Let's find a model for the given data using power regression, step by step.
Power regression is a method to find a relationship of the form [tex]\( y = a \cdot x^b \)[/tex], where constants [tex]\( a \)[/tex] and [tex]\( b \)[/tex].
The provided data points are:
[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & 2 & 3 & 5 & 6 & 8 \\
\hline
y & 96 & 27 & 6 & 3 & 1 \\
\hline
\end{array}
\][/tex]
To use power regression, we undertake the following steps:
1. Log-transform the data: Both [tex]\( x \)[/tex] and [tex]\( y \)[/tex] values are transformed using the natural logarithm:
[tex]\[
\ln(x) \text{ and } \ln(y)
\][/tex]
2. Fit a linear model to the transformed data: This involves finding the best fitting line for the transformed data points [tex]\((\ln(x_i), \ln(y_i))\)[/tex].
The natural logs of the [tex]\( x \)[/tex] values [tex]\((2, 3, 5, 6, 8)\)[/tex] are:
[tex]\[
\begin{align*}
\ln(2) &= 0.6931 \\
\ln(3) &= 1.0986 \\
\ln(5) &= 1.6094 \\
\ln(6) &= 1.7918 \\
\ln(8) &= 2.0794 \\
\end{align*}
\][/tex]
The natural logs of the [tex]\( y \)[/tex] values [tex]\((96, 27, 6, 3, 1)\)[/tex] are:
[tex]\[
\begin{align*}
\ln(96) &= 4.5643 \\
\ln(27) &= 3.2958 \\
\ln(6) &= 1.7918 \\
\ln(3) &= 1.0986 \\
\ln(1) &= 0 \\
\end{align*}
\][/tex]
3. Find the linear regression line [tex]\( \ln(y) = \ln(a) + b \ln(x) \)[/tex], resulting in the coefficients for [tex]\( \ln(a) \)[/tex] and [tex]\( b \)[/tex].
After fitting a linear model, we find:
[tex]\[
\ln(y) = -3.2 \ln(x) + \ln(955.3)
\][/tex]
4. Exponentiate to find the original scale:
- [tex]\( \ln(a) \)[/tex] corresponds to [tex]\( 6.8590473 \)[/tex] (which is calculated as [tex]\( 955.3 \)[/tex] when exponentiated),
- [tex]\( b \)[/tex] is simply the slope value, i.e., [tex]\( b \)[/tex].
So the final model is given by:
[tex]\[
y = 955.3 \cdot x^{-3.2}
\][/tex]
Therefore, the model for the given data using power regression is:
[tex]\[
y = 955.3 \cdot x^{-3.2}
\][/tex]
