Questions: Find the sample size needed to estimate the percentage of adults who can wiggle their ears. Use a margin of error of 3 percentage points and use a confidence level of 99%. Complete parts (a) and (b) below. a. Assume that p̂ and q̂ are unknown. n= (Round up to the nearest integer.) b. Assume that 23% of adults can wiggle their ears. n= (Round up to the nearest integer.)

Solution Steps

To find the sample size needed to estimate the percentage of adults who can wiggle their ears when the proportions \( \hat{p} \) and \( \hat{q} \) are unknown, we use the formula:

\[ n = \frac{z^2 \cdot \hat{p} \cdot \hat{q}}{E^2} \]

where:

• \( z \) is the z-score corresponding to the desired confidence level,
• \( \hat{p} = 0.5 \) (maximum variability),
• \( \hat{q} = 1 - \hat{p} = 0.5 \),
• \( E = 0.03 \) (margin of error).

For a confidence level of \( 99\% \), the z-score is approximately \( 2.576 \). Plugging in the values:

\[ n = \frac{(2.576)^2 \cdot (0.5) \cdot (0.5)}{(0.03)^2} \approx 1844 \]

Thus, rounding up, the sample size needed is:

\[ \boxed{n = 1844} \]

Next, we calculate the sample size assuming that \( 23\% \) of adults can wiggle their ears. Here, we set:

\[ \hat{p} = 0.23 \quad \text{and} \quad \hat{q} = 1 - \hat{p} = 0.77 \]

Using the same formula:

\[ n = \frac{z^2 \cdot \hat{p} \cdot \hat{q}}{E^2} \]

Substituting the values:

\[ n = \frac{(2.576)^2 \cdot (0.23) \cdot (0.77)}{(0.03)^2} \approx 1306 \]

Thus, rounding up, the sample size needed is:

\[ \boxed{n = 1306} \]

Final Answer