<PRE><B>Question 438470</B>
Chapter 11 bankruptcy in a certain year. Let p be the proportion of small businesses that declared Chapter 11 bankruptcy.
----------------------- 
a. If no preliminary sample is taken, how large a sample is needed to be 95% sure that sample proportion will be within 10% of p?
---
Since E = z*sqrt(pq/n)
--- 
sqrt(n) = z*sqrt(pq)/E
----
n = [z/E]^2(pq)
-----
Your Problem:
n = [1.96/0.10]^2*[(1/2)(1/2)
Notice: By using p=q=1/2 you avoid bias in the process.
---
n = [19.6]^2*(1/4)
n = 97 when rounded up
======================================

b. In a preliminary random sample of 38 businesses, it was found that six had declared Chapter 11. How many more small businesses should be included to be 95% sure that the sample proportion is within 10% of p?
-----
I guess you need 97-38 = 59 more
===============
cheers,
Stan H.

</PRE>