Question

For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.

Case studies showed that out of 10,562 convicts who escaped from certain prisons, only 7441 were recaptured.

(a) Let p represent the proportion of all escaped convicts who will eventually be recaptured. Find a point estimate for p. (Round your answer to four decimal places.)


(b) Find a 99% confidence interval for p. (Round your answers to three decimal places.)

lower limit
upper limit

Give a brief statement of the meaning of the confidence interval.

A.) 1% of all confidence intervals would include the true proportion of recaptured escaped convicts.

B.)1% of the confidence intervals created using this method would include the true proportion of recaptured escaped convicts.    

C.) 99% of all confidence intervals would include the true proportion of recaptured escaped convicts.99% of the confidence intervals created using this method would include the true proportion of recaptured escaped convicts.

(c) Is use of the normal approximation to the binomial justified in this problem? Explain.

A.) Yes; np < 5 and nq < 5.

B.) No; np > 5 and nq < 5.    

C.) Yes; np > 5 and nq > 5.

D.) No; np < 5 and nq > 5.

Answer 1

a)
sample proportion, pcap = 0.7045

b)
sample size, n = 10562
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.7045 * (1 - 0.7045)/10562) = 0.0044

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.7045 - 2.58 * 0.0044 , 0.7045 + 2.58 * 0.0044)
CI = (0.693 , 0.716)

lower limit = 0.693 ,
upper limit = 0.716

99% of the confidence intervals created using this method would include the true proportion of recaptured escaped convicts.

c)
C.) Yes; np > 5 and nq > 5.