To determine the minimum sample size needed to estimate the percentage of adults who can wiggle their ears, we can use the formula:
n = (Z^2 * p * q) / E^2
Where:
n = sample size
Z = Z-score corresponding to the desired confidence level (in this case, a confidence level of 95% corresponds to a Z-score of approximately 1.96)
p = estimated proportion of adults who can wiggle their ears (21% or 0.21)
q = 1 - p (proportion of adults who cannot wiggle their ears)
E = margin of error (5 percentage points or 0.05)
Let's plug in the values:
n = (1.96^2 * 0.21 * (1-0.21)) / (0.05^2)
n = (3.8416 * 0.21 * 0.79) / 0.0025
n = (0.646842816) / 0.0025
n ≈ 258.736
Therefore, the minimum sample size required to estimate the percentage of adults who can wiggle their ears with a margin of error of 5 percentage points and a confidence level of 95% is approximately 259.