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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 $90 \%$. Complete parts (a) and (b) below.
a. Assume that $\hat{p}$ and $\hat{q}$ are unknown
\[
\mathrm{n}=752
\]
(Round up to the nearest integer.)
b. Assume that $24 \%$ of adults can wiggle their ears
\[
n=\square
\]
(Round Up to the nearest integer.)
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$838 \mathrm{PM}$
$11 / 6 / 2023$
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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 \(\boxed{549}\).