Question

In Exercises 9.1 to \(9.4,\) use the computer output (from different computer packages) to estimate the intercept \(\beta_{0},\) the slope \(\beta_{1},\) and to give the equation for the least squares line for the sample. Assume the response variable is \(Y\) in each case. $$ \begin{aligned} &\text { The regression equation is } Y=29.3+4.30 \mathrm{X}\\\ &\begin{array}{lrrrr} \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 29.266 & 6.324 & 4.63 & 0.000 \\ \text { X } & 4.2969 & 0.6473 & 6.64 & 0.000 \end{array} \end{aligned} $$

Short answer

The intercept (\(\beta_{0}\)) is 29.266, the slope (\(\beta_{1}\)) is 4.2969, and the equation for the least squares line is \(Y = 29.266 + 4.2969X\).

Step-by-step solution

Interpret the Intercept (\(\beta_{0}\))

The coefficient for 'Constant' row under 'Coef' column in the table is the intercept (\(\beta_{0}\)). Here, the intercept (\(\beta_{0}\)) is 29.266.

Interpret the Slope (\(\beta_{1}\))

In the row for the predictor X, the coefficient under 'Coef' column is the slope (\(\beta_{1}\)). Here, the slope (\(\beta_{1}\)) is 4.2969.

Compose the least squares line equation

The least square line equation is given by \(Y = \beta_{0} + \beta_{1}X\). Substituting values from the given table, we get \(Y = 29.266 + 4.2969X\).

Key concepts

Intercept Estimation

When we talk about intercept estimation in regression analysis, we are referring to the process of determining the value at which the regression line will cross the y-axis. This point, known as the intercept, represents the expected value of the dependent variable (\texttt{\(Y\)}) when all independent variables (\texttt{\(X\)}) are set to zero.

In the context of our exercise, the intercept is identified from the output provided by statistical software. The row labeled 'Constant' in the table gives us the coefficient that serves as the intercept estimate. Here, it is listed as 29.266. This figure is key as it sets the starting point for the prediction of \texttt{\(Y\)} when \texttt{\(X\)} is zero, and it tells us that even without any influence from our independent variable, the response variable \texttt{\(Y\)} would start at 29.266.

Understanding the role of the intercept is crucial for interpreting the relationship between the variables, because it provides the baseline value that we use as a reference point for the predicted outcomes.

Slope Calculation

The slope of a regression line represents the change in the response variable for each one-unit increase in the predictor variable. It's an essential concept in understanding how the dependent and independent variables are related.

For our textbook exercise, the slope is extracted from the coefficient of the predictor (\texttt{\(X\)}) variable in the computer output table. We are given a 'Coef' value next to \texttt{\(X\)}, which is 4.2969. This means that for each one-unit increase in \texttt{\(X\)}, we expect \texttt{\(Y\)} to increase by approximately 4.2969 units, assuming a linear relationship exists between the two variables. So, if \texttt{\(X\)} changes from 0 to 1, \texttt{\(Y\)} will increase from the intercept value (29.266) to 29.266 plus 4.2969, and so on for subsequent units of \texttt{\(X\)}.

Calculating and interpreting the slope is fundamental in predictions; it shows the direction and strength of the relationship between variables. A positive slope, as in this case, indicates that as \texttt{\(X\)} increases, \texttt{\(Y\)} also increases.

Regression Analysis

Regression analysis is a statistical technique that estimates the relationships among variables. It allows us to understand how the typical value of the dependent variable changes when any one of the independent variables is varied while the others are held fixed. Essentially, it provides a prediction equation that describes these relationships.

In our given exercise, regression analysis helps us determine how well we can predict the outcome variable, \texttt{\(Y\)}, based on the predictor, \texttt{\(X\)}. The regression equation given is \texttt{\(Y = 29.3 + 4.30X\)}, which is based on the estimates obtained for the intercept and the slope. This equation is the result of the least squares method, which minimizes the sum of the squares of the differences between the observed values and the values predicted by the model.

With regression analysis, we can also assess the strength of the relationship through coefficients and determine the significance of the predictor variables using statistical tests, such as the t-test for coefficients. The provided 'T' and 'P' values in the output indicate that both the intercept and the slope are statistically significant predictors of \texttt{\(Y\)}. This analysis forms the backbone of predictions and inferences in various fields, from economics to engineering.