Question

According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft ('Consumer fraud and,' 2008). Assume a state had 283 complaints of identity theft out of 1400 consumer complaints . Calculate a 92% confidence interval for the proportion of all complaints that are complaints of identity theft in that state.

Answer 1

The 92% confidence interval for the proportion of identity theft complaints is calculated to be between 18.39% and 22.03% based on the sample data provided.

Step-by-step explanation:

To calculate the 92% confidence interval for the proportion of complaints that are for identity theft, we will use the formula for the confidence interval for a population proportion:

Confidence Interval = p ± z*(√(p(1-p)/n))

Where:

• p is the sample proportion (number of identity theft complaints divided by the total number of complaints),
• z is the z-score corresponding to the desired confidence level,
• n is the sample size (total number of complaints).

In this case, p is 283/1400, n is 1400, and the z-score for a 92% confidence level is approximately 1.75 (you would normally use a z-score table or statistical software to find this value).

Calculating the confidence interval:

• p = 283/1400 = 0.2021,
• Standard Error (SE) = √(0.2021*(1-0.2021)/1400) ≈ 0.0104,
• Margin of Error (ME) = 1.75 * SE ≈ 0.0182.

Therefore, the 92% confidence interval is:

0.2021 ± 0.0182 = (0.1839, 0.2203)

This means we are 92% confident that the true proportion of all complaints that are for identity theft is between 18.39% and 22.03%.