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True or false? a larger sample size produces a longer confidence interval for μ.

16/11/2020 Client: papadok01 Deadline: 3 days

True or false? Consider a random sample of size n from an x distribution. For such a sample, the margin of error for estimating μ is the magnitude of the difference between x and μ.

False. By definition, the margin of error is the magnitude of the difference between x and σ.

True. By definition, the margin of error is the magnitude of the difference between x and μ.

True. By definition, the margin of error is the magnitude of the difference between x and σ.

False. By definition, the margin of error is the magnitude of the difference between x and μ.

True or false? Every random sample of the same size from a given population will produce exactly the same confidence interval for μ.

True. Different random samples will produce the same x values, resulting in the same confidence intervals.

False. Different random samples may produce different x values, resulting in the same confidence intervals.

False. Different random samples may produce different x values, resulting in different confidence intervals.

True. Different random samples may produce different x values, resulting in different confidence intervals.

True or false? A larger sample size produces a longer confidence interval for μ.

False. As the sample size increases, the maximal error decreases, resulting in a shorter confidence interval.

True. As the sample size increases, the maximal error increases, resulting in a longer confidence interval.

True. As the sample size increases, the maximal error decreases, resulting in a longer confidence interval.

False. As the sample size increases, the maximal error increases, resulting in a shorter confidence interval.

Allen's hummingbird (Selasphorussasin) has been studied by zoologist Bill Alther.† Suppose a small group of 12 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.38 gram.

(a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. (Round your answers to two decimal places.)

lower limit 3.01

upper limit 3.29

margin of error 0.14

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

n is large

σ is unknown

uniform distribution of weights

normal distribution of weights

σ is known

(c) Give a brief interpretation of your results in the context of this problem.

There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's --- hummingbirds in this region. 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

. The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.

There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.

(d) Find the sample size necessary for an 80% confidence level with a maximal error of estimate E = 0.13 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)

15hummingbirds

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 taken nine blood tests for uric acid. The mean concentration was x = 5.35 mg/dl. The distribution of uric acid in healthy adult males can be assumed to be normal, with σ = 1.79 mg/dl.

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

lower limit 4.18

upper limit 6.52

margin of error 1.17

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

n is large

σ is known

normal distribution of uric acid

σ is unknown uniform distribution of uric acid

(c) Give a brief interpretation of your results in the context of this problem.

There is not enough information to make an interpretation.

The probability that this interval contains the true average uric acid level for this patient is 0.05.

There is a 95% chance that the confidence interval is one of the intervals containing the population average uric acid level for this patient.

The probability that this interval contains the true average uric acid level for this patient is 0.95

There is a 5% chance that the confidence interval is one of the intervals containing the population average uric acid level for this patient.

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

10blood tests

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 of 43 male firefighters are tested and that they have a plasma volume sample mean of x= 37.5 ml/kg (milliliters plasma per kilogram body weight). Assume that σ = 7.70 ml/kg for 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 limit 34.48

upper limit 40.52

margin of error 3.02

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

the distribution of weights is normal

the distribution of weights is uniform

σ is unknown

σ is known n is large

(c) Give a brief interpretation of 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.

99% of the intervals created using this method will contain the true average blood plasma volume in male firefighters.

The probability that this interval contains the true average blood plasma volume in male firefighters is 0.01.

1% of the intervals created using this method will contain the true average blood plasma volume in male firefighters.

(d) Find the sample size necessary for a 99% confidence level with maximal error of estimate E = 2.80 for the mean plasma volume in male firefighters. (Round up to the nearest whole number.)

51male firefighters

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

1320

1187

1278

1313

1268

1316

1275

1317

1275

(a) Use a calculator with mean and standard deviation keys to find the sample mean year x and sample standard deviation s. (Round your answers to the nearest whole number.)

X=A.D. 1283

s =yr42

(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

1257A.D.

upper limit

1309A.D.

How much does a sleeping bag cost? Let's say you want a sleeping bag that should keep you warm in temperatures from 20°F to 45°F. A random sample of prices ($) for sleeping bags in this temperature range is given below. Assume that the population of x values has an approximately normal distribution.

115

115

35

40

110

40

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 price x and sample standard deviation s. (Round your answers to two decimal places.)

X=$81.40

s =$33.07

(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 price μ of all summer sleeping bags. (Round your answers to two decimal places.)

lower limit

$68.61

upper limit

$94.19

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