In: Statistics and Probability
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 833 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.
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 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.
No; np < 5 and nq > 5.
No; np > 5 and nq < 5.
Yes; np > 5 and nq > 5.
Solution :
Given that,
a) n = 833
x = 36
Point estimate = sample proportion = = x / n = 36 / 833 = 0.0432
1 - = 1 - 0.0432 = 0.9568
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 (((0.0432 * 0.9568) / 833)
= 0.018
A 99% confidence interval for population proportion p is ,
± E
= 0.0432 ± 0.018
= (0.025, 0.061)
lower limit = 0.025
upper limit = 0.061
99% of the confidence intervals created using this method would include the true catch-and-release mortality rate.
c) Yes; np > 5 and nq > 5