Answer to Question #324581 in Statistics and Probability for Christina

Question

Consider a population consisting of the values (1, 3, 4). Solve for the following:

A.

a. Compute the population mean.

b. Compute the population variance.

c. Compute the population standard deviation.

B.

a. List all the possible samples of size 2 that can be drawn from the population with

replacement.

b. Compute the mean of the sampling distribution of the means.

c. Compute the variance of the sampling distribution of the means.

d. Compute the standard deviation of the sampling distribution of the means.

e. Construct the probability histogram of x̅ with replacements when n = 2.

C.

a. List all the possible samples of size 2 that can be drawn from the population without

replacement.

b. Compute the mean of the sampling distribution of the means.

c. Compute the variance of the sampling distribution of the means.

d. Compute the standard deviation of the sampling distribution of the means.

e. Construct the probability histogram of x̅ with replacements when n = 2.

Answer 1

$Aa:\\\mu =\frac{1+3+4}{3}=\frac{8}{3}\\b:\\\sigma ^2=\frac{\left( 1-\frac{8}{3} \right) ^2+\left( 3-\frac{8}{3} \right) ^2+\left( 4-\frac{8}{3} \right) ^2}{3}=\frac{14}{9}\\c:\\s=\sqrt{\sigma ^2}=\sqrt{\frac{14}{9}}=1.24722\\Ba:\\\left( 1,1 \right) ,\bar{x}=1\\\left( 1,3 \right) ,\bar{x}=2\\\left( 1,4 \right) ,\bar{x}=2.5\\\left( 3,1 \right) ,\bar{x}=2\\\left( 3,3 \right) ,\bar{x}=3\\\left( 3,4 \right) ,\bar{x}=3.5\\\left( 4,1 \right) ,\bar{x}=2.5\\\left( 4,3 \right) ,\bar{x}=3.5\\\left( 4,4 \right) ,\bar{x}=4\\b:\\\mu _{\bar{x}}=\frac{1+2+2.5+2+3+3.5+2.5+3.5+4}{9}=\frac{8}{3}\\c:\\{\sigma _{\bar{x}}}^2=\frac{\left( 1-\frac{8}{3} \right) ^2+\left( 2-\frac{8}{3} \right) ^2+\left( 2.5-\frac{8}{3} \right) ^2+\left( 2-\frac{8}{3} \right) ^2+\left( 3-\frac{8}{3} \right) ^2+\left( 3.5-\frac{8}{3} \right) ^2+\left( 2.5-\frac{8}{3} \right) ^2+\left( 3.5-\frac{8}{3} \right) ^2+\left( 4-\frac{8}{3} \right) ^2}{9}=\frac{7}{9}\\d:\\\sigma _{\bar{x}}=\sqrt{{\sigma _{\bar{x}}}^2}=\sqrt{\frac{7}{9}}=0.881917\\Ca:\\\left( 1,3 \right) ,\bar{x}=2\\\left( 1,4 \right) ,\bar{x}=2.5\\\left( 3,4 \right) ,\bar{x}=3.5\\b:\\\mu _{\bar{x}}=\frac{2+2.5+3.5}{3}=\frac{8}{3}\\c:\\{\sigma _{\bar{x}}}^2=\frac{\left( 2-\frac{8}{3} \right) ^2+\left( 2.5-\frac{8}{3} \right) ^2+\left( 3.5-\frac{8}{3} \right) ^2}{3}=\frac{7}{18}\\d:\\\sigma _{\bar{x}}=\sqrt{\frac{7}{18}}=0.62361$