What do you understand by unbiased estimate of population variance B) an infinite population consists of the…

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What do you understand by unbiased estimate of population variance B) an infinite population consists of the number 1,2,3 A) write down the population mean of the sample B) calculate the population variance C) random sample of size 2 are obtained the population D) write down all possible sample of sin 2 taken with replacement E) obtain the expected value of the mean and the population variance.

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Here are the answers to your questions:

A) What do you understand by unbiased estimate of population variance?
An unbiased estimate of population variance is a sample statistic that, on average, equals the true population variance. If you were to take many samples from a population and calculate this estimate for each sample, the average of all these estimates would be equal to the actual population variance. The formula for an unbiased estimate of population variance is $s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}$ .

B) An infinite population consists of the number 1, 2, 3.

A) Write down the population mean of the population.

List the population values.
The population consists of the numbers $1, 2, 3$.
Calculate the population mean ( $\mu$ ).
$\mu = \frac{\sum x_i}{N} = \frac{1 + 2 + 3}{3} = \frac{6}{3} = \boxed{2}$
B) Calculate the population variance.
Use the population mean from part A ( $\mu = 2$ ).
Calculate the squared differences from the mean.
$(1 - 2)^2 = (-1)^2 = 1$
$(2 - 2)^2 = (0)^2 = 0$
$(3 - 2)^2 = (1)^2 = 1$
Calculate the population variance ( $\sigma^2$ ).
$\sigma^2 = \frac{\sum (x_i - \mu)^2}{N} = \frac{1 + 0 + 1}{3} = \frac{2}{3} = \boxed{0.6667}$
C) Random sample of size 2 are obtained from the population.
This is an introductory statement for the following parts.

D) Write down all possible samples of size 2 taken with replacement.
The population is $\{1, 2, 3\}$ and the sample size is $n=2$ .
The possible samples taken with replacement are:
$\begin{array}{ccc} (1, 1) & (1, 2) & (1, 3) \\ (2, 1) & (2, 2) & (2, 3) \\ (3, 1) & (3, 2) & (3, 3) \end{array}$
E) Obtain the expected value of the mean and the variance of the sample means.
List the sample means for each sample from part D.
For each sample $(x_1, x_2)$ , the sample mean is $\bar{x} = \frac{x_1 + x_2}{2}$ .
$\begin{array}{cc} \text{Sample} & \text{Sample Mean } (\bar{x}) \\ (1, 1) & 1.0 \\ (1, 2) & 1.5 \\ (1, 3) & 2.0 \\ (2, 1) & 1.5 \\ (2, 2) & 2.0 \\ (2, 3) & 2.5 \\ (3, 1) & 2.0 \\ (3, 2) & 2.5 \\ (3, 3) & 3.0 \end{array}$
Calculate the expected value of the sample means ( $E(\bar{X})$ ).
This is the mean of all possible sample means.
$E(\bar{X}) = \frac{\sum \bar{x}}{N^n} = \frac{1.0 + 1.5 + 2.0 + 1.5 + 2.0 + 2.5 + 2.0 + 2.5 + 3.0}{9}$ $E(\bar{X}) = \frac{18.0}{9} = \boxed{2}$
(Note: This is equal to the population mean $\mu$ , as expected.)
Calculate the variance of the sample means ( $Var(\bar{X})$ ).
This is the variance of the sampling distribution of the mean.
$Var(\bar{X}) = \frac{\sum (\bar{x} - E(\bar{X}))^2}{N^n}$
Using $E(\bar{X}) = 2$ :
$\sum (\bar{x} - 2)^2 = (1.0-2)^2 + (1.5-2)^2 + (2.0-2)^2 + (1.5-2)^2 + (2.0-2)^2 + (2.5-2)^2 + (2.0-2)^2 + (2.5-2)^2 + (3.0-2)^2$ $= (-1)^2 + (-0.5)^2 + (0)^2 + (-0.5)^2 + (0)^2 + (0.5)^2 + (0)^2 + (0.5)^2 + (1)^2$ $= 1 + 0.25 + 0 + 0.25 + 0 + 0.25 + 0 + 0.25 + 1 = 3$ $Var(\bar{X}) = \frac{3}{9} = \frac{1}{3} = \boxed{0.3333}$
(Alternatively, $Var(\bar{X}) = \frac{\sigma^2}{n} = \frac{2/3}{2} = \frac{2}{6} = \frac{1}{3}$ .)

Step-by-step solution

List the population values.

Calculate the population mean ( $\mu$ ).

Use the population mean from part A ( $\mu = 2$ ).

Calculate the squared differences from the mean.

Calculate the population variance ( $\sigma^2$ ).

List the sample means for each sample from part D.

Calculate the expected value of the sample means ( $E(\bar{X})$ ).

Calculate the variance of the sample means ( $Var(\bar{X})$ ).

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Step-by-step solution

List the population values.

Calculate the population mean (μ\muμ).

Use the population mean from part A (μ=2\mu = 2μ=2).