16. A consumer advocacy group published a study of labeling of seafood sold in t
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16. A consumer advocacy group published a study of labeling of seafood sold in three U.S. states. The group purchased 235 pieces of seafood from various kinds of food stores and restaurants in the three states and genetically compared the pieces to standard gene fragments that can identify the species. The study found that 15 of the 26 "red snapper" packages tested were a different kind of fish. Assume that the study used a simple random sample.
Complete parts a through c below.
a) Are the conditions for creating a confidence interval satisfied? Explain.
A.
No, because the sample is a simple random sample, the sample is less than 10% of the population, and there are at least 10 expected "successes" and 10 expected "failures."
B.
No, because the sample is a simple random sample, the sample proportion is between 10% and 90%, and there are at least 20 expected "successes" and 20 expected "failures."
C.
Yes, because the sample is a simple random sample, the sample proportion is between 10% and 90%, and there are at least 20 expected "successes" and 20 expected "failures."
D.
Yes, because the sample is a simple random sample, the sample is less than 10% of the population, and there are at least 10 expected "successes" and 10 expected "failures."
b) Construct a
9595%
confidence interval for the proportion of "red snapper" packages that were a different kind of fish.
left parenthesis nothing comma nothing right parenthesis,
(Round to three decimal places as needed.)
c) Explain what the confidence interval from part (b) says about "red snapper" sold in these three states. Select the correct choice below and fill in the answer boxes within your choice.
(Round to one decimal place as needed.)
A.
One is
9595%
confident that between
nothing %
and
nothing %
of all "red snapper" purchased for the study in these three states was not actually red snapper.
B.
Ninety-five percent of the time, the true proportion of "red snapper" sold in these three states that is falsely labeled is between
nothing %
and
nothing %.
C.
There is a
9595%
chance that the probability of any given "red snapper" sold in these three states being actual red snapper is between
nothing %
and
nothing %.
D.
One is
9595%
confident that between
nothing %
and
nothing %
of all "red snapper" sold in food stores and restaurants in these three states is not actually red snapper.
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17. It the 1980s, it was generally believed that congenital abnormalities affected about 77% of a large nation's children. Some people believe that the increase in the number of chemicals in the environment has led to an increase in the incidence of abnormalities. A recent study examined 380 randomly selected children and found that 43 of them showed signs of an abnormality. Is this strong evidence that the risk has increased? (We consider a P-value of around 55% to represent reasonable evidence.) Complete parts a through f.
a) Write appropriate hypotheses. Let p be the proportion of children with genetic abnormalities. Choose the correct answer below.
A.
H0:
pequals=0.070.07
vs.
HA:
pnot equals0.070.07
B.
H0:
pequals=0.070.07
vs.
HA:
pgreater than>0.070.07
C.
H0:
pequals=0.11320.1132
vs.
HA:
pnot equals0.11320.1132
D.
H0:
pequals=0.11320.1132
vs.
HA:
pless than<0.11320.1132
E.
H0:
pequals=0.070.07
vs.
HA:
pless than<0.070.07
F.
H0:
pequals=0.11320.1132
vs.
HA:
pgreater than>0.11320.1132
b) Check the necessary assumptions. Which of the following are satisfied? Select all that apply.
A.
Less than 10% of the population was sampled.
B.
There are more than 10 successes and 10 failures.
C.
The sample is random.
D.
The data are independent.
c) Perform the mechanics of the test. What is the P-value?
P-valueequals=nothing
(Round to three decimal places as needed.)
d) Explain carefully what the P-value means in this context. Choose the correct answer below.
A.
The P-value is the chance of observing
4343
or more children with genetic abnormalities in a random sample of
380380
children.
B.
The P-value is the chance of observing
4343
or more children with genetic abnormalities in a random sample of
380380
children if
77%
of children actually have genetic abnormalities.
C.
The P-value is the chance of observing
77%
of children with genetic abnormalities.
D.
The P-value is the actual percentage of children who have genetic abnormalities.
e) What's your conclusion?
A.
RejectReject
H0.
There
isis
sufficient evidence that more than
77%
of the nation's children have genetic abnormalities.
B.
RejectReject
H0.
There
is notis not
sufficient evidence that more than
77%
of the nation's children have genetic abnormalities.
C.
Fail to rejectFail to reject
H0.
There
is notis not
sufficient evidence that more than
77%
of the nation's children have genetic abnormalities.
D.
Fail to rejectFail to reject
H0.
There
isis
sufficient evidence that more than
77%
of the nation's children have genetic abnormalities.
f) Do environmental chemicals cause congenital abnormalities?
A.
No, the conclusion of the hypothesis test shows that environmental chemicals do not cause genetic abnormalities.
B.
Yes, the conclusion of the hypothesis test shows that environmental chemicals cause genetic abnormalities.
C.
It is unknown if environmental chemicals cause genetic abnormalities, because the hypothesis test does not indicate the cause of any changes.
18. Data from a recent year showed that 63% of the tens of thousands of applicants to a certain program were accepted. A company that trains applicants claimed that 191 of the 290 students it trained that year were accepted. Assume these trainees were representative of the population of applicants. Has the company demonstrated a real improvement over the average? Complete parts a through c below.
a) What are the hypotheses?Which hypotheses below will test if the company demonstrated improvement over the average?
A.
Upper H 0H0:
pequals=0.65860.6586
Upper H Subscript Upper AHA:
pless than<0.65860.6586
B.
Upper H 0H0:
pequals=0.630.63
Upper H Subscript Upper AHA:
pless than<0.630.63
C.
Upper H 0H0:
pequals=0.630.63
Upper H Subscript Upper AHA:
pgreater than>0.630.63
D.
Upper H 0H0:
pequals=0.630.63
Upper H Subscript Upper AHA:
pnot equals0.630.63
E.
Upper H 0H0:
pequals=0.65860.6586
Upper H Subscript Upper AHA:
pgreater than>0.65860.6586
F.
Upper H 0H0:
pequals=0.65860.6586
Upper H Subscript Upper AHA:
pnot equals0.65860.6586
b) Verify that the conditions are satisfied and find the P-value.
Which assumptions and conditions below are met? Select all that apply.
A.The independence assumption is met.
B.The success/failure condition is met.
C.The 10% condition is met.
D.The randomization condition is met.
What is the P-value?
P-valueequals=nothing
(Round to four decimal places as needed.)
c) Would you recommend this company based on what you see here?
A.
The P-value provides
strong evidencestrong evidence
that the program is successful.
B.
The P-value provides
some indicationsome indication
that the program is successful.
C.
The P-value provides
insufficient evidenceinsufficient evidence
that the program is successful.
D.
Inference is not possible since the assumptions and conditions are not met.
____________________________________
19. The manufacturer of a metal stand for home TV sets must be sure that its product will not fail under the weight of the TV. Since some larger sets weigh nearly 300 pounds, the company's safety inspectors have set a standard of ensuring that the stand can support an average of over 500 pounds. Their inspectors regularly subject a random sample of the stands to increasing weight until they fail. They test the hypothesis Upper H 0H0: muequals=500500
against Upper H Subscript Upper AHA:
mugreater than>500500,
using the level of significance
alphaequals=0.01.
If the sample of stands fails to pass this safety test, the inspectors will not certify the product for sale to the general public. The manufacturer is thinking of revising its safety test. Complete parts a through c below.
a) If the company's lawyers are worried about being sued for selling an unsafe product, should they make the value of alpha smaller or larger? Explain. Choose the correct answer below.
A. They should make the value of alpha smaller. This means a smaller chance of declaring the stands unsafe when they are actually safe.
B.They should make the value of alpha smaller. This means a smaller chance of declaring the stands safe when they are not actually safe.
C.They should make the value of alpha larger. This means a smaller chance of declaring the stands unsafe when they are actually safe.
D.They should make the value of alpha larger. This means a smaller chance of declaring the stands safe when they are not actually safe.
b) In this context, what is meant by the power of the test?
A.
The probability of correctly detecting that the stands are not capable of holding more than
500 pounds.
B.The probability of incorrectly detecting that the stands are capable of holding more than 500
pounds.
C.The probability of correctly detecting that the stands are capable of holding more than 500
pounds.
D.
The probability of incorrectly detecting that the stands are not capable of holding more than 500
pounds.
c) If the company wants to increase the power of the test, what options does it have? Explain the advantages and disadvantages of each option. Select all that apply.
A.
The company could increase the "design load" to be well above
500500
pounds. This would probably be costly.
B.
The company could change the production procedure so that the standard deviation of the weights of the stands decreases. This would probably be costly.
C.
The company could increase the sample size. This would take more time for testing and be costly.
D.
The company could increase
alpha.
This would result in more Type I errors.
16. A consumer advocacy group published a study of labeling of seafood sold in three U.S. states. The group purchased 235 pieces of seafood from various kinds of food stores and restaurants in the three states and genetically compared the pieces to standard gene fragments that can identify the species. The study found that 15 of the 26 "red snapper" packages tested were a different kind of fish. Assume that the study used a simple random sample.
Complete parts a through c below.
a) Are the conditions for creating a confidence interval satisfied? Explain.
A.
No, because the sample is a simple random sample, the sample is less than 10% of the population, and there are at least 10 expected "successes" and 10 expected "failures."
B.
No, because the sample is a simple random sample, the sample proportion is between 10% and 90%, and there are at least 20 expected "successes" and 20 expected "failures."
C.
Yes, because the sample is a simple random sample, the sample proportion is between 10% and 90%, and there are at least 20 expected "successes" and 20 expected "failures."
D.
Yes, because the sample is a simple random sample, the sample is less than 10% of the population, and there are at least 10 expected "successes" and 10 expected "failures."
b) Construct a
9595%
confidence interval for the proportion of "red snapper" packages that were a different kind of fish.
left parenthesis nothing comma nothing right parenthesis,
(Round to three decimal places as needed.)
c) Explain what the confidence interval from part (b) says about "red snapper" sold in these three states. Select the correct choice below and fill in the answer boxes within your choice.
(Round to one decimal place as needed.)
A.
One is
9595%
confident that between
nothing %
and
nothing %
of all "red snapper" purchased for the study in these three states was not actually red snapper.
B.
Ninety-five percent of the time, the true proportion of "red snapper" sold in these three states that is falsely labeled is between
nothing %
and
nothing %.
C.
There is a
9595%
chance that the probability of any given "red snapper" sold in these three states being actual red snapper is between
nothing %
and
nothing %.
D.
One is
9595%
confident that between
nothing %
and
nothing %
of all "red snapper" sold in food stores and restaurants in these three states is not actually red snapper.
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Explanation / Answer
16. The statistical software output for this problem is:
One sample proportion summary confidence interval:
p : Proportion of successes
Method: Standard-Wald
95% confidence interval results:
Hence,
a) Option D
b) 95% confidence interval is:
(0.387, 0.767)
c) Interpretation: 38.7% and 76.7%
Proportion Count Total Sample Prop. Std. Err. L. Limit U. Limit p 15 26 0.57692308 0.096890668 0.38702086 0.7668253Related Questions
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