Question

A researcher wishes to estimate the proportion of adults who have​ high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.04 with 99​% confidence if ​
(a) she uses a previous estimate of 0.52​?

(b) she does not use any prior​ estimates?

Answer 1

Answer:

Step-by-step explanation:

Solution:-

- The sample size = n

- The Error of estimation, E = 0.04

- The confidence level, CI = 99%

a)

What size sample should be obtained when she uses previous estimate of p = 0.52​?

- We are given the sample proportion p = 0.52, the required sample size is a function of confidence value and error of estimation (E):

                 $n = p*( 1 - p ) * (\frac{Z-critical}{E})^2$

Where,

- The critical value of the confidence level = 99% would be:

           significance level ( α ) = 1 - CI = 1 - 0.99 = 0.01

           Z-critical = Z_α/2 = Z_0.005 = 2.575  

- The required sample size (n) can be calculated:

            $n = 0.52*( 1 - 0.52 ) * (\frac{2.575}{0.04})^2\\\\n = 0.2304*(51.5)^2 = 611.0784$

- Hence, the minimum required sample size (n) should be = 612 adults.  

b)

- If the preliminary estimate of proportion is missing or not given, we are to assume the proportion p = 0.5.

- Similarly, repeat the calculations for sample size (n) when p = 0.5

               $n = 0.5*( 1 - 0.5 ) * (\frac{2.575}{0.04})^2\\\\n = 0.25*(51.5)^2 = 663.0625$

- Hence, the minimum required sample size (n) should be = 664 adults.