Step 1 :Given that the margin of error (E) is 3 percentage points or 0.03, and the z-score (Z) for a 90% confidence level is 1.645.
Step 2 :For part a, we assume that the population proportion (p) and its complement (q) are unknown. In this case, we use p = q = 0.5. Substituting these values into the formula for the sample size when the population proportion is unknown, \(n = \frac{Z^2 * p * q}{E^2}\), we get \(n = \frac{(1.645)^2 * 0.5 * 0.5}{(0.03)^2}\), which rounds up to 752.
Step 3 :For part b, we assume that 24% of adults can wiggle their ears, so p = 0.24 and q = 1 - p = 0.76. Substituting these values into the formula for the sample size, we get \(n = \frac{(1.645)^2 * 0.24 * 0.76}{(0.03)^2}\), which rounds up to 549.
Step 4 :Final Answer: For part a, the sample size needed is \(\boxed{752}\). For part b, the sample size needed is \(\boxed{549}\).