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:

$\displaystyle

{\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.

**$\displaystyle We have {\sqrt{N-n}}/{\sqrt{N-1}} = {\sqrt{200-33}}/{\sqrt{200-1}}$**

=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!