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Chapter 9: Problem 28

The Gallup Organization conducts an annual survey on crime. It was reported that \(25 \%\) of all households experienced some sort of crime during the past year. This estimate was based on a sample of 1002 randomly selected households. The report states, "One can say with \(95 \%\) confidence that the margin of sampling error is \(\pm 3\) percentage points." Explain how this statement can be justified.

Short Answer

Expert verified
The statement can be justified by calculating the margin of error for a \(95%\) confidence interval using the sample size and proportion given. The calculated value of approximately 0.03, or \(3\%\), concurs with the reported margin of error.

Step by step solution

01

Identifying Given Variables

We have a sample size (n) of 1002 households, out of which \(25%\) have experienced some sort of crime. This constitutes our sample proportion (p虃), which equals 0.25. The margin of error (E) stated is \(3%\), or 0.03. We need to justify this value.
02

Understanding Confidence Interval Calculations

A \(95%\) confidence interval for a proportion is typically calculated as p虃 卤 z * (sqrt(( p虃 * (1 - p虃))/n)) where the z-score represents the number of standard deviations away from the mean a particular proportion is. In this case, for a \(95%\) confidence interval, the z-score is approximately 1.96.
03

Calculating and Comparing Margin of Error

Let's calculate the margin of error as E = z * sqrt(( p虃 * (1 - p虃))/n). Plugging in the given values, we get E = 1.96 * sqrt((0.25 * 0.75)/1002) which equals approximately 0.03. This is consistent with the reported margin of error of \( \pm 3% \), validating the claim.

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Most popular questions from this chapter

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