Confidence IntervalsDate: 06/09/99 at 00:12:50 From: Faizal Mohammed Subject: Confidence interval question Dear Sir, Can you please answer the following question for me? A fisheries research team caught 68 rainbow trout in a lake, tagged them, and released them. They waited a month for the fish to be evenly distributed and went fishing again. This time they landed 219 rainbow trout, 16 of which were tagged. (a) Find 90% confidence limits for the proportion of tagged fish in the lake. (b) Estimate the number of rainbow trout in the lake. (c) Establish a 90% confidence interval for the number of rainbow trout in the lake. Answers: (a) (0.0441, 0.1020) (b) 931 (c) (667, 1541) Thank you for your time. Date: 06/09/99 at 10:48:48 From: Doctor Anthony Subject: Re: Confidence interval question >(a) Find 90% confidence limits for the proportion of tagged fish in > the lake. Taking 16/219 as the proportion giving an estimate of the proportion in the whole population we have n = 219, p = 16/219, q = 203/219 For 90% confidence interval for p we have 16/219 - p 219(16/219 - p) --------------------- = ----------------- = +-1.645 sqrt[16 * 203/219^3] sqrt(16 * 203/219) 16 - 219p ------------ = +-1.645 3.8511 and so 219p = 16 +- 6.335 p = 16/219 +- 0.028927 p = 0.07306 +- 0.028927 So the confidence interval is 0.044132 < p < 0.10199 >(b) Estimate the number of rainbow trout in the lake. Suppose there are N fish in the lake. You tag 68 of them, so there are now 68 fish with tags and N-68 without tags. You then return and catch 219 more fish; 16 are found to have tags. The probability of this event is C(68,16) x C(N-68,203) ----------------------- C(N,219) We choose the value of N that makes this probability a maximum. We require the probability with N fish to be greater than the probability with N-1 fish. So this gives C(68,16) * C(N-68,203) C(68,16) * C(N-69,203) ---------------------- > --------------------- C(N,219) C(N-1,219) cancelling C(68,16) from each side this reduces to (N-68)!/[203!(N-271)!] (N-69)!/[203!(N-272)!] ---------------------- > ---------------------- N!/[219!(N-219)!] (N-1)!/[219!(N-220)!] cancelling as much as we can from one side with the corresponding term on the other side, we get: (N-68)/(N-271) -------------- > 1 N/(N-219) (N-68)/(N-271) > N/(N-219) (N-68)(N-219) > N(N-271) N^2 - 287N + 14892 > N^2 - 271N 14892 > 16N and so N < 14892/16 = 930.75 Similarly, comparing N with N+1 fish in the lake you will find that the probabilities increase while N < 931 and then decrease again, so that the optimum value is 931. So the probability of getting the second catch that we did is at a maximum if N is 931. >(c) Establish a 90% confidence interval for the number of rainbow >trout in the lake. If N = total number of fish in the lake, then from part (a) we had confidence interval is 0.044132 < p < 0.10199 which we could write 0.044132 < 68/N < 0.10199 this leads to 68/0.10199 < N < 68/0.044132 666.7 < N < 1540.8 and we have 667 < N < 1541 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/