Question

Light through water Some college students collected data on the intensity of light at various depths in a lake. Here is a scatterplot of their data:

a. At the top right is a scatterplot of the natural logarithm of light intensity versus depth. Based on this graph, explain why it would be reasonable to use an exponential model to describe the relationship between light intensity and depth.

b. Here is the computer output from a linear regression analysis of the transformed data. Give the equation of the least-squares regression line. Be sure to define any variables you use.

c. Use your model to predict the light intensity at a depth of $12$ meters.

Short answer

a). Scatter plot is not having strong curvature.

b). The equation of the least-squares regression line is $\ln \hat{y}=6.78910-0.333021x$.

c). The expected that the intensity is $16.3275$ lumens at depth of $12$ metre.

Step-by-step solution

Part (a) Step 1: Given Information

Given data:

Part (a) Step 2: Explanation

Presented the lack of substantial curvature in the given scatter plot, a linear model between the two scatter plot variables would be appropriate. As a result, a linear relationship between In(intensity) and depth is reasonable.

Expectations based on a general linear model Time and ln(intensity);

$\ln \left(\text{intensity}\right)=a+b\text{(depth)}$

Taking the exponential

$\text{intensity}=e^{\ln \left(\text{intensity}\right)}$

$=e^{a+b\left(\mathrm{depth}\right)}$$=e^a{e}^{b\left(\mathrm{depth}\right)}$

Part (b) Step 1: Given Information

Given data:

Part (b) Step 2: Explanation

Least square regression line's general equation

$\hat{y}=b_0+b_{1}x$

In the row "constant" and the column "Coef" of the computer output, the calculated constant $b_0$is mentioned.

$b_0=6.78910$

In the row "Depth" and the column "Coef" of the computer output, the calculated slope $b_1$ is mentioned.

$b_1=-0.333021$

Part (b) Step 3: Explanation

Substituting the value of $b_0$and $b_1$

$\hat{y}=b_0+b_{1}x$

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

Where$x$ represents the current time and $y$ is the ln (count)

role="math" localid="1654323193310" $\ln \hat{y}=6.78910-0.333021x$

Where $x$ is representing the depth and $\mathrm{y}$ is representing the intensity.

Part (c) Step 1: Given Information

Given data:

Part (c) Step 2: Explanation

Substituting the value of $x$

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

$\ln \hat{y}=6.78910-0.333021\left(12\right)$

$=2.792848$

Taking the exponential

$\hat{y}=e^{\ln \hat{y}}$

$=e^{2.792848}$

$=16.3275$