1. **State the problem:** We need to find the minimum sample size $n$ required to estimate the prevalence of Alzheimer's disease with a margin of error no more than 3.4% (0.034) at a conservative estimate. 2. **Formula used:** The margin of error (ME) for estimating a population proportion is given by: $$ ME = z \times \sqrt{\frac{p(1-p)}{n}} $$ where $z$ is the z-score corresponding to the desired confidence level, $p$ is the estimated proportion, and $n$ is the sample size. 3. **Important rule:** For a conservative estimate (maximum sample size), use $p = 0.5$ because $p(1-p)$ is maximized at $p=0.5$. 4. **Rearrange the formula to solve for $n$:** $$ n = \frac{z^2 \times p(1-p)}{ME^2} $$ 5. **Assuming a 95% confidence level, $z = 1.96$:** 6. **Plug in the values:** $$ n = \frac{(1.96)^2 \times 0.5 \times (1-0.5)}{(0.034)^2} $$ 7. **Calculate numerator:** $$ (1.96)^2 = 3.8416 $$ $$ 0.5 \times 0.5 = 0.25 $$ $$ \text{Numerator} = 3.8416 \times 0.25 = 0.9604 $$ 8. **Calculate denominator:** $$ (0.034)^2 = 0.001156 $$ 9. **Calculate $n$:** $$ n = \frac{0.9604}{0.001156} $$ 10. **Simplify with cancellation:** $$ n = \frac{\cancel{0.9604}}{\cancel{0.001156}} = 830.8 $$ 11. **Interpretation:** Since sample size must be a whole number and to ensure margin of error is not exceeded, round up: $$ n = 831 $$ **Final answer:** The minimum sample size required is **831**.