Question

Use a transformation to linearize this equation and then employ linear regression to determine parameters a and

Answer 1

Given the equation

$y=\left( \frac{a+\sqrt{x}}{b\sqrt{x}} \right)^2$

which models the data tabulated below:

$\begin{tabular} {|c|c|} x&y\\[1ex] 0.5&10.4\\ 1&5.8\\ 2&3.3\\ 3&2.4\\ 4&2 \end{tabular}$

The linear regression equation is given by

$y=a+bx$

where: $b= \frac{\Sigma xy-n\bar{x}\bar{y}}{\Sigma x^2-n\bar{x}^2}$ and $a=\bar{y}-b\bar{x}$

We extend the given table as follows:

$\begin{tabular} {|c|c|c|c|} x&y&x^2&xy\\[1ex] 0.5&10.4&0.25&5.2\\ 1&5.8&1&5.8\\ 2&3.3&4&6.6\\ 3&2.4&9&7.2\\ 4&2&16&8\\[1ex] \Sigma x=10.5&\Sigma y=23.9&\Sigma x^2=30.25&\Sigma xy=32.8 \end{tabular} \\ \\ \\ \bar{x}= \frac{\Sigma x}{n} = \frac{10.5}{5} =2.1 \\ \\ \bar{y}=\frac{\Sigma y}{n} = \frac{23.9}{5} =4.78$

Thus,

$b= \frac{32.8-5(2.1)(4.78)}{30.25-5(2.1)^2} \\ \\ = \frac{32.8-50.19}{30.25-22.05} = \frac{-17.39}{8.2} \\ \\ =-2.12$

and

$a=4.78-(-2.12)(2.1)=4.78+4.454=9.234$

Therefore, the linearlized form of the equation is y = 9.234 - 2.12x



Part B:

At x = 1.6,

$y=9.234-2.12(1.6)=9.234-3.392=5.842$