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 five percentage points? complete parts (a) and (b) below.
a. assume that nothing is known about the percentage of adults who gamble online.
n = 271
(round up to the nearest integer.)
b. assume that 18% of all adults gamble online.
n =
(round up to the nearest integer.)

Explanation:

Step1: Recall the formula for sample size

The formula for sample size \(n\) when estimating a proportion is \(n=\frac{z^{2}\cdot p(1 - p)}{E^{2}}\). For a \(90\%\) confidence level, the critical value \(z\) (from the standard normal distribution) is \(z = 1.645\), and the margin of error \(E=0.05\).

Part (a)

When nothing is known about the proportion \(p\), we use \(p = 0.5\) (this gives the maximum value of \(p(1 - p)\)).
Substitute \(z = 1.645\), \(p=0.5\), \(1 - p = 0.5\), and \(E = 0.05\) into the formula:

$$
LATEXBLOCK0
$$

Since we round up to the nearest integer, \(n = 271\).

Part (b)

Given \(p = 0.18\) and \(1 - p=0.82\)
Substitute \(z = 1.645\), \(p = 0.18\), \(1 - p = 0.82\), and \(E = 0.05\) into the formula:

$$
LATEXBLOCK1
$$

Rounding up to the nearest integer, \(n = 161\).

Answer:

a. \(n = 271\)
b. \(n = 161\)