Question

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.If no preliminary sample is taken to estimate p, how large a sample is necessary to be 95% sure that a point estimate will be within a distance of 0.08 from p? (Round your answer up to the nearest whole number.)

Answer 1

Answer:

n = 150.06

Explanation:

Since the confidence c = 95% = 0.95

α = 1 - 0.95 = 0.05

$\frac{\alpha }{2} = \frac{0.05}{2}=0.025$

z score of 0.025 is the same as the z score of 0.5 - 0.025 = 0.475

From the probability table, $z_{0.025}=z_{0.475}=1.96$

Also E = 0.08

Therefore the sample size n is  given by:

$n = \frac{1}{4}(\frac{z_{0.025}}{E} ) ^2=\frac{1}{4} *(\frac{1.96}{0.08})^2 =150.06$

n = 150.06

The sample must be at least 150.06  to be 95% sure that a point estimate will be within a distance of 0.08 from p