Linear Least-Squares Regression

Problem 14.14.

An investigator has reported the data tabulated below for an experiment to determine the growth rate of bacteria k (per d) as a function of oxygen concentration c (mg/L). It is known that such data can be modeled by the following equation: 

 Eq.(14.14(a))

where c_s and k_max are parameters. Use a transformation to linearize this equation. Then use linear regression to estimate c_s and k_max and predict the growth rate at c = 2 mg/L.

Solution:

Equation (14.14(a)), which is in the format of saturation-growth-rate model can be linearized by inverting it to give:

Thus, a plot of 1/k (y) versus 1/c² (x) will yield a straight line with a slope of c_s/k_max and an intercept of 1/k_max. The data can be set up in tabular form and the necessary sums computed as in table below.

The means can be computed as:

The slope and intercept can then be calculated with Eqs (14.15) and (14.16) as in textbook.

The least-squares fit of the transformed data is 1/k = 0.09666 + 0.20201(1/c²).

The model coefficients can then be calculated as:

k_max = 1/0.09666 = 10.34554

c_s = 0.20201 x 10.34554 = 2.0899

The growth rate at c = 2 mg/L can be predicted as:

1/k = 0.09666 + 0.20201(1/4) = 0.1471625

k = 6.7952 per day

*The linregr M-file can also be used to determine the least-squares fit:

>> format long

>> c = [0.5 0.8 1.5 2.5 4];

>> k = [1.1 2.5 5.3 7.6 8.9];

>> [a,r2,syx] = linregr(1./c.^2,1./k)

a =

   0.202005196520525   0.096665700618091

r2 =

   0.999569288346182

syx =

   0.007995513530533

The model coefficients can then be calculated as:

>> kmax = 1/a(2)

kmax =

  10.344930969370603

>> cs = kmax*a(1)

cs =

   2.089729813458969