Statistics, averages and deviations.

RossBrons · #1 $Old$ January 3rd, 2009, 09:22 PM

A population consists of the five numbers 2,3,6,8 and 11. Consider all possible samples of 2 that can be drawn with replacement from this population (i.e. as soon as soon as number has been drawn and its value noted it is replaced and can be drawn again). Find

i) The mean of the population.

ii) The standard deviation of the population.

iii) The mean of the sampling distibution of means.

iv) The standard deviation of the sampling distribution of means.

---------------------------------------

For i) I calculated the mean to be 6.0 however I found the standard deviation for part ii) to be 3.67 to which I am unsure of its reliability.

Can anyone please help with these please, I am keen to learn the method to solving these.

Regards

Aryth · #2 $Old$ January 4th, 2009, 12:27 AM

You have found the mean to be:

$\bar{x} = 6.0$

So, the standard deviation is:

$\sigma = \sqrt{\frac{1}{N}\sum_{i=1}^N (x_i - \bar{x})^2}$

Where N is the number of elements in the population, $x_i$ is the i-th element in the set, and $\bar{x}$ is the mean.

$\sigma = \sqrt{\frac{1}{5} \sum_{i = 1}^5 (x_i - 6.0)^2}$

$\sigma = \sqrt{\frac{1}{5}\left[ (2 - 6)^2 + (3 - 6)^2 + (6 - 6)^2 + (8 - 6)^2 + (11 - 6)^2 \right] }$

$\sigma = \sqrt{\frac{1}{5}\left[(-4)^2 + (-3)^2 + (0)^2 + (2)^2 + (5)^2 \right]}$

$\sigma = \sqrt{\frac{1}{5} (16 + 9 + 0 + 4 + 25)}$

$\sigma = \sqrt{\frac{1}{5} (54)}$

$\sigma = \sqrt{10.8}$

$\sigma = 3.29$

And there you go.

mr fantastic · #3 $Old$ January 4th, 2009, 01:24 AM

Quote:

Originally Posted by RossBrons $View Post$

A population consists of the five numbers 2,3,6,8 and 11. Consider all possible samples of 2 that can be drawn with replacement from this population (i.e. as soon as soon as number has been drawn and its value noted it is replaced and can be drawn again). Find

i) The mean of the population.

ii) The standard deviation of the population.

iii) The mean of the sampling distibution of means.

iv) The standard deviation of the sampling distribution of means.

---------------------------------------

For i) I calculated the mean to be 6.0 however I found the standard deviation for part ii) to be 3.67 to which I am unsure of its reliability.

Can anyone please help with these please, I am keen to learn the method to solving these.

Regards

For (iii) and (iv) read this: Sampling distribution - Wikipedia, the free encyclopedia

RossBrons · #4 $Old$ January 4th, 2009, 10:06 AM

Thanks, for the input Aryth, that was very helpful.

Is the standard deviation not...

$\sigma = \sqrt{\frac{1}{N-1}\sum_{i=1}^N (x_i - \bar{x})^2}$

And is the standard deviation of $\bar{X} = \frac{\sigma}{\sqrt{n}}$
Thereby meaning the standard deviation is not 3.29 but 1.47?
I'm lost, can you please untangle this mess?

Thanks

Mr Fantastic, I can't understand how to apply the math on that wiki article. Kind Regards,
Ross

vg284 · #5 $Old$ January 4th, 2009, 12:23 PM

the second one in an unbiased estimatior ?

Aryth · #6 $Old$ January 4th, 2009, 12:41 PM

Quote:

Originally Posted by RossBrons $View Post$

Thanks, for the input Aryth, that was very helpful.

Is the standard deviation not...

$\sigma = \sqrt{\frac{1}{N-1}\sum_{i=1}^N (x_i - \bar{x})^2}$

And is the standard deviation of $\bar{X} = \frac{\sigma}{\sqrt{n}}$
Thereby meaning the standard deviation is not 3.29 but 1.47?
I'm lost, can you please untangle this mess?

Thanks

Mr Fantastic, I can't understand how to apply the math on that wiki article. Kind Regards,
Ross

Both can be used, but the one I provided has a smaller error in estimation, but using the N-1 formula, the standard deviation is the 3.67 that you had initially, and not 1.47.

Your sampling distribution formula is the correct one.

RossBrons · #7 $Old$ January 4th, 2009, 12:44 PM

Quote:

Originally Posted by Aryth $View Post$

Both can be used, but the one I provided has a smaller error in estimation, but using the N-1 formula, the standard deviation is the 3.67 that you had initially, and not 1.47.

Your sampling distribution formula is the correct one.

Yeah, thanks for that. Greatly appreciated.

Has anyone got any ideas on how to tackle part iii and iv?

Cheers,
Ross

mr fantastic · #8 $Old$ January 4th, 2009, 02:46 PM

Quote:

Originally Posted by RossBrons $View Post$

Thanks, for the input Aryth, that was very helpful.

Is the standard deviation not...

$\sigma = \sqrt{\frac{1}{N-1}\sum_{i=1}^N (x_i - \bar{x})^2}$

[snip]

This is an unbiased estimator and is used to estimate the sd of a population from a sample taken from the population. You have only a sample (no mention of where it's come from) and you want the sd of only the sample. So the original formula you were given is the one to use.

mr fantastic · #9 $Old$ January 4th, 2009, 03:00 PM

Quote:

Originally Posted by RossBrons $View Post$

And is the standard deviation of $\bar{X} = \frac{\sigma}{\sqrt{n}}$ Mr F says: Yes.

Mr Fantastic, I can't understand how to apply the math on that wiki article. Kind Regards,
Ross

$\mu_{\overline{X}} = \mu$ .

$\sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}}$ .

Substitute $\mu = 6.0$ and $\sigma = 3.29$ and $n = 2$ .

It's that simple.