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)

Question

Accepted Answer

# 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:
$$
\begin{align*}
n&=\frac{(2.576)^{2}\times0.5\times0.5}{(0.02)^{2}}\
&=\frac{6.635776\times0.25}{0.0004}\
&=\frac{1.658944}{0.0004}\
& = 4147.36
\end{align*}
$$
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:
$$
\begin{align*}
n&=\frac{(2.576)^{2}\times0.22\times0.78}{(0.02)^{2}}\
&=\frac{6.635776\times0.1716}{0.0004}\
&=\frac{1.1387}{0.0004}\
&=2846.75
\end{align*}
$$
Round up to $n = 2847$

# Answer:
a. $n = 4148$
b. $n = 2847$

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QUESTION IMAGE

Question

Explanation:

Step1: Find the z - value

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

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

Answer: