1.

[-/4.42 Points]DETAILSBBUNDERSTAT12 7.1.015.

MY NOTES

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Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.Suppose a small group of12Allen's hummingbirds has been under study in Arizona. The average weight for these birds isx= 3.15 grams.Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with=0.32gram.(a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.)

lower limit
upper limit
margin of error

(b) What conditions are necessary for your calculations? (Select all that apply.)is knownnis largeuniform distribution of weightsnormal distribution of weightsis unknown

(c) Interpret your results in the context of this problem.We are 20% confident that the true average weight of Allen's hummingbirds falls within this interval.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20. We are 80% confident that the true average weight of Allen's hummingbirds falls within this interval.

(d) Find the sample size necessary for an 80% confidence level with a maximal margin of errorE=0.15for the mean weights of the hummingbirds. (Round up to the nearest whole number.) hummingbirds

2.

[-/4.42 Points]DETAILSBBUNDERSTAT12 7.1.016.MI.

MY NOTES

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Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an advance indication of illness such as gout, leukemia, or lymphoma.Over a period of months, an adult male patient has takennineblood tests for uric acid. The mean concentration wasx=5.35mg/dl.The distribution of uric acid in healthy adult males can be assumed to be normal, with=1.93mg/dl.

(a)

Find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. What is the margin of error? (Round your answers to two decimal places.)

lower limitupper limitmargin of error

(b)

What conditions are necessary for your calculations? (Select all that apply.)

normal distribution of uric acidis unknownis knownnis largeuniform distribution of uric acid

(c)

Interpret your results in the context of this problem.

The probability that this interval contains the true average uric acid level for this patient is 0.95.We are 5% confident that the true uric acid level for this patient falls within this interval. The probability that this interval contains the true average uric acid level for this patient is 0.05.We are 95% confident that the true uric acid level for this patient falls within this interval.

(d)

Find the sample size necessary for a 95% confidence level with maximal margin of errorE=1.14for the mean concentration of uric acid in this patient's blood. (Round your answer up to the nearest whole number.)

blood tests

3.

[-/4.42 Points]DETAILSBBUNDERSTAT12 7.1.017.MI.

MY NOTES

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Total plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma volume is influenced by the overall health and physical activity of an individual. Suppose that a random sample of48male firefighters are tested and that they have a plasma volume sample mean ofx= 37.5 ml/kg(milliliters plasma per kilogram body weight). Assume that=7.40ml/kgfor the distribution of blood plasma.

(a)

Find a 99% confidence interval for the population mean blood plasma volume in male firefighters. What is the margin of error? (Round your answers to two decimal places.)

lower limitupper limitmargin of error

(b)

What conditions are necessary for your calculations? (Select all that apply.)

nis largethe distribution of volumes is normalthe distribution of volumes is uniformis unknownis known

(c)

Interpret your results in the context of this problem.

The probability that this interval contains the true average blood plasma volume in male firefighters is 0.99.We are 99% confident that the true average blood plasma volume in male firefighters falls within this interval. We are 1% confident that the true average blood plasma volume in male firefighters falls within this interval.The probability that this interval contains the true average blood plasma volume in male firefighters is 0.01.

(d)

Find the sample size necessary for a 99% confidence level with maximal margin of errorE=2.60for the mean plasma volume in male firefighters. (Round up to the nearest whole number.)

male firefighters

4.

[-/4.42 Points]DETAILSBBUNDERSTAT12 7.2.013.S.

MY NOTES

ASK YOUR TEACHER

The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population ofxvalues has an approximately normal distribution.

1,271

1,208

1,250

1,271

1,268

1,316

1,275

1,317

1,275

(a) Use a calculator with mean and standard deviation keys to find the sample mean yearxand sample standard deviations. (Round your answers to four decimal places.)

x=	A.D.
s=	yr

(b) Find a 90% confidence interval for the mean of all tree ring dates from this archaeological site. (Round your answers to the nearest whole number.)

lower limit	A.D.
upper limit	A.D.

5.

[-/4.42 Points]DETAILSBBUNDERSTAT12 7.2.014.MI.S.

MY NOTES

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How much does a sleeping bag cost? Let's say you want a sleeping bag that should keep you warm in temperatures from20Fto45F.A random sample of prices ($) for sleeping bags in this temperature range is given below. Assume that the population ofxvalues has an approximately normal distribution.

120	35	80	110	60	55	30	23	100	110
105	95	105	60	110	120	95	90	60	70

(a)

Use a calculator with mean and sample standard deviation keys to find the sample mean pricexand sample standard deviations. (Round your answers to four decimal places.)

x=$ s=$

(b)

Using the given data as representative of the population of prices of all summer sleeping bags, find a 90% confidence interval for the mean priceof all summer sleeping bags. (Round your answers to two decimal places.)

lower limit$ upper limit$

6.

[-/4.42 Points]DETAILSBBUNDERSTAT12 7.2.016.S.

MY NOTES

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Do you want to own your own candy store? Wow! With some interest in running your own business and a decent credit rating, you can probably get a bank loan on startup costs for franchises such as Candy Express, The Fudge Company, Karmel Corn, and Rocky Mountain Chocolate Factory. Startup costs (in thousands of dollars) for a random sample of candy stores are given below. Assume that the population ofxvalues has an approximately normal distribution.

96

174

127

93

75

94

116

100

85

(a) Use a calculator with mean and sample standard deviation keys to find the sample mean startup costxand sample standard deviations. (Round your answers to four decimal places.)

x=	thousand dollars
s=	thousand dollars

(b) Find a 90% confidence interval for the population average startup costsfor candy store franchises. (Round your answers to one decimal place.)

lower limit	thousand dollars
upper limit	thousand dollars

7.

[-/4.42 Points]DETAILSBBUNDERSTAT12 7.3.011.S.

MY NOTES

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For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. In a random sample of64professional actors, it was found that43were extroverts.

(a) Letprepresent the proportion of all actors who are extroverts. Find a point estimate forp. (Round your answer to four decimal places.) (b) Find a 95% confidence interval forp. (Round your answers to two decimal places.)

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upper limit

Give a brief interpretation of the meaning of the confidence interval you have found.We are 5% confident that the true proportion of actors who are extroverts falls within this interval.We are 5% confident that the true proportion of actors who are extroverts falls above this interval. We are 95% confident that the true proportion of actors who are extroverts falls outside this interval.We are 95% confident that the true proportion of actors who are extroverts falls within this interval.

(c) Do you think the conditionsnp>5 andnq>5 are satisfied in this problem? Explain why this would be an important consideration.Yes, the conditions are satisfied. This is important because it allows us to say thatpis approximately binomial.No, the conditions are not satisfied. This is important because it allows us to say thatpis approximately normal. Yes, the conditions are satisfied. This is important because it allows us to say thatpis approximately normal.No, the conditions are not satisfied. This is important because it allows us to say thatpis approximately binomial.

8.

[-/4.42 Points]DETAILSBBUNDERSTAT12 7.3.012.MI.S.

MY NOTES

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For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.

In a random sample of506judges, it was found that295were introverts.

(a)

Letprepresent the proportion of all judges who are introverts. Find a point estimate forp. (Round your answer to four decimal places.)

(b)

Find a 99% confidence interval forp. (Round your answers to two decimal places.)

lower limitupper limit

Give a brief interpretation of the meaning of the confidence interval you have found.

We are 1% confident that the true proportion of judges who are introverts falls within this interval.We are 99% confident that the true proportion of judges who are introverts falls outside this interval. We are 99% confident that the true proportion of judges who are introverts falls within this interval.We are 1% confident that the true proportion of judges who are introverts falls above this interval.

(c)

Do you think the conditionsnp>5 andnq>5 are satisfied in this problem? Explain why this would be an important consideration.

Yes, the conditions are satisfied. This is important because it allows us to say thatpis approximately binomial.No, the conditions are not satisfied. This is important because it allows us to say thatpis approximately normal. No, the conditions are not satisfied. This is important because it allows us to say thatpis approximately binomial.Yes, the conditions are satisfied. This is important because it allows us to say thatpis approximately normal.

9.

[-/4.42 Points]DETAILSBBUNDERSTAT12 7.4.014.S.

MY NOTES

ASK YOUR TEACHER

Independent random samples of professional football and basketball players gave the following information. Assume that the weight distributions are mound-shaped and symmetric.Weights (in lb) of pro football players:x₁;n₁= 21

245	261	254	251	244	276	240	265	257	252	282
256	250	264	270	275	245	275	253	265	270

Weights (in lb) of pro basketball players:x₂;n₂= 19

204	200	220	210	193	215	223	216	228	207
225	208	195	191	207	196	182	193	201

(a) Use a calculator with mean and standard deviation keys to calculatex₁,s₁,x₂, ands₂. (Round your answers to four decimal places.)

x1=
s1=
x2=
s2=

(b) Let₁be the population mean forx₁and let₂be the population mean forx₂. Find a 99% confidence interval for₁₂.(Round your answers to one decimal place.)

lower limit
upper limit

(c) Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the 99% level of confidence, do professional football players tend to have a higher population mean weight than professional basketball players?Because the interval contains only negative numbers, we can say that professional football players have a lower mean weight than professional basketball players.Because the interval contains both positive and negative numbers, we cannot say that professional football players have a higher mean weight than professional basketball players. Because the interval contains only positive numbers, we can say that professional football players have a higher mean weight than professional basketball players.

(d) Which distribution did you use? Why?The Student'st-distribution was used because₁and₂are known.The standard normal distribution was used because₁and₂are known. The standard normal distribution was used because₁and₂are unknown.The Student'st-distribution was used because₁and₂are unknown.

10.

[-/4.42 Points]DETAILSBBUNDERSTAT12 7.4.016.MI.S.

MY NOTES

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The following data represent petal lengths (in cm) for independent random samples of two species of iris.

5.0	5.8	6.1	6.1	5.1	5.5	5.3	5.5	6.9	5.0	4.9	6.0	4.8	6.1	5.6	5.1
5.6	4.8	5.4	5.1	5.1	5.9	5.2	5.7	5.4	4.5	6.4	5.3	5.5	6.7	5.7	4.9
4.8	5.7	5.2	5.2

1.6	1.8	1.4	1.5	1.5	1.6	1.4	1.1	1.2	1.4	1.7	1.0	1.7	1.9	1.6	1.4
1.5	1.4	1.2	1.3	1.5	1.3	1.6	1.9	1.4	1.6	1.5	1.4	1.6	1.2	1.9	1.5
1.6	1.4	1.3	1.7	1.5	1.5

(a)

Use a calculator with mean and standard deviation keys to calculatex₁,s₁,x₂, ands₂. (Round your answers to four decimal places.)

x₁=s₁=x₂=s₂=

(b)

Let₁be the population mean forx₁and let₂be the population mean forx₂. Find a 99% confidence interval for₁₂.(Round your answers to two decimal places.)

lower limitupper limit

(c)

Explain what the confidence interval means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the 99% level of confidence, is the population mean petal length ofIris virginicalonger than that ofIris setosa?

Because the interval contains only positive numbers, we can say that the mean petal length ofIris virginicais longer.Because the interval contains only negative numbers, we can say that the mean petal length ofIris virginicais shorter. Because the interval contains both positive and negative numbers, we cannot say that the mean petal length ofIris virginicais longer.

(d)

Which distribution did you use? Why?

The Student'st-distribution was used because₁and₂are known.The Student'st-distribution was used because₁and₂are unknown. The standard normal distribution was used because₁and₂are unknown.The standard normal distribution was used because₁and₂are known.

Do you need information about the petal length distributions? Explain.

Both samples are large, so information about the distributions is needed.Both samples are small, so information about the distributions is not needed. Both samples are small, so information about the distributions is needed.Both samples are large, so information about the distributions is not needed.

11.

[-/4.42 Points]DETAILSBBUNDERSTAT12 7.4.017.S.

MY NOTES

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A random sample of370married couples found that282had two or more personality preferences in common. In another random sample of574married couples, it was found that only36had no preferences in common. Letp₁be the population proportion of all married couples who have two or more personality preferences in common. Letp₂be the population proportion of all married couples who have no personality preferences in common.

(a) Find a95%confidence interval forp₁-p₂. (Use 3 decimal places.)

lower limit
upper limit

(b) Explain the meaning of the confidence interval in part (a) in the context of this problem. Does the confidence interval contain all positive, all negative, or both positive and negative numbers? What does this tell you (at the95%confidence level) about the proportion of married couples with two or more personality preferences in common compared with the proportion of married couples sharing no personality preferences in common?We can not make any conclusions using this confidence interval.Because the interval contains both positive and negative numbers, we can not say that a higher proportion of married couples have two or more personality preferences in common. Because the interval contains only positive numbers, we can say that a higher proportion of married couples have two or more personality preferences in common.Because the interval contains only negative numbers, we can say that a higher proportion of married couples have no personality preferences in common.

12.

[-/4.51 Points]DETAILSBBUNDERSTAT12 7.4.019.

MY NOTES

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The U.S. Geological Survey compiled historical data about Old Faithful Geyser (Yellowstone National Park) from 1870 to 1987. Letx₁be a random variable that represents the time interval (in minutes) between Old Faithful eruptions for the years 1948 to 1952. Based on8860observations, the sample mean interval wasx₁=63.2minutes. Letx₂be a random variable that represents the time interval in minutes between Old Faithful eruptions for the years 1983 to 1987. Based on24,170observations, the sample mean time interval wasx₂=69.4minutes. Historical data suggest that₁=9.40minutes and₂=12.55minutes. Let₁be the population mean ofx₁and let₂be the population mean ofx₂.(a) Compute a99%confidence interval for₁-₂. (Use 2 decimal places.)

lower limit
upper limit

(b) Comment on the meaning of the confidence interval in the context of this problem. Does the interval consist of positive numbers only? negative numbers only? a mix of positive and negative numbers? Does it appear (at the99%confidence level) that a change in the interval length between eruptions has occurred? Many geologic experts believe that the distribution of eruption times of Old Faithful changed after the major earthquake that occurred in 1959.Because the interval contains only positive numbers, we can say that the interval length between eruptions has gotten shorter.Because the interval contains both positive and negative numbers, we can not say that the interval length between eruptions has gotten longer. We can not make any conclusions using this confidence interval.Because the interval contains only negative numbers, we can say that the interval length between eruptions has gotten longer.