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 10,001 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 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 outside this interval.     We are 1% confident that the true proportion of recaptured escaped convicts falls above 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.Yes; np < 5 and nq < 5.     Yes; np > 5 and nq > 5.No; np > 5 and nq < 5.

Answer 1

Solution :

Given that,

n = 10001

x = 7582

Point estimate = sample proportion = = x / n = 7582/10001=0.7581

1 -   = 1-0.7581 =0.2419

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576* (((0.7581*0.2419) /10001 )

= 0.011

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.7581-0.011 < p <0.7581+ 0.011

0.747< p < 0.769

The 99% confidence interval for the population proportion p is :lower limit= 0.747 , upper limit= 0.769

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