Question

Case studies showed that out of 10,154 convicts who escaped from certain prisons, only 7488 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.99% of the confidence intervals created using this method would include the true proportion of recaptured escaped convicts.     

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

d.99% of all confidence intervals 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.No; np > 5 and nq < 5.

b.Yes; np > 5 and nq > 5.     

c.Yes; np < 5 and nq < 5.

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

Answer 1

Solution :

Given that,

n = 10154

x = 7488

a) Point estimate = sample proportion = = x / n = 7488 / 10154 = 0.7374

1 - = 1 - 0.7374 = 0.2626

b) At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z_{0.005 = 2.576}

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

= 2.576 (((0.7374 * 0.2626) / 10154)

= 0.011

A 99% confidence interval for population proportion p is ,

± E

= 0.7374 ± 0.011

= (0.726, 0.748 )

lower limit = 0.726

upper limit = 0.748

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

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

yes,10154 * 0.7374 = 7487.55 > 5 and 10154 * 0.2626 = 2666.44 > 5