Solution: The power equation to fit the data for specific growth rate and available food is k=0.1479f0.4224.

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

fk70.2990.37150.48250.65400.80750.971000.991501.07

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

Here,

a=n∑i=1nxiyi−∑j=1nxj∑k=1nykn∑i=1nxi2−(∑i=1nxi)2 and b=1n(∑i=1nyi−a∑j=1nxj).

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

y=αxβ

Here, α=10b and β=a.

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

Consider logarithmic terms and evaluate the summationsas,

logflogklogf⋅logk(logf)20.85−0.54−0.450.710.95−0.43−0.410.911.18−0.32−0.371.381.40−0.19−0.261.951.60−0.10−0.162.571.88−0.01−0.023.522.000.00−0.014.002.180.030.064.74∑12.03−1.56−1.6319.78

Calculate the value of a as,

a=n∑i=1nxiyi−∑i=1nxi∑i=1nyin∑i=1nxi2−(∑i=1nxi)2

Here, x is logf and y is logk.

Thus,

a=n∑i=1nlogfi⋅logki−∑i=1nlogfi∑i=1nlogkin∑i=1nlogfi2−(∑i=1nlogfi)2

Simplify,

a=n∑logf⋅logk−∑logf∑logkn∑logf2−(∑logf)2

Substitute the values from the summation table and solve as,

a=8×(−1.63)−(12.03)(−1.56)8×(19.78)−(12.03)2=0.4224

Calculate the value of b as,

b=1n(∑i=1nyi−a∑j=1nxj)

Here, x is logf and y is logk.

Thus,

b=1n(∑i=1nlogki−a∑j=1nlogfj)

Simplify,

b=1n(∑logk−a∑logf)

Substitute the values from the summation table and solve as,

b=18(−1.56−(0.4224)(12.03))=−0.8300

Thus, the linear regression equation is,

logf=0.4224logk−0.83

Therefore, α=10−0.83 and β=0.4224.

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

k=0.1479f0.4224

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.