Adjusted Coefficient of Determination For Exercise 2, why is it better to use values of adjusted \({R^2}\)instead of simply using values of \({R^2}\)?

Short Answer

Expert verified

When more variables are added, the unadjusted\({R^2}\)increases (or stays the same). But the adjusted\({R^2}\)adjusts according to the number of variables and the sample size. It increases/decreases depending on whether the added variable explains the variation in the model better than the previous one.

The unadjusted \({R^2}\)incorrectly suggests that the best multiple regression equation is generated by adding all available variables. On the other hand, the adjusted \({R^2}\) is more helpful in eliminating factors that should not be included as they add nothing useful to the original model.

Step by step solution

01

Given information

A regression equation is computed to predict the weight of a bear (in lb) by using the variables “weight”, “length,” and “chest size” and the adjusted\({R^2}\)value is noted.

Further, another predictor variable “neck size” is added to the equation and the adjusted \({R^2}\)is noted again.

02

Step 2: \({R^2}\) vs. adjusted \({R^2}\)

It is always better to consider the value of adjusted \({R^2}\) as compared to the value of simple \({R^2}\) whenever another predictor variable is added to a given regression equation because of the following factors:

  • The unadjusted value of \({R^2}\) always increases if a new predictor variable is added to an equation, irrespective of whether the new variable significantly adds to the explanation of the response variable.
  • On the other hand, the value of adjusted \({R^2}\) increases for the new equation only when the new variable significantly adds to the explanation of variation in the response variable. It can decrease if the variable added is useless.

Since the value of adjusted \({R^2}\)considers the number of variables as well as the sample size, it helps in eliminating the variables that do not add anything to the model. Therefore, it helps to correctly identify the best available regression model.

Here, a new variable, “neck size,” is added to the given regression equation.

If the value of unadjusted \({R^2}\) is considered, it will either increase or remain the same. Thus, it cannot be determined whether the new variable significantly enhances the original model.

When the value of unadjusted \({R^2}\) is considered, it increases. If the new variable “neck size” is added, it explains the variation in the model better. Thus, it should be used for future purposes.

Therefore, the value of adjusted\({R^2}\)is better than that ofsimple\({R^2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Testing for a Linear Correlation. In Exercises 13–28, construct a scatterplot, and find the value of the linear correlation coefficient r. Also find the P-value or the critical values of r from Table A-6. Use a significance level of A = 0.05. Determine whether there is sufficient evidence to support a claim of a linear correlation between the two variables. (Save your work because the same data sets will be used in Section 10-2 exercises.)

POTUS Media periodically discuss the issue of heights of winning presidential candidates and heights of their main opponents. Listed below are those heights (cm) from severalrecent presidential elections (from Data Set 15 “Presidents” in Appendix B). Is there sufficient evidence to conclude that there is a linear correlation between heights of winning presidential candidates and heights of their main opponents? Should there be such a correlation?

President

178

182

188

175

179

183

192

182

177

185

188

188

183

188

Opponent

180

180

182

173

178

182

180

180

183

177

173

188

185

175

Interpreting the Coefficient of Determination. In Exercises 5–8, use the value of the linear correlation coefficient r to find the coefficient of determination and the percentage of the total variation that can be explained by the linear relationship between the two variables.

Weight , Waist r = 0.885 (x = weight of male, y = waist size of male)

The following exercises are based on the following sample data consisting of numbers of enrolled students (in thousands) and numbers of burglaries for randomly selected large colleges in a recent year (based on data from the New York Times).

Conclusion The linear correlation coefficient r is found to be 0.499, the P-value is 0.393, and the critical values for a 0.05 significance level are\( \pm 0.878\). What should you conclude?

The following exercises are based on the following sample data consisting of numbers of enrolled students (in thousands) and numbers of burglaries for randomly selected large colleges in a recent year (based on data from the New York Times).

Repeat the preceding exercise, assuming that the linear correlation coefficient is r= 0.997.

Interpreting a Graph The accompanying graph plots the numbers of points scored in each Super Bowl to the last Super Bowl at the time of this writing. The graph of the quadratic equation that best fits the data is also shown in red. What feature of the graph justifies the value of\({R^2}\)= 0.255 for the quadratic model?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free