College Statistics Quiz: R2 And Model Fit

20 questions · exam conditions

0:00

R2 And Model FitQuestion 1 of 20

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?

0.27

0.49

0.88

1.58

← Back to quizzes

College Statistics Quiz

College Statistics Quiz: R2 And Model Fit

Practice R2 And Model Fit in College Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on R2 And Model Fit, giving you a quick way to practice the rules, question types, and explanations that matter most for College Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

0.27
0.49 (correct answer)
0.88
1.58

Explanation: 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.

Question 2

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?

0.25
0.33
0.75 (correct answer)
0.87

Explanation: 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.

Question 3

0.27
0.49 (correct answer)
0.88
1.58

Question 4

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)?

The correlation r will increase (become more positive), and R² will increase.
The correlation r will decrease (become less positive), and R² will also decrease. (correct answer)
The correlation r will decrease (become less positive), but R² will increase.
The correlation r will become negative, but R² will remain positive and increase.

Explanation: 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.

Question 5

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?

R² will decrease.
R² will remain exactly 0.80.
R² will increase. (correct answer)
The effect on R² depends on whether the new point's y-value is positive or negative.

Explanation: 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.

Question 6

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?

R² will likely increase because an influential point was removed.
R² will likely decrease because a point that confirmed the trend at an extreme x-value has been removed. (correct answer)
R² will remain unchanged since the point was already close to the regression line.
The effect cannot be predicted without knowing the slope of the regression line.

Explanation: 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.

Question 7

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?

The new R² will be slightly greater than 0.70, while the new adjusted R² will be less than the original adjusted R². (correct answer)
Both the new R² and the new adjusted R² will be less than their original values.
Both the new R² and the new adjusted R² will be greater than their original values.
The new R² will be equal to 0.70, and the new adjusted R² will be less than the original adjusted R².

Explanation: 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.

Question 8

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?

The slope coefficient $c_1$ in Model 2 must be numerically larger than the slope coefficient $b_1$ in Model 1.
Model 1 is invalid for prediction because its R² value is less than 0.50.
The sum of squared residuals (SSE) is smaller for Model 2 than for Model 1. (correct answer)
The total sum of squares (SST) is larger for Model 2 than for Model 1.

Explanation: 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.

Question 9

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?

Model A is preferable because simple models are generally better, and an R² of 0.65 is adequate.
Model B is preferable because it has a higher R² and its residual plot indicates a better model fit. (correct answer)
Both models are equally good because their R² values are both statistically significant and greater than 0.50.
Neither model is adequate because a good model should have an R² value of at least 0.90.

Explanation: 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.

Question 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?

The slope coefficient $c_1$ in Model 2 must be numerically larger than the slope coefficient $b_1$ in Model 1.
Model 1 is invalid for prediction because its R² value is less than 0.50.
The sum of squared residuals (SSE) is smaller for Model 2 than for Model 1. (correct answer)
The total sum of squares (SST) is larger for Model 2 than for Model 1.

Question 11

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?

-0.80
0.41
0.64
0.80 (correct answer)

Explanation: 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.

Question 12

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?

The model's predictions for monthly sales will be correct 75% of the time.
For every $100 increase in advertising, monthly sales are predicted to increase by 75%.
75% of the variability in monthly sales can be explained by the linear relationship with the monthly advertising budget. (correct answer)
75% of the data points for monthly sales and advertising budget lie on the least-squares regression line.

Explanation: 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.

Question 13

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?

Exactly 1
Exactly 0 (correct answer)
Approximately 0.5
Negative

Explanation: 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.

Question 14

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?

0.250
0.500
0.750
0.866 (correct answer)

Explanation: 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.

Question 15

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?

The first model's slope must be substantially steeper than the second model's slope.
The first model will generally produce narrower prediction intervals for Y for a given value of X than the second model. (correct answer)
The first model provides strong evidence of a causal relationship, while the second does not.
The y-intercept of the first model will necessarily be closer to zero than the y-intercept of the second model.

Explanation: 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.

Question 16

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?

0.36
0.40
0.60
0.64 (correct answer)

Explanation: The coefficient of determination, R², can be interpreted as the proportional reduction in variance. It can be approximated by the formula R² ≈ 1 - (s_e² / s_y²), 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.

Question 17

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?

The relationship between X and Y is perfectly linear with a positive slope.
The mean of the response variable Y must be 0.
The correlation coefficient between X and Y is 0. (correct answer)
All data points (X, Y) must lie on a vertical line.

Explanation: 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.

Question 18

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)?

0.64
0.80
-0.60
-0.80 (correct answer)

Explanation: 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.

Question 19

-0.80
0.41
0.64
0.80 (correct answer)

Question 20

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?

R² will increase because the corrected point is no longer an extreme outlier, improving the overall fit of the linear model. (correct answer)
R² will decrease because the total variation in income (SST) is substantially reduced by correcting the outlier.
R² will remain unchanged because the sample size is large enough to absorb the effect of a single point.
R² will become negative because the original model was heavily biased by the erroneous data point.

Explanation: 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.