Question

Case studies showed that out of 10,409 convicts who escaped from certain prisons, only 7958 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

Answer 1

Solution :

Given that,

n = 10409

x = 7958

Point estimate = sample proportion = = x / n = 7958 /10409=0.7645

1 - = 1 - 0.7645=0.2355

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 (((0.0.7645*0.2355) /10409 )

= 0.011

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.7645 -0.011 < p < 0.7645 +0.011

0.754< p < 0.776

The 99% confidence interval for the population proportion p is : lower limit =0.754 upper limit=0.776