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In the previous exercise, \(r^{2}=0.88\) when \(x_{1}\) is the predictor and \(R^{2}=0.914\) when both \(x_{1}\) and \(x_{2}\) are predictors. Why do you think that the predictions of \(y\) don't improve much when \(x_{2}\) is added to the model? (The association of \(x_{2}\) with \(y\) is \(r=0.5692 .)\)

Short Answer

Expert verified
Adding \(x_2\) provides minimal predictive improvement because it does not contribute much unique information beyond \(x_1\).

Step by step solution

01

Understanding the Variables

The exercise involves two models for predicting the outcome variable, \(y\). The first model uses \(x_1\) as a predictor and achieves \(r^2 = 0.88\), indicating that 88% of the variance in \(y\) is explained by \(x_1\). The second model includes both \(x_1\) and \(x_2\) as predictors, achieving an \(R^2 = 0.914\), which means that 91.4% of the variance in \(y\) is explained when both predictors are used.
02

Calculating Increase in Explained Variance

When adding \(x_2\) to the model, the \(R^2\) increases from 0.88 to 0.914. The increase in explained variance is thus \(R^2 - r^2 = 0.914 - 0.88 = 0.034\), which indicates a minimal improvement of 3.4%.
03

Interpreting the Effect of Adding a Predictor

The small increase suggests that \(x_2\) contributes little additional information about \(y\) beyond what is already explained by \(x_1\). This can happen if \(x_1\) and \(x_2\) are correlated with each other, meaning \(x_2\) does not provide much unique predictive value in the presence of \(x_1\).
04

Analyzing Association between \(x_2\) and \(y\)

The correlation of \(x_2\) with \(y\) is \(r=0.5692\). While this indicates a moderate relationship between \(x_2\) and \(y\), the value is not sufficiently strong, or \(x_2\) might be overlapping in the variance that \(x_1\) already explains, resulting in a minimal increase in the overall explanatory power of the model.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Predictor Variables
Predictor variables, commonly known as independent variables, are essential in the field of regression analysis. Their key role is to explain or predict the variations in the dependent or outcome variable, denoted as \(y\). In any regression model, selecting effective predictor variables is crucial, as it determines the overall success of predicting \(y\).

When creating a predictive model, we often start with one predictor. This allows us to see the level of variance explained by that single variable. Adding more predictors can improve the model, but as we see in the exercise, adding another predictor variable, \(x_2\), increased the explained variance only slightly. This can happen if the new predictor doesn't provide much unique information beyond what's already offered by existing predictors.

Sometimes, new predictor variables don't significantly enhance the model's predictive power because they are highly correlated with other predictors. In regression, it's important that each predictor contributes uniquely to the prediction of \(y\). If they don't, the addition might not be worth the increase in model complexity.
Explained Variance
Explained variance is a fundamental concept in regression analysis, represented as \(R^2\). It measures the proportion of the total variance in the outcome variable \(y\) that is accounted for by the predictor variables. A higher \(R^2\) value indicates better predictive power of the model.

In the exercise's context, the initial model using \(x_1\) as the sole predictor states an \(r^2 = 0.88\), meaning \(88\%\) of \(y\)'s variance is explained by \(x_1\). When \(x_2\) is added, \(R^2\) increases to \(0.914\), translating to an overall explained variance of \(91.4\%\).

This change from \(0.88\) to \(0.914\) reflects a minimal increase of \(3.4\%\). Such a small increase suggests that while \(x_2\) is somewhat related to \(y's\) variance, it doesn't provide much additional context beyond what \(x_1\) offers. This illustrates the importance of carefully selecting additional variables that are not redundant but rather bring new, valuable insights to the prediction task.
Correlation Coefficient
The correlation coefficient, expressed as \(r\), quantifies the degree to which two variables are linearly related. It ranges between -1 and 1, where 1 indicates a perfect positive linear relationship, -1 signifies a perfect negative relationship, and 0 means no linear correlation at all.

In this scenario, the exercise states that \(x_2\) and \(y\) have a correlation coefficient of \(r=0.5692\). This suggests a moderate positive relationship between these two variables. However, this correlation doesn't necessarily imply that \(x_2\) will significantly enhance the predictive model when added.

Even if a variable exhibits a noticeable correlation with the dependent variable, it may overlap with what is already explained by other predictors, like \(x_1\). This overlap could lead to minimal improvement in the model's explanatory power, as seen in this exercise. Therefore, while correlation is a helpful tool in understanding linear relationships, it should be viewed in conjunction with other factors when assessing a variable's contribution to the model.

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Most popular questions from this chapter

For all students at Walden University, the prediction equation for \(y=\) college GPA (range \(0-4.0\) ) and \(x_{1}=\) high school GPA (range \(0-4.0\) ) and \(x_{2}=\) college board score (range \(200-800\) ) is \(\hat{y}=0.20+0.50 x_{1}+0.002 x_{2}\) a. Find the predicted college GPA for students having (i) high school GPA \(=4.0\) and college board score \(=800\) and (ii) \(x_{1}=2.0\) and \(x_{2}=200\). b. For those students with \(x_{2}=500\), show that \(\hat{y}=1.20+0.50 x_{1}\) c. For those students with \(x_{2}=600\), show that \(\hat{y}=1.40+0.50 x_{1}\). Thus, compared to part \(b\), the slope for \(x_{1}\) is still 0.50 , and increasing \(x_{2}\) by 100 (from 500 to 600 ) shifts the intercept upward by \(100 \times\left(\right.\) slope for \(\left.x_{2}\right)=100(0.002)=0.20\) units.

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