Imagine that you are doing an exhaustive study on the children in all of the daycares in your school district. You are particularly interested in how much time children spend in active play on weekends. You find that for this population of 2,431 children, the average number of minutes spent in active play on weekends is μ = 65.87, with a standard deviation of σ = 87.08. You select a random sample of 25 children of daycare age in this same school district. In this sample, you find that the average number of minutes the children spend in active play on weekends is M = 59.28, with a standard deviation of s = 95.79. The difference between M and μ is due to the . Suppose you compile all possible samples of 25 children of daycare age in your school district. If you calculate the mean of each sample (M) and create a frequency distribution of these means, this distribution is referred to as the . The mean of this distribution, that is, the mean of all the sample means (when n = 25), is the expected value of M and will be equal to . The standard deviation of this distribution is called the standard error of M and will be equal to . Suppose you compile all possible samples of 100 children of daycare age in your school district; the expected value of M (when n = 100) will be the expected value of M for all of the possible samples of 25 children of daycare age in your school district. The standard error of M for this distribution of samples when n = 100 will be equal to . The standard error of M for all the possible samples of 100 is the standard error of M for all of the possible samples of 25. You can predict the size of the standard error of M for a sample size of 100 relative to a sample size of 25 because of the . If you were interested in how much time children spend in active play on weekends among this population of 2,431 children, would you actually compile all possible samples of a certain size?

Imagine that you are doing an exhaustive study on the children in all of the daycares in your school district. You are particularly interested in how much time children spend in active play on weekends. You find that for this population of 2,431 children, the average number of minutes spent in active play on weekends is μ = 65.87, with a standard deviation of σ = 87.08. You select a random sample of 25 children of daycare age in this same school district. In this sample, you find that the average number of minutes the children spend in active play on weekends is M = 59.28, with a standard deviation of s = 95.79. The difference between M and μ is due to the . Suppose you compile all possible samples of 25 children of daycare age in your school district. If you calculate the mean of each sample (M) and create a frequency distribution of these means, this distribution is referred to as the . The mean of this distribution, that is, the mean of all the sample means (when n = 25), is the expected value of M and will be equal to . The standard deviation of this distribution is called the standard error of M and will be equal to . Suppose you compile all possible samples of 100 children of daycare age in your school district; the expected value of M (when n = 100) will be the expected value of M for all of the possible samples of 25 children of daycare age in your school district. The standard error of M for this distribution of samples when n = 100 will be equal to . The standard error of M for all the possible samples of 100 is the standard error of M for all of the possible samples of 25. You can predict the size of the standard error of M for a sample size of 100 relative to a sample size of 25 because of the . If you were interested in how much time children spend in active play on weekends among this population of 2,431 children, would you actually compile all possible samples of a certain size?

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Question

You find that for this population of 2,431 children, the average number of minutes spent in active play on weekends is μ = 65.87, with a standard deviation of σ = 87.08.

You select a random sample of 25 children of daycare age in this same school district. In this sample, you find that the average number of minutes the children spend in active play on weekends is M = 59.28, with a standard deviation of s = 95.79.

The difference between M and μ is due to the .

Suppose you compile all possible samples of 25 children of daycare age in your school district. If you calculate the mean of each sample (M) and create a frequency distribution of these means, this distribution is referred to as the .

The mean of this distribution, that is, the mean of all the sample means (when n = 25), is the expected value of M and will be equal to . The standard deviation of this distribution is called the standard error of M and will be equal to .

Suppose you compile all possible samples of 100 children of daycare age in your school district; the expected value of M (when n = 100) will be the expected value of M for all of the possible samples of 25 children of daycare age in your school district. The standard error of M for this distribution of samples when n = 100 will be equal to .

The standard error of M for all the possible samples of 100 is the standard error of M for all of the possible samples of 25. You can predict the size of the standard error of M for a sample size of 100 relative to a sample size of 25 because of the .

If you were interested in how much time children spend in active play on weekends among this population of 2,431 children, would you actually compile all possible samples of a certain size?

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here