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 10,358 convicts who escaped from certain prisons, only 7582 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.

We are 99% confident that the true proportion of recaptured escaped convicts falls outside this interval.We are 1% confident that the true proportion of recaptured escaped convicts falls above this interval.     We are 1% confident that the true proportion of recaptured escaped convicts falls within this interval.We are 99% confident that the true proportion of recaptured escaped convicts falls within this interval.

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

No; np > 5 and nq < 5.No; np < 5 and nq > 5.     Yes; np > 5 and nq > 5.Yes; np < 5 and nq < 5.

Answer 1

Solution :

Given that,

n = 10358

x = 7582

a) Point estimate = sample proportion = = x / n = 7582 / 10358 = 0.7320

1 - = 1 - 0.7320 = 0.2680

b) At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.05

Z/2 = Z0.005  = 2.576

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.576 (((0.7320 * 0.2680) / 10358)

= 0.0112

A 99% confidence interval for population proportion p is ,

± E

= 0.7320 ± 0.0112

= ( 0.721, 0.743 )

lower limit = 0.721

upper limit = 0.743

We are 99% confident that true proportion of recaptured escaped convicts falls within this interval.

c) Yes; np > 5 and nq > 5