Around 37 questions in statistics.Workable templates or links to online calculators
Brian_1234
12/12/2017 Section 10.4 HomeworkClayton Leach
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1.
Student: Clayton Leach Date: 12/12/17
Instructor: Ralph Leon Course: Math 110 Fall 2017 Assignment: Section 10.4 Homework
1: Critical Values for the ChiSquare Distribution
Determine the critical values for these tests of a population standard deviation.
(a) A righttailed test with degrees of freedom at the level of significance15 α = 0.01 (b) A lefttailed test for a sample of size n at the level of significance= 31 α = 0.05 (c) A twotailed test for a sample of size n at the level of significance= 25 α = 0.01
Click the icon to view a table a critical values for the ChiSquare Distribution.1
(a) The critical value for this righttailed test is .30.578 (Round to three decimal places as needed.)
(b) The critical value for this lefttailed test is . (Round to three decimal places as needed.)
(c) The critical values for this twotailed test are . (Round to three decimal places as needed. Use a comma to separate answers as needed.)
12/12/2017 Section 10.4 HomeworkClayton Leach
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12/12/2017 Section 10.4 HomeworkClayton Leach
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2.
3.
To test versus , a random sample of size is obtained from a population that is known to be normally distributed. Complete parts (a) through (d).
H : σ = 2.30 H : σ > 2.31 n = 15
(a) If the sample standard deviation is determined to be , compute the test statistic.s = 2.1
(Round to three decimal places as needed.) χ20 =
(b) If the researcher decides to test this hypothesis at the level of significance, determine the critical value.α = 0.01
(Round to three decimal places as needed.) χ20.01 =
(c) Draw a chisquare distribution and depict the critical region.
29.1411.67 29.1411.67 4.460 29.141 4.460 29.141
(d) Will the researcher reject the null hypothesis?
because .Reject H0 χ 2 0 > χ
2 0.01
because .Reject H0 χ 2 0 < χ
2 0.01
because .Do not reject H0 χ 2 0 < χ
2 0.01
because .Do not reject H0 χ 2 0 > χ
2 0.01
(1) less greater
(2) will will not
To test versus , a random sample of size is obtained from a population that is known to be normally distributed.
H : σ = 1.60 H : σ > 1.61 n = 22
(a) If the sample standard deviation is determined to be , compute the test statistic.s = 2.2 (b) If the researcher decides to test this hypothesis at the level of significance, use technology to determine the Pvalue.
α = 0.05
(c) Will the researcher reject the null hypothesis?
(a) The test statistic is .χ20 = (Round to two decimal places as needed.)
(b) The Pvalue is . (Round to three decimal places as needed.)
(c) Since the Pvalue is (1) than the level of significance, the researcher (2) reject the null hypothesis .H : σ = 1.60
12/12/2017 Section 10.4 HomeworkClayton Leach
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4.
5.
To test versus , a random sample of size is obtained from a population that is known to be normally distributed.
H : σ = 2.90 H : σ ≠ 2.91 n = 19
(a) If the sample standard deviation is determined to be , compute the test statistic.s = 1.9
(Round to three decimal places as needed.)=χ20
(b) If the researcher decides to test this hypothesis at the level of significance, determine the critical values.α = 0.10
The critical values are and .=χ20.05 =χ 2 0.95
(Round to three decimal places as needed.)
(c) Draw a chisquared distribution and depict the critical region. Choose the correct graph below.
A. B. C. D.
(d) Will the researcher reject the null hypothesis? Why? Choose the correct answer below.
A. , because the test statistic in the critical region.No is B. , because the test statistic in the critical region.No is not C. , because the test statistic in the critical region.Yes is not D. , because the test statistic in the critical region.Yes is
(1) less greater
(2) will not will
To test versus , a random sample of size is obtained from a population that is known to be normally distributed.
H : σ = 4.70 H : σ ≠ 4.71 n = 19
(a) If the sample standard deviation is determined to be , compute the test statistic.s = 6.6 (b) If the researcher decides to test this hypothesis at the level of significance, use technology to determine the Pvalue.
α = 0.05
(c) Will the researcher reject the null hypothesis?
(a) The test statistic is .χ20 = (Round to two decimal places as needed.)
(b) The Pvalue is . (Round to three decimal places as needed.)
(c) Since the Pvalue is (1) than the level of significance, the researcher (2) reject the null hypothesis .H : σ = 4.70
12/12/2017 Section 10.4 HomeworkClayton Leach
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6.
2: Chisquare distribution table
Suppose a professional golfing association requires that the standard deviation of the diameter of a golf ball be less than inch. Determine whether these randomly selected golf balls conform to this requirement at the level of significance. Assume that the population is normally distributed.
0.005 α = 0.05
Click the icon to view the chisquare distribution table.2
1.684 1.682 1.678 1.678 1.683 1.677 1.679 1.677 1.682 1.677 1.683 1.678
What are the correct hypotheses for this test?
H0: (1) (2) versus
H1: (3) (4) (Type integers or decimals. Do not round.)
Find the sample standard deviation.
s (Round to five decimal places as needed.)=
Use s to calculate the value of the test statistic.
(Round to two decimal places as needed.)χ20 =
Identify the critical value(s) for this test.
The critical value is . (Round to three decimal places as needed. Use a comma to separate answers as needed.)
What is the correct conclusion at the level of significance?α = 0.05
Since the test statistic is (5) than the critical value, (6) the null hypothesis. There
(7) sufficient evidence to conclude that these golf balls conform to the requirement at the level of significance.
0.05
12/12/2017 Section 10.4 HomeworkClayton Leach
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(1)
(2)
(3)
(4)
(5) less greater
(6) do not reject reject
(7) is is not
p σ
μ
≠ < > =
μ
σ
p
> < = ≠
12/12/2017 Section 10.3 HomeworkClayton Leach
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3.
3: Table of Critical tValues
A college entrance exam company determined that a score of on the mathematics portion of the exam suggests that a student is ready for collegelevel mathematics. To achieve this goal, the company recommends that students take a core curriculum of math courses in high school. Suppose a random sample of students who completed this core set of courses results in a mean math score of on the college entrance exam with a standard deviation of
. Do these results suggest that students who complete the core curriculum are ready for collegelevel mathematics? That is, are they scoring above on the math portion of the exam? Complete parts a) through d) below.
24
250 24.4
3.8 24
Click the icon to view the table of critical tvalues.3
a) State the appropriate null and alternative hypotheses. Fill in the correct answers below.
The appropriate null and alternative hypotheses are H0: (1) (2) versus
H1: (3) (4) .
b) Verify that the requirements to perform the test using the tdistribution are satisfied. Check all that apply.
A. The sample size is larger than 30. B. The students' test scores were independent of one another. C. The students were randomly sampled. D. None of the requirements are satisfied.
c) Use the classical approach at the level of significance to find the critical value and test the hypotheses. Identify the test statistic.
α = 0.10
t0 = (Round to two decimal places as needed.)
Identify the critical value. Select the correct choice below and fill in the answer box within your choice. (Round to two decimal places as needed.)
A. tα =
B. tα / 2 =
d) Write a conclusion based on the results. Choose the correct answer below.
(5) the null hypothesis and claim that there (6) sufficient evidence to conclude that
the population mean is (7) than .24
12/12/2017 Section 10.3 HomeworkClayton Leach
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12/12/2017 Section 10.3 HomeworkClayton Leach
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4.
(1)
(2)
(3)
(4)
(5) Do not reject Reject
(6) is not is
(7) less greater
μ
x = < ≤ ≥
> x μ
≤ > ≥ <
=
(1)
(2)
(3) Fail to reject Reject
(4) is is not
(5) equal to less than greater than
In a study, researchers wanted to measure the effect of alcohol on the hippocampal region, the portion of the brain responsible for longterm memory storage, in adolescents. The researchers randomly selected adolescents with alcohol use disorders to determine whether the hippocampal volumes in the alcoholic adolescents were less than the normal volume of . An analysis of the sample data revealed that the hippocampal volume is approximately normal with and s . Conduct the appropriate test at the level of significance.
18
9.02 cm3
x = 8.08 cm3 = 0.8 cm3 α = 0.01
State the null and alternative hypotheses.
: (1) H0 μ
: (2) H1 μ (Type integers or decimals. Do not round.)
Identify the tstatistic.
(Round to two decimal places as needed.)t0 =
Identify the Pvalue.
Pvalue (Round to three decimal places as needed.)=
Make a conclusion regarding the hypothesis.
(3) the null hypothesis. There (4) sufficient evidence to claim that the mean
hippocampal volume is (5) .cm3
≠ > < =
> = < ≠
12/12/2017 Section 10.3 HomeworkClayton Leach
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5.
(1)
(2)
(3)
(4)
The average daily volume of a computer stock in 2011 was million shares, according to a reliable source. A stock analyst believes that the stock volume in 2014 is different from the 2011 level. Based on a random sample of trading days in 2014, he finds the sample mean to be million shares, with a standard deviation of
s million shares. Test the hypotheses by constructing a % confidence interval. Complete parts (a) through (c) below.
μ = 35.1
30 31.5 = 14.7 95
(a) State the hypotheses for the test.
: (1) (2) million sharesH0 35.1
: (3) (4) million sharesH1 35.1
(b) Construct a % confidence interval about the sample mean of stocks traded in 2014.95
The lower bound is million shares. The upper bound is million shares. (Round to three decimal places as needed.)
(c) Will the researcher reject the null hypothesis?
A. the null hypothesis because million shares in the confidence interval.Reject μ = 35.1 falls B. the null hypothesis because million shares in the confidence
interval. Reject μ = 35.1 does not fall
C. the null hypothesis because million shares in the confidence interval. Do not reject μ = 35.1 falls
D. the null hypothesis because million shares in the confidence interval. Do not reject μ = 35.1 does not fall
p σ
μ
≠ > < =
σ
μ
p
= ≠ > <
12/12/2017 Section 10.3 HomeworkClayton Leach
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6. A math teacher claims that she has developed a review course that increases the scores of students on the math portion of a college entrance exam. Based on data from the administrator of the exam, scores are normally distributed with . The teacher obtains a random sample of students, puts them through the review class, and finds that the mean math score of the students is with a standard deviation of . Complete parts (a) through (d) below.
μ = 521 2200 2200 528 113
(a) State the null and alternative hypotheses. Let be the mean score. Choose the correct answer below.μ
A. , H : μ > 5210 H : μ ≠ 5211 B. , H : μ = 5210 H : μ > 5211 C. , H : μ = 5210 H : μ ≠ 5211 D. , H : μ < 5210 H : μ > 5211
(b) Test the hypothesis at the level of significance. Is a mean math score of statistically significantly higher than ? Conduct a hypothesis test using the Pvalue approach.
α = 0.10 528 521
Find the test statistic.
t0 = (Round to two decimal places as needed.)
Find the Pvalue.
The Pvalue is . (Round to three decimal places as needed.)
Is the sample mean statistically significantly higher?
No Yes
(c) Do you think that a mean math score of versus will affect the decision of a school admissions administrator? In other words, does the increase in the score have any practical significance?
528 521
No, because the score became only % greater.1.34 Yes, because every increase in score is practically significant.
(d) Test the hypothesis at the 0.10 level of significance with n students. Assume that the sample mean is still and the sample standard deviation is still . Is a sample mean of significantly more than ? Conduct a
hypothesis test using the Pvalue approach.
α = = 375 528 113 528 521
Find the test statistic.
t0 = (Round to two decimal places as needed.)
Find the Pvalue.
The Pvalue is . (Round to three decimal places as needed.)
Is the sample mean statistically significantly higher?
Yes No
What do you conclude about the impact of large samples on the Pvalue?
A. As n increases, the likelihood of not rejecting the null hypothesis increases. However, large samples tend to overemphasize practically significant differences.
B. As n increases, the likelihood of not rejecting the null hypothesis increases. However, large
12/12/2017 Section 10.3 HomeworkClayton Leach
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samples tend to overemphasize practically insignificant differences. C. As n increases, the likelihood of rejecting the null hypothesis increases. However, large
samples tend to overemphasize practically insignificant differences. D. As n increases, the likelihood of rejecting the null hypothesis increases. However, large
samples tend to overemphasize practically significant differences.
12/12/2017 Section 10.3 HomeworkClayton Leach
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7. Suppose the mean IQ score of people in a certain country is . Suppose the director of a college obtains a simple random sample of students from that country and finds the mean IQ is with a standard deviation of
. Complete parts (a) through (d) below.
101 39 104.4
13.1
(a) Consider the hypotheses : versus : . Explain what the director is testing. Perform the test at the level of significance. Write a conclusion for the test.
H0 μ = 101 H1 μ > 101 α = 0.01
Explain what the director is testing. Choose the correct answer below.
A. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually equal to .101
B. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually greater than .101
C. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually not greater than .101
D. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually not equal to .101
Find the test statistic for this hypothesis test.
(Round to two decimal places as needed.)
Find the Pvalue for this hypothesis test.
(Round to three decimal places as needed.)
Write a conclusion for the test. Choose the correct answer below.
A. . There sufficient evidence to conclude that the mean IQ score of people in the country is greater than at the level of significance. Do not reject H0 is
101 α = 0.01 B. . There sufficient evidence to conclude that the mean IQ score of people in
the country is greater than at the level of significance. Reject H0 is not
101 α = 0.01 C. . There sufficient evidence to conclude that the mean IQ score of
people in the country is greater than at the level of significance. Do not reject H0 is not
101 α = 0.01 D. . There sufficient evidence to conclude that the mean IQ score of people in the
country is greater than at the level of significance. Reject H0 is
101 α = 0.01
(b) Consider the hypotheses : versus : . Explain what the director is testing. Perform the test at the level of significance. Write a conclusion for the test.
H0 μ = 102 H1 μ > 102 α = 0.01
Explain what the director is testing. Choose the correct answer below.
A. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually not greater than .102
B. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually greater than .102
C. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually equal to .102
D. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually not equal to .102
Find the test statistic for this hypothesis test.
(Round to two decimal places as needed.)
Find the Pvalue for this hypothesis test.
(Round to three decimal places as needed.)
Write a conclusion for the test. Choose the correct answer below.
12/12/2017 Section 10.3 HomeworkClayton Leach
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A. . There sufficient evidence to conclude that the mean IQ score of people in the country is greater than at the level of significance. Do not reject H0 is
102 α = 0.01 B. . There sufficient evidence to conclude that the mean IQ score of
people in the country is greater than at the level of significance. Do not reject H0 is not
102 α = 0.01 C. . There sufficient evidence to conclude that the mean IQ score of people in the
country is greater than at the level of significance. Reject H0 is
102 α = 0.01 D. . There sufficient evidence to conclude that the mean IQ score of people in
the country is greater than at the level of significance. Reject H0 is not
102 α = 0.01
(c) Consider the hypotheses : versus : . Explain what the director is testing. Perform the test at the level of significance. Write a conclusion for the test.
H0 μ = 103 H1 μ > 103 α = 0.01
Explain what the director is testing. Choose the correct answer below.
A. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually greater than .103
B. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually equal to .103
C. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually not equal to .103
D. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually not greater than .103
Find the test statistic for this hypothesis test.
(Round to two decimal places as needed.)
Find the Pvalue for this hypothesis test.
(Round to three decimal places as needed.)
Write a conclusion for the test. Choose the correct answer below.
A. . There sufficient evidence to conclude that the mean IQ score of people in the country is greater than at the level of significance. Do not reject H0 is
103 α = 0.01 B. . There sufficient evidence to conclude that the mean IQ score of
people in the country is greater than at the level of significance. Do not reject H0 is not
103 α = 0.01 C. . There sufficient evidence to conclude that the mean IQ score of people in
the country is greater than at the level of significance. Reject H0 is not
103 α = 0.01 D. . There sufficient evidence to conclude that the mean IQ score of people in the
country is greater than at the level of significance. Reject H0 is
103 α = 0.01
(d) Based on the results of parts (a) (c), write a few sentences that explain the difference between "accepting" the statement in the null hypothesis versus "not rejecting" the statement in the null hypothesis.
–
One should (1) rather than (2) the null hypothesis. If one (3) the null
hypothesis, this indicates that the (4) is a specific value, such as , , or , and so the same
data have been used to conclude that the (5) is three different values. If one (6) the
null hypothesis, this indicates that the (7) could be , , or or even some other value; we are
simply not ruling them out as the value of the (8) Therefore, (9) the null hypothesis
can lead to contradictory conclusions, whereas (10) does not.
101 102 103
101 102 103
12/12/2017 Section 10.3 HomeworkClayton Leach
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8.
9.
(1) not reject accept
(2) not reject accept
(3) does not reject accepts
(4) sample mean population mean
(5) population mean sample mean
(6) accepts does not reject
(7) sample mean population mean
(8) sample mean. population mean.
(9) accepting not rejecting
(10) not rejecting accepting
The head of institutional research at a university believed that the mean age of fulltime students was declining. In 1995, the mean age of a fulltime student was known to be 27.4 years. After looking at the enrollment records of all 4934 fulltime students in the current semester, he found that the mean age was 27.1 years, with a standard deviation of 7.3 years. He conducted a hypothesis of : 27.4 years versus : 27.4 years and obtained a Pvalue of 0.0020. He concluded that the mean age of fulltime students did decline. Is there anything wrong with his research?
H0 μ = H1 μ <
Choose the correct answer below.
A. Yes, the head of institutional research stated the hypotheses incorrectly; a lefttailed hypothesis test was conducted instead of a twotailed test.
B. Yes, the head of institutional research has access to the entire population, inference is unnecessary. He can say with 100% confidence that the mean age has decreased.
C. Yes, a Pvalue only indicates the likelihood of getting a result as extreme or more extreme as the one found, the head of institutional research needs to include a confidence level.
D. No, the hypothesis test was conducted correctly, and the correct conclusion was made.
Explain the difference between statistical significance and practical significance.
Choose the correct answer below.
A. Statistical significance means that the hypothesis test being performed is useful for building theoretical foundations for other statistical work. Practical significance means that the particular application of the hypothesis test is of great importance to the real world.
B. Statistical significance means that the sample statistic is not likely to come from the population whose parameter is stated in the null hypothesis. Practical significance refers to whether the difference between the sample statistic and the parameter stated in the null hypothesis is large enough to be considered important in an application.
C. Statistical significance refers to the type of hypothesis test needed to analyze a population, with some tests being more important than Z tests. Practical significance refers to how difficult a desired hypothesis test is to perform in an application, with some tests being easier to perform than others.
D. Statistical significance refers to how an unusual event is unlikely to actually appear in a real world application, such as every entry in a sample of size 50 having the same value. Practical significance refers to how an unusual event is likely to actually appear in a real world application, such as a rejection of a null hypothesis using data that looks feasible.
12/12/2017 Section 10.2 HomeworkClayton Leach
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5.
(1) Do not reject Reject
(2) less greater
Test the hypothesis using the Pvalue approach. Be sure to verify the requirements of the test.
: p versus : pH0 = 0.7 H1 > 0.7 n ; x , = 200 = 150 α = 0.01
Is 10?np 1 − p0 0 ≥
No Yes
Use technology to find the Pvalue.
Pvalue = (Round to three decimal places as needed.)
(1) the null hypothesis, because the Pvalue is (2) than .α
12/12/2017 Section 10.2 HomeworkClayton Leach
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6.
(1)
(2) greater than less than
(3) is given to be random, can be reasonably assumed to be random,
cannot be reasonably assumed to be random, is given to not be random,
(4) are are not
(5)
(6)
(7)
(8)
Twenty years ago, % of parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. A recent survey found that of parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. Do parents feel differently today than they did twenty years ago? Use the level of significance.
51 222 750
α = 0.01
Because (1) 10, the sample size is (2) 5% of the
population size, and the sample (3) the requirements for testing the hypothesis
(4) satisfied.
np 1 − p0 0 =
(Round to one decimal place as needed.)
What are the null and alternative hypotheses?
: (5) (6) versus
: (7) (8)
H0
H1 (Type integers or decimals. Do not round.)
Find the test statistic.
(Round to two decimal places as needed.) z0 =
Find the Pvalue.
Pvalue (Round to three decimal places as needed.) =
Determine the conclusion for this hypothesis test. Choose the correct answer below.
A. Since Pvalue , the null hypothesis and conclude that there sufficient evidence that parents feel differently today.
< α reject is
B. Since Pvalue , the null hypothesis and conclude that there sufficient evidence that parents feel differently today.
> α reject is not
C. Since Pvalue , the null hypothesis and conclude that there sufficient evidence that parents feel differently today.
> α do not reject is not
D. Since Pvalue , the null hypothesis and conclude that there sufficient evidence that parents feel differently today.
< α do not reject is
> < = ≠
p σ
μ
> < = ≠
μ
σ
p
> = ≠ <
12/12/2017 Section 10.2 HomeworkClayton Leach
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7.
(1)
(2)
Several years ago, % of parents who had children in grades K12 were satisfied with the quality of education the students receive. A recent poll asked parents who have children in grades K12 if they were satisfied with the quality of education the students receive. Of the surveyed, indicated that they were satisfied. Construct a % confidence interval to assess whether this represents evidence that parents' attitudes toward the quality of
education have changed.
47 1,165
1,165 485 95
What are the null and alternative hypotheses?
: p (1) versus : p (2) H0 H1 (Round to two decimal places as needed.)
Use technology to find the % confidence interval.95
The lower bound is . The upper bound is . (Round to two decimal places as needed.)
What is the correct conclusion?
A. Since the interval the proportion stated in the null hypothesis, there is evidence that parents' attitudes toward the quality of education have changed.
contains insufficient
B. Since the interval the proportion stated in the null hypothesis, there is evidence that parents' attitudes toward the quality of education have changed.
does not contain sufficient
C. Since the interval the proportion stated in the null hypothesis, there is evidence that parents' attitudes toward the quality of education have changed.
does not contain insufficient
D. Since the interval the proportion stated in the null hypothesis, there is evidence that parents' attitudes toward the quality of education have changed.
contains sufficient
> < = ≠
= ≠ > <
12/12/2017 Section 10.2 HomeworkClayton Leach
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8.
1: Standard Normal Distribution Table (page 1)
In a survey, % of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so randomly selected pet owners and discovered that of them spoke to their pet on the telephone. Does the veterinarian have a right to be skeptical? Use the level of significance.
41 he 170 64
α = 0.1 Click here to view the standard normal distribution table (page 1).1 Click here to view the standard normal distribution table (page 2).2
Because (1) 10, the sample size is (2) 5% of the
population size, and the sample (3) the requirements for testing the hypothesis
(4) satisfied.
np0 1 − p0 =
(Round to one decimal place as needed.)
What are the null and alternative hypotheses?
: (5) (6) versus
: (7) (8)
H0
H1 (Type integers or decimals. Do not round.)
Determine the test statistic, .z0
(Round to two decimal places as needed.)z0 =
Determine the critical value(s). Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.)
A. zα =
B. ± zα / 2 = ±
Does the veterinarian have a right to be skeptical?
A. The veterinarian a right to be skeptical. There sufficient evidence to conclude that the true proportion of pet owners who talk to their pets on the telephone %.
has is is not 41
B. The veterinarian a right to be skeptical. There sufficient evidence to conclude that the true proportion of pet owners who talk to their pets on the telephone is %.
does not have is not 41
C. The veterinarian a right to be skeptical. There sufficient evidence to conclude that the true proportion of pet owners who talk to their pets on the answering machine or telephone is less than %.
does not have is not
41 D. The veterinarian a right to be skeptical. There sufficient evidence to conclude that the
true proportion of pet owners who talk to their pets on the telephone is less than %. has is
41
12/12/2017 Section 10.2 HomeworkClayton Leach
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2: Standard Normal Distribution Table (page 2)
12/12/2017 Section 10.2 HomeworkClayton Leach
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(1)
(2) less than greater than
(3) is given to not be random, can be reasonably assumed to be random,
is given to be random, cannot be reasonably assumed to be random,
(4) are are not
(5)
(6)
(7)
(8)
< = ≠ >
σ
μ
p
≠ = > <
μ
σ
p
≠ > = <
12/12/2017 Section 10.2 HomeworkClayton Leach
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9. A blind taste test is conducted to determine which of two colas, Brand A or Brand B, individuals prefer. Individuals are randomly asked to drink one of the two types of cola first, followed by the other cola, and then asked to disclose the drink they prefer. Results of the taste test indicate that of individuals prefer Brand A. Complete parts a through c.
57 100
(a) Conduct a hypothesis test (preferably using technology) : versus : p for , , , ..., , , at the 0.05 level of significance. For which values of do you not reject the null hypothesis?
What do each of the values of represent?
H0 p = p0 H1 ≠ p0 p0 = 0.46 0.47 0.48 0.66 0.67 0.68 α = p0
p0
Do not reject the null hypothesis for the values of between and , inclusively.p0 (Type integers or decimals as needed.)
(b) Construct a 95% confidence interval for the proportion of individuals who prefer Brand A.
The lower bound is . The upper bound is . (Round to three decimal places as needed.)
(c) Suppose you changed the level of significance in conducting the hypothesis test to 0.01. What would happen to the range of values for for which the null hypothesis is not rejected? Why does this make sense? Choose the correct answer below.
α = p0
A. The range of values would increase because the corresponding confidence interval would increase in size.
B. The range of values would decrease because the corresponding confidence interval would decrease in size.
C. The range of values would decrease because the corresponding confidence interval would increase in size.
D. The range of values would increase because the corresponding confidence interval would decrease in size.
12/12/2017 Section 10.2 HomeworkClayton Leach
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10. Two professors at a local college developed a new teaching curriculum designed to increase students' grades in math classes. In a typical developmental math course, % of the students complete the course with a letter grade of A, B, or C. In the experimental course, of the students enrolled, completed the course with a letter grade of A, B, or C. Is the experimental course effective at the level of significance? Complete parts (a) through (g).
51 18 13
α = 0.025
(a) State the appropriate null and alternative hypotheses.
: (1) (2) versus
: (3) (4)
H0
H1 (Type integers or decimals. Do not round.)
(b) Verify that the normal model may not be used to estimate the Pvalue.
Because (5) 10, the normal model (6) be used to approximate the Pvalue.
np 1 − p0 0 =
(Round to one decimal place as needed.)
(c) Explain why this is a binomial experiment.
There is a (7) number of trials with (8) The trials (9) independent and the probability of success is fixed at for each trial. (Type an integer or a decimal. Do not round.)
(d) Determine the Pvalue using the binomial probability distribution. State your conclusion to the hypothesis test.
First determine the Pvalue.
Pvalue (Round to three decimal places as needed.)=
Is there sufficient evidence to support the research that the experimental course is effective?
A. , the null hypothesis because the Pvalue is than . There evidence to conclude that the experimental course is effective. Yes reject less α is sufficient
B. , the null hypothesis because the Pvalue is than . There evidence to conclude that the experimental course is effective.
No do not reject greater α is insufficient
C. , the null hypothesis because the Pvalue is than . There evidence to conclude that the experimental course is effective. No reject less α is insufficient
D. , the null hypothesis because the Pvalue is than . There evidence to conclude that the experimental course is effective.
Yes do not reject greater α is sufficient
(e) Suppose the course is taught with students and complete the course with a letter grade of A, B, or C. Verify whether the normal model may now be used to estimate the Pvalue.
54 39
Because (10) 10, the sample size is (11) 5% of the
population size, and the sample (12) the normal model (13) be used to approximate the Pvalue.
np 1 − p0 0 =
(Round to one decimal place as needed.)
(f) Use the normal model to obtain and interpret the Pvalue. State your conclusion to the hypothesis test.
First find the test statistic, .z0
(Round to two decimal places as needed.)z0 =
Now determine the Pvalue.
Pvalue (Round to three decimal places as needed.)=
Is there sufficient evidence to support the research that the experimental course is effective?
12/12/2017 Section 10.2 HomeworkClayton Leach
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(1)
(2)
(3)
(4)
(5)
(6) may not may
(7) fixed variable
(8) one outcome. three outcomes. two mutually exclusive outcomes. two outcomes that can occur simultaneously.
(9) are are not
(10)
(11) less than greater than
(12) is given to not be random, cannot be reasonably assumed to be random, can be reasonably assumed to be random,
is given to be random,
(13) may may not
(14) insignificant. substantial.
(15) do not reject reject
A. , the null hypothesis because the Pvalue is than . There is evidence to conclude that the experimental course is effective.
No do not reject less α insufficient
B. , the null hypothesis because the Pvalue is than . There is evidence to conclude that the experimental course is effective. Yes reject less α sufficient
C. , the null hypothesis because the Pvalue is than . There is evidence to conclude that the experimental course is effective.
No do not reject greater α insufficient
D. , the null hypothesis because the Pvalue is than . There is evidence to conclude that the experimental course is effective. Yes reject greater α sufficient
(g) Explain the role that sample size plays in the ability to reject statements in the null hypothesis.
When there are small sample sizes, the evidence against the statement in the null hypothesis must be
(14) One should be wary of studies that (15) the null hypothesis when the test was conducted with a small sample size.
σ
p μ
> ≠ < =
p σ
μ
≠ < > =
< > = ≠
≠ > < =
12/12/2017 Section 10.2 HomeworkClayton Leach
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11.
(1)
(2)
(3)
(4)
(5)
(6) may may not
Previously, % of workers had a travel time to work of more than 60 minutes. An urban economist believes that the percentage has increased since then. She randomly selects workers and finds that of them have a travel time to work that is more than 60 minutes. Test the economist's belief at the level of significance.
6.6 55 13
α = 0.05
What are the null and alternative hypotheses?
: (1) (2) versus
: (3) (4)
H0
H1 (Type integers or decimals. Do not round.)
Because (5) 10, the normal model (6) be used to approximate the Pvalue.
np0 1 − p0 =
(Round to one decimal place as needed.)
Find the Pvalue.
Pvalue (Round to three decimal places as needed.)=
Is there sufficient evidence to support the economist's belief?
A. , the null hypothesis. There sufficient evidence because the Pvalue is than .
Yes do not reject is greater α
B. , the null hypothesis. There sufficient evidence because the Pvalue is than .
No do not reject is not greater α
C. , the null hypothesis. There sufficient evidence because the Pvalue is than . No reject is not less
α
D. , the null hypothesis. There sufficient evidence because the Pvalue is than .
Yes reject is less α
α
p μ
= > < ≠
p α
μ
≠ = < >
= < > ≠
12/12/2017 Section 9.3 HomeworkClayton Leach
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4. A simple random sample of size n is drawn from a population that is known to be normally distributed. The sample variance, , is determined to be . Complete parts (a) through (c).s2 12.8
(a) Construct a 90% confidence interval for if the sample size, n, is 20.σ2
The lower bound is . (Round to two decimal places as needed.)
The upper bound is . (Round to two decimal places as needed.)
(b) Construct a 90% confidence interval for if the sample size, n, is 30.σ2
The lower bound is . (Round to two decimal places as needed.)
The upper bound is . (Round to two decimal places as needed.)
How does increasing the sample size affect the width of the interval?
The width increases The width decreases The width does not change
(c) Construct a 98% confidence interval for if the sample size, n, is 20.σ2
The lower bound is . (Round to two decimal places as needed.)
The upper bound is . (Round to two decimal places as needed.)
Compare the results with those obtained in part (a). How does increasing the level of confidence affect the confidence interval?
The width decreases The width increases The width does not change
12/12/2017 Section 9.3 HomeworkClayton Leach
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5.
1: ChiSquare Distribution Critical Values Table
The following data represent the age (in weeks) at which babies first crawl based on a survey of mothers. The data are normally distributed and s weeks. Construct and interpret a % confidence interval for the population standard deviation of the age (in weeks) at which babies first crawl.
12 = 10.247 99
54 29 43 35 38 26 46 37 56 25 39 29
Click the icon to view the table of critical values of the chisquare distribution.1
Select the correct choice below and fill in the answer boxes to complete your choice. (Use ascending order. Round to three decimal places as needed.)
A. There is a % chance that the true population standard deviation is between and .
99
B. If repeated samples are taken, % of them will have the sample standard deviation between and .
99
C. There is % confidence that the population standard deviation is between and .
99
12/12/2017 Section 9.3 HomeworkClayton Leach
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12/12/2017 Section 9.3 HomeworkClayton Leach
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6.
2: Peanut jar data
3: Chi Square Distribution Table
A jar of peanuts is supposed to have ounces of peanuts. The filling machine inevitably experiences fluctuations in filling, so a qualitycontrol manager randomly samples 12 jars of peanuts from the storage facility and measures their contents. She obtains the accompanying data. Complete parts (a) through (d) below.
20
Click here to view the peanut jar data.2 Click here to view the table of critical values of the chisquare distribution.3 Click here to view the standard normal distribution table (page 1).4 Click here to view the standard normal distribution table (page 2).5
(a) Verify that the data are normally distributed by constructed a normal probability plot. Choose the correct normal probability plot below.
A.
19.2 20.6 1.75
1.75
observation
expected zscore
B.
19.2 20.6 1.75
1.75
observation
expected zscore
C.
1.75 1.75 19.2
20.6
observation
expected zscore
D.
19.2 20.6 1.75
1.75
observation
expected zscore
Are the data normally distributed?
Yes No
(b) Determine the sample standard deviation.
s (Round to three decimal places as needed.)=
(c) Construct a % confidence interval for the population standard deviation of the number of ounces of peanuts. Select the correct choice below and fill in the answer boxes to complete your choice.
95
(Use ascending order. Round to three decimal places as needed.)
A. If repeated samples are taken, % of them will have the sample standard deviation between and .
95
B. There is % confidence that the population standard deviation is between and .
95
C. There is a % chance that the true population standard deviation is between and .
95
(d) The quality control manager wants the machine to have a population standard deviation below ounce. Does the confidence interval validate this desire?
0.20
A. Yes the lower bound of the confidence interval is greater than .— 0.20 B. Yes the lower bound of the confidence interval is less than .— 0.20 C. No the lower bound of the confidence interval is less than .— 0.20 D. No the lower bound of the confidence interval is greater than .— 0.20
19.95 19.74 20.23 19.29 19.82 19.82 19.54 20.15 19.81 19.54 20.28 20.59
12/12/2017 Section 9.3 HomeworkClayton Leach
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4: Standard Normal Distribution Table (page 1)
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5: Standard Normal Distribution Table (page 2)
12/12/2017 Section 9.3 HomeworkClayton Leach
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12/12/2017 Section 9.2 HomeworkClayton Leach
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8. A researcher wishes to estimate the average blood alcohol concentration (BAC) for drivers involved in fatal accidents who are found to have positive BAC values. He randomly selects records from such drivers in 2009 and determines the sample mean BAC to be g/dL with a standard deviation of g/dL. Complete parts (a) through (d) below.
60 0.16 0.080
(a) A histogram of blood alcohol concentrations in fatal accidents shows that BACs are highly skewed right. Explain why a large sample size is needed to construct a confidence interval for the mean BAC of fatal crashes with a positive BAC.
A. Since the distribution of blood alcohol concentrations is normally distributed, the sample must be large so that the distribution of the sample mean will be approximately normal.
B. Since the distribution of blood alcohol concentrations is normally distributed, the sample must be large to ensure that the sample size is greater than 5% of the population.
C. Since the distribution of blood alcohol concentrations is not normally distributed (highly skewed right), the sample must be large to ensure that the sample size is greater than 5% of the population.
D. Since the distribution of blood alcohol concentrations is not normally distributed (highly skewed right), the sample must be large so that the distribution of the sample mean will be approximately normal.
(b) Recently there were approximately 25,000 fatal crashes in which the driver had a positive BAC. Explain why this, along with the fact that the data were obtained using a simple random sample, satisfies the requirements for constructing a confidence interval.
A. The sample size is likely greater than 10% of the population. B. The sample size is likely less than 10% of the population. C. The sample size is likely greater than 5% of the population. D. The sample size is likely less than 5% of the population.
(c) Determine and interpret a 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC. (Use ascending order. Round to three decimal places as needed.)
A. The lower bound is and the upper bound is . The researcher is 10% confident that the population mean BAC is in the confidence interval for drivers involved in fatal accidents who have a positive BAC value.
B. The lower bound is and the upper bound is . The researcher is 90% confident that the population mean BAC is not in the confidence interval for drivers involved in fatal accidents who have a positive BAC value.
C. The lower bound is and the upper bound is . The researcher is 90% confident that the population mean BAC is in the confidence interval for drivers involved in fatal accidents who have a positive BAC value.
(d) All areas of the country use a BAC of g/dL as the legal intoxication level. Is it possible that the mean BAC of all drivers involved in fatal accidents who are found to have positive BAC values is less than the legal intoxication level? Explain.
0.08
A. No, it is not possible that the mean BAC is less than g/dL, because it is possible that the true mean is not captured in the confidence interval.
0.08
B. Yes, it is possible that the mean BAC is less than g/dL, because it is possible that the true mean is not captured in the confidence interval, but it is not likely.
0.08
C. Yes, it is possible that the mean BAC is less than g/dL, because it is possible that the true mean is not captured in the confidence interval and it is highly probable.
0.08
D. No, it is not possible that the mean BAC is less than g/dL, because it is possible that the true mean is not captured in the confidence interval, but it is not likely.
0.08
12/12/2017 Section 9.2 HomeworkClayton Leach
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10. The data shown below represent the repair cost for a lowimpact collision in a simple random sample of mini and microvehicles. Complete parts (a) through (d) below.
$3148 $1016 $743 $663 $773 $1758 $3345 $2027 $2633 $1386 Click here to view the table of critical tvalues. 6 Click here to view page 1 of the standard normal distribution table.7 Click here to view page 2 of the standard normal distribution table.8
(a) Draw a normal probability plot to determine if it is reasonable to conclude the data come from a population that is normally distributed. Choose the correct answer below.
A.
0 2000 4000 2
1
0
1
2
Repair Cost ($)
E xp ec te d z sc or e
B.
0 2000 4000 2
1
0
1
2
Repair Cost ($)
E xp ec te d z sc or e
C.
0 2000 4000 4
2
0
2
4
Repair Cost ($) E xp ec te d z sc or e
D.
0 2000 4000 4
2
0
2
4
Repair Cost ($)
E xp ec te d z sc or e
Is it reasonable to conclude that the data come from a population that is normally distributed?
A. No, because the plotted values are not linear. B. Yes, because the plotted values are not linear. C. No, because there are not enough values to make a determination. D. Yes, because the plotted values are approximately linear.
(b) Draw a boxplot to check for outliers. Choose the correct answer below.
A.
0 2000 4000
B.
0 2000 4000
C.
0 2000 4000
D.
0 2000 4000
Does the boxplot suggest that there are outliers?
A. Yes, there is at least one point that is outside of the 1.5(IQR) boundary. B. No, there are no points that are outside of the 1.5(IQR) boundary. C. No, there are no points that are greater than the third quartile or less than the first quartile. D. Yes, there is at least one point that is greater than the third quartile or less than the first
quartile.
(c) Construct and interpret a % confidence interval for population mean cost of repair. Select the correct choice and fill in the answer boxes to complete your choice.
95
(Round to one decimal place as needed.)
A. The lower bound is $ and the upper bound is $ . We are % confident that the mean cost of repair is within the confidence interval.
95
B. The lower bound is $ and the upper bound is $ . We are % confident that the mean cost of repair is outside of the confidence interval.
95
(d) Suppose you obtain a simple random sample of size n of a specific type of minivehicle that was in a lowimpact collision and determine the cost of repair. Do you think a % confidence interval would be wider or narrower? Explain.
= 10 95
12/12/2017 Section 9.2 HomeworkClayton Leach
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6: Table of Critical tValues
A. Wider, because taking a second random sample will always lead to a wider confidence interval.
B. Narrower, because taking a second random sample will always lead to a narrower confidence interval.
C. Narrower, because there is less variability in the data because any variability caused by the different types of vehicles has been removed.
D. Wider, because there is more variability in the data because variability in the repair cost of the car has been added.
12/12/2017 Section 9.2 HomeworkClayton Leach
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7: Standard Normal Distribution (Page 1)
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8: Standard Normal Distribution (Page 2)
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11.
9: First Sample Volumes (millions of shares traded)
The trade volume of a stock is the number of shares traded on a given day. The data in the first accompanying table, in millions (so that represents shares traded), represent the volume of a certain stock traded for a random sample of trading days in a certain year. A second random sample of days in the same year resulted in the data in second accompanying table. Complete parts (a) through (d) below.
6.06 6,060,000 40 40
Click here to view the data for the first sample.9 Click here to view the data for the second sample. 10 Click here to view the table of critical tvalues.11
(a) Use the data from the first sample to compute a point estimate for the population mean number of shares traded per day in the year.
A point estimate for the population mean number of shares traded per day in the year is million. (Round to three decimal places as needed.)
(b) Using the data from the first sample, construct a % confidence interval for the population mean number of shares traded per day in the certain year. Interpret the confidence interval.
95
Select the correct choice below and fill in the answer boxes to complete your choice. (Round to three decimal places as needed.)
A. There is a % probability that the mean number of shares of stock traded per day in the specified year is between million and million.
B. One can be % confident that the mean number of shares of the stock traded per day in the specified year is between million and million.
C. One can be % confident that the mean number of shares of the stock traded in all days of the specified year are between million and million.
D. The number of shares of the stock traded per day is between million and million for % of all days in the specified year.
(c) Using the data from the second sample, construct another % confidence interval for the population mean number of shares traded per day in the year. Interpret the confidence interval.
95
Select the correct choice below and fill in the answer boxes to complete your choice. (Round to three decimal places as needed.)
A. The number of shares of the stock traded per day is between million and million for % of all days in the specified year
B. One can be % confident that the mean number of shares of the stock traded in all days of the specified year are between million and million.
C. There is a % probability that the mean number of shares of stock traded per day in the specified year is between million and million.
D. One can be % confident that the mean number of shares of the stock traded per day in the specified year is between million and million.
(d) Explain why the confidence intervals obtained in parts (b) and (c) are different.
A. They are different because the samples have different sizes. Because one sample was larger than another, the confidence from the larger sample is narrower.
B. They are different because they represent different populations. The populations have different means and standard deviations that lead to different confidence intervals.
C. They are different because there is an error in at least one of the samples. If the samples were obtained correctly, then the confidence intervals would be the same.
D. They are different because of variation in sampling. The samples have different means and standard deviations that lead to different confidence intervals.
12/12/2017 Section 9.2 HomeworkClayton Leach
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10: Second Sample Volumes (millions of shares traded)
11: Table of Critical tValues
6.06 4.18 6.22 5.92 3.84 2.88 4.62 2.98 3.24 8.98 4.60 2.94 6.52 5.78 3.32 4.56 3.16 3.90 3.34 5.38 2.94 2.20 6.12 6.08 5.66 3.60 3.00 9.10 5.68 3.94 4.04 3.50 3.74 3.92 4.06 5.38 3.62 3.70 3.28 1.64
2.86 5.70 6.86 9.12 8.98 3.84 3.04 7.16 10.38 4.76 6.64 4.88 4.44 5.48 2.74 3.20 3.42 3.28 4.40 2.86 2.96 4.10 7.50 6.60 5.38 4.88 3.34 2.76 6.24 3.38 3.58 4.10 7.18 3.66 4.40 3.08 1.68 4.38 3.38 3.54
12/12/2017 Section 9.2 HomeworkClayton Leach
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12/12/2017 Section 9.2 HomeworkClayton Leach
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12.
12: Table of Critical tValues
A study asked respondents, "If ever married, how old were you when you first married?" The results are summarized in the technology excerpt that follows. Complete parts (a) through (d) below.
OneSample T: AGEWED Variable N Mean StDev SE Mean .0% CI90 AGEWED 26690 022.52 4.798 0.029 ( , )22.472 22.568
Click the icon to view the table of critical tvalues.12
(a) Use the summary to determine the point estimate of the population mean and margin of error for the confidence interval.
(Type an integer or a decimal. Do not round.)x =
E (Type an integer or a decimal. Do not round.)=
(b) Interpret the confidence interval.
A. There is a % probability that the mean age of people when first married is 0 years.90 22.52 B. There is a % probability that the mean age of people when first married is between
and years. 90 22.472
22.568 C. One can be % confident that the mean age of people when first married is 0 years.90 22.52 D. One can be % confident that the mean age of people when first married is between
and years. 90
22.472 22.568
(c) Verify the results by computing a % confidence interval with the information provided.90
Begin with the lower bound.
(1) • 4.798
=
(Round to three decimal places as needed.)
Next find the upper bound.
(2) • 4.798
=
(Round to three decimal places as needed.)
(d) Why is the margin of error for this confidence interval so small?
A. The mean is very small. B. The standard deviation is very large. C. The standard deviation is very small. D. The sample size used is very large.
12/12/2017 Section 9.2 HomeworkClayton Leach
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13.
14.
(1)
(2)
− +
+ −
A doctor wants to estimate the mean HDL cholesterol of all 20 to 29yearold females. How many subjects are needed to estimate the mean HDL cholesterol within points with confidence assuming based on earlier studies? Suppose the doctor would be content with confidence. How does the decrease in confidence affect the sample size required?
3 99% s = 17.3 95%
A 99% confidence level requires subjects. (Round up to the nearest subject.)
A confidence level requires subjects. (Round up to the nearest subject.)95%
How does the decrease in confidence affect the sample size required?
A. The sample size is the same for all levels of confidence. B. Decreasing the confidence level decreases the sample size needed. C. Decreasing the confidence level increases the sample size needed.
People were polled on how many books they read the previous year. Initial survey results indicate that s books. Complete parts (a) through (d) below.
= 16.7
(a) How many subjects are needed to estimate the mean number of books read the previous year within books with % confidence?
four 95
This confidence level requires subjects. (Round up to the nearest subject.)95%
(b) How many subjects are needed to estimate the mean number of books read the previous year within books with % confidence?
two 95
This confidence level requires subjects. (Round up to the nearest subject.)95%
(c) What effect does doubling the required accuracy have on the sample size?
A. Doubling the required accuracy nearly halves the sample size. B. Doubling the required accuracy nearly quarters the sample size. C. Doubling the required accuracy nearly quadruples the sample size. D. Doubling the required accuracy nearly doubles the sample size.
(d) How many subjects are needed to estimate the mean number of books read the previous year within books with % confidence?
four 99
This % confidence level requires subjects. (Round up to the nearest subject.)99
Compare this result to part (a). How does increasing the level of confidence in the estimate affect sample size? Why is this reasonable?
A. Increasing the level of confidence decreases the sample size required. For a fixed margin of error, greater confidence can be achieved with a larger sample size.
B. Increasing the level of confidence increases the sample size required. For a fixed margin of error, greater confidence can be achieved with a smaller sample size.
C. Increasing the level of confidence decreases the sample size required. For a fixed margin of error, greater confidence can be achieved with a smaller sample size.
D. Increasing the level of confidence increases the sample size required. For a fixed margin of error, greater confidence can be achieved with a larger sample size.
12/12/2017 Section 9.2 HomeworkClayton Leach
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15.
(1) greater than or equal to 0. less than or equal to 1. between 0 and 1, inclusive. greater than 0 and less than 1.
In the following probability distribution, the random variable x represents the number of activities a parent of a to grade student is involved in. Complete parts (a) through (f) below.
6th- 8th
x 0 1 2 3 4 P(x) 0.091 0.263 0.333 0.139 0.174
(a) Verify that this is a discrete probability distribution.
This is a discrete probability distribution because the sum of the probabilities is and each probability
is (1)
(b) Graph the discrete probability distribution. Choose the correct graph below.
A.
0 1 2 3 4 0
0.1
0.2 0.3
0.4 0.5
B.
0 1 2 3 4 0
0.1
0.2 0.3
0.4 0.5
C.
0 1 2 3 4 0
0.1
0.2 0.3
0.4 0.5
D.
0 1 2 3 4 0
0.1
0.2 0.3
0.4 0.5
(c) Compute and interpret the mean of the random variable x.
The mean is activities. (Type an integer or a decimal.)
Which of the following interpretations of the mean is correct?
A. As the number of experiments increases, the mean of the observations will approach the mean of the random variable.
B. The observed value of an experiment will be less than the mean of the random variable in most experiments.
C. As the number of experiments decreases, the mean of the observations will approach the mean of the random variable.
D. The observed value of an experiment will be equal to the mean of the random variable in most experiments.
(d) Compute the standard deviation of the random variable x.
The standard deviation is activities. (Round to one decimal place as needed.)
(e) What is the probability that a randomly selected student has a parent involved in three activities?
The probability is . (Type an integer or a decimal.)
(f) What is the probability that a randomly selected student has a parent involved in three or four activities?
The probability is . (Type an integer or a decimal.)
12/12/2017 Section 8.2 HomeworkClayton Leach
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3. Suppose a simple random sample of size n is obtained from a population whose size is and whose population proportion with a specified characteristic is Complete parts (a) through (c) below.
= 50 N= 25,000 p = 0.6.
(a) Describe the sampling distribution of .p Choose the phrase that best describes the shape of the sampling distribution below.
A. Not normal because and n ≤ 0.05N np(1 − p) ≥ 10. B. Approximately normal because and n ≤ 0.05N np(1 − p) ≥ 10. C. Not normal because and n ≤ 0.05N np(1 − p) < 10. D. Approximately normal because and n ≤ 0.05N np(1 − p) < 10.
Determine the mean of the sampling distribution of .p
(Round to one decimal place as needed.)μ = p
0.6
Determine the standard deviation of the sampling distribution of .p
(Round to six decimal places as needed.)σ p = 0.069282
(b) What is the probability of obtaining x or more individuals with the characteristic? That is, what is P( )?= 32 p ≥ 0.64
P( ) (Round to four decimal places as needed.)p ≥ 0.64 =
(c) What is the probability of obtaining x or fewer individuals with the characteristic? That is, what is P( )?= 27 p ≤ 0.54
P( ) (Round to four decimal places as needed.)p ≤ 0.54 =
12/12/2017 Section 8.2 HomeworkClayton Leach
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4. According to a study conducted by a statistical organization, the proportion of people who are satisfied with the way things are going in their lives is . Suppose that a random sample of people is obtained. Complete parts (a) through (e) below.
0.78 100
(a) Suppose the random sample of people is asked, "Are you satisfied with the way things are going in your life?" Is the response to this question qualitative or quantitative? Explain.
100
A. The response is qualitative because the responses can be classified based on the characteristic of being satisfied or not.
B. The response is quantitative because the responses can be classified based on the characteristic of being satisfied or not.
C. The response is qualitative because the responses can be measured numerically and the values added or subtracted, providing meaningful results.
D. The response is quantitative because the responses can be measured numerically and the values added or subtracted, providing meaningful results.
(b) Explain why the sample proportion, , is a random variable. What is the source of the variability?p
A. The sample proportion is a random variable because the value of varies from sample to sample. The variability is due to the fact that different people feel differently regarding their satisfaction.
p p
B. The sample proportion is a random variable because the value of varies from sample to sample. The variability is due to the fact that people may not be responding to the question truthfully.
p p
C. The sample proportion is a random variable because the value of represents a random person included in the sample. The variability is due to the fact that different people feel differently regarding their satisfaction.
p p
D. The sample proportion is a random variable because the value of represents a random person included in the sample. The variability is due to the fact that people may not be responding to the question truthfully.
p p
(c) Describe the sampling distribution of , the proportion of people who are satisfied with the way things are going in their life. Be sure to verify the model requirements.
p
Since the sample size is no (1) than 5% of the population size and
np(1 p) 10, the distribution of is (2) with and
.
− = ≥ p μ = p
σ = p
(Round to three decimal places as needed.)
(d) In the sample obtained in part (a), what is the probability that the proportion who are satisfied with the way things are going in their life exceeds 0. ?80
The probability that the proportion who are satisfied with the way things are going in their life exceeds 0. is .
80
(Round to four decimal places as needed.)
(e) Using the distribution from part (c), would it be unusual for a survey of 100 people to reveal that or fewer people in the sample are satisfied with their lives?
70
The probability that or fewer people in the sample are satisfied is , which
(3) unusual because this probability (4) less than (5) %.
70
(Round to four decimal places as needed.)
12/12/2017 Section 8.2 HomeworkClayton Leach
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5.
6.
(1) more less
(2) approximately normal skewed left uniform
skewed right
(3) is is not
(4) is is not
(5) 5 0.5 50 0.05
Exit polling is a popular technique used to determine the outcome of an election prior to results being tallied. Suppose a referendum to increase funding for education is on the ballot in a large town (voting population over 100,000). An exit poll of voters finds that voted for the referendum. How likely are the results of your sample if the population proportion of voters in the town in favor of the referendum is ? Based on your result, comment on the dangers of using exit polling to call elections.
200 102 0.48
How likely are the results of your sample if the population proportion of voters in the town in favor of the referendum is ?0.48
The probability that than people voted for the referendum is .more 102 (Round to four decimal places as needed.)
Comment on the dangers of using exit polling to call elections. Choose the correct answer below.
A. The result unusual because the probability that is equal to or more extreme than the sample proportion is than 5%. Thus, it unusual for a wrong call to be made in an election if exit polling alone is considered.
is not p greater is not
B. The result unusual because the probability that is equal to or more extreme than the sample proportion is than 5%. Thus, it unusual for a wrong call to be made in an election if exit polling alone is considered.
is p greater is not
C. The result unusual because the probability that is equal to or more extreme than the sample proportion is than 5%. Thus, it unusual for a wrong call to be made in an election if exit polling alone is considered.
is p less is
D. The result unusual because the probability that is equal to or more extreme than the sample proportion is than 5%. Thus, it unusual for a wrong call to be made in an election if exit polling alone is considered.
is not p less is
A researcher studying public opinion of proposed Social Security changes obtains a simple random sample of adult Americans and asks them whether or not they support the proposed changes. To say that the distribution of the sample proportion of adults who respond yes, is approximately normal, how many more adult Americans does the researcher need to sample in the following cases?
25
(a) % of all adult Americans support the changes15 (b) % of all adult Americans support the changes20
(a) The researcher must ask more American adults. (Round up to the nearest integer.)54
(b) The researcher must ask more American adults. (Round up to the nearest integer.)38
12/12/2017 Section 8.1 HomeworkClayton Leach
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1.
2.
Student: Clayton Leach Date: 12/12/17
Instructor: Ralph Leon Course: Math 110 Fall 2017 Assignment: Section 8.1 Homework
A simple random sample of size n is obtained from a population with and . Does the population need to be normally distributed for the sampling distribution of to be approximately normally distributed? Why? What is the sampling distribution of ?
= 56 μ = 65 σ = 8 x
x
Does the population need to be normally distributed for the sampling distribution of to be approximately normally distributed? Why?
x
A. No because the Central Limit Theorem states that regardless of the shape of the underlying population, the sampling distribution of becomes approximately normal as the sample size, n, increases.
x
B. No because the Central Limit Theorem states that only if the shape of the underlying population is normal or uniform does the sampling distribution of become approximately normal as the sample size, n, increases.
x
C. Yes because the Central Limit Theorem states that only for underlying populations that are normal is the shape of the sampling distribution of normal, regardless of the sample size, n.x
D. Yes because the Central Limit Theorem states that the sampling variability of nonnormal populations will increase as the sample size increases.
What is the sampling distribution of ? Select the correct choice below and fill in the answer boxes within your choice.x (Type integers or decimals rounded to three decimal places as needed.)
A. The sampling distribution of is skewed left with and
.
x μx =
σx =
B. The sampling distribution of follows Student's tdistribution with and
.
x μx =
σx =
C. The sampling distribution of is normal or approximately normal with
and .
x μx =
σx =
D. The sampling distribution of is uniform with and .x μx = σx =
Determine and from the given parameters of the population and sample size.μx σx
, , nμ = 76 σ = 27 = 81
μx = 76
σx = 3
12/12/2017 Section 8.1 HomeworkClayton Leach
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5. The reading speed of second grade students in a large city is approximately normal, with a mean of words per minute (wpm) and a standard deviation of 10 wpm. Complete parts (a) through (f).
91
(a) What is the probability a randomly selected student in the city will read more than words per minute?97
The probability is . (Round to four decimal places as needed.)
Interpret this probability. Select the correct choice below and fill in the answer box within your choice.
A. If 100 different students were chosen from this population, we would expect to read more than words per minute.97
B. If 100 different students were chosen from this population, we would expect to read less than words per minute.97
C. If 100 different students were chosen from this population, we would expect to read exactly words per minute.97
(b) What is the probability that a random sample of second grade students from the city results in a mean reading rate of more than words per minute?
10 97
The probability is . (Round to four decimal places as needed.)
Interpret this probability. Select the correct choice below and fill in the answer box within your choice.
A. If 100 independent samples of n students were chosen from this population, we would expect sample(s) to have a sample mean reading rate of less than words per minute.
= 10 97
B. If 100 independent samples of n students were chosen from this population, we would expect sample(s) to have a sample mean reading rate of more than words per minute.
= 10 97
C. If 100 independent samples of n students were chosen from this population, we would expect sample(s) to have a sample mean reading rate of exactly words per minute.
= 10 97
(c) What is the probability that a random sample of second grade students from the city results in a mean reading rate of more than words per minute?
20 97
The probability is . (Round to four decimal places as needed.)
Interpret this probability. Select the correct choice below and fill in the answer box within your choice.
A. If 100 independent samples of n students were chosen from this population, we would expect sample(s) to have a sample mean reading rate of exactly words per minute.
= 20 97
B. If 100 independent samples of n students were chosen from this population, we would expect sample(s) to have a sample mean reading rate of less than words per minute.
= 20 97
C. If 100 independent samples of n students were chosen from this population, we would expect sample(s) to have a sample mean reading rate of more than words per minute.
= 20 97
(d) What effect does increasing the sample size have on the probability? Provide an explanation for this result.
A. Increasing the sample size increases the probability because decreases as n increases.σx B. Increasing the sample size decreases the probability because increases as n increases.σx C.
σ
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Increasing the sample size increases the probability because increases as n increases.σxD. Increasing the sample size decreases the probability because decreases as n increases.σx
(e) A teacher instituted a new reading program at school. After 10 weeks in the program, it was found that the mean reading speed of a random sample of second grade students was wpm. What might you conclude based on this result? Select the correct choice below and fill in the answer boxes within your choice.
19 93.3
(Type integers or decimals rounded to four decimal places as needed.)
A. A mean reading rate of wpm is unusual since the probability of obtaining a result of wpm or more is . This means that we would expect a mean reading rate of
or higher from a population whose mean reading rate is in of every 100 random samples of size n students. The new program is abundantly more effective than the old program.
93.3 93.3
93.3 91 = 19
B. A mean reading rate of wpm is not unusual since the probability of obtaining a result of wpm or more is . This means that we would expect a mean reading rate
of or higher from a population whose mean reading rate is in of every 100 random samples of size n students. The new program is not abundantly more effective than the old program.
93.3 93.3 93.3 91
= 19
(f) There is a 5% chance that the mean reading speed of a random sample of second grade students will exceed what value?
25
There is a 5% chance that the mean reading speed of a random sample of second grade students will exceed wpm. (Round to two decimal places as needed.)
25
12/12/2017 Section 8.1 HomeworkClayton Leach
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8.
(1) skewed right. approximately normal. skewed left.
In the game of roulette, a wheel consists of 38 slots numbered 0, 00, 1, 2,..., 36. To play the game, a metal ball is spun around the wheel and is allowed to fall into one of the numbered slots. If the number of the slot the ball falls into matches the number you selected, you win $35; otherwise you lose $1. Complete parts (a) through (g) below.
(a) Construct a probability distribution for the random variable X, the winnings of each spin.
x P(x) 35 1−
(Type integers or decimals rounded to four decimal places as needed.)
(b) Determine the mean and standard deviation of the random variable X. Round your results to the nearest penny.
μ = σ =
(c) Suppose that you play the game times so that n . Describe the sampling distribution of , the mean amount won per game.
70 = 70 x
The sample mean is (1) x
What are the mean and standard deviation of the sampling distribution of ? Round your results to the nearest penny.x
μx =
σx =
(d) What is the probability of being ahead after playing the game times? That is, what is the probability that the sample mean is greater than 0 for n ?
70 = 70
P( 0)x > = (Type an integer or decimal rounded to four decimal places as needed.)
(e) What is the probability of being ahead after playing the game times?140
P( 0)x > = (Type an integer or decimal rounded to four decimal places as needed.)
(f) What is the probability of being ahead after playing the game times?700
P( 0)x > = (Type an integer or decimal rounded to four decimal places as needed.)
(g) Compare the results of parts (d) through (f). What lesson does this teach you?
A. The probability of being ahead decreases as the number of games played increases. B. The probability of being ahead decreases as the number of games played decreases. C. The probability of being ahead increases as the number of games played increases. D. The probability of being ahead remains the same as the number of games played varies.
12/12/2017 Section 7.1 HomeworkClayton Leach
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1.
2.
3.
Student: Clayton Leach Date: 12/12/17
Instructor: Ralph Leon Course: Math 110 Fall 2017 Assignment: Section 7.1 Homework
Fill in the blank to complete the statement.
The area under the normal curve to the right of equals _______.μ
The area under the normal curve to the right of equals μ . 1 2
The graph to the right is the uniform probability density function for a friend who is x minutes late. (a) Find the probability that the friend is between and minutes late.10 30 (b) It is 10 A.M. There is a % probability the friend will arrive within how many minutes?50
0 10 20 30
1/30
0
Time (min)
D en
si ty
X
(a) The probability that the friend is between and minutes late is .10 30 (Type an integer or a decimal. Round to three decimal places as needed.)
(b) There is a % probability the friend will arrive within minutes.50 (Type a whole number.)
The randomnumber generator on calculators randomly generates a number between 0 and 1. The random variable X, the number generated, follows a uniform probability distribution. (a) Identify the graph of the uniform density function. (b) What is the probability of generating a number between and ?0.63 0.83 (c) What is the probability of generating a number greater than ?0.88
(a) Choose the correct graph of the uniform density function below.
A.
0 0.2 0.4 0.6 0.8 1 1.2 0
0.2 0.4 0.6 0.8 1
1.2
x
Density
B.
0 0.2 0.4 0.6 0.8 1 1.2 0
0.2 0.4 0.6 0.8 1
1.2
x
Density
C.
0 0.2 0.4 0.6 0.8 1 1.2 0
0.2 0.4 0.6 0.8 1
1.2
x
Density
(b) The probability is . (Simplify your answer.)
(c) The probability is . (Simplify your answer.)
12/12/2017 Section 7.1 HomeworkClayton Leach
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9.
1: Golf Drive Distances
Michael went to the driving range with his range finder and hit 75 golf balls with his pitching wedge and measured the distance each ball traveled (in yards). The accompanying table shows his data. Complete parts a and b below. Click the icon to view the table of drive distances.1
a. Use technology to construct a relative frequency histogram. Comment on the shape of the distribution. Draw a normal density function on the relative frequency histogram. Choose the correct relative frequency histogram below.
A.
93 96 99 102 105 0
12.5
25
Distance (yards)
R el at iv e Fr eq
. B.
93 96 99 102 105 0
12.5
25
Distance (yards)
R el at iv e Fr eq
.
C.
93 96 99 102 105 0
0.16
0.32
Distance (yards)
R el at iv e Fr eq
. D.
93 96 99 102 105 0
0.16
0.32
Distance (yards)
R el at iv e Fr eq
. Comment on the shape of the distribution.
A. The shape of the distribution is roughly uniform. B. The shape of the distribution is roughly normal. C. The shape of the distribution seems to have no pattern.
Draw a normal density function on the relative frequency histogram. Choose the correct answer below.
A.
93 96 99 102 105 0
12.5
25
Distance (yards)
R el at iv e Fr eq
.
B.
93 96 99 102 105 0
0.16
0.32
Distance (yards)
R el at iv e Fr eq
.
C.
93 96 99 102 105 0
0.16
0.32
Distance (yards)
R el at iv e Fr eq
.
D.
93 96 99 102 105 0
12.5
25
Distance (yards)
R el at iv e Fr eq
.
b. Do you think the normal density function accurately describes the distance Michael hits with a pitching wedge? Why?
A. Yes, because the distance Michael hits the ball is a random variable, so it is normally distributed.
B. Yes, because the histogram shape resembles a normal curve, and the area of each bar is roughly equal to the area under the normal curve for the same region.
C. No, because the bars in the histogram are sometimes higher than the normal curve and sometimes lower.
D. No, because the normal distribution is continuous, but the distance Michael hits the ball is a discrete variable.
12/12/2017 Section 7.1 HomeworkClayton Leach
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100 98 102 100 102 100 101 100 99 100 101 99 100 101 97 99 102 98 101 99 99 98 102 101 102 100 99 99 95 101 100 101 100 99 101 100 100 96 101 100 102 101 102 99 98 99 102 99 100 103 103 101 101 95 104 99 100 98 100 99 97 100 101 98 100 101 99 99 101 100 99 97 95 98 96
12/12/2017 Section 6.3 HomeworkClayton Leach
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1.
2.
3.
Student: Clayton Leach Date: 12/12/17
Instructor: Ralph Leon Course: Math 110 Fall 2017 Assignment: Section 6.3 Homework
State the conditions required for a random variable X to follow a Poisson process.
Select all that apply.
A. The probability of two or more successes in any sufficiently small subinterval is 0. B. The experiment is performed a fixed number of times. C. The number of successes in any interval is independent of the number of successes in any
other interval provided the intervals are not overlapping. D. A sample of size n is obtained from the population of size N without replacement. E. The probability of success is the same for any two intervals of equal length.
The random variable X follows a Poisson process with the given mean. Assuming compute the following.μ = 9, (a) P( )4 (b) P(X )< 4 (c) P(X )≥ 4 (d) P( X )5 ≤ ≤ 7
(a) P( )4 ≈ 0.0337 (Do not round until the final answer. Then round to four decimal places as needed.)
(b) P(X )< 4 ≈ 0.0212 (Do not round until the final answer. Then round to four decimal places as needed.)
(c) P(X )≥ 4 ≈ 0.9788 (Do not round until the final answer. Then round to four decimal places as needed.)
(d) P( X )5 ≤ ≤ 7 ≈ 0.2689 (Do not round until the final answer. Then round to four decimal places as needed.)
The random variable X follows a Poisson process with the given value of and t. Assuming and t , compute the following.
λ λ = 0.13 = 13 (a) P( )4 (b) P(X )< 4 (c) P(X )≥ 4 (d) P( X )3 ≤ ≤ 7 (e) μ and σX X
(a) P( )4 ≈ (Do not round until the final answer. Then round to four decimal places as needed.)
(b) P(X )< 4 ≈ (Do not round until the final answer. Then round to four decimal places as needed.)
(c) P(X )≥ 4 ≈ (Do not round until the final answer. Then round to four decimal places as needed.)
(d) P( X )3 ≤ ≤ 7 ≈ (Do not round until the final answer. Then round to four decimal places as needed.)
(e) (Round to two decimal places as needed.)μX ≈
(Round to three decimal places as needed.)σX ≈
12/12/2017 Section 6.3 HomeworkClayton Leach
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4.
(1) 14 16 13 19
(2) at least more than fewer than exactly
(3) 74 77 78 80
(4) exactly at least
fewer than more than
(5) 22 23 26 24
(6) exactly fewer than at least more than
(7) 62 58 57 59
(8) at least exactly
fewer than more than
(9) be not be
(10) 2 2 4 1
A government entity sets a Food Defect Action Level (FDAL) for the various foreign substances that inevitably end up in the foods we eat. The FDAL level for insect filth in peanut butter is insect fragment (larvae, eggs, body parts, and so on) per gram. Suppose that a supply of peanut butter contains insect fragment per gram. Compute the probability that the number of insect fragments in a gram sample of peanut butter is
0.1 0.1
9 (a) exactly . Interpret the results.two (b) fewer than . Interpret the results.two (c) at least . Interpret the results.two (d) at least one. Interpret the results. (e) Would it be unusual for a gram sample of this supply of peanut butter to contain or more insect fragments?9 four
(a) P( ) (Round to four decimal places as needed.)2 =
Fill in the blanks to complete the statement below.
About (1) of every 100 gram samples of this supply will contain (2) insect fragments.
9 2
(b) P(X ) (Round to four decimal places as needed.)< 2 =
Fill in the blanks to complete the statement below.
About (3) of every 100 gram samples of this supply will contain (4) insect fragments.
9 2
(c) P(X ) (Round to four decimal places as needed.)≥ 2 =
Fill in the blanks to complete the statement below.
About (5) of every 100 gram samples of this supply will contain (6) insect fragments.
9 2
(d) P(X 1) (Round to four decimal places as needed.)≥ =
Fill in the blanks to complete the statement below.
About (7) of every 100 gram samples of this supply will contain (8) 1 insect fragment.
9
(e) It would (9) unusual. About (10) of every 100 gram samples of this supply will contain at least insect fragments.
9 4
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5. For the past years, a certain state suffered direct hits from major (category 3 to 5) hurricanes. Assume that this was typical and the number of hits per year follows a Poisson distribution. Complete parts (a) through (d).
111 24
(a) What is the probability that the state will not be hit by any major hurricanes in a single year?
The probability is . (Round to four decimal places as needed.)
(b) What is the probability that the state will be hit by at least one major hurricane in a single year?
The probability is . (Round to four decimal places as needed.)
Is this unusual?
No Yes
(c) What is the probability that the state will be hit by at least three major hurricanes in a single year, as happened last year?
The probability is . (Round to four decimal places as needed.)
Does this indicate that the 2004 hurricane season in this state was unusual?
No Yes
(d) What is the probability that the state will be hit by at least two major hurricanes in a single year, as happened a couple of years ago?
The probability is . (Round to four decimal places as needed.)
Does this indicate that the 2005 hurricane season in this state was unusual?
Yes No
12/12/2017 Section 6.3 HomeworkClayton Leach
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6. Cars arrive at a certain restaurant's drivethrough at a rate of 0.2 cars per minute between the hours of 11:00 P.M. and on Saturday evening. The restaurant begins an advertising blitz that touts its latenight service. After one week of advertising, the restaurant's officials count the number of cars, X, arriving at the restaurant's drive through between the hours of 12:00 midnight and 12:30 A.M. for of its restaurants. The results are shown in the following table. Use the table to answer parts (a) through (c) to the right.
1:00 A.M.
196
Click the icon to view the data table.1
(a) Construct a probability distribution for the random variable X, assuming it follows a Poisson process with
0.2 and t 30. This is the probability distribution of X before the advertising. Remember that x 0,1,2,3,...,16. λ = =
=
Calculate the probabilities below.
x (# of cars arriving) P(x)
x (# of cars arriving) P(x)
0 9
1 10
2 11
3 12
4 13
5 14
6 15
7 16
8 (Type integers or decimals rounded to four decimal places as needed.)
(b) Compute the expected number of restaurants that will have 0 arrivals, 1 arrival, and so on.
Compute the number of restaurants that expect to have x cars. x (# of cars)
Number of restaurants
x (# of cars)
Number of restaurants
0 9
1 10
2 11
3 12
4 13
5 14
6 15
7 16
8 (Type whole numbers.)
(c) Compare these results with the number of arrivals after the advertising. Does it appear the advertising was effective? Why?
What is the observed total number of arrivals?
What is the expected total number of arrivals?
Was the advertising campaign effective?
A. The advertising campaign does not appear to be effective because the total number of arrivals is the same as the
12/12/2017 Section 6.3 HomeworkClayton Leach
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1: Restaurant car arrival frequency
number of arrivals is the same as the expected total number of arrivals.
B. The advertising campaign appears to be effective because the observed total number of arrivals is greater than the expected total number of arrivals.
C. The advertising campaign does not appear to be effective because the total number of arrivals is less than the expected total number of arrivals.
x (number of cars arriving) Frequency 1 4 2 4 3 13 4 22 5 24 6 28 7 24 8 26 9 20 10 15 11 6 12 2 13 3 14 2 15 0 16 3
12/12/2017 Section 6.3 HomeworkClayton Leach
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7.
2: Deaths in the Army
An army monitored 10 cavalry corps for 20 years and recorded the number X of annual fatalities due to horse kicks for the 200 observations. The accompanying table shows the data. Complete parts (a) through (d). Click the icon to view the data table.2
(a) Compute the proportion of years in which there were 0 deaths, 1 death, 2 deaths, 3 deaths, and 4 deaths. Complete the table below.
Number of deaths, x Proportion of years 0 1 2 3 4
(Type an integer or a decimal.)
(b) From the data in the table, what was the mean number of deaths per year?
The mean number of deaths per year is . (Type an integer or a decimal.)
(c) Use the mean number of deaths per year found in part (b) and the Poisson probability law to determine the theoretical proportion of years that 0 deaths should occur. Repeat this for 1, 2, 3, and 4 deaths. Complete the table below.
Number of deaths, x P(x) 0 1 2 3 4
(Round to four decimal places as needed.)
(d) Compare the observed proportions to the theoretical proportions. Do you think the data can be modeled by the poisson probability law?
A. The observed proportions and the theoretical proportions have close values, therefore the data can be modeled by the poisson probability law.
B. The observed proportions and the theoretical proportions have different values, therefore the data can not be modeled by the poisson probability law.
C. Comparison of these values can not give the information about whether the data can be modeled by the poisson probability law.
Number of deaths, x Number of times x deaths were observed
0 105 1 64 2 24 3 5 4 2