some states now allow online gambling. as a marketing manager for a casino, you need to determine the…

some states now allow online gambling. as a marketing manager for a casino, you need to determine the percentage of adults in the state you survey in order to be 95% confident that your estimate is in error by no more than three percentage points? complete parts a and b.\n a. assume that nothing is known about the percentage of adults who gamble online.\n n = (round up to the nearest integer)\n b. assume that 17% of all adults gamble online.\n n = (round up to the nearest integer.)
Answer
Explanation:
Step1: Recall sample - size formula for proportion
The formula for sample size $n$ in estimating a proportion $p$ is $n=\frac{z^{2}\cdot p(1 - p)}{E^{2}}$, where $z$ is the z - score corresponding to the confidence level, $p$ is the estimated proportion, and $E$ is the margin of error. For a 95% confidence level, the z - score $z = 1.96$, and the margin of error $E=0.03$.
Step2: Calculate $n$ when $p$ is unknown
When nothing is known about the proportion $p$, we use $p = 0.5$ (this gives the maximum value of $p(1 - p)$). Substitute $z = 1.96$, $p = 0.5$, and $E=0.03$ into the formula: [ \begin{align*} n&=\frac{(1.96)^{2}\times0.5\times(1 - 0.5)}{(0.03)^{2}}\ &=\frac{3.8416\times0.5\times0.5}{0.0009}\ &=\frac{3.8416\times0.25}{0.0009}\ &=\frac{0.9604}{0.0009}\ &\approx1067.11 \end{align*} ] Round up to the nearest integer, so $n = 1068$.
Step3: Calculate $n$ when $p = 0.17$
Substitute $z = 1.96$, $p = 0.17$, and $E = 0.03$ into the formula $n=\frac{z^{2}\cdot p(1 - p)}{E^{2}}$: [ \begin{align*} n&=\frac{(1.96)^{2}\times0.17\times(1 - 0.17)}{(0.03)^{2}}\ &=\frac{3.8416\times0.17\times0.83}{0.0009}\ &=\frac{3.8416\times0.1411}{0.0009}\ &=\frac{0.54215}{0.0009}\ &\approx602.39 \end{align*} ] Round up to the nearest integer, so $n = 603$.
Answer:
a. 1068 b. 603