Question

Acreage and Value. The data from Exercise $14.37$for lot size (in acres) and assessed value (in thousands of dollars) for a sample of homes in a particular area are on the WeissStats site.

Short answer

(a) The regression t-test is reasonable to apply for the given data.

(b) The data do not support the conclusion that the predictor variable "lot size" is beneficial in forecasting home "values" at the $5\%$level.

Step-by-step solution

Part (a) Step 1: Given information

Given in the question that, Acreage and Value. The data from Exercise 14.37 for lot size (in acres) and assessed value (in thousands of dollars) for a sample of homes in a particular area are on the WeissStats site.We need to decide that whether we can reasonably apply the regression t-lest. If so, then also do part (b).

Part (a) Step 2: Explanation

Given:

MINITAB is used to create the residual plot.

Procedure for MINITAB:

Step 1: Select Stat > Regression > Regression from the drop-down menu.

Step 2: Fill in the Value column in Response.

Step 3: Fill in the columns Lot Size in Predictors.

Step 4: In Graphs, under Residuals versus the variables, enter the columns Lot Size.

Step 5: Click the OK button.

OUTPUT FROM MINITAB:

Part(a) Step 3: Construct the residual plot

MINITAB is used to create a normal probability plot of residuals.

Procedure for MINITAB:

Step 1: Select Stat > Regression > Regression from the drop-down menu.

Step 2: Fill in the Value column in Response.

Step 3: In Predictors, fill in the columns Lot Size and Lot Value.

Step 4: Select Normal probability plot of residuals from the Graphs menu.

Step 5: Click the OK button.

OUTPUT FROM MINITAB:

The following is the assumption for regression inferences:

Line of population regression:

For each value $X$of the predicator variable, the conditional mean of the response variable $\left(Y\right)$is${\beta }_0+{\beta }_{1}X$ .

Standard deviations are equal:

The response variable's $\left(Y\right)$standard deviation is the same as the explanatory variable's (X) standard deviation. The standard deviation is represented by the symbol $\sigma$.

Typical populations include:

The response variable follows a normal distribution.

Observations made independently:

The response variable observations are unrelated to one another.

Examine whether the graph shows a violation of one or more of the regression inference assumptions.

- It is obvious from the residual plot that the residuals fall into the horizontal band.

- It is obvious from the normal probability plot of residuals that

Part (b) Step 1: Given information

Given in the question that , Acreage and Value. The data from Exercise 14.37 for lot size (in acres) and assessed value (in thousands of dollars) for a sample of homes in a particular area are on the WeissStats site.We need to decide, at the signiflcance level, whether the data provide sufflcient evidence to conclude that the predictor variable is useful for predicting the response variable.

Part (b) Step 2: Explanation

The following are the suitable hypotheses:

Hypothesis of nullity:

$H_0:{\beta }_1=0$

In other words, the predictor variable "lot size" is useless for predicting "value."

Another possibility:

$H_{\alpha }:{\beta }_1\ne 0$

In other words, the predictor variable "lot size" can help forecast "value."

Rule of Rejection:

If the p-value $\le a(=0.05)$, reject the null hypothesis $\left(H_0\right)$.

MINITAB can be used to find the test statistic and $p$-value.

Part (b) Step 3 : Procedure for MINITAB 

Procedure for MINITAB:

Step 1: Select Stat > Regression > Regression from the drop-down menu.

Step 2: Fill in the Value column in Response.

Step 3: In Predictors, fill in the columns Lot Size and Lot Value.

Step 4: Click the OK button.

OUTPUT FROM MINITAB:

Value vs. LOT SIZE: Regression Analysis

Model Overview

The test statistic value is $1.12$, and the p-value is $0.269$, according to the MINITAB result.

Conclusion: Use the$\alpha =0.05$significance threshold.

The p-value is higher than the level of significance in this case.

That is, p-value $(=0.269)>\alpha (=0.05)$.

According to the rejection rule, there is insufficient evidence to reject the null hypothesis$\left(H0\right)$at $\alpha =0.05$.