College Statistics Quiz: Least Squares Regression
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Least Squares RegressionQuestion 1 of 20

A student is trying to find the best-fit line for a dataset. Instead of minimizing the sum of the squared vertical distances (∑(yi−y^i)2\sum(y_i - \hat{y}_i)^2), they mistakenly minimize the sum of the squared horizontal distances (∑(xi−x^i)2\sum(x_i - \hat{x}_i)^2). Which of the following is a necessary consequence of the student's procedure?

The resulting line will be the true least-squares regression line of y on x.
The student's line will have a smaller sum of squared vertical residuals than the true least-squares regression line.
The student's line will have a slope that is the reciprocal of the slope of the true least-squares regression line.
The sum of the horizontal residuals, ∑(xi−x^i)\sum(x_i - \hat{x}_i), for the student's line will be zero.
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College Statistics Quiz

College Statistics Quiz: Least Squares Regression

Practice Least Squares Regression 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 Least Squares Regression, 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

A student is trying to find the best-fit line for a dataset. Instead of minimizing the sum of the squared vertical distances (∑(yi−y^i)2\sum(y_i - \hat{y}_i)^2), they mistakenly minimize the sum of the squared horizontal distances (∑(xi−x^i)2\sum(x_i - \hat{x}_i)^2). Which of the following is a necessary consequence of the student's procedure?

  1. The resulting line will be the true least-squares regression line of y on x.
  2. The student's line will have a smaller sum of squared vertical residuals than the true least-squares regression line.
  3. The student's line will have a slope that is the reciprocal of the slope of the true least-squares regression line.
  4. The sum of the horizontal residuals, ∑(xi−x^i)\sum(x_i - \hat{x}_i), for the student's line will be zero. (correct answer)
Explanation: When you encounter regression problems, it's crucial to understand what minimizing different types of distances actually accomplishes. The student's approach fundamentally changes the nature of the regression problem. The correct answer is D because when you minimize horizontal distances ∑(xi−x^i)2\sum(x_i - \hat{x}_i)^2, you're essentially finding the regression line of x on y (treating x as the dependent variable). Just like in standard least squares regression where ∑(yi−y^i)=0\sum(y_i - \hat{y}_i) = 0, this procedure guarantees that the sum of horizontal residuals equals zero. This is a mathematical consequence of the minimization process - the residuals must sum to zero when you minimize their squared values. Option A is incorrect because minimizing horizontal distances produces the regression of x on y, not y on x. These are completely different lines unless there's perfect correlation. Option B is wrong because the student's line wasn't designed to minimize vertical residuals, so it will actually have a larger sum of squared vertical residuals than the true least squares line. Option C presents a tempting misconception - while the slopes of "y on x" and "x on y" regressions are related, they're not simple reciprocals. The relationship involves correlation: if the y-on-x slope is bb, then the x-on-y slope is r2/br^2/b, where rr is the correlation coefficient. Remember this pattern: whatever type of residual you minimize (vertical or horizontal) will always sum to zero. This is a fundamental property of least squares estimation that applies regardless of which variable you treat as dependent.

Question 2

A dataset has a least-squares regression line of y^=50−2x\hat{y} = 50 - 2x. The point of means, (xˉ,yˉ)(\bar{x}, \bar{y}), lies on this line. One of the points in the dataset is (10,35)(10, 35). Which of the following statements is true?

  1. The residual for the point (10, 35) is +5. (correct answer)
  2. The mean of the x-values, xˉ\bar{x}, must be 10, and the mean of the y-values, yˉ\bar{y}, must be 30.
  3. The sum of all the residuals for this dataset must be equal to 35.
  4. The correlation between x and y must be positive because the residual is positive.
Explanation: First, calculate the residual for the point (10,35)(10, 35). The predicted value at x=10x=10 is y^=50−2(10)=50−20=30\hat{y} = 50 - 2(10) = 50 - 20 = 30. The observed value is y=35y=35. The residual is the observed value minus the predicted value: y−y^=35−30=5y - \hat{y} = 35 - 30 = 5. Statement A is correct. Let's check the distractors. B is incorrect; while (xˉ,yˉ)(\bar{x}, \bar{y}) lies on the line, we cannot assume that the single point (10, 35) is the point of means. C is incorrect; the sum of all residuals for a least-squares regression line must be 0. D is incorrect; the slope of the regression line is -2, which indicates a negative correlation, and the sign of a single residual has no bearing on the overall correlation.

Question 3

For a given dataset, Line M is the least-squares regression line. Line N is another line passing through the data such that the sum of its residuals is also zero (∑(yi−y^Ni)=0\sum(y_i - \hat{y}_{Ni}) = 0). However, the sum of squared residuals for Line N is greater than for Line M. Which of the following statements must be true?

  1. Line N must be parallel to Line M, but with a different y-intercept.
  2. The y-intercept of Line N must be greater than the y-intercept of Line M.
  3. Line N cannot be a valid linear model for the data because its SSE is not minimized.
  4. Line N must also pass through the point of means, (xˉ,yˉ)(\bar{x}, \bar{y}). (correct answer)
Explanation: When you encounter questions about regression lines and residuals, focus on the fundamental properties that define the least-squares regression line versus other lines that might fit the data. The key insight here involves understanding what it means for residuals to sum to zero. Any line where ∑(yi−y^i)=0\sum(y_i - \hat{y}_i) = 0 must pass through the point of means (xˉ,yˉ)(\bar{x}, \bar{y}). This is a mathematical requirement: if you sum all the residuals and get zero, the line must balance above and below the center point of the data cloud. The least-squares regression line always has this property, but it's not unique to that line—other lines can also achieve this balance while having larger sums of squared residuals. Option A is incorrect because lines with zero-sum residuals don't need to be parallel; they just need to pass through (xˉ,yˉ)(\bar{x}, \bar{y}). Option B is wrong because the relationship between y-intercepts depends on the specific slopes and data—there's no requirement that Line N's y-intercept exceeds Line M's. Option C represents a common misconception: Line N can absolutely be a valid linear model even though it's not optimal. Many valid models exist for any dataset; the least-squares line is simply the best among linear models using the SSE criterion. Option D must be true because any line with residuals summing to zero mathematically must pass through the centroid of the data. Remember this principle: the sum of residuals equaling zero is equivalent to the line passing through the point of means—this connects algebraic properties to geometric intuition about regression lines.

Question 4

An analyst fits two different linear models, Line A and Line B, to the same set of bivariate data. For Line A, the sum of the residuals, ∑(y−y^)\sum(y - \hat{y}), is 0, and the sum of the squared residuals, ∑(y−y^)2\sum(y - \hat{y})^2, is 45.2. For Line B, the sum of the residuals is 2.1, and the sum of the squared residuals is 42.8. Which of the following statements is the most valid conclusion?

  1. Line A must be the least-squares regression line because the sum of its residuals is 0.
  2. Line B is a better fit than Line A in the least-squares sense, as it has a smaller sum of squared residuals. (correct answer)
  3. Neither line can be the least-squares regression line because for the LSRL, both the sum of residuals and the sum of squared residuals are minimized.
  4. It is impossible to determine which line provides a better fit without knowing the correlation coefficient for the data.
Explanation: The least-squares regression line (LSRL) is defined as the line that minimizes the sum of squared residuals (SSE). Line B has a smaller SSE (42.8) than Line A (45.2), meaning it is a better fit according to the least-squares criterion. While it is a property of the LSRL that the sum of its residuals is 0, this is a consequence of the minimization process, not the primary definition. The defining characteristic is the minimum possible SSE. Therefore, Line B provides a better fit. Distractor A states a true property of the LSRL but misses the more fundamental definition. C is incorrect because only SSE is minimized. D is incorrect because the LSRL is determined by minimizing SSE directly.

Question 5

A scatterplot of yy versus xx shows a strong positive linear association. A least-squares regression line is fitted to the data. One data point, (x1,y1)(x_1, y_1), has a large positive residual. If this single data point were shifted vertically downward so that its new residual (relative to the original line) became zero, and then a new least-squares line was fit to the modified data, which of the following would most likely occur?

  1. The slope of the new regression line would increase, and the sum of squared residuals would decrease.
  2. The slope of the new regression line would decrease, and the sum of squared residuals would decrease. (correct answer)
  3. The slope of the new regression line would increase, and the sum of squared residuals would increase.
  4. The slope of the new regression line would decrease, and the sum of squared residuals would increase.
Explanation: The point (x1,y1)(x_1, y_1) is above the regression line (large positive residual). Shifting it vertically downward reduces its y-value. This change will 'pull' the regression line toward the new, lower point. For a point with a positive residual in a positive association, this pull will typically cause the slope to decrease. The sum of squared residuals (SSE) for the new line will decrease because a point that was a major source of error has been moved to be less of an error source, reducing the overall variability of points around the fitted line. Distractor A gets the effect on the slope wrong. Distractors C and D incorrectly state that the SSE would increase, which is incorrect because the data has been modified to be less scattered.

Question 6

The least-squares regression line y^=10.5−2.1x\hat{y} = 10.5 - 2.1x is calculated for a dataset. One of the data points in the set is (x=5,y=0)(x=5, y=0). Which of the following statements about this data point is true?

  1. The removal of this point would significantly increase the slope of the line.
  2. The residual for this point is zero. (correct answer)
  3. This point must be an influential outlier that negatively affects the model fit.
  4. The correlation coefficient must be -1.0 because this point is predicted perfectly.
Explanation: To find the residual, we first calculate the predicted value y^\hat{y} for x=5x=5. y^=10.5−2.1(5)=10.5−10.5=0\hat{y} = 10.5 - 2.1(5) = 10.5 - 10.5 = 0. The observed value is y=0y=0. The residual is y−y^=0−0=0y - \hat{y} = 0 - 0 = 0. A residual of zero means the point lies exactly on the regression line. A is not necessarily true; since the point lies on the line, its removal may have little effect. C is incorrect; a point on the line is perfectly predicted and is the opposite of an outlier in terms of fit error. D is incorrect; for the correlation to be -1.0, all points must lie on the line, not just one.

Question 7

A least-squares regression model is fitted to a dataset with 10 observations. The residuals for the first 9 observations are: 2.5, -1.3, 0.8, -3.1, 4.0, -0.5, 1.7, -2.2, and 1.9. What is the residual for the 10th observation?

  1. -3.8 (correct answer)
  2. 0.38
  3. 3.8
  4. It cannot be determined without the original data or the regression equation.
Explanation: A fundamental property of the least-squares regression line is that the sum of all residuals is equal to zero. Let the 10th residual be e10e_{10}. We must have ∑i=110ei=0\sum_{i=1}^{10} e_i = 0. Summing the first 9 residuals gives: 2.5 - 1.3 + 0.8 - 3.1 + 4.0 - 0.5 + 1.7 - 2.2 + 1.9 = 3.8. To make the total sum zero, 3.8+e10=03.8 + e_{10} = 0, which means e10=−3.8e_{10} = -3.8. Distractor B is a calculation error (dividing the sum by 10). Distractor C is a sign error. D is incorrect because this property holds for any LSRL regardless of the data.

Question 8

For a set of 20 data points, the mean of the explanatory variable is xˉ=10\bar{x} = 10 and the mean of the response variable is yˉ=25\bar{y} = 25. The least-squares regression line for this data is given by y^=a+1.5x\hat{y} = a + 1.5x. What is the value of the y-intercept, aa?

  1. 1.5
  2. 10 (correct answer)
  3. 15
  4. 40
Explanation: A required property of the least-squares regression line is that it must pass through the point of means, (xˉ,yˉ)(\bar{x}, \bar{y}). This means the coordinates of this point must satisfy the equation of the line. We can substitute xˉ=10\bar{x} = 10 and yˉ=25\bar{y} = 25 into the equation y^=a+1.5x\hat{y} = a + 1.5x. Since the predicted value at the mean of x is the mean of y, we get 25=a+1.5(10)25 = a + 1.5(10). This simplifies to 25=a+1525 = a + 15. Solving for aa, we find a=25−15=10a = 25 - 15 = 10. Distractor A confuses the slope with the intercept. C is the value of the term 1.5x1.5x. D is a calculation error (25+1525+15).

Question 9

A researcher is modeling the price of a used car (yy, in dollars) based on its age (xx, in years). They calculate the least-squares regression line, y^=25000−2000x\hat{y} = 25000 - 2000x, and find the sum of squared residuals to be 5,400,000. An alternative model, proposed by a colleague, is a simple rule of thumb: Price = 24000 - 1900x. If the researcher calculates the sum of squared residuals for the colleague's model using the same data, which of the following statements must be true?

  1. The sum of squared residuals for the colleague's model will be less than 5,400,000.
  2. The sum of squared residuals for the colleague's model will be equal to 5,400,000.
  3. The sum of squared residuals for the colleague's model will be greater than or equal to 5,400,000. (correct answer)
  4. The sum of the residuals (not squared) for the colleague's model will be zero.
Explanation: The least-squares regression line is, by definition, the unique line that minimizes the sum of the squared residuals (SSE). The problem states that y^=25000−2000x\hat{y} = 25000 - 2000x is the least-squares line and its SSE is 5,400,000. Therefore, any other linear model, such as the colleague's, must have an SSE that is greater than or, in the extremely unlikely event the colleague guessed the exact same line, equal to 5,400,000. A and B contradict the definition of the LSRL. D is not necessarily true; only the LSRL is guaranteed to have a sum of residuals equal to zero.

Question 10

A real estate agent uses a least-squares regression model to predict the selling price of houses (yy, in thousands of dollars) based on their square footage (xx). The resulting model is y^=50+0.15x\hat{y} = 50 + 0.15x. A particular house with 2,000 square feet sold for $340,000. What is the interpretation of the residual for this house?

  1. The house sold for $10,000 more than its predicted price based on its size.
  2. The house sold for $10,000 less than its predicted price based on its size. (correct answer)
  3. The house's price was $10 per square foot lower than the average price per square foot for all houses.
  4. The model's prediction was off by $340,000 for this particular house.
Explanation: First, calculate the predicted price for a 2,000 sq ft house. The variable xx represents square footage, so x=2000x=2000. The predicted price (in thousands) is y^=50+0.15(2000)=50+300=350\hat{y} = 50 + 0.15(2000) = 50 + 300 = 350. This corresponds to a predicted price of $350,000. The actual selling price was y=340y=340 (in thousands). The residual is y−y^=340−350=−10y - \hat{y} = 340 - 350 = -10. Since the units are in thousands of dollars, a residual of -10 means the actual price was $10,000 less than the predicted price. Distractor A has the wrong sign. C misinterprets the residual as a rate. D misinterprets the calculation entirely.

Question 11

An analyst fits a least-squares regression line to predict sales based on advertising spending. The sum of squared residuals is found to be 400 (in squared thousands of dollars). A new data point, which lies exactly on the calculated regression line, is added to the dataset. What will be the sum of squared residuals for a new least-squares line fit to this augmented dataset?

  1. Less than 400.
  2. Equal to 400. (correct answer)
  3. Greater than 400.
  4. It could be greater or less than 400 depending on the x-value of the new point.
Explanation: Let the original least-squares regression line be L1L_1. The sum of squared residuals for the original data with respect to L1L_1 is 400. The new data point lies exactly on L1L_1, which means its residual relative to L1L_1 is 0. If we were to use L1L_1 as the model for the new, augmented dataset, the sum of squared residuals would be the original 400 plus the new point's contribution of 020^2, which is still 400. Because the new point perfectly conforms to the original trend, it does not pull the line in any new direction. Therefore, the best-fit line for the augmented dataset, L2L_2, will be identical to L1L_1. Since the line does not change and the new point has a zero residual, the sum of squared residuals remains 400.

Question 12

The least-squares regression line for a dataset is y^=10+3x\hat{y} = 10 + 3x, and the corresponding sum of squared residuals (SSE) is 50. If the slope of the line is changed to 3.1 while the y-intercept is kept at 10, creating a new line y^∗=10+3.1x\hat{y}^* = 10 + 3.1x, what can be said about the SSE calculated for this new line?

  1. The new SSE will be less than 50.
  2. The new SSE will be equal to 50.
  3. The new SSE will be greater than 50. (correct answer)
  4. The new SSE could be greater or smaller than 50, depending on the data's correlation.
Explanation: The least-squares regression line (in this case, y^=10+3x\hat{y} = 10 + 3x) is the unique line that produces the minimum possible sum of squared residuals (SSE) for a given dataset. The problem states this minimum SSE is 50. Any other line, including the slightly altered line y^∗=10+3.1x\hat{y}^* = 10 + 3.1x, is not the optimal line in the least-squares sense. Therefore, when applied to the same dataset, it must result in a larger SSE. Distractors A, B, and D all contradict the fundamental definition of the least-squares line.

Question 13

A least-squares regression line is fit to a scatterplot of data that shows a moderate positive linear association. A new point is then added to the data. This new point has an x-value that is much larger than the other x-values in the dataset and a y-value that gives it a large positive residual with respect to the original regression line. How will a new least-squares line, fit to the augmented data, likely compare to the original line?

  1. The slope will increase and the y-intercept will also increase.
  2. The slope will decrease and the y-intercept will also decrease.
  3. The slope will decrease and the y-intercept will increase.
  4. The slope will increase and the y-intercept will decrease. (correct answer)
Explanation: When you encounter questions about how outliers affect regression lines, focus on understanding leverage and influence. Points with extreme x-values have high leverage, meaning they can strongly pull the regression line toward themselves. Let's analyze this scenario step by step. The new point has a much larger x-value than existing data, giving it high leverage. Since it has a large positive residual from the original line, it sits well above where the original line would predict. This high-leverage point above the line will pull the new regression line upward toward itself. When the line gets pulled up toward this outlier, two things happen: the line rotates to have a steeper slope (increase), and since the line pivots around the center of the data while being pulled up on the right side, the left side drops down, decreasing the y-intercept. Think of it like a seesaw - when the right side goes up, the left side goes down. Looking at the wrong answers: Choice A incorrectly suggests both slope and y-intercept increase, missing that the line pivots downward on the left. Choice B suggests both decrease, which would happen if the outlier were below the original line. Choice C suggests the slope decreases while y-intercept increases, which contradicts how high-leverage points above the line behave. The correct answer is D - slope increases and y-intercept decreases. Study tip: Remember that high-leverage outliers pull regression lines toward themselves like magnets. Always visualize how the line would tilt: if the outlier is above the line on the right, the slope increases and y-intercept decreases.

Question 14

An analyst fits two different linear models, Line A and Line B, to the same set of bivariate data. For Line A, the sum of the residuals, ∑(y−y^)\sum(y - \hat{y}), is 0, and the sum of the squared residuals, ∑(y−y^)2\sum(y - \hat{y})^2, is 45.2. For Line B, the sum of the residuals is 2.1, and the sum of the squared residuals is 42.8. Which of the following statements is the most valid conclusion?

  1. Line A must be the least-squares regression line because the sum of its residuals is 0.
  2. Line B is a better fit than Line A in the least-squares sense, as it has a smaller sum of squared residuals. (correct answer)
  3. Neither line can be the least-squares regression line because for the LSRL, both the sum of residuals and the sum of squared residuals are minimized.
  4. It is impossible to determine which line provides a better fit without knowing the correlation coefficient for the data.
Explanation: The least-squares regression line (LSRL) is defined as the line that minimizes the sum of squared residuals (SSE). Line B has a smaller SSE (42.8) than Line A (45.2), meaning it is a better fit according to the least-squares criterion. While it is a property of the LSRL that the sum of its residuals is 0, this is a consequence of the minimization process, not the primary definition. The defining characteristic is the minimum possible SSE. Therefore, Line B provides a better fit. Distractor A states a true property of the LSRL but misses the more fundamental definition. C is incorrect because only SSE is minimized. D is incorrect because the LSRL is determined by minimizing SSE directly.

Question 15

A least-squares regression model is fitted to a dataset with 10 observations. The residuals for the first 9 observations are: 2.5, -1.3, 0.8, -3.1, 4.0, -0.5, 1.7, -2.2, and 1.9. What is the residual for the 10th observation?

  1. -3.8 (correct answer)
  2. 0.38
  3. 3.8
  4. It cannot be determined without the original data or the regression equation.
Explanation: A fundamental property of the least-squares regression line is that the sum of all residuals is equal to zero. Let the 10th residual be e10e_{10}. We must have ∑i=110ei=0\sum_{i=1}^{10} e_i = 0. Summing the first 9 residuals gives: 2.5 - 1.3 + 0.8 - 3.1 + 4.0 - 0.5 + 1.7 - 2.2 + 1.9 = 3.8. To make the total sum zero, 3.8+e10=03.8 + e_{10} = 0, which means e10=−3.8e_{10} = -3.8. Distractor B is a calculation error (dividing the sum by 10). Distractor C is a sign error. D is incorrect because this property holds for any LSRL regardless of the data.

Question 16

The relationship between the number of hours a student studies per week (xx) and their score on a final exam (yy) is modeled by the least-squares regression line y^=45+5.2x\hat{y} = 45 + 5.2x. One student in the study, who studied for 8 hours, scored an 82 on the exam. What is the residual for this student, and what does it indicate?

  1. -4.6; the student scored 4.6 points lower than the model predicted for 8 hours of study. (correct answer)
  2. 4.6; the student scored 4.6 points higher than the model predicted for 8 hours of study.
  3. -4.6; for each hour studied, the student's score was an average of 4.6 points below the prediction.
  4. 8.2; the student's score was 8.2 points above the y-intercept of the regression line.
Explanation: First, calculate the predicted score (y^\hat{y}) for x=8x=8: y^=45+5.2(8)=45+41.6=86.6\hat{y} = 45 + 5.2(8) = 45 + 41.6 = 86.6. The student's actual score (yy) was 82. The residual is the actual value minus the predicted value: y−y^=82−86.6=−4.6y - \hat{y} = 82 - 86.6 = -4.6. A negative residual indicates that the observed value is below the predicted value. Therefore, the student scored 4.6 points lower than predicted by the model. Distractor B calculates the residual correctly but gets the sign wrong in the interpretation. C misinterprets the meaning of a single residual, incorrectly describing it as a rate. D is nonsensical.

Question 17

A real estate agent uses a least-squares regression model to predict the selling price of houses (yy, in thousands of dollars) based on their square footage (xx). The resulting model is y^=50+0.15x\hat{y} = 50 + 0.15x. A particular house with 2,000 square feet sold for $340,000. What is the interpretation of the residual for this house?

  1. The house sold for $10,000 more than its predicted price based on its size.
  2. The house sold for $10,000 less than its predicted price based on its size. (correct answer)
  3. The house's price was $10 per square foot lower than the average price per square foot for all houses.
  4. The model's prediction was off by $340,000 for this particular house.
Explanation: First, calculate the predicted price for a 2,000 sq ft house. The variable xx represents square footage, so x=2000x=2000. The predicted price (in thousands) is y^=50+0.15(2000)=50+300=350\hat{y} = 50 + 0.15(2000) = 50 + 300 = 350. This corresponds to a predicted price of $350,000. The actual selling price was y=340y=340 (in thousands). The residual is y−y^=340−350=−10y - \hat{y} = 340 - 350 = -10. Since the units are in thousands of dollars, a residual of -10 means the actual price was $10,000 less than the predicted price. Distractor A has the wrong sign. C misinterprets the residual as a rate. D misinterprets the calculation entirely.

Question 18

The least-squares regression line for a dataset is y^=10+3x\hat{y} = 10 + 3x, and the corresponding sum of squared residuals (SSE) is 50. If the slope of the line is changed to 3.1 while the y-intercept is kept at 10, creating a new line y^∗=10+3.1x\hat{y}^* = 10 + 3.1x, what can be said about the SSE calculated for this new line?

  1. The new SSE will be less than 50.
  2. The new SSE will be equal to 50.
  3. The new SSE will be greater than 50. (correct answer)
  4. The new SSE could be greater or smaller than 50, depending on the data's correlation.
Explanation: The least-squares regression line (in this case, y^=10+3x\hat{y} = 10 + 3x) is the unique line that produces the minimum possible sum of squared residuals (SSE) for a given dataset. The problem states this minimum SSE is 50. Any other line, including the slightly altered line y^∗=10+3.1x\hat{y}^* = 10 + 3.1x, is not the optimal line in the least-squares sense. Therefore, when applied to the same dataset, it must result in a larger SSE. Distractors A, B, and D all contradict the fundamental definition of the least-squares line.

Question 19

For a given dataset, Line M is the least-squares regression line. Line N is another line passing through the data such that the sum of its residuals is also zero (∑(yi−y^Ni)=0\sum(y_i - \hat{y}_{Ni}) = 0). However, the sum of squared residuals for Line N is greater than for Line M. Which of the following statements must be true?

  1. Line N must be parallel to Line M, but with a different y-intercept.
  2. The y-intercept of Line N must be greater than the y-intercept of Line M.
  3. Line N cannot be a valid linear model for the data because its SSE is not minimized.
  4. Line N must also pass through the point of means, (xˉ,yˉ)(\bar{x}, \bar{y}). (correct answer)
Explanation: When you encounter questions about regression lines and residuals, focus on the fundamental properties that define the least-squares regression line versus other lines that might fit the data. The key insight here involves understanding what it means for residuals to sum to zero. Any line where ∑(yi−y^i)=0\sum(y_i - \hat{y}_i) = 0 must pass through the point of means (xˉ,yˉ)(\bar{x}, \bar{y}). This is a mathematical requirement: if you sum all the residuals and get zero, the line must balance above and below the center point of the data cloud. The least-squares regression line always has this property, but it's not unique to that line—other lines can also achieve this balance while having larger sums of squared residuals. Option A is incorrect because lines with zero-sum residuals don't need to be parallel; they just need to pass through (xˉ,yˉ)(\bar{x}, \bar{y}). Option B is wrong because the relationship between y-intercepts depends on the specific slopes and data—there's no requirement that Line N's y-intercept exceeds Line M's. Option C represents a common misconception: Line N can absolutely be a valid linear model even though it's not optimal. Many valid models exist for any dataset; the least-squares line is simply the best among linear models using the SSE criterion. Option D must be true because any line with residuals summing to zero mathematically must pass through the centroid of the data. Remember this principle: the sum of residuals equaling zero is equivalent to the line passing through the point of means—this connects algebraic properties to geometric intuition about regression lines.

Question 20

A least-squares regression line is fit to a scatterplot of data that shows a moderate positive linear association. A new point is then added to the data. This new point has an x-value that is much larger than the other x-values in the dataset and a y-value that gives it a large positive residual with respect to the original regression line. How will a new least-squares line, fit to the augmented data, likely compare to the original line?

  1. The slope will increase and the y-intercept will also increase.
  2. The slope will decrease and the y-intercept will also decrease.
  3. The slope will decrease and the y-intercept will increase.
  4. The slope will increase and the y-intercept will decrease. (correct answer)
Explanation: When you encounter questions about how outliers affect regression lines, focus on understanding leverage and influence. Points with extreme x-values have high leverage, meaning they can strongly pull the regression line toward themselves. Let's analyze this scenario step by step. The new point has a much larger x-value than existing data, giving it high leverage. Since it has a large positive residual from the original line, it sits well above where the original line would predict. This high-leverage point above the line will pull the new regression line upward toward itself. When the line gets pulled up toward this outlier, two things happen: the line rotates to have a steeper slope (increase), and since the line pivots around the center of the data while being pulled up on the right side, the left side drops down, decreasing the y-intercept. Think of it like a seesaw - when the right side goes up, the left side goes down. Looking at the wrong answers: Choice A incorrectly suggests both slope and y-intercept increase, missing that the line pivots downward on the left. Choice B suggests both decrease, which would happen if the outlier were below the original line. Choice C suggests the slope decreases while y-intercept increases, which contradicts how high-leverage points above the line behave. The correct answer is D - slope increases and y-intercept decreases. Study tip: Remember that high-leverage outliers pull regression lines toward themselves like magnets. Always visualize how the line would tilt: if the outlier is above the line on the right, the slope increases and y-intercept decreases.