Confidence Intervals

Date: 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/