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 = \square \\) (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 \(\alpha/2 = 0.05\).
From the standard normal distribution table, \(z_{\alpha/2}=z_{0.05}\approx1.645\).

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

When \(\hat{p}\) and \(\hat{q}\) are unknown, we use \(\hat{p}=\hat{q} = 0.5\) (this maximizes the value of \(\hat{p}\hat{q}\)).
The formula for sample size \(n\) is \(n=\frac{z_{\alpha/2}^{2}\hat{p}\hat{q}}{E^{2}}\), where \(E = 0.02\) (since \(2\) percentage points \(=0.02\)).
Substitute the values: \(n=\frac{(1.645)^{2}\times0.5\times0.5}{(0.02)^{2}}\).
First, calculate \((1.645)^{2}=2.706025\), and \(0.5\times0.5 = 0.25\), \((0.02)^{2}=0.0004\).
Then \(n=\frac{2.706025\times0.25}{0.0004}=\frac{0.67650625}{0.0004}=1691.265625\).

Answer:

\(1692\)