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 32 out of 815 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.

We are 99% confident that the true catch-and-release mortality rate falls within this interval.

We are 1% confident that the true catch-and-release mortality rate falls above this interval.     

We are 1% confident that the true catch-and-release mortality rate falls within this interval.

We are 99% confident that the true catch-and-release mortality rate falls outside this interval.

(c) Is 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

Solution :

Given that,

n = 815

x = 32

a) Point estimate = sample proportion = = x / n = 32 / 815 = 0.0393

1 - = 1 - 0.0393 = 0.9607

b) At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005  = 2.576

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

= 2.576 (((0.0393 * 0.9607) / 815)

= 0.0175

A 99% confidence interval for population proportion p is ,

± E

= 0.0393 ± 0.0175

= ( 0.022, 0.057 )

lower limit = 0.022

upper limit = 0.057

We are 99% confident that the true catch-and-release mortality rate falls within this interval.

c) Yes; np > 5 and nq > 5