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When we use \(R^{2}\) for a random sample to estimate a population \(R^{2}\), it's a bit biased. It tends to be a bit too large, especially when \(n\) is small. Some software also reports Adjusted \(R^{2}=R^{2}-\\{p /[n-(p+1)]\\}\left(1-R^{2}\right)\) where \(p=\) number of predictor variables in the model. This is slightly smaller than \(R^{2}\) and is less biased. Suppose \(R^{2}=0.500\) for a model with \(p=2\) predictors. Calculate adjusted \(R^{2}\) for the following sample sizes: 10,100,1000 . Show that the difference between adjusted \(R^{2}\) and \(R^{2}\) diminishes as \(n\) increases.

Short Answer

Expert verified
As sample size increases, the difference between Adjusted \( R^{2} \) and \( R^{2} \) diminishes.

Step by step solution

01

Understand the Adjusted R² Formula

The formula for Adjusted \( R^{2} \) is given by:\[\text{Adjusted } R^{2} = R^{2} - \frac{p}{n - (p + 1)} (1 - R^{2})\]where \( p \) is the number of predictor variables, \( R^{2} \) is the coefficient of determination, and \( n \) is the sample size.
02

Define Known Values

We are given that \( R^{2} = 0.500 \) and \( p = 2 \). We need to compute Adjusted \( R^{2} \) for three different sample sizes: \( n = 10 \), \( n = 100 \), and \( n = 1000 \).
03

Calculate Adjusted R² for n = 10

Substitute \( R^{2} = 0.500 \), \( p = 2 \), and \( n = 10 \) into the formula:\[\text{Adjusted } R^{2} = 0.500 - \frac{2}{10 - (2 + 1)} (1 - 0.500)\]This simplifies to:\[\text{Adjusted } R^{2} = 0.500 - \frac{2}{7} \times 0.500 = 0.500 - 0.143 = 0.357\]
04

Calculate Adjusted R² for n = 100

Substitute \( n = 100 \) into the formula:\[\text{Adjusted } R^{2} = 0.500 - \frac{2}{100 - (2 + 1)} (1 - 0.500)\]This simplifies to:\[\text{Adjusted } R^{2} = 0.500 - \frac{2}{97} \times 0.500 = 0.500 - 0.0103 = 0.4897\]
05

Calculate Adjusted R² for n = 1000

Substitute \( n = 1000 \) into the formula:\[\text{Adjusted } R^{2} = 0.500 - \frac{2}{1000 - (2 + 1)} (1 - 0.500)\]This simplifies to:\[\text{Adjusted } R^{2} = 0.500 - \frac{2}{997} \times 0.500 = 0.500 - 0.0010 = 0.4990\]
06

Compare Results and Conclude

The Adjusted \( R^{2} \) values we calculated are:- For \( n = 10 \): Adjusted \( R^{2} = 0.357 \)- For \( n = 100 \): Adjusted \( R^{2} = 0.4897 \)- For \( n = 1000 \): Adjusted \( R^{2} = 0.4990 \)As the sample size \( n \) increases, the difference between \( R^{2} \) and Adjusted \( R^{2} \) becomes smaller, confirming that the bias diminishes with larger sample sizes.

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Key Concepts

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

Coefficient of Determination
Understanding the coefficient of determination, commonly represented as \(R^2\), is fundamental in statistics, especially in regression analysis. This metric assesses how well a statistical model explains the variance of the dependent variable. Simply put, \(R^2\) indicates the proportion of the total variation in the outcome variable that is captured by the model using the predictor variables.

For example, if you have an \(R^2\) value of 0.75, it means 75% of the variation can be explained by the model's inputs. However, it is crucial to remember that \(R^2\) does not indicate the correctness of the model; it may sometimes be misleading as values can artificially inflate, particularly when more predictors are added.
Sample Size Effect
The sample size \(n\) plays a crucial role in the reliability and bias of \(R^2\) values. In smaller samples, \(R^2\) tends to be biased upwards, making it appear as though the model is explaining more variance than it truly is. This is where adjusted \(R^2\) becomes valuable.

As you compute adjusted \(R^2\) for different sample sizes, a pattern emerges: the larger the sample size, the closer the adjusted \(R^2\) value is to the original \(R^2\) value. This change happens because the adjustment formula compensates for the number of predictors in relation to the sample size, reducing the bias in smaller samples. This effect vividly demonstrates the power of having more data to accurately describe the model's predictive ability.
Predictor Variables
Predictor variables are the independent variables in your model, which you use to predict the dependent variable. The number \(p\) of these variables directly affects the calculation of adjusted \(R^2\).

Incorporating more predictor variables can inflate \(R^2\), as the model will capture more variation, but this doesn't mean it improves predictive accuracy. This potential overfitting is curbed by adjusted \(R^2\), which reduces \(R^2\) by considering the number of predictors.

This is critical in model selection, ensuring your model is both simple and effective, avoiding unnecessary complexity that doesn't translate into real predictive improvement.
Bias Reduction
Bias in regression analysis occurs when there is a systematic error in estimation. Regular \(R^2\) is susceptible to bias, particularly with small sample sizes and numerous predictors. Adjusted \(R^2\) is designed to reduce this bias.

By factoring in the sample size and the number of predictors, adjusted \(R^2\) lowers the inflation caused by these variables. The correction diminishes as the sample size grows, making adjusted \(R^2\) align closely with \(R^2\) with larger datasets.

This bias reduction is crucial for making realistic and reliable inferences from your model. It ensures the interpretability of results stays consistent, offering a more truthful performance metric.

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Most popular questions from this chapter

Let's use multiple regression to predict total body weight (in pounds) using data from a study of University of Georgia female athletes. Possible predictors are \(\mathrm{HGT}=\) height (in inches), \(\% \mathrm{BF}=\) percent body fat, and age. The display shows correlations among these explanatory variables. a. Which explanatory variable gives by itself the best predictions of weight? Explain. b. With height as the sole predictor, \(\hat{y}=-106+3.65\) \((\mathrm{HGT})\) and \(r^{2}=0.55 .\) If you add \(\% \mathrm{BF}\) as a predictor, you know that \(R^{2}\) will be at least \(0.55 .\) Explain why. c. When you add \% body fat to the model, \(\hat{y}=-121+3.50(\mathrm{HGT})+1.35(\% \mathrm{BF})\) and \(R^{2}=0.66 .\) When you add age to the model, \(\hat{y}=-97.7+3.43(\mathrm{HGT})+1.36(\% \mathrm{BF})-0.960(\mathrm{AGE})\) and \(R^{2}=0.67\). Once you know height and \(\%\) body fat, does age seem to help you in predicting weight? Explain, based on comparing the \(R^{2}\) values.

A study of horseshoe crabs found a logistic regression equation for predicting the probability that a female crab had a male partner nesting nearby using \(x=\) width of the carapace shell of the female crab (in centimeters). The results were Predictor Coef Constant -12.351 Width \(\quad 0.497\) a. For width, \(Q 1=24.9\) and \(Q 3=27.7\). Find the estimated probability of a male partner at \(\mathrm{Q} 1\) and at Q3. Interpret the effect of width by estimating the increase in the probability over the middle half of the sampled widths. b. At which carapace shell width level is the estimated probability of a male partner (i) equal to 0.50 , (ii) greater than 0.50 , and (iii) less than \(0.50 ?\)

In \(100-200\) words, explain to someone who has never studied statistics the purpose of multiple regression and when you would use it to analyze a data set or investigate an issue. Give an example of at least one application of multiple regression. Describe how multiple regression can be useful in analyzing complex relationships.

Suppose you fit a straight-line regression model to \(x=\) age of subjects and \(y=\) driving accident rate. Sketch what you would expect to observe for (a) the scatterplot of \(x\) and \(y\) and (b) a plot of the residuals against the values of age.

For the 59 observations in the Georgia Student Survey data file on the text CD, the result of regressing college GPA on high school GPA and study time follows. a. Explain in nontechnical terms what it means if the population slope coefficient for high school GPA equals \(0 .\) b. Show all steps for testing the hypothesis that this slope equals \(0 .\)

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