A researcher wishes to estimate the proportion of adults who have high-speed Internet access what size sample should be obtained if she wishes to the estimate to be within 0.01 with 95% confidence if
(a) she uses a prevlous estimate of 0.56 ?
(b) she does not use any prior estimates?
Click the icon to view the table of critical values.
(a) $n=\square$ (Round up to the nearest integer.)
Table of critical values
\begin{tabular}{|ccc|}
\hline \begin{tabular}{c}
Level of Confidence, \\
$(\mathbf{1}-\boldsymbol{\alpha}) \cdot \mathbf{1 0 0 \%}$
\end{tabular} & Area in Each Tail, $\frac{\boldsymbol{\alpha}}{\mathbf{2}}$ & Critical Value, $\mathrm{z}_{\mathbf{1}}$ \\
\hline $\mathbf{9 0 \%}$ & 0.05 & 1645 \\
\hline $95 \%$ & 0.025 & 1.96 \\
\hline $99 \%$ & 0.005 & 2575 \\
\hline
\end{tabular}
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Step 1 ：The problem is asking for the sample size needed to estimate the proportion of adults who have high-speed Internet access with a certain level of confidence and precision. The formula for sample size in this case is given by: \(n = \frac{{Z^2 \cdot p \cdot (1-p)}}{{E^2}}\), where: n is the sample size, Z is the z-score, which corresponds to the desired confidence level (in this case, 1.96 for 95% confidence), p is the estimated proportion (in this case, 0.56), E is the desired margin of error (in this case, 0.01).

Step 2 ：Plugging these values into the formula, we get: \(n = \frac{{(1.96)^2 \cdot 0.56 \cdot (1-0.56)}}{{(0.01)^2}}\).

Step 3 ：Calculating the above expression, we find that the sample size needed for part (a) is 9466. This means that the researcher would need to sample 9466 adults in order to estimate the proportion of adults who have high-speed Internet access with a margin of error of 0.01 and a confidence level of 95%, assuming a previous estimate of 0.56.

Step 4 ：For part (b) of the question, if the researcher does not use any prior estimates, we use a value of 0.5 for p (since this gives the maximum variability and thus the largest sample size). We'll use the same formula and values for Z and E.

Step 5 ：Plugging these values into the formula, we get: \(n = \frac{{(1.96)^2 \cdot 0.5 \cdot (1-0.5)}}{{(0.01)^2}}\).

Step 6 ：Calculating the above expression, we find that the sample size needed for part (b) is 9604. This means that the researcher would need to sample 9604 adults in order to estimate the proportion of adults who have high-speed Internet access with a margin of error of 0.01 and a confidence level of 95%, without using any prior estimates.

Step 7 ：Final Answer: For part (a), the required sample size is \(\boxed{9466}\). For part (b), the required sample size is \(\boxed{9604}\).