Imagine that you are doing an exhaustive study on the children in all of the ele

Question

Imagine that you are doing an exhaustive study on the children in all of the elementary schools in your school district. You are particularly interested in how much time children spend reading on weekdays. You find that for this population of 2, 431 children, the average number of minutes spent reading on weekdays is mu = 11.03, with a standard deviation of sigma = 19.64. You select a random sample of 25 children of elementary school age in this same school district. In this sample, you find that the average number of minutes the children spend reading on weekdays is M = 9.93, with a standard deviation of s = 21.60. The difference between M and, mu is due to the sampling error. Suppose you compile all possible samples of 25 children of elementary school 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 sampling distribution of M 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 11.03. The standard deviation of this distribution is called the standard error of M and will be equal to 3.93. Suppose you compile all possible samples of 100 children of elementary school 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 25 children of elementary school 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

Explanation / Answer

1) Equal. Because the sample mean is equal to the population mean ir respective of the sample size.

2) SE = standard deviation / square root of the sample size = 19.64 / square root of 100 = 19.64 / 10 = 1.964.

3) half. Because S10/S5 = 1/2 where S10 and S5 are standard error for sample size 100 and 25 respectively.

4) Central limit theorem.

Thanks.

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