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 90% confident that your estimate is in error by no more than one percentage point? complete parts (a) and (b) below.
a. assume that nothing is known about the percentage of adults who gamble online.
n = 6766
(round up to the nearest integer.)
b. assume that 18% of all adults gamble online.
n = □
(round up to the nearest integer.)

Explanation:

Step1: Find the z - score

For a 90% confidence level, the significance level \(\alpha=1 - 0.90 = 0.10\), and \(\alpha/2=0.05\). Using the standard normal distribution table or a calculator, \(z_{\alpha/2}=z_{0.05}\approx1.645\). The margin of error \(E = 0.01\) (since 1 percentage point \(=0.01\)).

Step2: Use the formula for sample size when \(p\) is known

The formula for sample size \(n\) is \(n=\frac{z_{\alpha/2}^{2}\times p\times(1 - p)}{E^{2}}\). Given \(p = 0.18\) and \(1-p=1 - 0.18=0.82\), \(z_{\alpha/2}=1.645\), and \(E = 0.01\).

Substitute the values into the formula:

$$
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$$

Answer:

\(n = 3995\) (rounded up to the nearest integer)