Two-Variable Data: Models and Scatterplots Flashcards

A researcher plots the population of a bacterial colony (in thousands) versus time (in hours) and fits both a linear model ŷ = 3.2x + 1.4 and an exponential model ŷ = 1.4(2.1)^x. At hour 5, the linear model predicts 17,400 bacteria and the exponential model predicts 52,600. The actual count at hour 5 is 51,800. Which conclusion is best supported?

Answer: The exponential model is more appropriate because the residual from the linear model is far larger than from the exponential model.

The residual (actual − predicted) for the linear model is 51,800 − 17,400 = 34,400, whereas for the exponential model it is 51,800 − 52,600 = −800. The exponential model's residual is dramatically smaller in absolute value, indicating a much better fit. Whether a model over- or underestimates at one point is less important than the magnitude of the residual. 'Simpler is always better' is not a valid statistical principle without comparing fit quality.

A scatterplot of 40 data points shows a strong positive linear association with r = 0.91. A student removes the two points with the highest x-values and recomputes the correlation, obtaining r = 0.63. What is the most accurate interpretation?

Answer: The two removed points were likely high-leverage influential points that were strengthening the apparent linear relationship.

Points with extreme x-values (high leverage) can dramatically influence the correlation coefficient. When removing them drops r from 0.91 to 0.63, those points were acting as influential points that pulled the regression line and inflated r. This does not automatically mean the original r was 'wrong'—it means the relationship may be weaker among the bulk of the data, and those extreme points deserve further investigation. We cannot conclude from this alone that 0.63 is more 'reliable,' nor is it true that any two points would cause such a change.

The scatterplot below (described in text): a dataset of 12 points has a least-squares regression line ŷ = −0.5x + 80. One point is located at (60, 50). The mean of all x-values is 30 and the mean of all y-values is 65. Which of the following correctly describes the point (60, 50)?

Answer: It is a high-leverage point because its x-value is far from x̄, and its residual is exactly 0.

The predicted value at x = 60 is ŷ = −0.5(60) + 80 = 50, which exactly equals the observed y = 50, so the residual is 50 − 50 = 0. Since x = 60 is far from x̄ = 30, this is a high-leverage point. However, a residual of zero means the line passes through it perfectly—it is NOT pulling the line away from the other points in a distorting way. High leverage + zero residual = the point sits exactly where the line predicts, so while it has potential influence, its actual influence on the slope is minimal. Option A incorrectly calls it a y-outlier; option C ignores leverage; option D overstates influence when residual is 0.

A data analyst fits a linear regression to data and produces a residual plot. The residual plot shows residuals that are negative for low x-values, positive for middle x-values, and negative again for high x-values, forming a curved (U-shaped inverted) pattern. What does this pattern indicate?

Answer: A linear model is not the best fit; the curved residual pattern suggests a non-linear (e.g., quadratic) model would be more appropriate.

A well-fitting linear model produces a residual plot with points scattered randomly around zero with no pattern. A systematic curved (inverted U) pattern in residuals—negative, then positive, then negative—indicates that the relationship between x and y is itself curved (non-linear). This means a linear model is systematically over- or under-predicting across ranges of x, and a quadratic or other curved model would better capture the true relationship. Heteroscedasticity would appear as a funnel shape (changing spread), not a curved pattern. The residual plot conveys nothing about the sign of the correlation.

Two variables x and y have a correlation coefficient of r = −0.78. A researcher takes the log (base 10) of all y-values and recomputes the correlation between x and log(y), obtaining r = −0.43. What is the correct interpretation?

Answer: The original linear model between x and y fits better than a model between x and log(y), since |−0.78| > |−0.43|.

Correlation measures the strength of a LINEAR association. Since |r| = 0.78 for the original (x, y) pair is greater than |r| = 0.43 for (x, log y), the straight-line relationship between x and y is actually stronger than the straight-line relationship between x and log(y). This means a simple linear model y = mx + b fits better than an exponential model. A weaker correlation after transformation does NOT indicate improvement—it indicates the transformation made the linear relationship weaker. The log transformation is not universally appropriate; it helps only when the original relationship is exponential.

A least-squares regression line is computed for a dataset: ŷ = 4x − 12. The standard deviation of x is 5, and the standard deviation of y is 25. What is the correlation coefficient r?

Answer: 0.80

The slope of the least-squares regression line is related to the correlation coefficient by the formula: b = r · (s_y / s_x), where b is the slope, s_y is the standard deviation of y, and s_x is the standard deviation of x. Here, b = 4, s_y = 25, s_x = 5, so: 4 = r · (25/5) = r · 5, which gives r = 4/5 = 0.80. Since the slope is positive, r is positive. This question requires knowing the algebraic relationship between slope and correlation rather than reading r directly from a plot.