6. Dogs vary in size a great deal more than cats do. Suppose the mean weight of dogs in a certain city is 28 pounds with a standard deviation of 14 pounds, and the mean weights of cats is 9.5 pounds with a standard deviation of 2 pounds. Cat weights are roughly Normally distributed, but dog weights are skewed right. Suppose you take a random sample of 10 dogs and 8 cats. Which of the following is an accurate description of the center, spread, and shape of the sampling distribution of the difference in the mean weight dogs and the mean weight of cats? (a) Mean = 18.5 pounds, standard deviation = 4.48 pounds, shape approximately Normal (b) Mean = 18.5 pounds, standard deviation = 14.1 pounds, shape non-Normal (c) Mean = 18.5 pounds, standard deviation = 14.1 pounds, shape approximately Normal (d) Mean = 18.5 pounds, standard deviation = 4.48 pounds, shape non-Normal (e) The difference of means is 18.5 pounds and the shape is non-Normal, but the standard deviation cannot be determined since the sample size is small and the distribution of dog weights is non-Normal.

6. Dogs vary in size a great deal more than cats do. Suppose the mean weight of dogs in a certain city is 28 pounds with a standard deviation of 14 pounds, and the mean weights of cats is 9.5 pounds with a standard deviation of 2 pounds. Cat weights are roughly Normally distributed, but dog weights are skewed right. Suppose you take a random sample of 10 dogs and 8 cats. Which of the following is an accurate description of the center, spread, and shape of the sampling distribution of the difference in the mean weight dogs and the mean weight of cats? (a) Mean = 18.5 pounds, standard deviation = 4.48 pounds, shape approximately Normal (b) Mean = 18.5 pounds, standard deviation = 14.1 pounds, shape non-Normal (c) Mean = 18.5 pounds, standard deviation = 14.1 pounds, shape approximately Normal (d) Mean = 18.5 pounds, standard deviation = 4.48 pounds, shape non-Normal (e) The difference of means is 18.5 pounds and the shape is non-Normal, but the standard deviation cannot be determined since the sample size is small and the distribution of dog weights is non-Normal.

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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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Concept explainers

Inverse Normal Distribution

The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of two parameters.

Mean, Median, Mode

It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency or measure of central tendency is a central or typical value for a probability distribution.

Z-Scores

A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.

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Question

6. Dogs vary in size a great deal more than cats do. Suppose the mean weight of dogs in a certain city is 28 pounds with
a standard deviation of 14 pounds, and the mean weights of cats is 9.5 pounds with a standard deviation of 2 pounds.
Cat weights are roughly Normally distributed, but dog weights are skewed right. Suppose you take a random sample
of 10 dogs and 8 cats. Which of the following is an accurate description of the center, spread, and shape of the
sampling distribution of the difference in the mean weight dogs and the mean weight of cats?
(a) Mean = 18.5 pounds, standard deviation = 4.48 pounds, shape approximately Normal
(b) Mean = 18.5 pounds, standard deviation = 14.1 pounds, shape non-Normal
(c) Mean = 18.5 pounds, standard deviation = 14.1 pounds, shape approximately Normal
(d) Mean = 18.5 pounds, standard deviation = 4.48 pounds, shape non-Normal
(e) The difference of means is 18.5 pounds and the shape is non-Normal, but the standard deviation cannot be
determined since the sample size is small and the distribution of dog weights is non-Normal.

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