R² & Model Fit - College Statistics

1. In a simple linear regression analysis with 50 data points, the total sum of squares (SST) is found to be 1200, and the sum of squared errors (SSE) is 300. What is the coefficient of determination for this model?
   Answer: 0.75 The coefficient of determination (R²) can be calculated using the formula R² = 1 - (SSE / SST), where SSE is the sum of squared errors and SST is the total sum of squares. Substituting the given values: R² = 1 - (300 / 1200) = 1 - 0.25 = 0.75. Alternatively, one could calculate the regression sum of squares (SSR) as SST - SSE = 1200 - 300 = 900, and then use the formula R² = SSR / SST = 900 / 1200 = 0.75.
2. An economist studies the relationship between GDP per capita (X) and life expectancy (Y) for two different groups of countries. * Group A: 20 highly developed countries. The relationship is found to be positive and roughly linear, with R² = 0.25. * Group B: 150 countries from all development levels. The relationship is also found to be positive and roughly linear, with R² = 0.75. Assuming a linear model is appropriate for both groups, what is the best statistical explanation for the large difference in R² values?
   Answer: The range of GDP per capita values is much wider in Group B than in Group A, making the underlying trend more apparent. The value of R² is highly sensitive to the range of the predictor variable. Group A consists of countries with similar, high levels of GDP per capita, representing a restricted range of X values. In this narrow range, the variation in life expectancy may be dominated by other factors, resulting in a low R². Group B includes countries from all development levels, representing a much wider range of X values. Over this broad range, the strong underlying relationship between X and Y becomes much more pronounced relative to the random error, leading to a much higher R².
3. A linear regression model is fit to a dataset. The resulting analysis shows that the proportion of total variation in the response variable that is unexplained by the model is 0.36. The slope of the regression line is negative. What is the value of the correlation coefficient (r)?
   Answer: -0.80 The proportion of unexplained variation is given by SSE/SST. We are told this value is 0.36. The coefficient of determination, R², is the proportion of explained variation, calculated as R² = 1 - SSE/SST. So, R² = 1 - 0.36 = 0.64. The correlation coefficient, r, is the square root of R², and its sign matches the sign of the slope. So, r = ±√0.64 = ±0.80. Since the slope is negative, we must choose the negative root. Thus, r = -0.80.
4. A scientist models the relationship between temperature in Celsius (X) and the metabolic rate of a certain insect (Y). The resulting least-squares regression line has a coefficient of determination of R² = 0.49. The scientist then decides to convert the temperature measurements from Celsius to Fahrenheit using the formula F = (9/5)C + 32. If the regression is re-run using temperature in Fahrenheit as the predictor, what will the new R² be?
   Answer: 0.49 The correlation coefficient (r) and the coefficient of determination (R²) are invariant under linear transformations of the variables. Converting from Celsius to Fahrenheit is a linear transformation (multiplying by a constant and adding a constant). This transformation changes the scale and origin of the predictor variable but does not change the strength of the linear association between the variables. Therefore, the R² value will remain 0.49.
5. A scatterplot of data for two variables, X and Y, shows a perfect U-shaped parabolic relationship, where the data points are perfectly symmetric about the line X = 5. A simple linear regression model is fit to this data. What will the coefficient of determination, R², be?
   Answer: Exactly 0 Because the data is perfectly symmetric in a U-shape, for every point (x, y) on one side of the axis of symmetry, there is a corresponding point with the same y-value on the other side. The least-squares regression line that best fits this data will be a horizontal line at the mean of Y. A horizontal line has a slope of 0. When the slope of the regression line is 0, the model explains none of the variation in Y. Thus, the regression sum of squares (SSR) is 0, and the coefficient of determination, R² = SSR/SST, is also 0.
6. A researcher fits a simple linear regression model and obtains an R² value of 0.95. Another researcher fits a model to a different dataset and obtains R² = 0.25. Both models satisfy the assumptions of linear regression. Which statement correctly compares the predictive power of the two models?
   Answer: The first model will generally produce narrower prediction intervals for Y for a given value of X than the second model. A higher R² value indicates that the data points are, on average, closer to the regression line. The width of a prediction interval depends on the standard error of the estimate, which is a measure of the typical distance of data points from the line. A smaller standard error (resulting from a higher R²) leads to narrower, more precise prediction intervals. The other options are incorrect: R² is independent of the slope's magnitude, does not imply causation, and provides no information about the y-intercept.
7. A market analyst develops a linear model to predict monthly sales of a product (in thousands of dollars) based on the monthly advertising budget (in hundreds of dollars). The model has an R² value of 0.75. Which of the following is the most accurate interpretation of this value?
   Answer: 75% of the variability in monthly sales can be explained by the linear relationship with the monthly advertising budget. The coefficient of determination, R², measures the proportion of the total variation in the response variable (Y) that is explained by the linear regression model with the predictor variable (X). Therefore, an R² of 0.75 means that 75% of the variation in monthly sales is accounted for by the linear model based on advertising budget.
8. A linear regression is performed on a set of 20 data points. The resulting model has R² = 0.80. A 21st data point, whose x-value is different from the mean of the original x-values, is then added to the dataset. This new point lies exactly on the original least-squares regression line. How will the addition of this point affect the value of R²?
   Answer: R² will increase. A point that lies exactly on the regression line has a residual of zero, so it does not add to the sum of squared errors (SSE). However, a point whose x-value is not the mean of x will increase the total sum of squares (SST) because it adds variation in the y-direction. Since R² = 1 - SSE/SST, and SSE remains constant while SST increases, the fraction SSE/SST will decrease. Consequently, R² will increase. Such a point reinforces the existing trend, strengthening the measured association.
9. In a simple linear regression, a researcher finds that the sum of squared errors (SSE) is equal to the total sum of squares (SST). Which of the following must be true?
   Answer: The coefficient of determination R² is 0. The coefficient of determination is calculated as R² = 1 - (SSE/SST). If SSE = SST, then the ratio SSE/SST is equal to 1. Substituting this into the formula gives R² = 1 - 1 = 0. This means that the linear model explains none of the variability in the response variable, which occurs when the predictor and response variables are uncorrelated (r=0).
10. A researcher attempts to model crop yield (Y) based on rainfall (X) using data from 100 farms. Two models are considered: * Model 1: A simple linear model, $\hat{Y} = b_0 + b_1 X$, yields R² = 0.40. * Model 2: A logarithmic transformation is applied to the predictor, $\hat{Y} = c_0 + c_1 \ln(X)$, yielding R² = 0.70. What can be concluded by comparing these two models?
    Answer: The sum of squared residuals (SSE) is smaller for Model 2 than for Model 1. Both models are predicting the same response variable, crop yield (Y). This means that the total sum of squares (SST), which measures the total variation in Y, is identical for both models. The coefficient of determination is defined as R² = 1 - SSE/SST. Since SST is the same for both, a model with a higher R² must have a lower sum of squared errors (SSE). As Model 2 has a higher R² (0.70 vs 0.40), its SSE must be smaller, indicating a better fit to the data.
11. A researcher is modeling the relationship between years of education and annual income. The initial analysis of 200 individuals yields an R² of 0.30. Upon review, it is discovered that a data entry error was made for one individual, who was recorded as having 8 years of education and an annual income of $500,000. The correct income for this individual is $50,000. All other individuals in the sample have incomes below $150,000. How will correcting this single data point most likely affect the R² value?
    Answer: R² will increase because the corrected point is no longer an extreme outlier, improving the overall fit of the linear model. The original point ($500,000 income) is a significant outlier in the y-direction. Such outliers dramatically increase both the total sum of squares (SST) and the sum of squared errors (SSE). While correcting the error reduces both SST and SSE, the reduction in SSE (the measure of error) is proportionally much larger than the reduction in SST. Since R² = 1 - SSE/SST, a large proportional decrease in SSE relative to SST will cause R² to increase, reflecting a better fit for the corrected data.
12. A simple linear regression model is built to predict Y from X. The standard deviation of the Y values is $s_y = 10$. The standard error of the estimate (which measures the typical size of a residual) is $s_e = 6$. Assuming the model assumptions are met and the sample size is reasonably large, what is the approximate coefficient of determination, R²?
    Answer: 0.64 The coefficient of determination, R², can be interpreted as the proportional reduction in variance. It can be approximated by the formula R² ≈ 1 - ($s_e$² / $s_y$²), where $s_e$ is the standard error of the estimate and $s_y$ is the standard deviation of the response variable. In this case, R² ≈ 1 - (6² / 10²) = 1 - (36 / 100) = 1 - 0.36 = 0.64.
13. For a given simple linear regression model, the sum of squared errors (SSE) is found to be one-fourth of the total sum of squares (SST). What is the absolute value of the correlation coefficient, |r|, for this model?
    Answer: 0.866 The coefficient of determination is given by R² = 1 - SSE/SST. We are given that SSE = (1/4)SST, so SSE/SST = 1/4 = 0.25. Plugging this into the formula, we get R² = 1 - 0.25 = 0.75. The absolute value of the correlation coefficient, |r|, is the square root of R². Therefore, |r| = √0.75 = √(3/4) = √3 / 2 ≈ 0.866.
14. A regression analysis is performed on 30 data points that show a moderately strong positive linear association, with R² ≈ 0.60. A new, 31st data point is added. This point has an x-value that is extremely large compared to the other data points, and it lies far below the trend established by the original 30 points. What is the most likely effect of adding this influential point on the correlation coefficient (r) and the coefficient of determination (R²)?
    Answer: The correlation r will decrease (become less positive), and R² will also decrease. The added point is an influential point with high leverage (extreme x-value) that deviates substantially from the existing linear pattern. Such a point will pull the regression line towards itself, which in this case means it will flatten the slope. A flatter slope corresponds to a weaker linear relationship. Therefore, the positive correlation coefficient (r) will decrease, moving closer to zero. Since R² = r², a decrease in the magnitude of r will cause R² to decrease as well.
15. A multiple regression model with 3 predictor variables is used to predict a response variable Y, resulting in R² = 0.70. A fourth predictor variable, which is known to be completely unrelated to Y, is added to the model. Which of the following statements about the new R² and the new adjusted R² is most likely to be true?
    Answer: The new R² will be slightly greater than 0.70, while the new adjusted R² will be less than the original adjusted R². Adding any predictor variable to a model can only cause the standard R² to increase or stay the same; it can never decrease. Even a useless predictor will explain a small amount of variance by chance. However, the adjusted R² penalizes the model for adding predictors that do not contribute meaningfully to explaining the response. Since the fourth predictor is unrelated to Y, it will not improve the model fit enough to offset the penalty, causing the adjusted R² to decrease.
16. In a simple linear regression analysis, the regression sum of squares (SSR) is calculated to be 0. What must be true about the relationship between the predictor variable X and the response variable Y?
    Answer: The correlation coefficient between X and Y is 0. The regression sum of squares, SSR = Σ(ŷᵢ - ȳ)², measures the variation in Y that is explained by the regression model. If SSR = 0, it means that the predicted value for every data point is simply the mean of Y (ŷᵢ = ȳ). This occurs when the regression line is a horizontal line with a slope of 0. A slope of 0 indicates that there is no linear relationship between X and Y. Therefore, the correlation coefficient (r) and the coefficient of determination (R²) must both be 0.
17. An analyst fits two different models to predict house prices. Model A is a simple linear regression using square footage as the predictor, yielding R² = 0.65. Its residual plot shows a clear curved (parabolic) pattern. Model B is a quadratic regression using square footage and square footage squared, yielding R² = 0.85. Its residual plot shows no obvious pattern. Which of the following is the best conclusion?
    Answer: Model B is preferable because it has a higher R² and its residual plot indicates a better model fit. Model evaluation depends on both the coefficient of determination (R²) and an assessment of model assumptions, often done via a residual plot. The curved pattern in the residual plot for Model A indicates that the linear model is not appropriate for the data. Model B has a substantially higher R² and a residual plot with no pattern, which suggests that it provides a much better fit to the data by capturing the non-linear relationship. Therefore, Model B is the superior model.
18. A linear regression model is used to predict a student's final exam score based on their hours of study. The model yields a coefficient of determination of 0.64. A residual analysis shows no discernible pattern. The covariance between hours of study and exam score is positive. What is the correlation coefficient between hours of study and exam score?
    Answer: 0.80 The coefficient of determination, R², is the square of the correlation coefficient, r. Therefore, r = ±√R². Given R² = 0.64, r = ±√0.64 = ±0.80. The sign of the correlation coefficient is the same as the sign of the slope of the regression line. Since the covariance between the two variables is positive, the slope of the regression line must also be positive. Thus, the correlation coefficient r must be positive, so r = 0.80.
19. A regression model is built from 50 data points. The R² is 0.85. One of the data points is an influential point with high leverage (its x-value is far from the mean of x). This point, however, lies very close to the least-squares regression line determined by the other 49 points. If this single influential point is removed from the dataset, what is the most likely effect on R²?
    Answer: R² will likely decrease because a point that confirmed the trend at an extreme x-value has been removed. A high-leverage point that conforms to the existing trend (i.e., lies near the regression line) disproportionately increases the regression sum of squares (SSR) and the total sum of squares (SST) while adding very little to the error sum of squares (SSE). This has the effect of inflating the R² value. Removing such a point removes its substantial contribution to SSR and SST. This typically leads to a decrease in R², as the model's apparent strength was heavily dependent on that one confirming, high-leverage point.

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