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Suppose that in a random sample of 200 adults, 75 were victims of a crime. Estimate the true proportion of adults who were victims of a crime using a 99% confidence interval.

User Underyx
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Final answer:

To create a 99% confidence interval for the proportion of adults who were victims of a crime, we can use the formula CI = p ± Z * √((p*(1-p))/n). Given that 75 out of 200 adults were victims of a crime, the 99% confidence interval is approximately 0.315 to 0.435.

Step-by-step explanation:

To create a 99% confidence interval for the true proportion of adults who were victims of a crime, we can use the formula:

CI = p ± Z * √((p*(1-p))/n)

Where:

CI is the confidence interval

p is the proportion of the sample who were victims of a crime

Z is the z-value corresponding to the desired confidence level

n is the sample size

Given that 75 out of 200 adults were victims of a crime, the point estimate for the proportion is 75/200 = 0.375. With a 99% confidence level, the corresponding z-value is approximately 2.576.

Plugging these values into the formula, we get:

CI = 0.375 ± 2.576 * √((0.375*(1-0.375))/200)

Simplifying, we find that the 99% confidence interval for the true proportion of adults who were victims of a crime is approximately 0.315 to 0.435.

User Nghauran
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