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 30 out of 869 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 the confidence intervals created using this method 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. 99% of all confidence intervals would include the true catch-and-release mortality rate. 1% of all confidence intervals would include the true catch-and-release mortality rate. (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:

    We are given that: In a combined study of northern pike, cutthroat trout, rainbow trout, and lake trout, it was found that 30 out of 869 fish died when caught and released using barbless hooks on flies or lures. Thus x = 30 and n = 869

Part 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.

Part b) Find a 99% confidence interval for p.

Formula:

where

Z_c is z critical value for c = 0.99 confidence level.

Find Area = ( 1+c)/2 = ( 1 + 0.99 ) / 2 = 1.99 /2 = 0.9950

Thus look in z table for Area = 0.9950 or its closest area and find corresponding z critical value.

From above table we can see area 0.9950 is in between 0.9949 and 0.9951 and both are at same distance from 0.9950, Hence corresponding z values are 2.57 and 2.58

Thus average of both z values is 2.575

Thus Z_c = 2.575

Thus

Thus lower and upper limits are:

Give a brief explanation of the meaning of the interval.

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

Part c) Is the normal approximation to the binomial justified in this problem?

n*p = 869 * 0.0345 = 29.98 = 30 > 5

n*q = n*(1-p) = 869 * ( 1 - 0.0345) = 869 * 0.9655= 839.02 = 839 > 5

Thus correct option is fourth option: Yes; np > 5 and nq > 5.