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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)
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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.
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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
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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
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b
Devore, Jay and Berk, Kenneth,
Modern Mathematical Statistics with Applications,
Thomson Brooks/Cole, 2007.
4.
An engineer selected 30 components at random and measure their strengths.
34,54,73,38,89,52,75,33,50,39,42,42,40,66,72,85,28,71,52,47,41,36,33,38,49,51,55,63,72,78
Does this data suggest that the population mean strength of the components differs from 50? Set up
an appropriate hypothesis test to answer this question.
Given:
n
=
30:
As
the
population
standard
deviation
is
not
given
use
t-test
claim:
the
population
mean
strength
of
the
components
diggers
from
to
·
Ho:
M
=
50;
Hail
+
50
Input
R-studio:
data
=
c
(34,
1,76,38
....
-
63,72,78)
o
r
to
test
(data,
n
50)
mean
(dota)
=
53.26667
·
t
=
1,0457
5
d
(data)
=
17.11006
t
=
xj
-
M
53.26667
58
=
1.04572
a
17.
11006
30
This
is
two-tailed
test
->ICT-1,047(or
TC,
1,04572)
=
2*pt(+,n
-
1)
-
2Ppt
710472,29)
=
0.3043288c
=
0.05
·
fail
to
reject
the
will
hypothesis.
There
is
no
sufficient
evidence
to
support
the
claim
that
population
mean
strength
of
the
components
digers
from
to
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b
Devore, Jay and Berk, Kenneth,
Modern Mathematical Statistics with Applications,
Thomson Brooks/Cole, 2007.
5.
An experiment on the side effects of pain relievers assigned arthritis patients to take one of several
over-the-counter pain medications.
Of the 440 patients who took one brand of pain reliever, 23
suffered some “adverse symptom.”
(a)
If 10% of all patients suffer adverse symptoms, what would be the sampling distribution of the
proportion with adverse symptoms in a sample of 440 patients?
(b)
Does the experiment provide strong evidence that fewer than 10% of patients who take this
medication have adverse symptoms?
Verify that the conditions for inference are met.
State the
hypothesis, calculate the test statistic, then obtain and interpret the P-value.
a)
Given:
n
=
440;
p
=
0.1
up
=
440.0.1
=
44>10
n(1
p)
=
440.0.9
=
396>10
·
We
have
n
=
440
>
30
and
up
>
10,n(1
-
p)
>
10
·
We
can
use
normal
difibution
with:
·
the
mean
of
sampling
distribution
of
the
sample
statistic
is:
up
=
p
=
0.1
·
standard
deviation
of
sampling
distribution
of
the
sample
statistic
is
.
p(1
-
p)
0.1.0.9
=
0.0143
Op
=
m
=
440
b
Ho:
p
=
0.1,
Haip<0.1;
p
231440
=
0.0532
test
statistic:
z
=
0
P
0.0532-0.1
=
-
3.3352
0.0143
value=R(t
c
-3.3337)
=
pnorm
(3.3351)
=
0.004
<1=0.05
·
Reject
the
will
hypothesis.
*
The
experiment
provides
strong
evidence
that
fewer
than
10%
of
patients
who
take
this
medication
have
adverse
symptoms.
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b
Devore, Jay and Berk, Kenneth,
Modern Mathematical Statistics with Applications,
Thomson Brooks/Cole, 2007.
6.
In a random sample of 100 registered voters, 54 said that they might vote for Candidate A. Test the
research hypothesis that Candidate A’s approval rating is greater than 50% against the null hypothesis
that it is 50%
, with α
=0.05.
(a)
Check the assumption for the hypothesis testing.
(b)
State the null hypothesis and alternative hypothesis.
(c)
Determine the rejection region.
(d)
Calculate the test statistics.
(e)
Obtain the p-value for this hypothesis testing.
(f)
Make your decision and state your conclusion.
as
Assumption
for
the
hypothesis:
Candidate
A's
approval
rating
is
equal
to
50%
b
flo:
p
=
0.5;
+a:
p>0.5
c)
Rejection
region:
gnorm
(0.95)
=
1.645;
reject
the
wall
hypothesis
is
23
1.695
5440
-
0.5
dz
=
T
=
0.5(1
-
0.53
=
0.8
10
2)
Brale=1((
>
0.8)
=
1-phorm
(0.8)
=
0.2219
8)
Dvalue
=
0.2219>
x
=
0.05
·
Fail
to
reject
the
wall
hypothesis
·
there
is
no
evidence
that
candidate
A's
approval
rating
is
greater
than
70%
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b
Devore, Jay and Berk, Kenneth,
Modern Mathematical Statistics with Applications,
Thomson Brooks/Cole, 2007.
7.
b
Some college students did a study of textbook pricing.
They compared prices at the campus
bookstore and Amazon.com for the same price.
To be fair, they included the sales tax for the local
store and the added shipping form Amazon.
Here are the prices for a sample of 10 books.
Campus
Amazon
99.34
113.94
51.53
61.44
20.45
31.59
97.22
108.29
61.89
78.44
58.17
65.74
61.63
63.49
44.63
40.39
96.69
117.99
48.88
58.94
(a)
We want to determine if there is a significant difference in the price of textbooks from the campus
bookstore and from Amazon.com.
Which type of test would we use for this data?
(b)
Determine a 95% confidence interval for the difference of the population means.
(c)
Interpret your results.
Is there a substantial difference between the two ways to be textbook?
Assuming that the populations remain unchanged and you have just these two sources, where
would you buy?
&
=campus
-
amazon
-
14.6
9.91
-
11.14
-
11.07
-
16.55
7.57
1.86
4.24
21.3
-
10.6
snam:
-gg.82
as
the
prices
of
same
textbook
from
the
campus
and
Amazon
are
given
·
the
samples
are
paired
and
we
would
use
paired
t
test
b)
Rstudio:
>d
=
c(
-
14.6,
-
9.91
.....
-
(1.3,
-10.6)
-
mn
=
mean
(d)
=
-
9,982
->
Sd(d)
=
7.22
For
95%
confidence
level,
at
df=n
-
1
=
g
d
1
+
+
(dv
=
-
9.g8z1qt(155,9).7.1
2
=
9.982
15.16
=>
(F
=
(
-
15.151
-
4.82)
c)
We
are
95%
confident
that
the
price
of
textbook
from
cample
bookstore
cheaper
than
from
Amazon
between
$
4.82
and
15.15
Since
the
confidence
interval
does
not
contain
othere
is
a
significant
diferent
in
the
price
from
campus
bookstore
and
from
Amazon
I
would
buy
in
Campus
bookstore.
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b
Devore, Jay and Berk, Kenneth,
Modern Mathematical Statistics with Applications,
Thomson Brooks/Cole, 2007.
8.
Concentrations of a pollutant were measured at 25 locations in a lake after rains caused heavy runoff
from surrounding areas. After a period of dry weather, they were measured again at the same
locations. The data are in the file “runoff
.txt
” from the textbook.
(a)
Perform an appropriate hypothesis testing to see if there is any effect of the runoff on pollutant
concentrations.
(b)
Find a 90% confidence interval for the effect of runoff on pollutant concentrations.
a)
Ho:
M
=
Mr
Ha:
Mr
FM2
>
t
test
(X.rainy,
X.
dry
-
I
wong.level
=
0.9)
t
=
2.7158;
df=42.899;
p-value
=
0.009491
·
P-value
<x=0.1.
Reject
the
hypothesis
b)
>
t
test
(X.wainy,
X.dry
.,
conf.
level
=
0.3,
paired
=
T)
+
=
3.8073,
df
=24,
p-value
=
0.000852
90%
confidence
interval
(0.017,
0.045)
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b
Devore, Jay and Berk, Kenneth,
Modern Mathematical Statistics with Applications,
Thomson Brooks/Cole, 2007.
9.
b
An article wanted to look at the compression strength of aluminum cans filled with strawberry drink
and another sample filled with cola.
(a)
Does the following data suggest that the extra carbonation of cola results in a higher average
compression strength?
Base your answer on a P-value.
What assumptions are necessary for your
analysis?
(b)
Develop a 90% confidence interval of the difference of compression strength between the two
beverages.
Beverage
Sample Size
Sample Mean
Sample SD
Strawberry drink
15
540
21
Cola
15
554
15
se
=
29.4,2
=
15
a)
Ho:
Mn
=
M2
M
=
n2
=
15;
x1=
5310;
s1=21
Mr
Ha:
Mr
<Mc
x2
=
354;62
=
15
(
+
sir
an
229.4
+
15)
~
25
·
df
=
(m)
(Sc
nz)
=
29.42
+
152
+
14
M
1
12
1
.
t
=
E
=
590
-
55
=
-
2.101
t
29.4
+
15
Ne
N2
A
value
=
1((-2,101)
=
pt.(2.101,
25)
=
0.0229
<0.05
Reject
the
wall
hypothes.
The
data
suggests
that
cola
has
a
higher
average
compression
strength
than
strawberry
drink
the
distribution
of
compression
strength
care
approximately
normal
b)
for
80%
confidence
interval
at
df=15;F=
a
+
E
=
547
54719+(E,(5)*11
+
152
15
547
1
11.98191
->
(I
=
(535.62,58.38)
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