91Ó°ÊÓ

$\begin{array}{cc}\text{Depth (meters)}& \text{Light intensity (lumens)}\\ 5& 168.00\\ 6& 120.42\\ 7& 86.31\\ 8& 61.87\\ 9& 44.34\\ 10& 31.78\\ 11& 22.78\end{array}$

The scatter plot of the data is:

As a result, we can see from the scatterplot that,

Because the scatterplot slopes downhill, the direction is negative.

Because the points do not lie in a straight line, the shape is curved.

Strength: Strong because all points in the same pattern are relatively near together.

The values for light intensity and depth are now provided in the question. Because the corresponding scatterplot exhibits a broadly linear pattern with no apparent strong outliers, we can infer that using an exponential model to represent the relationship between light intensity and depth would be plausible.

Minitab output from a linear regression analysis on the transformed data is shown below.

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& \text{T}& \text{P}\\ \text{Constant}& 6.78910& 0.00009& 78575.46& 0.000\\ \text{Depth}\left(\mathrm{m}\right)& -0.333021& 0.000010& -31783.44& 0.000\end{array}$

$\mathrm{S}=0.000055\mathrm{R}-\mathrm{Sq}=100.0\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=100.0\%$

Now, $x$represents depth and $y$represents light intensity. As a result, the transformation is $\ln y$, which is the natural logarithm of the light intensity.

As we already know, the regression line's general equation is as follows:

role="math" localid="1652851293597" $\ln \hat{y}=a+bx$

The slope and constant in the computer output are as follows:

$a=6.78910$

$b=-0.333021$

As a result, the regression line looks like this:

role="math" localid="1652851670453" $$\begin{gathered} \ln \hat{y}=a+bx \\ =6.78910-0.333021x \end{gathered}$$

Minitab output from a linear regression analysis on the transformed data is shown below.

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& \text{T}& \text{P}\\ \text{Constant}& 6.78910& 0.00009& 78575.46& 0.000\\ \text{Depth}\left(\mathrm{m}\right)& -0.333021& 0.000010& -31783.44& 0.000\end{array}$

$\mathrm{S}=0.000055\mathrm{R}-\mathrm{Sq}=100.0\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=100.0\%$

The regression line is

$\ln \hat{y}=6.78910-0.333021x$

Calculating the equation as:

$$\begin{gathered} \ln \hat{y}=a+bx \\ =6.78910-0.333021x \\ =6.78910-0.333021\left(12\right) \\ =2.792848 \\ =e^{2.792848} \\ =16.3275 \end{gathered}$$

The residual is:

role="math" localid="1652851853646" $$\begin{gathered} \text{Residual}=\ln y-\ln \hat{y} \\ =\ln 16.2-2.792848 \\ =-0.007837 \end{gathered}$$

We are surprised by the outcome because the residual is more than $s=0.000055$.