91Ó°ÊÓ

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.