Stat. 6304 (Fall, 2011) HW 1 Solutions
6.1 No, the sample mean does not have to correspond to a data point. For example, suppose the data set
is cfw_0, 1. The sample mean is 0.5, which is not part of the data set.
6.2 No. Using the Jordan data set from cl
Homework 6 Solutions
6.5-1
a. p = 0.03738318
b. (0.02270901, 0.05205734)
c. (0.02524793, 0.0550219)
d. (0.02504435, 0.055451)
p
e. p + 1.645 p(1 p)/642 = 0.04969899.
6.5-3
a. p = 690/1000 = .69.
b. There are varying acceptable answers depending on which f
Homework 5 Solutions
6.3-12
For this data, xd = 0.07875 and sd = 0.2549222.
a. A point estimate of d is xd = 0.07875.
b. t.05;23 = 1.714, so the confidence interval is (0.01043932, ).
c. Since the lower bound is negative, there is not enough evidence to s
Stat. 6304 (Fall, 2011) Homework Assignment 3
1. (Problem 4-12) If X is a continuous random variable, argue that P (a < X b) = P (a < X b) =
P (a X < b) = P (a < X < b).
2. (Problem 4-28) Suppose f (x) = x/8 for 0 < x < 4.
a. Show that f (x) is a probabil
Homework 1 Solutions
5.1-4
Parts (c) and (d) will vary.
a.
0, x 0
x2 , 0 < x < 1
FX (x) =
1, x 1
b. We can simulate X by setting Y = FX (X) = X 2 X = Y .
We can then generate random U(0, 1) realizations and plug them in for
Y in the above to get realizat
Homework 3 Solutions
6.1-1
Let X1 , X2 , . . . , Xn be a random sample from the N (, 2 ), where
= cfw_ : < < .
Then the likelihood function is given by
P
(xi )2
2 n/2
L() = (2 )
exp
,
2 2
.
Then, the log-likelihood function is
P
2
`() = (n/2) ln(2 )
Homework 2 Solutions
5.6-1
Assuming n = 12 is large enough for the CLT to apply,
1
N 1/2,
.
X
12 12
Therefore
2/3) = P
P (1/2 X
1/2 1/2
2/3 1/2
Z
1/12
1/12
= P (0 Z 2) = 0.4472.
5.6-8
= E(X) = 24.43
a. E(X)
= V ar(X)/n = 2.20/30 = 0.0733
b. V ar(X)
c.
Stat. 6304 (Fall, 2011) HW 8 Solutions
10-55. Use Stat > Basic Statistics > 2 Variances for this problem.
a. H0 : men = women . Ha : men 6= women . = 0.02. Assuming that the individual values in
both groups are normal, F = 0.92 and the p-value = 0.843. Th
Stat. 6304 (Fall, 2011) HW 6 Solutions
9-99. This problem can be done in MINITAB, using STAT > Tables > Chi-Square goodness of fit (1
Var). Put the observed data in a column and the probabilities below (the probabilities X equals the
given values, except
Homework 5 Solutions
3.5-1
a. Here, X Gamma( = 10, = 3/2), so the p.d.f. is
f (x) =
b.
M (t) =
x9 e2x/3
(10)( 23 )10
1
1
=
,
(1 t)
(1 (3/2)t)10
t < 2/3
X = = 10(3/2) = 15
2
X
= 2 = 10(3/2)2 = 45/2
3.5-8
Here, X P ois( = 16).
a. W Gamma( = 7, = 1/16).
b. H
Homework 4 Solutions
2.5-1
a. This is Binomial
with n = 1 and p = 2/3. Thus, = 2/3, 2 = 2/9
q
and = 29
b. Thisq
is Binomial with n = 12 and p = 3/4. Thus, = 9, 2 = 9/4 and
= 94
2.5-5
This is an example of the Geometric distribution with p = 1/5.
P (X = 4
Stat. 6304 (Fall, 2011) HW 3 Solutions
1. P (a X b) = P (a < X b) + P (X = a) = P (a < X b) + 0 = P (a < X b). P (a < X b) =
P (a < X < b) + P (X = b)P (a < X < b) + 0 = P (a < X < b). P (a X b) = P (a X < b) + P (X =
b) = P (a X < b) + 0 = P (a X < b).
2
Stat. 6304 (Fall, 2011) HW 2 Solutions
3.3 The possible values are cfw_0, 1, 2, ., 99999.
3.4 The possible values are cfw_0, 1, 2, ., 10.
3.24 The probability mass function is
0.3
p(x) = 0.5
0.2
0
if x=$10,000,000
if x=$5,000,000
if x = $1, 000, 000
othe
Homework 2 Solutions
1.3-4
a. There are 4 5 2 = 40 possible experiments.
b. There are 2 2 2 = 8 possible experiments.
1.3-14
(n 1)!
n1
n1
(n 1)!
+
+
=
r!(n r 1)! (r 1)!(n 1 (r 1)!
r
r1
(n 1)!(n r) + r(n 1)!
=
r!(n r)!
n!
=
r!(n r)!
n
=
r
1.3-20
a. For