# Thread: Statistics! One more that needs to be checked!

1. ## Statistics! One more that needs to be checked!

Hi all:

thanks for all your help - especially Mr. Fantastic! You have all been great help! I have one more question - I have done my part and just want to make sure that what I did was correct!

As always am putting my response in BOLD!! Lets hope my response is right!

When obtaining a confidence interval for a population mean in the case of a finite population of size N and a sample size n which is greater than 0.05N, the margin of error is multiplied by the following finite population correction factor:

$
{\sqrt{N-n}}/{\sqrt{N-1}}
$

a. Find the 95% confidence interval for the mean of 200 weights if a sample of 33 of those weights yields a mean of 147.7 pounds and a standard deviation of 26.7 pounds.
$We have {\sqrt{N-n}}/{\sqrt{N-1}} = {\sqrt{200-33}}/{\sqrt{200-1}}$
=0.916076405
Now, going back to the data and using the table we have the confidence score as 1.96.
147.7 - 1.96(26.7)/sqrt(13) < mu < 147.7 + 1.96(26.7)/sqrt(13)
simplifying
=> 133.19 < mu < 162.21
Now I assume that we multiply each of these by .9160... ?? Am i right?
122.0122 < mu < 148.596

b. Assume the sample of 33 weights is selected from a population of 2000 weights. Would you use the same technique as in part a? Why or why not? (Base your decision and explanation on the criteria for using this correction factor that is given above.)

No we cannot use the same method as 2000 x 0.05 = 100 . 100 is the minimum number of samples we need before we can proceed along the same lines. The result will be skewed if we proceed in a similar fashion as above.

Thanks in advance
Darren!

2. Originally Posted by darren_a1
Hi all:

thanks for all your help - especially Mr. Fantastic! You have all been great help! I have one more question - I have done my part and just want to make sure that what I did was correct!

As always am putting my response in BOLD!! Lets hope my response is right!

When obtaining a confidence interval for a population mean in the case of a finite population of size N and a sample size n which is greater than 0.05N, the margin of error is multiplied by the following finite population correction factor:

$
{\sqrt{N-n}}/{\sqrt{N-1}}
$

a. Find the 95% confidence interval for the mean of 200 weights if a sample of 33 of those weights yields a mean of 147.7 pounds and a standard deviation of 26.7 pounds.
$We have {\sqrt{N-n}}/{\sqrt{N-1}} = {\sqrt{200-33}}/{\sqrt{200-1}}$
=0.916076405
Now, going back to the data and using the table we have the confidence score as 1.96.
147.7 - 1.96(26.7)/sqrt(13) < mu < 147.7 + 1.96(26.7)/sqrt(13)
simplifying
=> 133.19 < mu < 162.21
Now I assume that we multiply each of these by .9160... ?? Am i right?
122.0122 < mu < 148.596

b. Assume the sample of 33 weights is selected from a population of 2000 weights. Would you use the same technique as in part a? Why or why not? (Base your decision and explanation on the criteria for using this correction factor that is given above.)

No we cannot use the same method as 2000 x 0.05 = 100 . 100 is the minimum number of samples we need before we can proceed along the same lines. The result will be skewed if we proceed in a similar fashion as above.

Thanks in advance
Darren!
b. looks OK. In a., the population sd is unknown - you're using the sample sd as an estimate - and n is small. So I think you'd want to use the t-statistic, NOT the z-statistic .....

3. ## Thanks!

Mr F:

thanks for the quick reply.

My question would be do I still multiply by .9160.. with the answers that I get? Is that the correct thing to do?

Cheers

Darren

4. ## Final answer check

Mr. F:

I finally did the math on that

here is what i get before i multiply with the factor of 0.916076405

162.37 < mu < 133.03

after multiplying by 0.916076405 we have

148.74 < mu < 121.86

Hope some one can help me out with this!

Thanks

Darren

5. Originally Posted by darren_a1
Mr. F:

I finally did the math on that

here is what i get before i multiply with the factor of 0.916076405

162.37 < mu < 133.03

after multiplying by 0.916076405 we have

148.74 < mu < 121.86
Mr F says: Yes, this is the interval you want.

Hope some one can help me out with this!

Thanks

Darren
And I hope you see that as n gets larger the FPCF gets smaller. This makes sense since you become more confident in your estimate as n increases.

When n = N the FPCF = 0 and your confidence interval consists of the single point ${\color{blue}\bar{x}}$. This makes sense since when n = N you've obviously sampled the entire population and so ${\color{blue}\bar{x} = \mu}$.