: there is not enough evidence to conclude the manufacturers
claim is false.
24.
Homework Set #9 Solutions, part 1
Stat 430 Spring 2011
1. (a) Let R be the average functional improvement for patients receiving radon spa
treatment, and let O be the average functional improvement for patients receiving
ordinary bath treatment. Then
H0 :
Homework #10 Solutions: Stat 430, Spring 2011
Chapter 11 problems:
8. The summary statistics are:
= 15.505,
= 9.448, Sxy = 1546.459, Sxx = 2359.929. Thus,
= 0.712 + 0.655x. When x = 12, the best estimate of y is
= .712 + 0.655(12) = 7.148.
15.
Homework #11 Solutions: Stat 430, Spring 2011
Chapter 11 problems:
36. From Ex. 11.13 and 11.24, when x* = 5,
= 452.119 29.402(5) = 305.11 so that V( ) is
estimated to be 402.98. Thus, a 90% CI for E(Y) is 305.11 1.782
= 305.11 35.773.
42. Simil
Homework #1 Solutions: Stat 430, Spring 2011
4. a. They are equal.
b.
8. a.
Note that
,
.
,
and
are simple linear combinations of Y1, Y2, and Y3. So, it is
easily shown that all four of these estimators are unbiased. From Ex. 6.81 it was shown that
has an