Explanation Given: The sample mean is 39.8. The sample standard deviation is 17.2. The sample size is 50. Formula used: The margin of error is given by Z α 2 s n . Calculations: Since, sample mean is the unbiased estimator of population mean. Therefor, μ ^ = x ¯ = 39

Introduction to Probability and Statistics

14th Edition

ISBN: 9781133103752

Author: Mendenhall, William

Publisher: Cengage Learning

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Chapter 8.4, Problem 8.15E

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Chapter 8.4, Problem 8.14E

Chapter 8.4, Problem 8.16E

Chapter 8 Solutions

Introduction to Probability and Statistics

Ch. 8.4 - Prob. 8.3E Ch. 8.4 - Prob. 8.4E Ch. 8.4 - Prob. 8.5E Ch. 8.4 - Prob. 8.6E Ch. 8.4 - Prob. 8.7E Ch. 8.4 - Prob. 8.8E Ch. 8.4 - Prob. 8.9E Ch. 8.4 - Prob. 8.10E Ch. 8.4 - Prob. 8.11E Ch. 8.4 - Prob. 8.12E

Ch. 8.4 - Prob. 8.13E Ch. 8.4 - Prob. 8.14E Ch. 8.4 - Prob. 8.15E Ch. 8.4 - Prob. 8.16E Ch. 8.4 - Prob. 8.18E Ch. 8.4 - Prob. 8.19E Ch. 8.4 - Hotel Costs Even within a particular chain of...Ch. 8.4 - Prob. 8.21E Ch. 8.5 - Find and interpret a 95% confidence interval for a...Ch. 8.5 - Find a 90% confidence interval for a population...Ch. 8.5 - Prob. 8.26E Ch. 8.5 - Prob. 8.27E Ch. 8.5 - Prob. 8.28E Ch. 8.5 - Prob. 8.29E Ch. 8.5 - Compare the confidence intervals in Exercise 8.29....Ch. 8.5 - Prob. 8.31E Ch. 8.5 - Prob. 8.32E Ch. 8.5 - Prob. 8.33E Ch. 8.5 - Prob. 8.34E Ch. 8.5 - Prob. 8.35E Ch. 8.5 - Prob. 8.36E Ch. 8.5 - Prob. 8.38E Ch. 8.5 - What’s Normal? What is normal, when it comes to...Ch. 8.5 - Prob. 8.40E Ch. 8.6 - Independent random samples were selected from...Ch. 8.6 - Prob. 8.42E Ch. 8.6 - Prob. 8.43E Ch. 8.6 - Prob. 8.44E Ch. 8.6 - Prob. 8.45E Ch. 8.6 - Prob. 8.46E Ch. 8.6 - Prob. 8.47E Ch. 8.6 - Prob. 8.48E Ch. 8.6 - Prob. 8.49E Ch. 8.6 - Prob. 8.50E Ch. 8.6 - Hotel Costs Refer to Exercise 8.20. The means and...Ch. 8.6 - Noise and Stress To compare the effect of stress...Ch. 8.6 - Prob. 8.53E Ch. 8.7 - Prob. 8.54E Ch. 8.7 - Prob. 8.55E Ch. 8.7 - Prob. 8.56E Ch. 8.7 - Prob. 8.58E Ch. 8.7 - Prob. 8.61E Ch. 8.7 - Prob. 8.64E Ch. 8.7 - Prob. 8.65E Ch. 8.7 - Prob. 8.66E Ch. 8.9 - Prob. 8.67E Ch. 8.9 - Prob. 8.68E Ch. 8.9 - Prob. 8.69E Ch. 8.9 - Prob. 8.70E Ch. 8.9 - Prob. 8.71E Ch. 8.9 - Prob. 8.72E Ch. 8.9 - Prob. 8.73E Ch. 8.9 - Prob. 8.75E Ch. 8.9 - Prob. 8.77E Ch. 8.9 - Prob. 8.78E Ch. 8.9 - Prob. 8.80E Ch. 8.9 - Prob. 8.81E Ch. 8.9 - Prob. 8.83E Ch. 8 - Prob. 8.86SE Ch. 8 - Prob. 8.87SE Ch. 8 - Prob. 8.88SE Ch. 8 - Prob. 8.89SE Ch. 8 - Prob. 8.90SE Ch. 8 - Prob. 8.91SE Ch. 8 - Prob. 8.92SE Ch. 8 - Prob. 8.93SE Ch. 8 - Prob. 8.94SE Ch. 8 - Prob. 8.95SE Ch. 8 - Prob. 8.96SE Ch. 8 - Prob. 8.98SE Ch. 8 - Prob. 8.99SE Ch. 8 - Prob. 8.100SE Ch. 8 - Prob. 8.101SE Ch. 8 - Prob. 8.103SE Ch. 8 - Prob. 8.107SE Ch. 8 - Prob. 8.108SE Ch. 8 - Prob. 8.109SE Ch. 8 - Prob. 8.110SE Ch. 8 - Prob. 8.111SE Ch. 8 - Prob. 8.112SE Ch. 8 - Prob. 8.113SE Ch. 8 - Prob. 8.114SE Ch. 8 - Prob. 8.115SE Ch. 8 - Prob. 8.116SE Ch. 8 - Prob. 8.117SE

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Planetary Velocity The following table gives the mean velocity of planets in their orbits versus their mean distance from the sun. Note that 1AU astronomical unit is the mean distance from Earth to the sun, abut 93 million miles. Planet d=distance AU v=velocity km/sec Mercury 0.39 47.4 Venus 0.72 35.0 Earth 1.00 29.8 Mars 1.52 24.1 Jupiter 5.20 13.1 Saturn 9.58 9.7 Uranus 19.20 6.8 Neptune 30.05 5.4 Astronomers tell us that it is reasonable to model these data with a power function. a Use power regression to express velocity as a power function of distance from the sun. b Plot the data along with the regression equation. c An asteroid orbits at a mean distance of 3AU from the sun. According to the power model you found in part a, what is the mean orbital velocity of the asteroid?

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The estimate of population mean and the margin of error.

Solution Summary: The author explains how the best estimate of population mean is 39.8 and the margin of error is 4.7676. The level of significance is taken as 5%.

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To find: The estimate of population mean and the margin of error.

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Chapter 8 Solutions