Answered: 1.In a trial for the university…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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1.In a trial for the university athletics competition, one aims to estimate the mean chest size of the male athletes. A random sample of 70 male athletes gave rise to an average chest size of 42 inches with a standard deviation of 16 inches. What would be the distribution of the statistic that we need to estimate the mean chest size?

a. it is normal distribution with mean 42 and standard deviation 16 inches.

b. it is a student's t distribution with degree freedom 69

c. it is a student's t distribution with mean 42 inches and standard deviation 16 inches.

d. It is a normal distribution with degree of freedom 69.

2.What would be the margin of error for a 95% confidence interval of the true chest size in problem 1?

a. 0.46

b. 0.52

c. 3.82

d. 4.38

3.How do you conclude using the 95% confidence interval?

a. With 95% confidence, the sample mean chest size is between 38.18 inches and 45.82 inches.

b. With 95% confidence, the sample mean chest size is between 37.62 and 46.38.

c. With 95% confidence, the true mean chest size is between 37.62 inches and 46.38 inches.

d. With 95% confidence, the true mean chest size is between 38.18 inches and 45.82 inches.

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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