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Homework 6- Math 3339
Statistics for the Sciences (University of Houston)
Studocu is not sponsored or endorsed by any college or university
Homework 6- Math 3339
Statistics for the Sciences (University of Houston)
Downloaded by Antonia Marinoni (antoniamarinoni2@gmail.com)
lOMoARcPSD|9388477
a
Probability And Statistics for Engineers And Scientists
, Hayter, A. (2007)
Math 3339
Homework 6 (Chapters 8, 10)
Name:__________________________________ PeopleSoft ID:_______________
Instructions:
•
Homework will NOT be accepted through email or in person. Homework must be submitted through
CourseWare BEFORE the deadline.
•
Print out this file and complete the problems or you can complete it using your computer.
•
Use blue or black ink or a dark pencil if completing this by hand.
•
Write your solutions in the space provided.
You must show all work for full credit.
•
Submit this assignment at
http://www.casa.uh.edu
under “
Assignments" and choose
hw6
.
•
Total possible points:
15.
1.
a
An experimenter is interested in the hypothesis testing problem with
H
0:
µ = 420
versus
H
a
:
µ <
420
, where
µ
is the average radiation level in a research laboratory. Suppose that a sample of 49
radiation level measurements is obtained and that the experimenter wishes to use
σ=10.0 for the
standard deviation of the radiation levels.
(a)
What is the rejection
region when α=0.01?
(b)
Suppose that the sample mean is 415.7. What would be the experimenter’s decision of the
hypothesis test based on the rejection region?
(c)
What is the p-value for this hypothesis test?
Given:
Ho:
M
=420;
HaiM
<420;
n
=
49;0
=
10
a)
(
0.01
=>
z
=
gnorm
(0.01)
=
-
2.326
·
the
rejection
regions
when
z
<-2.323
or
&value
<a
=0.01
57
suppose
sample
mean
x
=
415.7
·
Using
normal
distribution
the
statistic
2)
X
-
1
=
415.7
-
428
t
=
=
-
3.01
o
10
N
49
1(<z)
=
prorm
(3.01)
=
0.001306
<
x
=
0.01
·
Reject
the
will
hypothes
at
0.01
significant
level.
The
population
mean
is
less
than
400.
Downloaded by Antonia Marinoni (antoniamarinoni2@gmail.com)
lOMoARcPSD|9388477
b
Devore, Jay and Berk, Kenneth,
Modern Mathematical Statistics with Applications,
Thomson Brooks/Cole, 2007.
2.
b
Water samples are taken from water used for cooling as it is being discharged from a power plant
into a river.
It has been determined that as long as the mean temperature of the discharged water is at
most 15
0̊
F, there will be no negative effects on the river
’
s ecosystem. To investigate whether the
plant is in compliance with regulations that prohibit a mean discharge-water temperature above 15
0̊
,
50 water sample will be tank at randomly selected times, and the temperature of each sample will be
recorded.
(a)
Determine an appropriate null and alternative hypothesis for this test.
(b)
In the context of this situation, describe type I and type II errors.
a)
Ho:
M
=
150;
HaM>
150
b)
Type
I:
Here
the
mean
of
discharged
water
is
above
1500,
but
actually
it
is
1500F
Type
I:
Here
the
mean
of
discharged
unter
is
equal
to
list,
but
actually
it
is
above
1500
Downloaded by Antonia Marinoni (antoniamarinoni2@gmail.com)
lOMoARcPSD|9388477
b
Devore, Jay and Berk, Kenneth,
Modern Mathematical Statistics with Applications,
Thomson Brooks/Cole, 2007.
3.
For each of the following scenarios, state whether the null hypothesis should be rejected or not. State
any assumptions that you make beyond the information that is given.
(a)
H
0:
µ = 6
,
H
a
:
µ ≠ 6
, n=25,
x̄
=5.6, s=2.5,
α=0.05
(b)
H
0:
µ = 6
,
H
a
:
µ < 6
, n=25,
x̄
=5.6, s=2.5,
α=0.05
(c)
H
0:
µ = 25
,
H
a
:
µ > 25
, n=81,
x̄
=26.4, s=3.2,
α
=0.01
(d)
H
0:
µ = 50
,
H
a
:
µ ≠ 50
, n=55, p-value=0.053
a)
The
sample
size
n
=25
<
30
·
Normal
distribution,
use
t-test
x
-
u
=
5.6
-
6
t
=
=
-
0.8
Svr
2.525
-value
=
1(T
=
-
0.80r
T(,0.8)
=
a+pt(t,n
-
1)
=
2*pt
(
-
0.8,24)
=
0.43156
Since
R-value
=
0.43156>
=
0.05
·
fail
to
reject
the
mill
hypothesis
*
the
mean
of
the
sample
is
equal
to
b
=
=
=
0.8
P-value
=
1(T
=
0.8)
=
pt(-0.8,24)
=
0.21578c
=
0.05
-
Do
fail
to
reject
the
will
hypothesis
-
the
mean
of
the
sample
is
equal
to
c)
Sample
size
1
=
81
-
Normall
distribution,
apply
the
central
limit
theorem
Uset-test:
A
=
I
-
r
=
26425
=
3.9375
n
3.281
B.value
=
f(T
>
5.9375)
1
-
8(T
(5.9375)
1
-
pt
(3.9375,80)
=
8.79.105
Since
P-value
=
8.74.105
<c
=0.01
Reject
the
mull
hypothesis
*
the
mean
of
the
sample
is
above
at
d)
If
we
assume
2
=
0.05,
we
would
fail
to
reject
the
mall
hypothesis.
Because
I
value
0.053
x
=
0.05
·
The
mean
of
the
sample
is
equal
to
50
Downloaded by Antonia Marinoni (antoniamarinoni2@gmail.com)
lOMoARcPSD|9388477
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