Answer to Question #186675 in Statistics and Probability for Acierlynx

Question

Solve the given problems.

1. A population consists of the four numbers 1, 2, 4 and 5. List all the possible

samples of size n = 3 which can be drawn with replacement from the population.

Find the following:

a. Population mean

b. Population variance

c. Population standard deviation.

d. Mean of the sampling distribution of sample means

e. Variance of the sampling distribution of sample means

f. Standard deviation of the sampling distribution of sample means

2. A population consists of the four numbers 1, 2, 8, and 9. List all the possible

samples of size n = 2 which can be drawn without replacement from the population. 

Find the following:

a. Population mean

b. Population variance

c. Population standard deviation

d. Mean of the sampling distribution of sample means

e. Variance of the sampling distribution of sample means

f. Standard deviation of the sampling distribution of sample means

Answer 1

We have given the population,

1, 2, 4 and 5

Sample size = 3

a.) Population mean $= \dfrac{1+2+4+5}{4} = \dfrac{12}{4} = 3$

b.)Population variance $= \dfrac{(1-3)^2+(2-3)^2+(4-3)^2+(5-3)^2}{4} = 2.5$

c.)Population standard deviation $= \sqrt{Population \hspace{3mm} variance} = 1.58$

d.)Mean of the sampling distribution of sample means $= \dfrac{11.65}{4} = 2.9125$

e.)Variance of the sampling distribution of sample means

$= \dfrac{(2.33-2.91)^2+(3.33-2.91)^2+(3.66-2.91)^2+(2.33-2.91)^2}{4} = 0.225$

f.)Standard deviation of the sampling distribution of sample means $= \sqrt{75.135} =0.46$

We have given the population,

1, 2, 8, and 9

Sample size = 2

a.)Population mean $= \dfrac{1+2+8+9}{4} = \dfrac{20}{4} = 5$

b.)Population variance $= \dfrac{(1-5)^2+(2-5)^2+(8-5)^2+(9-5)^2}{4} = 12.5$

c.)Population standard deviation $= \sqrt{12.5} = 3.53$

d.)Mean of the sampling distribution of sample means = $\dfrac{30.3}{6} = 5.05$

e.)Variance of the sampling distribution of sample means

$= \dfrac{(1.5-5.05)^2+(4.5-5.05)^2+(5-5.05)^2+(5-5.05)^2+(5.5-5.05)^2+(8.8-5.05)^2}{6} = 4.16$

f.)Standard deviation of the sampling distribution of sample means $= \sqrt{4.16} = 2.03$