Question

American Medical Association wants to estimate the proportion of medical doctors who had a solo practice. How many doctors should be included in the study if the association wants 99% confident that the estimate is within 5 percentage point of the true population proportion?

Answer 1

Answer:

The number of doctors 'n' = 6.76

Step-by-step explanation:

Explanation:-

Given Estimate of the true population proportion = 5 % = 0.05

Given level of significance ∝ = 99% or 1%

Z₀.₀₁ = 2.576

The margin of error is determined by

$M.E = \frac{2.576 X \sqrt{p(1-p)} }{\sqrt{n} }$

we know that $\sqrt{p(1-p)} \leq \frac{1}{2}$

$0.05 = \frac{2.576 X \frac{1}{2} }{\sqrt{n} }$

Cross multiplication , we get

$\sqrt{n} = \frac{2.576 X \frac{1}{2} }{5}$

√n = 0.2576≅0.26

squaring on both sides,we get

n = 0.0676 X 100

n=  6.76

Final answer:-

The number of doctors 'n' = 6.76