Ramin Shamshiri ABE6981 HW_04

7/30/2019 Ramin Shamshiri ABE6981 HW_04

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Ramin Shamshiri ABE 6986, HW #4 Due 09/24/08

Homework #4

Due 09/24/08

ABE 6986

Ramin Shamshiri

UFID # 90213353

Population dynamics

Data on population growth of snails in a fish aquarium are given in Table 1 for surface areas of 4500 and

9300 cm2.

t (wk) P Z P Z 4500 cm

29300 cm

2

0 100 100

2 533 595

4 1594 2503

6 2299 49388 2467 5807

10 2499 5994

12 2500 5997

14 2500 5999

16 2500 5999

18 2500 5999

A 2500 6000

b

c, wk-1r

Table 1: Data source, Snail population (P) with elapsed time (t) for a fish aquarium.

Assume that population growth follows the logistic model

= + Where t is elapsed time since snails were introduced, wk; P is the number of snails at time t; A is

the maximum number for large time; b is the intercept parameter; and c is the response coefficient,

wk-1. Now Eq. (1) can be rearranged to the linear form:

= = Calculation of values of Z requires an estimate of parameter A.


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1. Plot population (P) vs. time (t) on linear-linear graph paper.Answer:

Figure 1: Plot of P vs. t for surface areas of 4500 cm2.

Figure 2: Plot of P vs. t for surface areas of 9300 cm2.


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2. From #1 estimate A=2500 and A=6000 for the two areas, respectively, and use Eq. (2) to estimate Zfor each t.

Answer:Using the below set of calculations, the answer is given in Table.

t P Z P Z

A=2500, area=4500 cm2

A=6000, area=9300 cm2

0 100 -3.17805 100 -4.07754

2 533 -1.30574 595 -2.20652

4 1594 0.564963 2503 -0.33442

6 2299 2.436925 4938 1.536806

8 2467 4.314251 5807 3.404129

10 2499 7.823646 5994 6.906755

12 2500 Math Error 5997 7.600402

14 2500 Math Error 5999 8.699348

16 2500 Math Error 5999 8.699348

18 2500 Math Error 5999 8.699348

Table 2: Estimation of Z values for each t.

A=2500 for area of 4500 cm2:

t=0 => P=100

=> = 1 = ln 2500

100 1 = 3.17805

t=2 => P=533

=> = 1 = ln 2500

533 1 = 1.30574

.

.

.t=18 => P=2500

=> = 1 = ln 2500

2500 1 = Math Error

A=6000 for area of 9300 cm2

t=0 => P=100

=> = 1 = ln 6000

100 1 = 4.07754

t=2 => P=595

=>

=

1

=

ln

6000

595 1

=

2.20652

.

.

.t=18 => P=5999

=> = 1 = ln 6000

595 1 = 8.699348


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3. Perform linear regression of Z vs. t to obtain estimates of b and c as well as the linear correlationcoefficient (r). Omit values at t>8 wk from regression.

Answer:

t Z Z

A=2500, area=4500 cm2 A=6000, area=9300 cm2

0 -3.17805 -4.07754

2 -1.30574 -2.20652

4 0.564963 -0.33442

6 2.436925 1.536806

8 4.314251 3.404129

Table 3: Values of Z for different t, (Omitting values at t>8 wk)

Performing linear regression for the Z vs. t data corresponding to A=2500 and area=4500 cm2

Linear model: Z = ln AP 1 = ct bCoefficients (with 95% confidence bounds):

c = 0.9364 (0.9353, 0.9374)b = 3.179 (3.174, 3.184)

Z = 0.9364t 3.179R-square: 0.999999601

r= 0.9999998

Performing linear regression for the Z vs. t data corresponding to A=6000 and area=4500 cm2

Linear model: =

= Coefficients (with 95% confidence bounds):

c = 0.9353 (0.9345, 0.9361)

b = 4.077 (4.073, 4.081)

Z = 0.9353t 4.077R-square: 0.999999789r=0.999999894


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4. Plot values of Z vs. t on linear-linear graph paper. Show the regression lines as well.Answer:

Plotting all the values of Z vs. t corresponding to A=2500 and area=4500cm2:

Linear model:

=

=

Coefficients (with 95% confidence bounds):

c = 1.053 (0.8658, 1.241)

b = 3.491 (-2.355,-4.628)

= ..R-square: 0.9838

r=0.991866

Figure 3: Plot of Z vs. t with regression line corresponding to data in Table above.

t Z

A=2500, area=4500 cm2

0 -3.17805

2 -1.30574

4 0.564963

6 2.436925

8 4.314251

10 7.823646


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Plotting the values of Z vs. t (Omitting values at t>8 wk from regression) corresponding to A=2500 and

area=4500cm2:

Linear model: Z = ln AP 1 = ct b


c = 0.9364 (0.9353, 0.9374)b = 3.179 (3.174, 3.184)

Z = 0.9364t 3.179R-square: 0.999999601

r= 0.9999998

Figure 4: Plot of Z vs. t (omitting values at t>8). Data: Table above.

t Z

A=2500, area=4500 cm2

0 -3.17805

2 -1.30574

4 0.564963

6 2.436925

8 4.314251


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Plotting all the values of Z vs. t corresponding to A=6000 and area=9300cm2:

Linear model: Z = ln AP 1 = ct b


c = 0.7824 (0.6114, 0.9534)b = 3.149 (1.324,4.974)

Z = 0.7824t 3.149R-square: 0.933

r=0.96591

Figure 5: Plot of Z vs. t with regression line corresponding to data in Table above.

t Z

A=6000, area=9300 cm2

0 -4.07754

2 -2.20652

4 -0.33442

6 1.536806

8 3.404129

10 6.906755

12 7.600402

14 8.699348

16 8.699348

18 8.699348


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Plotting the values of Z vs. t (Omitting values at t>8 wk from regression) corresponding to A=6000 and

area=9300cm2:

Linear model: = = Coefficients (with 95% confidence bounds):

c = 0.9353 (0.9345, 0.9361)

b = 4.077 (4.073, 4.081) Z = 0.9353t 4.077

R-square: 0.999999789r=0.999999894

Figure 6: Plot of Z vs. t (omitting values at t>8). Data: Table above.

t Z

A=6000, area=9300 cm2

0 -4.07754

2 -2.206524 -0.33442

6 1.536806

8 3.404129


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5. Estimates values of for each t from Eq. (1) with your parameters.Answer:

= + Eq. (1)

For the data corresponding to A=2500 and area=4500 cm2, the coefficients b and c are 3.179 and 0.9364

respectively. Therefore Eq. (1) can be written as: = 25001 + 3.179 0.9364

t=0 => = 25001+ 3.1790.9364(0) = 99.9092

.

.

t=18 => = 25001+ 3.1790.9364(18) = 2499.997

For the data corresponding to A=6000 and area=930000 cm2, the coefficients b and c are 4.077 and0.9353 respectively. Therefore Eq. (1) can be written as:

= 60001 + 4.077 0.9353t=0 => = 2500

1+ 4.0770.9353(0) = 100.0529..

t=18 => = 25001+ 4.0770.9353(18) = 5999.983

t (wk) P Z P Z 4500 cm2 9300 cm2

0 100 -3.17805 99.90921 100 -4.07754 100.0529

2 533 -1.30574 532.8085 595 -2.20652 595.06344 1594 0.564963 1594.946 2503 -0.33442 2500.98

6 2299 2.436925 2299.457 4938 1.536806 4936.245

8 2467 4.314251 2466.933 5807 3.404129 5807.237

10 2499 7.823646 2494.86 5994 6.906755 5969.479

12 2500 Math Error 2499.209 5997 7.600402 5995.278

14 2500 Math Error 2499.878 5999 8.699348 5999.272

16 2500 Math Error 2499.981 5999 8.699348 5999.888

18 2500 Math Error 2499.997 5999 8.699348 5999.983

A 2500 6000b 3.179 4.077

c, wk-1 0.9364 0.9353

r 0.9999998 0.999999894


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6. Plot the prediction curves on #1.Answer:

Figure 7: Plot of the prediction curve for P values corresponding to A=2500 and area=4500cm2

Figure 8: Plot of the prediction curve for P values corresponding to A=6000 and area=9300cm2


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Figure 9: Plot of = /( + (..) ) alone.

Figure 10: Plot of P=2500/(1+exp(3.179-0.9364t) ) alone.


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7. Discuss your results. How well does the model describe the data? Is the linearizationprocedure justified? How could estimates of the parameters be improved? Explain the

effect of surface area on the parameters.

Answer:

The model is a very strong one in predicting the response variable and describing data. Yes, the

linearization procedure is justified because it helped to accurately estimate the b and c coefficientsthrough a simple procedure. This estimation is very accurate as it leaded to a very high correlation

coefficient, very close to 1, however the estimation of the parameters could be improved through non-

linear procedure or with fitting more data.

As the area has increased from 4500 cm2 to 6000 cm2, the values of P have increased too. In the other

word, the surface area has a positive effect on the values of P but a negative effect on the values of Z for

t

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