Statistics Help: Sample Variability

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prettylittlepixels

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Consider a box containing a POPULATION of 5 identical blocks numbered 2, 4, 6, 8, and 10. Samples of size n = 2 are drawn WITH REPLACEMENT. Each draw is independent of other draws (the first block is drawn, observed, and returned to the box; then the second block is drawn and observed).

How would you describe the shape of the distribution of the POPULATION of five blocks?

A. Skewed to the right.

B. Normal or triangular.

C. Skewed to the left.

D. Uniform or rectangular.

E. None of these describes the population distribution.


What is the mean of the POPULATION?


What is the standard deviation of the POPULATION?


How many different samples of size two are possible?


What is the mean of the sample means for all samples of size n = 2?


What is the standard deviation of the sample means for all samples of size n = 2?


What conclusion do you reach when you compare the population standard deviation to the standard deviation of the means of all possible samples of size n = 2?

A. They are exactly the same.

B. Sample means are less variable than individual x values.

C. Sample means are more variable than individual x values.


Of the 20 statistics problems this week, this is the only one I am stuck on. I'm not looking for someone to do my homework for me, but I have read the required reading 3 times and I have no idea how to do this problem. Any explanation of formulas or how to do this problem is greatly appreciated. Thanks!
 
Where are you getting stuck? For instance, the mean of the listed values is easy to compute. Which parts are proving not to be so easy?

When you reply, please include your thoughts and efforts so far. Thank you! ;)
 
prettylittlepixels said:
Consider a box containing a POPULATION of 5 identical blocks numbered 2, 4, 6, 8, and 10. Samples of size n = 2 are drawn WITH REPLACEMENT. Each draw is independent of other draws (the first block is drawn, observed, and returned to the box; then the second block is drawn and observed).

How would you describe the shape of the distribution of the POPULATION of five blocks?

A. Skewed to the right.

B. Normal or triangular.

C. Skewed to the left.

D. Uniform or rectangular.

E. None of these describes the population distribution.


What is the mean of the POPULATION?


What is the standard deviation of the POPULATION?


How many different samples of size two are possible?


What is the mean of the sample means for all samples of size n = 2?


What is the standard deviation of the sample means for all samples of size n = 2?


What conclusion do you reach when you compare the population standard deviation to the standard deviation of the means of all possible samples of size n = 2?

A. They are exactly the same.

B. Sample means are less variable than individual x values.

C. Sample means are more variable than individual x values.


Of the 20 statistics problems this week, this is the only one I am stuck on. I'm not looking for someone to do my homework for me, but I have read the required reading 3 times and I have no idea how to do this problem. Any explanation of formulas or how to do this problem is greatly appreciated. Thanks!
Click to expand...
Do you think the POPULATION means the probability for a single block? Then you would be comparing the population distribution to the distribution of SAMPLE MEANS when n=2. You could start by defining the sample space and actually COUNTING how many ways there are to get each possible result when the sample size is n=2.
 
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Where I'm stuck

This is where I am on this problem:

How would you describe the shape of the distribution of the POPULATION of five blocks?

B. Normal or triangular. (Not sure if this is right though)

What is the mean of the POPULATION?

6 (Does this mean that they are looking for the mean of the values 2, 4, 6, 8 and 10 and not the mean of each 1 block?)

What is the standard deviation of the POPULATION?

2.82843 (Calculated this with the values 2, 4, 6, 8, and 10 on my calculator).


How many different samples of size two are possible?

I know 5!= 5x4x3x2x1= 120 but because n=2 and not 5, is it 120/2 = 60? When I wrote them all out below, I get 25, but since it is with replacement, is it 25x2= 50 because the block order can be reversed?

2-2 2-4 2-6 2-8 2-10
4-2 4-4 4-6 4-8 4-10
6-2 6-4 6-6 6-8 6-10
8-2 8-4 8-6 8-8 8-10
10-2 10-4 10-6 10-8 10-10


What is the mean of the sample means for all samples of size n = 2?

Not sure what this is asking... do I take each of the values above 2, 2, 2, 4, 2, 6, 2, 8, 2, 10... 10, 8, 10, 10 and find the mean?

What is the standard deviation of the sample means for all samples of size n = 2?

Same as above, do I use the values I wrote out?


What conclusion do you reach when you compare the population standard deviation to the standard deviation of the means of all possible samples of size n = 2?

A. They are exactly the same. Thinking this is correct if I use the values I wrote out because each value (2, 4, 6, 8, 10) is each written 10 times, making the mean 6 still?
 
prettylittlepixels said:
This is where I am on this problem:

How would you describe the shape of the distribution of the POPULATION of five blocks?

B. Normal or triangular. (Not sure if this is right though) X
All probabilities are equal: P(2) = P(4) = P(6) = P(8) = P(10) = 1/5

What is the mean of the POPULATION?

6 (Does this mean that they are looking for the mean of the values 2, 4, 6, 8 and 10 Yes. Mean=6 and not the mean of each 1 block?)

What is the standard deviation of the POPULATION?

2.82843 (Calculated this with the values 2, 4, 6, 8, and 10 on my calculator). Yes. Exact value is sqrt(8)


How many different samples of size two are possible?

I know 5!= 5x4x3x2x1= 120 [= permutations, or ways to arrange the 5 blocks] but because n=2 and not 5, is it 120/2 = 60? When I wrote them all out below, I get 25, but since it is with replacement, is it 25x2= 50 because the block order can be reversed? You have already included all the reversed block orders, for instance 2-10 and 10-2. 5 choices of 1st block times 5 choices of 2nd block = 25

2-2 2-4 2-6 2-8 2-10
4-2 4-4 4-6 4-8 4-10
6-2 6-4 6-6 6-8 6-10
8-2 8-4 8-6 8-8 8-10
10-2 10-4 10-6 10-8 10-10


What is the mean of the sample means for all samples of size n = 2? The sample mean for any of the 25 possible samples is the sum divided by 2:
(2+2)/2 = 1, (2+4)/2 = 3, (2+6)/2 = 4, . . . (10+10)/2 = 10
Each of the 25 samples has a mean between 2 and 10.

Not sure what this is asking... do I take each of the values above 2, 2, 2, 4, 2, 6, 2, 8, 2, 10... 10, 8, 10, 10 and find the mean? Find the mean of the 25 sample means.

What is the standard deviation of the sample means for all samples of size n = 2?

Same as above, do I use the values I wrote out? Find the standard deviation of the 25 sample means.



What conclusion do you reach when you compare the population standard deviation to the standard deviation of the means of all possible samples of size n = 2?

A. They are exactly the same. Thinking this is correct if I use the values I wrote out because each value (2, 4, 6, 8, 10) is each written 10 times, making the mean 6 still?
Click to expand...
You need to work out the mean and the standard deviation of the distribution of sample means. Your reasoning that the MEAN should be the same is appropriate, BUT you can't make the same argument for the Standard Deviation. It WILL be different, and you should be able to find the ratio.

EDIT: It may be useful for you to work out the actual distribution of sample means. If you write the 25 values in increasing order, you should get something like this:
2, 3, 3, 4, 4, 4, . . . 9, 9, 10
How would you describe that shape?

ANOTHER EDIT: What this exercise has led you through is a "simple" example of the effects of taking samples from a population distribution. When you take samples of size n, from ANY population distribution, and then look at the distribution of sample means,
a) what happens to the SHAPE of the distribution?
b) what happens to the MEAN of the distribution?
c) what happens to the STANDARD DEVIATION of the distribution? [You will soon be learning a theorem that tells you how the standard deviation of the distribution of sample means depends on sample size, n.]
 
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Thanks!

DrPhil said:
You need to work out the mean and the standard deviation of the distribution of sample means. Your reasoning that the MEAN should be the same is appropriate, BUT you can't make the same argument for the Standard Deviation. It WILL be different, and you should be able to find the ratio.

EDIT: It may be useful for you to work out the actual distribution of sample means. If you write the 25 values in increasing order, you should get something like this:
2, 3, 3, 4, 4, 4, . . . 9, 9, 10
How would you describe that shape?

ANOTHER EDIT: What this exercise has led you through is a "simple" example of the effects of taking samples from a population distribution. When you take samples of size n, from ANY population distribution, and then look at the distribution of sample means,
a) what happens to the SHAPE of the distribution?
b) what happens to the MEAN of the distribution?
c) what happens to the STANDARD DEVIATION of the distribution? [You will soon be learning a theorem that tells you how the standard deviation of the distribution of sample means depends on sample size, n.]
Click to expand...


Thank you so much for helping me with this! I only missed:

What is the standard deviation of the sample means for all samples of size n = 2?

I put 2.0412 as the answer, but the correct answer was 2. Now, I see what you were saying. Will the answer always be equal to the value of n?
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prettylittlepixels said:
Thank you so much for helping me with this! I only missed:

What is the standard deviation of the sample means for all samples of size n = 2?

I put 2.0412 as the answer, but the correct answer was 2. Now, I see what you were saying. Will the answer always be equal to the value of n?
spacer.gif
Click to expand...
It I take the sum of squares of the 25 sample means, divide by 25, and take the square root, I get about 1.96. If on the other hand I use the "unbiased" formula for the standard deviation, in which I divide by (N-1) instead of N, then I get 2.00.

Recalling that the standard deviation of the population distribution was sqrt(8), the ratio is

\(\displaystyle \sigma(\text{sample means}) / \sigma(\text{population}) = 2 / \sqrt 8 = 1 / \sqrt 2 = 1 / \sqrt n \)

The general rule is that for samples of size n, the distribution of sample means is normal, with the same mean as the population, and wth standard deviation = population / sqrt(n).
 
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