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 five 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 confidence level and margin of error

The confidence level is \(99\%\), so the significance level \(\alpha = 1 - 0.99=0.01\). Then \(\alpha/2=0.005\). From the standard normal distribution table, \(z_{\alpha/2}=z_{0.005} = 2.576\). The margin of error \(E = 0.05\) (since \(5\%\) points \(= 0.05\)). When no prior estimate of the proportion \(p\) is known, we use \(p = 0.5\) (this maximizes the value of \(p(1 - p)\)).

Step2: Use the formula for sample size in proportion

The formula for sample size \(n\) in estimating a proportion is \(n=\frac{z_{\alpha/2}^{2}\times p(1 - p)}{E^{2}}\).
Substitute \(z_{\alpha/2}=2.576\), \(p = 0.5\), \(1-p=0.5\) and \(E = 0.05\) into the formula:

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Answer:

\(n = 664\)