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Data Analysis - Finding Sample Sizes for Confidence Intervals

INSTRUCTIONS:
Include a "summary statement" for each CI.

FACTS:
Elected officials in a small Florida town are preparing for the annual budget for their community. Specifically, they would like to estimate how much their constituents living in this town are typically paying each year in real estate taxes. Given that there are over 3000 homeowners in this small community, officials have decided to sample a representative subset of taxpayers and thoroughly study their tax payments. The latest frame of homeowners is given in the file P8_49.XLS.

Suppose that elected officials in this community would like to estimate the proportion of taxpayers whose annual real estate tax payments exceed $2.000.

PROBLEM: (part a, b only)

(a) What sample size would be required to generate a 99% confidence interval for the proportion with a half-length of 0.10? Assume for now that the relevant population proportion p is close to 0.50.

(b) Assume now that the officials discover old tax records that suggest that approximately 30% of all property owners in this community pay more than $2000 annually in real estate taxes. What sample size would now be required to generate a 99% confidence interval for this population with a half-length of 0.10?

CL 0.99 0.99
Z 2.5758 2.5758
p 0.5 0.3
B 0.1 0.1

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Solution Preview

(a) What sample size would be required to generate a 99% confidence interval for the proportion with a half-length of 0.10? Assume for now that the relevant population proportion p is close to 0.50.

In this problem, we want to use the sample proportion, to estimate the population proportion. We want to be 99% confident that the difference between the estimate and the actual value of the population proportion is at most 0.100 in absolute value.

The formula for the sample size needed to estimate under these ...

Solution Summary

The solution finds sample sizes for confidence intervals. A summary statement for each different confidence interval is given for a small town in Florida.

$2.19