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 90%. complete parts (a) and (b) below.
a. assume that \\( \hat { p } \\) and \\( \hat { q } \\) are unknown.
n = 1692
(round up to the nearest integer.)
b. assume that 24% of adults can wiggle their ears
n =
(round up to the nearest integer.)

Explanation:

Step1: Determine the value of \(z_{\alpha/2}\)

For a 90% confidence level, \(\alpha=1 - 0.90=0.10\), and \(\alpha/2 = 0.05\). Using the standard normal distribution table, \(z_{\alpha/2}=z_{0.05}\approx1.645\).
The margin of error \(E = 0.02\) (since 2 percentage points \(=2\%=0.02\)).

Step2: Use the formula for sample size when \(\hat{p}\) is known

The formula for sample size \(n\) in estimating a proportion is \(n=\frac{z_{\alpha/2}^{2}\hat{p}\hat{q}}{E^{2}}\). Given \(\hat{p}=0.24\), then \(\hat{q}=1-\hat{p}=1 - 0.24 = 0.76\).
Substitute the values into the formula:

$$
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Answer:

\(n = 1234\) (rounded up to the nearest integer)