For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. The National Council of Small Businesses is interested in the proportion of small businesses that declared Chapter 11 bankruptcy last year. Since there are so many small businesses, the National Council intends to estimate the proportion from a random sample. Let p be the proportion of small businesses that declared Chapter 11 bankruptcy last year. (a) If no preliminary sample is taken to estimate p, how large a sample is necessary to be 99% sure that a point estimate p̂ will be within a distance of 0.08 from p? (Round your answer up to the nearest whole number.) ?? small businesses (b) In a preliminary random sample of 30 small businesses, it was found that eleven had declared Chapter 11 bankruptcy. How many more small businesses should be included in the sample to be 99% sure that a point estimate p̂ will be within a dista

Given Information : 

The National Council of Small Businesses is interested in the proportion of small businesses that declared Chapter 11 bankruptcy last year. Since there are so many small businesses, the National Council intends to estimate the proportion from a random sample.

Let p be the proportion of small businesses that declared Chapter 11 bankruptcy last year. 

(a)

At 99% confidence level , $\alpha$ = 1 - 0.98 = 0.02 . 

Critical value of z : z_c = z_{(1-$\alpha$)/2} = z_0.01 . 

Using standard normal table z_0.01  = 2.32  

Provided margin of error = 0.08 . It's formula is given as : 

$M.E.=z·\sqrt{\frac{p(1-p)}{n}}$ 

For no preliminary , take standard proportion p = 0.05 . 

Therefore , sample size n computed as : 

$n=\frac{z^2·p(1-p)}{E^2}$=$\frac{2.{32}^2·0.5(1-0.5)}{0.{08}^2}$

n = 210.25 

Largest sample size : n = 211 small businesses