Chapter 20, Problem 1P

Perform the same computation as in Sec. 20.1, but use linear regression and transformations to fit the data with a power equation. Assess the result.

To determine

To calculate: A power equation fit to the data for specific growth rate and available food the data using linear regression equation and transformation.

Answer to Problem 1P

Solution: The power equation to fit the data for specific growth rate and available food is $k=0.1479f^{0.4224}$.

Explanation of Solution

Given Information: Data for specific growth rate $k$ and available food $f$ is given in the form of a table as,

$\boxed{\begin{array}{cc}f& k\\ 7& 0.29\\ 9& 0.37\\ 15& 0.48\\ 25& 0.65\\ 40& 0.80\\ 75& 0.97\\ 100& 0.99\\ 150& 1.07\end{array}}$

Formula used: A linear regression equation is given as $y=ax+b$.

Here,

$a=\frac{n{\displaystyle \sum _{i=1}^n{x}_i{y}_i}-{\displaystyle \sum _{j=1}^n{x}_j}{\displaystyle \sum _{k=1}^n{y}_k}}{n{\displaystyle \sum _{i=1}^n{x}_i^2}-{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2}$ and $b=\frac{1}{n}\left({\displaystyle \sum _{i=1}^n{y}_i}-a{\displaystyle \sum _{j=1}^n{x}_j}\right)$.

Power equation can be obtained from the linear regression equation as,

$y=\alpha x^{\beta }$

Here, $\alpha ={10}^b$ and $\beta =a$.

Calculation: For the given data, $n=8$ as there are eight observations.

Consider logarithmic terms and evaluate the summationsas,

$\boxed{\begin{array}{ccccc}& \log f& \log k& \log f\cdot \log k& {\left(\log f\right)}^2\\ & 0.85& -0.54& -0.45& 0.71\\ & 0.95& -0.43& -0.41& 0.91\\ & 1.18& -0.32& -0.37& 1.38\\ & 1.40& -0.19& -0.26& 1.95\\ & 1.60& -0.10& -0.16& 2.57\\ & 1.88& -0.01& -0.02& 3.52\\ & 2.00& 0.00& -0.01& 4.00\\ & 2.18& 0.03& 0.06& 4.74\\ {\displaystyle \sum }& 12.03& -1.56& -1.63& 19.78\end{array}}$

Calculate the value of $a$ as,

$a=\frac{n{\displaystyle \sum _{i=1}^n{x}_i{y}_i}-{\displaystyle \sum _{i=1}^n{x}_i}{\displaystyle \sum _{i=1}^n{y}_i}}{n{\displaystyle \sum _{i=1}^n{x}_i^2}-{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2}$

Here, $x$ is $\log f$ and $y$ is $\log k$.

Thus,

$a=\frac{n{\displaystyle \sum _{i=1}^n\log f_i\cdot \log k_i}-{\displaystyle \sum _{i=1}^n\log f_i}{\displaystyle \sum _{i=1}^n\log k_i}}{n{\displaystyle \sum _{i=1}^n\log f_i^2}-{\left({\displaystyle \sum _{i=1}^n\log f_i}\right)}^2}$

Simplify,

$a=\frac{n{\displaystyle \sum \log f\cdot \log k}-{\displaystyle \sum \log f}{\displaystyle \sum \log k}}{n{\displaystyle \sum \log f^2}-{\left({\displaystyle \sum \log f}\right)}^2}$

Substitute the values from the summation table and solve as,

$\begin{array}{c}a=\frac{8\times \left(-1.63\right)-\left(12.03\right)\left(-1.56\right)}{8\times \left(19.78\right)-{\left(12.03\right)}^2}\\ =0.4224\end{array}$

Calculate the value of $b$ as,

$b=\frac{1}{n}\left({\displaystyle \sum _{i=1}^n{y}_i}-a{\displaystyle \sum _{j=1}^n{x}_j}\right)$

Here, $x$ is $\log f$ and $y$ is $\log k$.

Thus,

$b=\frac{1}{n}\left({\displaystyle \sum _{i=1}^n\log k_i}-a{\displaystyle \sum _{j=1}^n\log f_j}\right)$

Simplify,

$b=\frac{1}{n}\left({\displaystyle \sum \log k}-a{\displaystyle \sum \log f}\right)$

Substitute the values from the summation table and solve as,

$\begin{array}{c}b=\frac{1}{8}\left(-1.56-\left(0.4224\right)\left(12.03\right)\right)\\ =-0.8300\end{array}$

Thus, the linear regression equation is,

$\log f=0.4224\log k-0.83$

Therefore, $\alpha ={10}^{-0.83}$ and $\beta =0.4224$.

Thus, the power equation fit to the datais written as,

$k=0.1479f^{0.4224}$

The plot for both the linear and the power models is obtained as,

Hence, the power equation fit gives better approximation than the linear regression fit.