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 0.56 ? (b) she does not use any prior estimates? E Click the icon to view the table of critical values. (a) $n=1022$ (Round up to the nearest integer) (b) $n=\square$ (Round up to the nearest integer) Table of critical values \begin{tabular}{|ccc|} \hline \begin{tabular}{c} Level of Confidence, \\ $(1-\alpha) \cdot \mathbf{1 0 0} \%$ \end{tabular} & Area in Each Tail, $\frac{\boldsymbol{\alpha}}{\mathbf{2}}$ & Critical Value, $z_{1}$ \\ \hline $90 \%$ & 0.05 & 1645 \\ \hline $95 \%$ & 0.005 & 196 \\ \hline $99 \%$ & 0.005 & 2.575 \\ \hline \end{tabular} Print Done w an example Get more help - Clear all Check answer

Step 1 ：The sample size needed for a proportion estimate can be calculated using the formula: \(n = \frac{{Z^2 \cdot p \cdot (1-p)}}{{E^2}}\), where \(n\) is the sample size, \(Z\) is the z-score, \(p\) is the estimated proportion, and \(E\) is the margin of error.

Step 2 ：For part (a), we are given a previous estimate of \(p = 0.56\), a margin of error \(E = 0.04\), and a confidence level of 99%, which corresponds to a z-score \(Z = 2.575\).

Step 3 ：For part (b), if no prior estimate is given, we use \(p = 0.5\) as this maximizes the product \(p \cdot (1-p)\) and thus gives the largest possible sample size.

Step 4 ：Let's calculate the sample size for each part.

Step 5 ：For part (a), substituting the given values into the formula, we get \(n_a = \frac{{(2.575)^2 \cdot 0.56 \cdot (1-0.56)}}{{(0.04)^2}} = 1022\).

Step 6 ：For part (b), substituting the given values into the formula, we get \(n_b = \frac{{(2.575)^2 \cdot 0.5 \cdot (1-0.5)}}{{(0.04)^2}} = 1037\).

Step 7 ：Final Answer: (a) The required sample size when using a previous estimate of 0.56 is \(\boxed{1022}\). (b) The required sample size when not using any prior estimates is \(\boxed{1037}\).