Question

For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. In a combined study of northern pike, cutthroat trout, rainbow trout, and lake trout, it was found that 38 out of 821 fish died when caught and released using barbless hooks on flies or lures. All hooks were removed from the fish. (a) Let p represent the proportion of all pike and trout that die (i.e., p is the mortality rate) when caught and released using barbless hooks. 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 explanation of the meaning of the interval. 99% of all confidence intervals would include the true catch-and-release mortality rate. 1% of the confidence intervals created using this method would include the true catch-and-release mortality rate. 1% of all confidence intervals would include the true catch-and-release mortality rate. 99% of the confidence intervals created using this method would include the true catch-and-release mortality rate. (c) Is the normal approximation to the binomial justified in this problem? Explain. Yes; np < 5 and nq < 5. Yes; np > 5 and nq > 5. No; np > 5 and nq < 5. No; np < 5 and nq > 5

Answer 1

Given that,

n = 821

x = 38

Point estimate = sample proportion = = x / n = 38/821=0.0463

1 - = 1-0.0463=0.9537

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 (((0.0463*0.9537) /821 )

E = 0.019

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.0463 - 0.019< p <0.0463+ 0.019

0.027< p < 0.065

The 99% confidence interval for the population proportion p is :lower limit = 0.027, upper limit =0.065

(B)99% of all confidence intervals would include the true catch-and-release mortality rate.

(C)np > 5 and nq > 5