4th Edition

ISBN: 9780073401331

Author: William Navidi Prof.

Publisher: McGraw-Hill Education

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Chapter 5.1, Problem 6E

The article “Application of Surgical Navigation to Total Hip Arthroplasty” (T. Ecker and S. Murphy, Journal of Engineering in Medicine, 2007:699–712) reports that in a sample of 123 hip surgeries of a certain type, the average surgery time was 136.9 minutes with a standard deviation of 22.6 minutes.

a. Find a 95% confidence interval for the mean surgery time for this procedure.
b. Find a 99.5% confidence interval for the mean surgery time for this procedure.
c. A surgeon claims that the mean surgery time is between 133.9 and 139.9 minutes. With what level of confidence can this statement be made?
d. Approximately how many surgeries must be sampled so that a 95% confidence interval will specify the mean to within ±3 minutes?
e. Approximately how many surgeries must be sampled so that a 99% confidence interval will specify the mean to within ±3 minutes?

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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Chapter 5.1, Problem 5E

Chapter 5.1, Problem 7E

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The article “Application of Surgical Navigation to Total Hip Arthroplasty” (T. Ecker and S.Murphy, Journal of Engineering in Medicine, 2007:699–712) reports that in a sample of 123 hipsurgeries of a certain type, the average surgery time was 136.9 minutes with a standard deviationof 22.6 minutes.a) Find a 95% confidence interval for the mean surgery time for this procedure.b) Find a 99.5% confidence interval for the mean surgery time for this procedure.c) A surgeon claims that the mean surgery time is between 133.9 and 139.9 minutes. Withwhat level of confidence can this statement be made?d) Approximately how many surgeries must be sampled so that a 95%confidence intervalwill specify the mean to within ±3 minutes?e) Approximately how many surgeries must be sampled so that a 99%confidence intervalwill specify the mean to within ±3 minutes?

The article “Tibiofemoral Cartilage Thickness Distribution and its Correlation with Anthropometric Variables” (A. Connolly, D. FitzPatrick, et al., Journal of Engineering in Medicine, 2008:29–39) reports that in a sample of 11 men, the average volume of femoral cartilage (located in the knee) was 18.7 cm3 with a standard deviation of 3.3 cm3 and the average volume in a sample of 9 women was 11.2 cm3 with a standard deviation of 2.4 cm2. Find a 95% confidence interval for the difference in mean femoral cartilage volume between men and women.

When 89 people used the Atkins diet for one year, their mean weight change was 6.4 lb(based on data from "Comparison of the Atkins, Ornish, Weight Watchers, and Zone Diets for Weight Loss and Heart Disease Reduction," by Dansinger, et al., Journal of the American Medical Association, Vol. 293, No. 1).The standard deviation of weight change was s = 6.8 lb and calculate a 69% confidence interval for the mean weight change.Use 3 decimal places

Chapter 5 Solutions

Statistics for Engineers and Scientists

Ch. 5.1 - Find the value of Z/2 to use in expression (5.1)...Ch. 5.1 - Find the levels of the confidence intervals that...Ch. 5.1 - As the confidence level goes up, the reliability...Ch. 5.1 - The article Modeling Arterial Signal Optimization...Ch. 5.1 - In a sample of 100 steel wires the average...Ch. 5.1 - The article Application of Surgical Navigation to...Ch. 5.1 - The capacities (in ampere-hours) were measured for...Ch. 5.1 - Oven thermostats were tested by setting them to...Ch. 5.1 - In a sample of 80 ten-penny nails, the average...Ch. 5.1 - In a sample of 60 electric motors, the average...

Ch. 5.1 - The sugar content in a one-cup serving of a...Ch. 5.1 - Refer to Exercise 5. a. Find a 95% lower...Ch. 5.1 - Prob. 13E Ch. 5.1 - Prob. 14E Ch. 5.1 - Prob. 15E Ch. 5.1 - Refer to Exercise 9. a. Find a 90% upper...Ch. 5.1 - Refer to Exercise 10. a. Find a 98% lower...Ch. 5.1 - Refer to Exercise 11. a. Find a 95% upper...Ch. 5.1 - Prob. 19E Ch. 5.1 - A 95% confidence interval for a population mean is...Ch. 5.1 - Based on a large sample of capacitors of a certain...Ch. 5.1 - Sixty-four independent measurements were made of...Ch. 5.1 - A large box contains 10,000 ball bearings. A...Ch. 5.1 - Prob. 24E Ch. 5.1 - Prob. 25E Ch. 5.1 - Prob. 26E Ch. 5.2 - In a simple random sample of 70 automobiles...Ch. 5.2 - During a recent drought, a water utility in a...Ch. 5.2 - A soft-drink manufacturer purchases aluminum cans...Ch. 5.2 - The article HIV-positive Smokers Considering...Ch. 5.2 - The article The Functional Outcomes of Total Knee...Ch. 5.2 - Refer to Exercise 1. 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