Question

1. In a study of crime, the FBI found that 13.2% of all Americans had been victims of crime during a one-year period. This result was based on a sample of 1,105 adults. (a) Estimate the percentage of US adults who were victims at the 90% confidence level. (b) Estimate the percentage of US adults who were victims at the 99% confidence level.

Answer 1

Answer:

(a) The estimate the percentage of US adults who were victims at the 90% confidence level is (12%, 15%).

(b) The estimate the percentage of US adults who were victims at the 99% confidence level is (11%, 16%).

Explanation:

The percentage of Americans who had been victims of crime during a one-year period is, $\hat p=0.132$.

The sample selected was of size, n = 1105.

The (1 - α)% confidence interval for the population proportion is:

$CI=\hat p\pm z_{\alpha/2}\cdot \sqrt{\frac{\hat p(1-\hat p)}{n}}$

(a)

The confidence level is 90%.

The critical value of z for 90% confidence level is, z = 1.645.

Compute the 90% confidence interval for the percentage of US adults who were victims as follows:

$CI=\hat p\pm z_{\alpha/2}\cdot \sqrt{\frac{\hat p(1-\hat p)}{n}}$

    $=0.132\pm 1.645\times\sqrt{\frac{0.132(1-0.132)}{1105}}\\\\=0.132\pm 0.0168\\\\=(0.1152, 0.1488)\\\\\approx (0.12, 0.15)$

Thus, the estimate the percentage of US adults who were victims at the 90% confidence level is (12%, 15%).

(b)

The confidence level is 99%.

The critical value of z for 99% confidence level is, z = 2.58.

Compute the 99% confidence interval for the percentage of US adults who were victims as follows:

$CI=\hat p\pm z_{\alpha/2}\cdot \sqrt{\frac{\hat p(1-\hat p)}{n}}$

    $=0.132\pm 2.58\times\sqrt{\frac{0.132(1-0.132)}{1105}}\\\\=0.132\pm 0.0263\\\\=(0.1057, 0.1583)\\\\\approx (0.11, 0.16)$

Thus, the estimate the percentage of US adults who were victims at the 99% confidence level is (11%, 16%).