Question: Bordeaux wine sold at auction. The uncertainty of the weather during the growing season, the phenomenon that wine tastes better with age, and the fact that some vineyards produce better wines than others encourage speculation concerning the value of a case of wine produced by a certain vineyard during a certain year (or vintage). The publishers of a newsletter titled Liquid Assets: The International Guide to Fine Wine discussed a multiple regression approach to predicting the London auction price of red Bordeaux wine. The natural logarithm of the price y (in dollars) of a case containing a dozen bottles of red wine was modelled as a function of weather during growing season and age of vintage. Consider the multiple regression results for hypothetical data collected for 30 vintages (years) shown below.

1. Conduct a t-test (at$\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$ ) for each of the $\mathbf{\beta }$parameters in the model. Interpret the results.
2. When the natural log of y is used as a dependent variable, the antilogarithm of a b coefficient minus 1—that is ebi - 1—represents the percentage change in y for every 1-unit increase in the associated x-value. Use this information to interpret each of the b estimates.
3. Interpret the values of ${\mathbf{R}}^{\mathbf{2}}$and s. Do you recommend using the model for predicting Bordeaux wine prices? Explain

For ${\mathbf{\beta }}_{\mathbf{1}}$:

${\mathbf{H}}_{\mathbf{0}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{=}\mathbf{0}$, whereas, ${\mathbf{H}}_{\mathbf{a}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{1}}\ne \mathbf{0}$

Here, the test statistic $\mathbf{=}\frac{{\hat{\mathbf{\beta }}}_{\mathbf{1}}}{{\mathbf{s\beta }}_{\mathbf{1}}}$

Test statistic $$\begin{gathered} \mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{03}}{\mathbf{0}\mathbf{.}\mathbf{006}} \\ \mathbf{=}\mathbf{5} \end{gathered}$$

For $\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$the critical value of ${\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{05}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{699}$using the formulae table

${\mathbf{H}}_{\mathbf{0}}$is rejected if$\left|\mathbf{t}\right|\mathbf{>}{\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{05}}$. Since 5 > 1.699,

Reject the null hypothesis at a 95% significance level

Therefore, the value of ${\mathbf{\beta }}_{\mathbf{1}}\ne \mathbf{0}$

For ${\mathbf{\beta }}_{\mathbf{2}}$:

${\mathbf{H}}_{\mathbf{0}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{=}\mathbf{0}$whereas${\mathbf{H}}_{\mathbf{a}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{2}}\ne \mathbf{0}$

Here, the test statistic $\mathbf{=}\frac{{\hat{\mathbf{\beta }}}_{\mathbf{2}}}{{\mathbf{s\beta }}_{\mathbf{2}}}$

$$\begin{gathered} Teststatistic\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{60}}{\mathbf{0}\mathbf{.}\mathbf{120}} \\ \mathbf{=}\mathbf{5} \end{gathered}$$

For $\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$, the critical value of ${\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{05}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{699}$using the formulae table

${\mathbf{H}}_{\mathbf{0}}$is rejected if $\left|\mathbf{t}\right|\mathbf{>}{\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{05}}$. Since 5 > 1.699,

Reject the null hypothesis at a 95% significant level

Thus, the value of ${\mathbf{\beta }}_{\mathbf{2}}\ne \mathbf{0}$

For${\mathbf{\beta }}_{\mathbf{3}}$:

${\mathbf{H}}_{\mathbf{0}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{3}}\mathbf{=}\mathbf{0}$, whereas, ${\mathbf{H}}_{\mathbf{a}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{3}}\ne \mathbf{0}$

Here, the test statistic $\mathbf{=}\frac{{\hat{\mathbf{\beta }}}_{\mathbf{3}}}{{\mathbf{s\beta }}_{\mathbf{3}}}$

Test statistic $$\begin{gathered} \mathbf{=}\frac{\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{004}}{\mathbf{0}\mathbf{.}\mathbf{001}} \\ \mathbf{=}\mathbf{-}\mathbf{4} \end{gathered}$$

For $\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$, the critical value of ${\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{05}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{699}$using the formulae table

${\mathbf{H}}_{\mathbf{0}}$is rejected if $\left|\mathbf{t}\right|\mathbf{>}{\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{05}}$. Since, -4 < 1.699,

Do not reject the null hypothesis at a 95% significance level.

Hence, the value of${\mathbf{\beta }}_{\mathbf{3}}\mathbf{=}\mathbf{0}$

For ${\mathbf{\beta }}_{\mathbf{4}}$:

${\mathbf{H}}_{\mathbf{0}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{4}}\mathbf{=}\mathbf{0}$, whereas, ${\mathbf{H}}_{\mathbf{a}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{4}}\ne \mathbf{0}$

Here, the test statistic $\mathbf{=}\frac{{\hat{\mathbf{\beta }}}_{\mathbf{4}}}{{\mathbf{s\beta }}_{\mathbf{4}}}$

Test statistic$$\begin{gathered} \mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{0015}}{\mathbf{0}\mathbf{.}\mathbf{0005}} \\ \mathbf{=}\mathbf{3} \end{gathered}$$

For $\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$, the critical value of ${\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{05}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{699}$using the formulae table

$H_0$is rejected if $\left|\mathbf{t}\right|\mathbf{>}{\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{05}}$. Since 3 > 1.699,

Reject the null hypothesis at a 95% significance level

Wherefore, the value of${\mathbf{\beta }}_{\mathbf{4}}\ne \mathbf{0}$

Antilogarithm formulae to be used to explain the percentage change in y for every 1-unit increase in the x-value is ${\mathbf{e}}^{\mathbf{\beta t}}\mathbf{-}\mathbf{1}$

For ${\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{\beta }}_{\mathbf{1}}$ will have the effect of $\mathbf{(}{\mathbf{e}}^{\mathbf{0}\mathbf{.}\mathbf{03}}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{03045}$

This means that for every 1- unit change in $X_1$, y changes by 0.0345 units.

For ${\mathbf{x}}_{\mathbf{2}}\mathbf{,}{\mathbf{\beta }}_{\mathbf{2}}$will have the effect of $\mathbf{(}{\mathbf{e}}^{\mathbf{0}\mathbf{.}\mathbf{60}}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8221}$

According to, the values of ${\mathbf{R}}^{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{85}$and the value of s = 0.30

Value of $R^2$ greater than 0.70 denoted that the model is available to explain the variations in the data. The value of s is 0.30 which indicates that there is less variability between the data points and that the data is not very spread.

The lower value of s and high value of makes the model a good fit for the data.