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 10,069 convicts who escaped from certain prisons, only 7817 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. 1% of the confidence intervals created using this method would include the true proportion of recaptured escaped convicts. 1% 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. 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. 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

Given that,

n = 10069

x = 7817

Point estimate = sample proportion = = x / n = 7817 /10069=0.7763

1 - = 1-0.7763=0.2237

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 ( Using z table )

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

= 2.576 (((0.7763*0.2237) / 10069)

E = 0.011

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.7763 -0.011 < p < 0.7763+0.011

0.7653< p < 0.7873

The 99% confidence interval for the population proportion p is :lower limit  0.765, upper limit 0.787

(b)99% of all confidence intervals would include the true proportion of recaptured escaped convicts.