Question

Past surveys reveal that 30% of tourists going to Las Vegas to gamble spend more than $1,000. The Visitor's Bureau of Las Vegas wants to update this percentage. How many tourists should be randomly surveyed to estimate the population proportion with a 99.9% level of confidence and a 1% margin of error?

Answer 1

To estimate the population proportion with a 99.9% level of confidence and a 1% margin of error, the Visitor's Bureau of Las Vegas should randomly survey 2,687 tourists.

Step-by-step explanation:

To estimate the population proportion with a 99.9% level of confidence and a 1% margin of error, we need to use the formula for sample size:

n = (Z^2 * p * (1 - p)) / E^2

Where:

• n is the sample size
• Z is the z-score for the desired level of confidence
• p is the estimated proportion
• E is the margin of error

In this case, the estimated proportion is 0.3 (30%), the margin of error is 0.01 (1%), and the z-score for a 99.9% confidence level is approximately 3.29.

Plugging these values into the formula:

n = (3.29^2 * 0.3 * (1 - 0.3)) / 0.01^2

Solving this equation gives us a sample size of approximately 2,687. Therefore, the Visitor's Bureau of Las Vegas should randomly survey 2,687 tourists to estimate the population proportion with a 99.9% level of confidence and a 1% margin of error.