Question

The FBI wants to determine the effectiveness of their 10 Most Wanted list. To do so, they need to find out the fraction of people who appear on the list that are actually caught. Suppose a sample of 716 suspected criminals is drawn. Of these people, 286 were captured. Using the data, estimate the proportion of people who were caught after being on the 10 Most Wanted list. Enter your answer as a fraction or a decimal number rounded to three decimal places. Also, construct the 90% confidence interval for the population proportion of people who are captured after appearing on the 10 Most Wanted list. Round your answers to three decimal places.

Answer 1

Answer:

1.) .399

2.) {.369, .429}

Step-by-step explanation:

The proportion of criminals in this sample that were caught was $\frac{286}{716}$=.39944, .399 rounded to three decimal places.

To construct the confidence interval, you need three pieces of information: the statistic, the critical value, and the standard error.

The statistic is given in the first part of the problem with the proportion of criminals in the sample that were caught was .399.

The critical value, we are told, is the z equivalent of 90%, or 1.645. You can find this value using a z table or with the inverse normal function on a calculator.

Finally, we need the standard error. The formula for standard error for a proportion with a single population is $\sqrt{\frac{(proportion)(1-proportion)}{sample size} }$so in this situation it would be $\sqrt{\frac{(.399)(.601)}{716} }$=.0183.

The confidence interval would be .399±.018×1.645 or .399±.030 {.369, .429}