Question

find the sample size needed to estimate the percentage of adults who can wiggle their ears. use a margin of error of 2 percentage points and use a confidence level of 99%. complete parts (a) and (b) below.
a. assume that \\( \hat { p } \\) and \\( \hat { q } \\) are unknown
\\( n = \square \\)
(round up to the nearest integer)
b. assume that 22% of adults can wiggle their ears.
\\( n = \square \\)
(round up to the nearest integer)

Explanation:

Step1: Find the z - value

For a 99% confidence level, the z - value \(z_{\alpha/2}\) is 2.576. The margin of error \(E = 0.02\) (since 2 percentage points \(= 0.02\)).

Step2: Formula for sample size when \(\hat{p}\) and \(\hat{q}\) are unknown

When \(\hat{p}\) and \(\hat{q}\) are unknown, we use \(\hat{p}=\hat{q} = 0.5\). The formula for sample size \(n=\frac{z_{\alpha/2}^{2}\hat{p}\hat{q}}{E^{2}}\).
Substitute \(z_{\alpha/2}=2.576\), \(\hat{p} = 0.5\), \(\hat{q}=0.5\), and \(E = 0.02\) into the formula:

$$
LATEXBLOCK0
$$

Round up to \(n = 4148\)

Step3: Formula for sample size when \(\hat{p}\) is known

Given \(\hat{p}=0.22\), then \(\hat{q}=1 - \hat{p}=0.78\).
Use the formula \(n=\frac{z_{\alpha/2}^{2}\hat{p}\hat{q}}{E^{2}}\)
Substitute \(z_{\alpha/2}=2.576\), \(\hat{p} = 0.22\), \(\hat{q}=0.78\), and \(E = 0.02\) into the formula:

$$
LATEXBLOCK1
$$

Round up to \(n = 2847\)

Answer:

a. \(n = 4148\)
b. \(n = 2847\)