Question 1110446: 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 36 out of 839 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.
99% 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.
1% 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 by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 36/839 is point estimate probability or 0.0429
99% CI has z of 2.576
the interval width is 2.576* sqrt (p*(1-p))/n
this is 2.576*sqrt (.0429*.9571)/839; sqrt term is 0.0700
the interval width is 0.0180
the interval is (0.025, 0.061)
The best choice of meaning is the second--that 99% of all CIs created this way would include the true mortality rate. It wouldn't include all confidence intervals.
Yes, both np and nq are greater than 5.
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