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 four percentage points? Complete parts (a) and (b) below.
a. Assume that nothing is known about the percentage of adults who gamble online.
\[
\mathrm{n}=
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(Round $u_{n}$ to the nearest integer.)
Answer
\(\boxed{1037}\) is the number of adults you must survey in order to be 99% confident that your estimate is in error by no more than four percentage points.
Step 1 ：We are given that we want to be 99% confident, which corresponds to a z-score of 2.576. The margin of error E is 4%, or 0.04. Since we don't know anything about the proportion of adults who gamble online, we'll use the most conservative estimate, which is p = 0.5. This gives us the maximum possible sample size, ensuring that our estimate will be within the desired margin of error.
Step 2 ：We use the formula for sample size in a proportion, which is \(n = \frac{Z^2 * p * (1-p)}{E^2}\), where n is the sample size, Z is the z-score, p is the estimated proportion of the population, and E is the desired margin of error.
Step 3 ：Substitute the given values into the formula: Z = 2.576, p = 0.5, E = 0.04.
Step 4 ：Calculate the sample size, n = 1037.
Step 5 ：\(\boxed{1037}\) is the number of adults you must survey in order to be 99% confident that your estimate is in error by no more than four percentage points.
