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.


Step 1 of 2: Suppose a sample of 369 suspected criminals is drawn. Of these people, 118 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.


Step 2 of 2: Suppose a sample of 369 suspected criminals is drawn. Of these people, 118 were captured. Using the data, 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.


Lower Endpoint:_____ Upper Endpoint:_____

Answer 1

Answer:

(1) The proportion of people who were caught after being on the 10 Most Wanted list is 0.319.

(2) 90% confidence interval for the population proportion of people who are captured after appearing on the 10 Most Wanted list is [0.279 , 0.359].

Step-by-step explanation:

We are given that the FBI wants to determine the effectiveness of their 10 Most Wanted list.

Suppose a sample of 369 suspected criminals is drawn. Of these people, 118 were captured.

(1) The proportion of people who were caught after being on the 10 Most Wanted list  =  $\frac{118}{369}$  = 0.319

(2) Firstly, the pivotal quantity for 90% confidence interval for the population proportion is given by;

                          P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample proportion of criminals captured = 0.319

           n = sample of criminals = 369

           p = population proportion

Here for constructing 90% confidence interval we have used One-sample z proportion test statistics.

So, 90% confidence interval for the population proportion, p is ;

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5%

                                           level of significance are -1.645 & 1.645}  

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p = [$\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$]

 = [ $0.319-1.645 \times {\sqrt{\frac{0.319(1-0.319)}{369} } }$ , $0.319+1.645 \times {\sqrt{\frac{0.319(1-0.319)}{369} } }$ ]

 = [0.279 , 0.359]

Therefore, 99% confidence interval for the population proportion of passing test scores is [0.279 , 0.359].

Lower Endpoint = 0.279

Upper Endpoint = 0.359