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The following data has been obtained from https://www.mcgi,state.ni.us/fishpop/#

ID: 3050918 • Letter: T

Question

The following data has been obtained from https://www.mcgi,state.ni.us/fishpop/# for brook trout in the Iron River of Upper Michigan. In this problem, you will use the data to develop two different modeling approaches Table 1: Iron River brook trout - Number of fish/acre 1996 1997 1998 | 1999 |2000 2001 |2002 20032004 2005 2006 909 Year Fish/Acre 609 736 841 7221066 841550 272518 and compare results on validity of these as well as possible long term behaviors predicted. (i) Let n denote the year after 1996 and r(n) denote the number of fish/acre corresponding to that year. Plot the data (n, r(n)) and obtain a least squares best fit 5th degree (quintic) polynomial approximation r(n)5(n) Determine the range of validity of the model according to qs(t) 20 and calculate the errors between data and q Then calculate the long term behavior from such a model. (ii) For each data point, the quantity [r(n+1)-x(n)|/x(n) can be thought of as approximating the relative growth rate at time n. Plot (x(n), [x(n +1) - r(n)|/x(n)) for all the values of n where data is available and determine a quadratic, q2 that best fits the data according to a least squares approach. Determine the range of validity of the model according to [r(n + 1)-x(n)]/2(n) = g2(z(n), rewrite this as 2(n + 1) = f(x(n), and calculate the long term behavior from such a model.

Explanation / Answer

i) As mentioned, we obtain the below table:

On fitting a least square linear regression model, we obtain the below results:

In order to check the validity, we analyse the residuals

The values fluctuate drastically post 2004 and may not be a good representation of the trend in the long term.

ii) For the sake of simplicity, please assume in 2005 , x(2005) = (x(2004)+x(2006))/2 [Replacing with mean value]

We obtain the below table:

On building a least square regression model and ignoring the last data point since x(n+1) is missing, we obtain:

Below are the residuals:

y x x^2 x^3 x^4 x^5 609 1996 3984016 7952095936 1.58724E+13 3.16813E+16 736 1997 3988009 7964053973 1.59042E+13 3.17607E+16 841 1998 3992004 7976023992 1.59361E+13 3.18403E+16 722 1999 3996001 7988005999 1.5968E+13 3.19201E+16 1066 2000 4000000 8000000000 1.6E+13 3.2E+16 841 2001 4004001 8012006001 1.6032E+13 3.20801E+16 550 2002 4008004 8024024008 1.60641E+13 3.21603E+16 272 2003 4012009 8036054027 1.60962E+13 3.22407E+16 518 2004 4016016 8048096064 1.61284E+13 3.23213E+16 0 2005 4020025 8060150125 1.61606E+13 3.2402E+16 909 2006 4024036 8072216216 1.61929E+13 3.24829E+16
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