Question

Random samples of size n = 2 are drawn from a finite population consisting of the numbers 5,6,7,8, and 9.
a. Find the mean of the population.
b. Find the standard deviation of the population.
c. Find the mean of the sampling distribution of the sample means.
d. Find the standard deviation of the sampling distribution of the sample means.

Answer 1

Answer:

Step-by-step explanation:

Mean =(2+4+6+8)/4

=20/4

=5

Standard deviation():

We know that,

σ? Σε – μ)? η –1

$\sigma^2=\frac{ (2-5 )^{2}+(4-5)^{2}+(6-5 )^{2}+(8-5 )^{2}}{4-1}$

$\sigma^2=\frac{ 9+1+1+9}{3}$

$\sigma^2=\frac{ 20}{3}$

$\sigma^2=6.6667$

$\sigma=\sqrt{6.6667}$

$\sigma=2.5819$

(b).

List the six possible random samples of size n = 2 that can be drawn from this population and calculate their means:

Samples are given by:

Samples Mean

(2,4) 3

(2,6) 4

(2,8) 5

(4,6) 5

(4,8) 6

(6,8) 7

(c).

Use the results of part (b) to construct the sampling distribution of the mean for random samples of size n =2 from the given population:

Sampling distribution of the mean is:

$\mu _{\bar{x}}=\mu =5$

(d).

Calculate the standard deviation of the sampling distribution obtained in part c) and verify the result by substituting n = 2, N = 4, and the value of σ obtained in part (a):

We know that,

$\sigma _{\bar{x}}=\frac{\sigma }{\sqrt{n}}$

$=\frac{2.5819 }{\sqrt{5}}$

$=2.5819/2.2361$

$=1.1546$

For n=2:

$\sigma _{\bar{x}}=\frac{\sigma }{\sqrt{n}}\times \sqrt{\frac{N-n}{N-1}}$

$=\frac{2.5819 }{\sqrt{2}}\times \sqrt{\frac{4-2}{4-1}}$

$=\frac{2.5819 }{1.4142}\times \sqrt{0.6667}$

$=1.8257\times 0.8165$

$\sigma _{\bar{x}} =1.4907$(Approximately)

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Answer 2

Given sample size $n=2$

The population consists of numbers 5,6,7,8 and 9.

a. Mean =$\mu=\frac{(2+4+6+8)}{4}=5$

b. Standard deviation $\sigma=\sqrt{\frac{\Sigma(x_{i}-\mu)^{2}}{N}$

$\sigma=\sqrt{\frac{(5-5)^{2}+(6-5)^{2}+(7-5)^{2}+(8-5)^{2}+(9-5)^{2}}{2}$$= \sqrt{\frac{1+4+9+16}{2}$$=\sqrt{15}=3.87$

c. Mean of the sampling distribution of the sample means

$\mu_{m}=\mu=5$

d. Standard deviation of the sampling distribution of the sample means

$\sigma_{m}=\frac{\sigma}{\sqrt{n}}=\frac{\sqrt{15}}{\sqrt{2}}=2.74$

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(b).

List the six possible random samples of size n = 2 that can be drawn from this population and calculate their means:

Samples are given by:

Samples Mean

(2,4) 3

(2,6) 4

(2,8) 5

(4,6) 5

(4,8) 6

(6,8) 7

(c).

Use the results of part (b) to construct the sampling distribution of the mean for random samples of size n =2 from the given population:

Sampling distribution of the mean is:

(d).

Calculate the standard deviation of the sampling distribution obtained in part c) and verify the result by substituting n = 2, N = 4, and the value of σ obtained in part (a):

We know that,

For n=2:

(Approximately)

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