MSPE PROGRAM
ECON 506  FALL 2011
Solution to HW7
(The problem numbers refer to numbers in 7
th
edition)
Part B:
Use likelihood ratio test.
The observed test statistics is –2 ln (24/196) = 4.2
The sampling distribution is
χ
2
with 1 degree of freedom. At for example
α
=0.05 the
critical value would be 3.84. Thus we reject the null.
Part C:
10.2
Note that
Y
is binomial with parameters
n
= 20 and
p
.
a.
If the experimenter concludes that less than 80% of insomniacs respond to the
drug when actually the drug induces sleep in 80% of insomniacs, a type I error
has occurred.
b.
α =
P
(reject
H
0

H
0
true) =
P
(
Y
≤ 12 
p
= .8) = .032 (using Appendix III).
c.
If the experimenter does not reject the hypothesis that 80% of insomniacs
respond to the drug when actually the drug induces sleep in fewer than 80% of
insomniacs, a type II error has occurred.
d.
β(.6) =
P
(fail to reject
H
0

H
a
true) =
P
(
Y
> 12 
p
= .6) = 1 –
P
(
Y
≤ 12 
p
= .6)
= .416.
e.
β(.4) =
P
(fail to reject
H
0

H
a
true) =
P
(
Y
> 12 
p
= .4) = .021.
10.9
a.
The simulation is performed with a known
p
= .5, so rejecting
H
0
is a type I error.
b.

e.
Answers vary.
f.
This is because of part a.
g.

h.
Answers vary.
10.18
H
0
: μ = 13.20,
H
a
: μ < 13.20.
Using the large sample test for a mean,
z
= –2.53, and with
α = .01, –
z
.01
= –2.326.
So,
H
0
is rejected: there is evidence that the company is paying
substandard wages.
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10.19
H
0
: μ = 130,
H
a
: μ < 130.
Using the large sample test for a mean,
z
=
40
/
1
.
2
130
6
.
128

= – 4.22 and
with –
z
.05
= –1.645,
H
0
is rejected: there is evidence that the mean output voltage is less
than 130.
10.20
H
0
: μ ≥ 64,
H
a
: μ < 64.
Using the large sample test for a mean,
z
= –1.77, and w/ α = .01,
–
z
.01
= –2.326.
So,
H
0
is not rejected: there is not enough evidence to conclude the
manufacturer’s claim is false.
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 Spring '08
 Staff
 Null hypothesis, Hypothesis testing, Statistical hypothesis testing, H0, Type I and type II errors

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