Pricey diamonds Here is a scatterplot showing the relationship between the
weight (in carats) and price (in dollars) of round, clear, internally flawless diamonds with excellent cuts:

a. Explain why a linear model is not appropriate for describing the relationship between price and weight of diamonds.
b. We used software to transform the data in hopes of achieving linearity. The output shows the results of two different transformations. Would an exponential model or a power model describe the relationship better? Justify your answer.

c. Use each model to predict the price for a diamond of this type that weighs 2 carats. Which prediction do you think will be better? Explain your reasoning.

To explain a linear model is not appropriate for describing the relationship between price and weight of diamonds.

The follow - up activities a scatterplot depicting the link between the weight and price of round circle internally flawless diamonds with exceptional cuts.
Because the above scatterplot of price vs weight contains a lot of curvature and so looks to be a curved relationship between the weight and the price, a liner model would not be appropriate for expressing the relationship between price and weight of diamonds.
As a result,a linear model will not be appropriate.

To explain that the exponential model or a power model describe the relationship better.

The follow - up activities a scatterplot depicting the link between the weight and price of round circle internally flawless diamonds with exceptional cuts. The question specifies the two transformations.
The transformation 2 as the Exponential model:
The above scatterplot of In(price) versus weight has a lot of curvature, so using a linear model between the two variables of the scatterplot isn't a good idea. Therefore, a linear relationship between the weight and is insufficient (price). So, to predict In(price) from weight, the general linear model is as follows:

$I\hat{n}\left(\text{Price}\right)=a+b\text{(Weight})$
Taking the exponential on both sides now,
$\hat{P}$rice $=e^{\hat{\ln }\text{(Price)}}$
$=e^{a+b\left(\text{Weight}\right)}$
$=e^a{e}^{b\left(\text{Weight}\right)}$

Transformation 1 as the Power model:
Since the scatterplot of $ln$ (weight) and $ln$ (price) does not have any considerable curvature, a linear model between the two variables of the scatterplot is not appropriate. As a result, a linear model between $ln$ (weight) and $ln$ (length) will be acceptable (price).
As a result, the general linear model for predicting this is as follows:
$l\hat{n}\text{(Price)}=a+b\ln \text{(Weight})$
Using the exponential on both sides as follows:

$\tilde{\mathrm{P}}\text{rice}=e^{\ln \left(\text{Price}\right)}$
$=e^{a+b\ln \left(\text{Weight}\right)}$

$=e^a{e}^{b\ln \left(\text{Weight}\right)}$
As a result, conclude that a power model better describes the relationship.

To use each model to predict the price for a diamond of the type that weighs 2 carats.

The scatterplot depicting the link between the weight and price of round circle internally flawless diamonds with exceptional cuts. The question specifies the two transformations. As a result,
Transformation 2: Exponential model,
The least square regression line's general equation is:
$\hat{y}=b_0+b_{1}x$
As a result of the computer output, the constant estimate is presented in the row "Constant" and the column "Coef" as:
$b_0=8.2709$
In the row "Weight" and the column "Coef" of the given computer output, the slope $b_1$ is presented as:
$b_1=1.3791$
Then, substitute the values in the equation:
$\hat{y}=b_0+b_{1}x$
$=8.2709+1.3791x$
Now, solve for the logarithm in the equation as follows:
$\ln \hat{y}=8.2709+1.3791x$

Substitute 2 for x:
$\ln \hat{y}=8.2709+1.3791x$
$=8.2709+1.3791\left(2\right)$
$=11.0291$
$\ln \hat{y}=8.2709+1.3791x$
$-8.2709+1.3791\left(2\right)$
$=11.0291$
Then taking exponential on both sides:
$\hat{y}=e^{\ln \hat{y}}$
$=e^{11.0291}$
$=61642.1$
The predicted price is $\$61642.1$.

Transformation 1: Power model,
The least square regression line's general equation is:
$\hat{y}=b_0+b_{1}x$
As a result of the computer output, we can see that the constant estimate is presented in the row "Constant" and the column "Coef" as:
$b_0=9.7062$
In the row "InWeight" and the column "Coef" of the given computer output, the slope $b_1$ is presented as:
$b_1=2.2913$

Then substitute the values in the equation:
$\hat{y}=b_0+b_{1}x$
$=9.7062+2.2913x$
Then, solve for the logarithm in the equation as follows:
$\ln \hat{y}=9.7062+2.2913x$
Substitute 2 for x:
$\ln \hat{y}=9.7062+2.2913x$
$=9.7062+2.2913\left(2\right)$
$=11.2944$
Taking exponential on both sides:
$\hat{y}=e^{\ln \hat{y}}$
$=e^{11.2944}$
$=80370.30$
The predicted price is $\$80370.30$.
As a result, predicting using the power model would be better based on the above results.