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MTH 233
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1
For the given claim, complete parts (a) and (b) below.
Claim: High school teachers have incomes with a standard deviation that is
moremore
than
$19 comma 00019,000.
A recent study of
133133
high school teacher incomes showed a standard deviation of
$21 comma 00021,000.
- Express the original claim in symbolic form.
- Identify the null and the alternative hypotheses that should be used to arrive at a conclusion that supports the claim.
Upper H 0H0:
Upper H 1H1:
2
For the given claim, complete parts (a) and (b) below.
Claim:
At mostAt most
2222%
of Internet users pay bills online. A recent survey of
397397
Internet users indicated that
1818%
pay their bills online.
- Express the original claim in symbolic form.
- Identify the null and the alternative hypotheses.
Upper H 0H0:
Upper H 1H1:
3
The claim is that the proportion of adults who smoked a cigarette in the past week is less than
0.250.25,
and the sample statistics include
nequals=17631763
subjects with
494494
saying that they smoked a cigarette in the past week. Find the value of the test statistic.
The test statistic is
(Round to two decimal places as needed.)
4
The claim is that the white blood cell counts of adult females are normally distributed, with a standard deviation equal to
2.052.05.
A random sample of
4444
adult females has white blood cell counts with a mean of
7.727.72
and a standard deviation of
3.923.92.
Find the value of the test statistic.
The test statistic is
(Round to three decimal places as needed.)
5
Assume that the significance level is
alpha equals 0.05α=0.05.
Use the given information to find the P-value and the critical value(s).
The test statistic of
zequals=1.471.47
is obtained when testing the claim that
p greater than 0.4p>0.4.
Click here to view page 1 of the Normal table.
LOADING…
Click here to view page 2 of the Normal table.
LOADING…
P-valueequals=
(Round to four decimal places as needed.)
The critical value(s) is/are
zequals=
(Round to two decimal places as needed. Use a comma to separate answers as needed.)
6
Assume that the significance level is
alpha equals 0.05α=0.05.
Use the given statement and find the P-value and critical value(s).
The test statistic of
zequals=negative 1.94−1.94
is obtained when testing the claim that
p equals one fifthp=15.
Click here to view page 1 of the Normal table.
LOADING…
Click here to view page 2 of the Normal table.
LOADING…
P-valueequals=
(Round to four decimal places as needed.)
The critical value(s) is/are
(Round to three decimal places as needed. Use a comma to separate answers as needed.)
7
Assume a significance level of
alpha equals 0.05α=0.05
and use the given information to complete parts (a) and (b) below.
Original claim: The proportion of male golfers is
lessless
than
0.50.5.
The hypothesis test results in a P-value of
0.1770.177.
- State a conclusion about the null hypothesis. (Reject
Upper H 0H0
or fail to reject
Upper H 0H0.)
Choose the correct answer below.
A.
Fail to rejectFail to reject
Upper H 0H0
because the P-value is
greatergreater
than
alphaα.
B.
RejectReject
Upper H 0H0
because the P-value is
lessless
than
alphaα.
C.
Fail to rejectFail to reject
Upper H 0H0
because the P-value is
lessless
than
alphaα.
D.
RejectReject
Upper H 0H0
because the P-value is
greatergreater
than
alphaα.
- Without using technical terms, state a final conclusion that addresses the original claim. Which of the following is the correct conclusion?
A.
There
is notis not
sufficient evidence to reject the claim that the proportion of male golfers is
lessless
than
0.50.5.
B.
There
isis
sufficient evidence to reject the claim that the proportion of male golfers is
lessless
than
0.50.5.
C.
There
isis
sufficient evidence to support the claim that the proportion of male golfers is
lessless
than
0.50.5.
D.
There
is notis not
sufficient evidence to support the claim that the proportion of male golfers is
lessless
than
0.50.5.
8
Assume a significance level of
alpha equals 0.01α=0.01
and use the given information to complete parts (a) and (b) below.
Original claim: Women have heights with a mean equal to
161.7161.7
- The hypothesis test results in a P-value of
0.19850.1985.
- State a conclusion about the null hypothesis. (Reject
Upper H 0H0
or fail to reject
Upper H 0H0.)
Choose the correct answer below.
A.
Fail to rejectFail to reject
Upper H 0H0
because the P-value is
greatergreater
than
alphaα.
B.
RejectReject
Upper H 0H0
because the P-value is
greatergreater
than
alphaα.
C.
RejectReject
Upper H 0H0
because the P-value is
lessless
than
alphaα.
D.
Fail to rejectFail to reject
Upper H 0H0
because the P-value is
lessless
than
alphaα.
- Without using technical terms, state a final conclusion that addresses the original claim. Which of the following is the correct conclusion?
A.
There
isis
sufficient evidence to support the claim that the mean height of women is equal to
161.7161.7
cm.
B.
There
is notis not
sufficient evidence to warrant rejection of the claim that the mean height of women is equal to
161.7161.7
cm.
C.
There
is notis not
sufficient evidence to support the claim that the mean height of women is equal to
161.7161.7
cm.
D.
There
isis
sufficient evidence to warrant rejection of the claim that the mean height of women is equal to
161.7161.7
cm.
9
A
0.10.1
significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is
lessless
than 0.5. Assume that sample data consists of
5555
girls in
121121
births, so the sample statistic of
five elevenths511
results in a z score that is 1 standard deviation
belowbelow
- Complete parts (a) through (h) below.
Click here to view page 1 of the Normal table.
LOADING…
Click here to view page 2 of the Normal table.
LOADING…
- Identify the null hypothesis and the alternative hypothesis. Choose the correct answer below.
A.
Upper H 0H0:
pequals=0.5
Upper H 1H1:
pnot equals≠0.5
B.
Upper H 0H0:
pnot equals≠0.5
Upper H 1H1:
pless than<0.5
C.
Upper H 0H0:
pequals=0.5
Upper H 1H1:
pgreater than>0.5
D.
Upper H 0H0:
pequals=0.5
Upper H 1H1:
pless than<0.5
- What is the value of
alphaα?
alphaαequals=
(Type an integer or a decimal.)
- What is the sampling distribution of the sample statistic?
Normal distribution
chi squaredχ2
Student (t) distribution
- Is the test two-tailed, left-tailed, or right-tailed?
Two-tailed
LeftLeft-tailed
RightRight-tailed
- What is the value of the test statistic?
The test statistic is
(Type an integer or a decimal.)
- What is the P-value?
The P-value is
(Round to four decimal places as needed.)
- What are the critical value(s)?
The critical value(s) is/are
(Round to two decimal places as needed. Use a comma to separate answers as needed.)
- What is the area of the critical region?
The area is
(Round to two decimal places as needed.)
Question is complete.
10
In 1997, a survey of
920920
households showed that
152152
of them use e-mail. Use those sample results to test the claim that more than 15% of households usee-mail. Use a 0.05 significance level. Use this information to answer the following questions.
- Which of the following is the hypothesis test to be conducted?
A.
Upper H 0 : p equals 0.15H0: p=0.15
Upper H 1 : p less than 0.15H1: p<0.15
B.
Upper H 0 : p not equals 0.15H0: p≠0.15
Upper H 1 : p equals 0.15H1: p=0.15
C.
Upper H 0 : p greater than 0.15H0: p>0.15
Upper H 1 : p equals 0.15H1: p=0.15
D.
Upper H 0 : p equals 0.15H0: p=0.15
Upper H 1 : p not equals 0.15H1: p≠0.15
E.
Upper H 0 : p equals 0.15H0: p=0.15
Upper H 1 : p greater than 0.15H1: p>0.15
F.
Upper H 0 : p less than 0.15H0: p<0.15
Upper H 1 : p equals 0.15H1: p=0.15
- What is the test statistic?
zequals=
(Round to two decimal places as needed.)
- What is the P-value?
P-valueequals=
(Round to three decimal places as needed.)
- What is the conclusion?
There
is notis not
sufficient evidence to support the claim that more than 15% of households use e-mail.
There
isis
sufficient evidence to support the claim that more than 15% of households use e-mail.
- Is the conclusion valid today? Why or why not?
A.
Yes, the conclusion is valid today because the requirements to perform the test are satisfied.
B.
No, the conclusion is not valid today because the population characteristics of the use of e-mail are changing rapidly.
C.
You can make no decisions about the validity of the conclusion today.
Question is complete.
11
Consider a flight to be on time if it arrives no later than 15 minutes after the scheduled arrival time. Negative arrival times correspond to flights arriving earlier than their scheduled arrival time. Use the sample data to test the claim that
78.378.3%
of flights are on time. Use a
0.0250.025
significance level and the P-value method to answer the following questions.
LOADING…
Click on the icon to view the table of arrival delay times.
What are the null and alternative hypotheses?
A.
Upper H 0H0:
pequals=0.7830.783
Upper H 1H1:
pgreater than>0.7830.783
B.
Upper H 0H0:
pequals=0.7830.783
Upper H 1H1:
pless than<0.7830.783
C.
Upper H 0H0:
pnot equals≠0.7830.783
Upper H 1H1:
pequals=0.7830.783
D.
Upper H 0H0:
pgreater than>0.7830.783
Upper H 1H1:
pequals=0.7830.783
E.
Upper H 0H0:
pless than<0.7830.783
Upper H 1H1:
pequals=0.7830.783
F.
Upper H 0H0:
pequals=0.7830.783
Upper H 1H1:
pnot equals≠0.7830.783
What is the test statistic?
zequals=
(Round to two decimal places as needed.)
What is the P-value?
P-valueequals=
(Round to four decimal places as needed.)
What is the conclusion about the null hypothesis?
A.
Fail to rejectFail to reject
the null hypothesis because the P-value is
greater thangreater than
the significance level,
alphaα.
B.
RejectReject
the null hypothesis because the P-value is
greater thangreater than
the significance level,
alphaα.
C.
Fail to rejectFail to reject
the null hypothesis because the P-value is
less thanless than
the significance level,
alphaα.
D.
RejectReject
the null hypothesis because the P-value is
less thanless than
the significance level,
alphaα.
What is the final conclusion?
A.
There is
insufficientinsufficient
evidence to warrant rejection of the claim that
78.378.3%
of flights are on time.
B.
There is
sufficientsufficient
evidence to support the claim that
78.378.3%
of flights are on time.
C.
There is
sufficientsufficient
evidence to warrant rejection of the claim that
78.378.3%
of flights are on time.
D.
There is
insufficientinsufficient
evidence to support the claim that
78.378.3%
of flights are on time.
12
Use technology to find the P-value for a right-tailed test with
nequals=2424
and test statistic t equals 3.193 .t=3.193.
P-valuealmost equals≈
(Round to four decimal places as needed.)
13
Assume that a simple random sample has been selected and test the given claim. Use the P-value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
The ages of actresses when they won an acting award is summarized by the statistics
nequals=7878,
xequals=35.835.8
years, and
sequals=11.511.5
years. Use a
0.010.01
significance level to test the claim that the mean age of actresses when they win an acting award is
3232
years.
What are the hypotheses?
A.
Upper H 0H0:
muμequals=3232
years
Upper H 1H1:
muμless than<3232
years
B.
Upper H 0H0:
muμequals=3232
years
Upper H 1H1:
muμnot equals≠3232
years
C.
Upper H 0H0:
muμnot equals≠3232
years
Upper H 1H1:
muμequals=3232
years
D.
Upper H 0H0:
muμequals=3232
years
Upper H 1H1:
muμgreater than or equals≥3232
years
Identify the test statistic.
tequals=
(Round to three decimal places as needed.)
Identify the P-value.
The P-value is
(Round to four decimal places as needed.)
State the final conclusion that addresses the original claim. Choose the correct answer below.
A.
Fail to rejectFail to reject
Upper H 0H0.
There is
insufficientinsufficient
evidence to warrant rejection of the claim that the mean age of actresses when they win an acting award is
3232
years.
B.
Fail to rejectFail to reject
Upper H 0H0.
There is
sufficientsufficient
evidence to warrant rejection of the claim that the mean age of actresses when they win an acting award is
3232
years.
C.
RejectReject
Upper H 0H0.
There is
sufficientsufficient
evidence to warrant rejection of the claim that the mean age of actresses when they win an acting award is
3232
years.
D.
RejectReject
Upper H 0H0.
There is
insufficientinsufficient
evidence to warrant rejection of the claim that the mean age of actresses when they win an acting award is
3232
years.
14
Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify the null and alternativehypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
A simple random sample of
2525
filtered 100 mm cigarettes is obtained, and the tar content of each cigarette is measured. The sample has a mean of
19.619.6
mg and a standard deviation of
3.243.24
- Use a
0.050.05
significance level to test the claim that the mean tar content of filtered 100 mm cigarettes is less than
21.121.1
mg, which is the mean for unfiltered king size cigarettes. What do the results suggest, if anything, about the effectiveness of the filters?
What are the hypotheses?
A.
Upper H 0H0:
muμless than<21.121.1
mg
Upper H 1H1:
muμgreater than or equals≥21.121.1
mg
B.
Upper H 0H0:
muμequals=21.121.1
mg
Upper H 1H1:
muμless than<21.121.1
mg
C.
Upper H 0H0:
muμequals=21.121.1
mg
Upper H 1H1:
muμgreater than or equals≥21.121.1
mg
D.
Upper H 0H0:
muμgreater than>21.121.1
mg
Upper H 1H1:
muμless than<21.121.1
mg
Identify the test statistic.
tequals=
(Round to three decimal places as needed.)
Identify the P-value.
The P-value is
(Round to four decimal places as needed.)
State the final conclusion that addresses the original claim. Choose the correct answer below.
A.
Fail to rejectFail to reject
Upper H 0H0.
There is
insufficientinsufficient
evidence to support the claim that the mean tar content of filtered 100 mm cigarettes is less than
21.121.1
mg.
B.
RejectReject
Upper H 0H0.
There is
insufficientinsufficient
evidence to support the claim that the mean tar content of filtered 100 mm cigarettes is less than
21.121.1
mg.
C.
Fail to rejectFail to reject
Upper H 0H0.
There is
sufficientsufficient
evidence to support the claim that the mean tar content of filtered 100 mm cigarettes is less than
21.121.1
mg.
D.
RejectReject
Upper H 0H0.
There is
sufficientsufficient
evidence to support the claim that the mean tar content of filtered 100 mm cigarettes is less than
21.121.1
mg.
What do the results suggest, if anything, about the effectiveness of the filters?
A.
The results suggest that the filters increase the tar content.
B.
The results suggest that the filtered cigarettes have the same tar content as unfiltered king size cigarettes.
C.
The results
do not suggestdo not suggest
that the filters are effective.
D.
The results are inconclusive because the sample size is less than 30.
E.
The results
suggestsuggest
that the filters are effective.
15
The accompanying data table lists the magnitudes of
5050
earthquakes measured on the Richter scale. Test the claim that the population of earthquakes has a mean magnitude greater than
1.001.00.
Use a
0.050.05
significance level. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, and conclusion for the test. Assume this is a simple random sample.
LOADING…
Click the icon to view the sample data.
What are the hypotheses?
A.
Upper H 0H0:
muμequals=1.001.00
in magnitude
Upper H 1H1:
muμgreater than>1.001.00
in magnitude
B.
Upper H 0H0:
muμequals=1.001.00
in magnitude
Upper H 1H1:
muμless than<1.001.00
in magnitude
C.
Upper H 0H0:
muμnot equals≠1.001.00
in magnitude
Upper H 1H1:
muμequals=1.001.00
in magnitude
D.
Upper H 0H0:
muμequals=1.001.00
in magnitude
Upper H 1H1:
muμnot equals≠1.001.00
in magnitude
Identify the test statistic.
tequals=
(Round to two decimal places as needed.)
Identify the P-value.
The P-value is
(Round to three decimal places as needed.)
Choose the correct answer below.
A.
RejectReject
Upper H 0H0.
There is
insufficientinsufficient
evidence to conclude that the population of earthquakes has a mean magnitude greater than
1.001.00.
B.
Fail to rejectFail to reject
Upper H 0H0.
There is
sufficientsufficient
evidence to conclude that the population of earthquakes has a mean magnitude greater than
1.001.00.
C.
Fail to rejectFail to reject
Upper H 0H0.
There is
insufficientinsufficient
evidence to conclude that the population of earthquakes has a mean magnitude greater than
1.001.00.
D.
RejectReject
Upper H 0H0.
There is
sufficientsufficient
evidence to conclude that the population of earthquakes has a mean magnitude greater than
1.001.00.
16
Which of the following is not a strategy for finding P-values with the Student t distribution?
Choose the correct answer below.
A.
Use the table in the book with the appropriate number of degrees of freedom to find a range of values containing the P-value.
B.
Use a TI-83/84 Plus calculator to find the P-value rounded to at least 4 decimal places.
C.
Use software such as Minitab, Excel, or STATDISK to find the P-value rounded to at least 4 decimal places.
D.
Use the table in the book to find the P-value rounded to at least 4 decimal places.
17
Which of the following is not true when using the confidence interval method for testing a claim about
muμ
when
sigmaσ
is unknown?
Choose the correct answer below.
A.
The P-value method, the traditional method, and the confidence interval method are equivalent and yield the same results.
B.
For a one-tailed hypothesis test with a 0.05 significance level, one must construct a 90% confidence interval.
C.
The P-value method and the classical method are not equivalent to the confidence interval method in that they may yield different results.
D.
For a two-tailed hypothesis test with a 0.05 significance level, one must construct a 95% confidence interval.