Question

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 1,002 randomly selected adults. The report states, "One can say with \(95 \%\) confidence that the margin of sampling error is ±3 percentage points." Explain how this statement can be justified.

Short answer

The statement can be justified if the calculated margin of error is approximately the same as the given one. This requires understanding and applying the formula for the margin of error in a confidence interval calculation.

Step-by-step solution

Understand the terms

First, we need to understand what each term means. The 25% is the sample estimate, or in other words, the point estimate, which is calculated based on the given sample. The ±3 percentage points is the margin of error, representing the range that the actual population parameter is expected to fall within, from the point estimate with 95% confidence level.

Calculating margin of error

The margin of error in a confidence interval calculation can be found using the formula: \(E = Z \cdot \sqrt{{p(1-p)}/n}\), where E is the margin of error, Z is the Z-value, p is the sample proportion and n is the size of the sample. The Z-value corresponds directly to the chosen confidence level. For a confidence level of 95%, the corresponding Z-value is approximately 1.96 (from Z-table or standard statistical programs). Now, substituting the given values, \(E = 1.96 \cdot \sqrt{{0.25 \cdot 0.75}/1002}\), will get you the margin of error.

Justify the statement

To justify the statement, compare the calculated margin of error with the given margin of error in the problem (±3 percentage points). If they are approximately the same, then the statement can be justified.