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 $90\%$ confident that your estimate is in error by no more than one percentage point? Complete parts (a) and (b) below. $\newline$ (a). Assume that nothing is known about the percentage of adults who gamble online. $\newline$ $n=6766$ $\newline$ (Round up to the nearest integer.) $\newline$ (b). Assume that $18\%$ of all adults gamble online. $\newline$ $n=\square$ $\newline$ (Round up to the nearest integer.)

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Q. 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

90\%

confident that your estimate is in error by no more than one percentage point? Complete parts (a) and (b) below.

\newline

(a). Assume that nothing is known about the percentage of adults who gamble online.

\newline

n=6766

\newline

(Round up to the nearest integer.)

\newline

(b). Assume that

18\%

of all adults gamble online.

\newline

n=\square

\newline

(Round up to the nearest integer.)

Sample Size Formula: To determine the sample size needed for a given confidence level and margin of error, we use the formula for sample size in a proportion: $\newline$ $n = \left(\frac{Z_{\alpha/2} \cdot \sqrt{p(1-p)}}{E}\right)^2$ $\newline$ where: $\newline$ - $n$ is the sample size, $\newline$ - $Z_{\alpha/2}$ is the z-score corresponding to the desired confidence level, $\newline$ - $p$ is the estimated proportion of the population that has the attribute of interest, $\newline$ - $E$ is the desired margin of error.
Part (a) Scenario: For part (a), since nothing is known about the percentage of adults who gamble online, we use $p = 0.5$ ( $50$ %) for the worst-case scenario, which provides the maximum possible sample size.
Confidence Level and Z-Score: The desired confidence level is $90\%$ , so we look up the z-score that corresponds to the $90\%$ confidence level. The z-score for a $90\%$ confidence interval is approximately $1.645$ (since $90\%$ confidence corresponds to an alpha level of $0.10$ , and the z-score that leaves $5\%$ in each tail is $1.645$ ).
Margin of Error: The desired margin of error $E$ is $1$ percentage point, or $0$ . $01$ when expressed as a proportion.
Sample Size Calculation for Part (a): Substitute the values into the formula to calculate the sample size for part (a): $\newline$ $n = \left(\frac{1.645 \cdot \sqrt{0.5(1-0.5)}}{0.01}\right)^2$ $\newline$ $n = \left(\frac{1.645 \cdot \sqrt{0.25}}{0.01}\right)^2$ $\newline$ $n = \left(\frac{1.645 \cdot 0.5}{0.01}\right)^2$ $\newline$ $n = \left(\frac{0.8225}{0.01}\right)^2$ $\newline$ $n = 82.25^2$ $\newline$ $n = 6766.0625$
Rounding Up: Since we cannot survey a fraction of a person, we round up to the nearest whole number. Therefore, the sample size for part (a) is $6767$ .
Part (b) Scenario: For part (b), we are given that $18$ % of all adults gamble online, so $p = 0.18$ .
Sample Size Calculation for Part (b): We use the same formula as before, but with $p = 0.18$ and $1-p = 0.82$ : $\newline$ $n = \left(\frac{1.645 \cdot \sqrt{0.18(0.82)}}{0.01}\right)^2$ $\newline$ $n = \left(\frac{1.645 \cdot \sqrt{0.1476}}{0.01}\right)^2$ $\newline$ $n = \left(\frac{1.645 \cdot 0.3842}{0.01}\right)^2$ $\newline$ $n = \left(\frac{0.6321}{0.01}\right)^2$ $\newline$ $n = 63.21^2$ $\newline$ $n = 3995.6641$
Sample Size Calculation for Part (b): We use the same formula as before, but with $p = 0.18$ and $1-p = 0.82$ : $\newline$ $n = \left(\frac{1.645 \cdot \sqrt{0.18(0.82)}}{0.01}\right)^2$ $\newline$ $n = \left(\frac{1.645 \cdot \sqrt{0.1476}}{0.01}\right)^2$ $\newline$ $n = \left(\frac{1.645 \cdot 0.3842}{0.01}\right)^2$ $\newline$ $n = \left(\frac{0.6321}{0.01}\right)^2$ $\newline$ $n = 63.21^2$ $\newline$ $n = 3995.6641$ Again, we round up to the nearest whole number. Therefore, the sample size for part (b) is $3996$ .