Question

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 99% 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 = □
(round up to the nearest integer.)

Explanation:

Step1: Determine the formula for sample size

When nothing is known about the proportion \(p\), we use \(p = 0.5\) (this gives the maximum value of \(p(1 - p)\)). The formula for sample size \(n\) for a proportion is \(n=\frac{z^{2}_{\alpha/2}\times p(1 - p)}{E^{2}}\). For a \(99\%\) confidence interval, \(\alpha=1 - 0.99=0.01\), and \(\alpha/2=0.005\). The \(z\) - score \(z_{\alpha/2}=z_{0.005}\). From the standard normal table, \(z_{0.005} = 2.576\). The margin of error \(E = 0.02\) (since \(2\%\) points \(=0.02\)).

Step2: Substitute values into the formula

Substitute \(z_{\alpha/2}=2.576\), \(p = 0.5\), \(1 - p=0.5\), and \(E = 0.02\) into the formula \(n=\frac{z^{2}_{\alpha/2}\times p(1 - p)}{E^{2}}\).

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Step3: Round up the value

Since we need to round up to the nearest integer, \(n = 4148\).

Answer:

\(4148\)