Question

for the following situation, find the mean and standard deviation of the population. list all samples (with replacement) of the given size from that population and find the mean of each. find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population. the word counts of five essays are 508, 639, 552, 612, and 575. use a sample size of 2. the mean of the sampling distribution is 577.2 (round to two decimal places as needed.) the standard deviation of the sampling distribution is 32.34 (round to two decimal places as needed.) choose the correct comparison of the population and sampling distribution below. a. the means are equal but the standard deviation of the sampling distribution is smaller. b. the means and standard deviations are equal c. the means are not equal and the standard deviation of the sampling distribution is larger. d. the means are equal but the standard deviation of the sampling distribution is larger. e. the means are not equal and the standard deviation of the sampling distribution is smaller.

Explanation:

Step1: Calculate population mean

Let the population data be $x_1 = 508,x_2=639,x_3 = 552,x_4=612,x_5 = 575$. The population mean $\mu=\frac{\sum_{i = 1}^{n}x_i}{n}$, where $n = 5$. So $\mu=\frac{508 + 639+552+612+575}{5}=\frac{2886}{5}=577.2$.

Step2: Calculate population standard - deviation

First, calculate the deviations from the mean: $(508 - 577.2)^2=(-69.2)^2 = 4788.64$, $(639 - 577.2)^2=(61.8)^2=3819.24$, $(552 - 577.2)^2=(-25.2)^2 = 635.04$, $(612 - 577.2)^2=(34.8)^2 = 1211.04$, $(575 - 577.2)^2=(-2.2)^2 = 4.84$. The sum of squared deviations $\sum_{i = 1}^{n}(x_i-\mu)^2=4788.64 + 3819.24+635.04+1211.04 + 4.84=10458.8$. The population standard - deviation $\sigma=\sqrt{\frac{\sum_{i = 1}^{n}(x_i - \mu)^2}{n}}=\sqrt{\frac{10458.8}{5}}\approx\sqrt{2091.76}\approx45.74$.

Step3: Analyze the relationship

We are given that the mean of the sampling distribution is $577.2$ and the standard deviation of the sampling distribution is $32.34$. The mean of the population is $577.2$ and the standard deviation of the population is approximately $45.74$. The means are equal and the standard deviation of the sampling distribution is smaller.

Answer:

A. The means are equal but the standard deviation of the sampling distribution is smaller.