You are a travel agent and wish to estimate, within $98 \%$ confidence, the proportion of vacationers who use an online service or the Internet to make reservations for lodging. Your estimate must be accurate to within $2 \%$ of the population proportion.
a. Find the minimum sample size necessary if no preliminary estimate is available.
b. Find the minimum sample size necessary if a previous study indicated that $91 \%$ of vacationers said they used an online service or the internet to make their lodging reservations.

Step 1 ：First, we need to use the formula for the sample size in a proportion estimation problem, which is \(n = \frac{{Z^2 * p * (1-p)}}{{E^2}}\), where n is the sample size, Z is the Z-score corresponding to the desired confidence level, p is the estimated proportion of the population, and E is the desired margin of error.

Step 2 ：For part a, since no preliminary estimate is available, we use \(p = 0.5\) to get a conservative estimate. The Z-score for a 98% confidence level is approximately 2.33, and the desired margin of error E is 0.02. Substituting these values into the formula, we get \(n_a = \frac{{(2.33)^2 * 0.5 * (1-0.5)}}{{(0.02)^2}}\), which simplifies to \(n_a = 3394\).

Step 3 ：For part b, we use the given estimate \(p = 0.91\). Again, the Z-score for a 98% confidence level is approximately 2.33, and the desired margin of error E is 0.02. Substituting these values into the formula, we get \(n_b = \frac{{(2.33)^2 * 0.91 * (1-0.91)}}{{(0.02)^2}}\), which simplifies to \(n_b = 1112\).

Step 4 ：So, the minimum sample size necessary if no preliminary estimate is available is \(\boxed{3394}\).

Step 5 ：And the minimum sample size necessary if a previous study indicated that 91% of vacationers said they used an online service or the internet to make their lodging reservations is \(\boxed{1112}\).