some states now allow online gambling. as a marketing manager for a casino, you need to determine the percentage of adults in those states who gamble online. how many adults must you survey, in order to be 95% confident that your estimate is in error by no more than two percentage points? complete parts (a) and (b) below.
a. assume that nothing is known about the percentage of adults who gamble online.
( n=square )
(round up to the nearest integer.)
b. assume that ( 10 % ) of all adults gamble online.
( n=square )
(round up to the nearest integer.)

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Accepted Answer

# Explanation:
## Step1: Recall the formula for sample size in proportion
The formula for sample size $n$ when estimating a proportion is $n=\frac{z^{2}\cdot p(1 - p)}{E^{2}}$. For a $95\%$ confidence level, $z = 1.96$ (from standard normal distribution), and the margin of error $E=0.02$ (since $2\%$ = $0.02$).
## Step2: Calculate $n$ when $p$ is unknown
When no prior information about $p$ is known, we use $p=0.5$ (this maximizes the value of $p(1 - p)$).
$$
\begin{align*}
n&=\frac{z^{2}\cdot p(1 - p)}{E^{2}}\
&=\frac{(1.96)^{2}\times0.5\times(1 - 0.5)}{(0.02)^{2}}\
&=\frac{3.8416\times0.25}{0.0004}\
&=\frac{0.9604}{0.0004}\
& = 2401
\end{align*}
$$
## Step3: Calculate $n$ when $p = 0.19$
Substitute $p = 0.19$ into the formula $n=\frac{z^{2}\cdot p(1 - p)}{E^{2}}$
$$
\begin{align*}
n&=\frac{(1.96)^{2}\times0.19\times(1 - 0.19)}{(0.02)^{2}}\
&=\frac{3.8416\times0.19\times0.81}{0.0004}\
&=\frac{3.8416\times0.1539}{0.0004}\
&=\frac{0.5912}{0.0004}\
&=1478
\end{align*}
$$

# Answer:
a. $n = 2401$
b. $n = 1478$

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Question

Explanation:

Step1: Recall the formula for sample size in proportion

Step2: Calculate \(n\) when \(p\) is unknown

Step3: Calculate \(n\) when \(p = 0.19\)

Answer: