Bluebook SAT Test Two-Variable Data: Models and Scatterplots 5

Question 1

A researcher models a dataset with the linear equation ŷ = 2.4x + 5.1. The residual plot shows residuals clustered tightly near zero when x is small, but spreading widely (both positive and negative) as x increases. What is the most appropriate conclusion?

Accepted Answer

The linear model is likely inappropriate; the increasing spread suggests a non-constant variance that a different model or transformation may address

Answer

A fan-shaped residual plot—where spread increases with x—indicates heteroscedasticity (non-constant variance). This violates a key assumption of linear regression and suggests the linear model is not the best fit. A transformation (such as a square root or log on y) or a nonlinear model should be considered. The fact that residuals average near zero does not mean the model is appropriate.

Question 2

A scientist measures a colony's population y (in thousands) at time x (in days). Plotting log₁₀(y) versus x produces a linear pattern described by log₁₀(ŷ) = 0.12x + 0.9. Which equation best models the original population data?

Accepted Answer

y = 10^0.9 · (10^0.12)^x

Answer

If log₁₀(y) = 0.12x + 0.9, then y = 10^(0.12x + 0.9) = 10^0.9 · 10^(0.12x) = 10^0.9 · (10^0.12)^x. This is the exponential form y = a · b^x, where a = 10^0.9 ≈ 7.94 and b = 10^0.12 ≈ 1.32. The key insight is that a linear relationship between log(y) and x signals an underlying exponential model.

Question 3

For 18 students, the correlation between hours of social media use per day (x) and GPA (y) is r = −0.52. One student uses social media 0 hours per day but has the lowest GPA in the class (1.8). If this student's data point is removed, what is the most likely effect on r?

Accepted Answer

r will become more negative (stronger) because this point weakens the negative association

Answer

This student (low social media, low GPA) is an outlier that contradicts the negative trend—the pattern suggests less social media corresponds to higher GPA, but this student breaks that pattern. Removing a point that goes against the general trend strengthens the correlation in that direction. Therefore r will become more negative, moving away from 0.

Question 4

A linear model predicting monthly electricity cost (y, in dollars) from average daily temperature (x, in °F) yields R² = 0.58. A quadratic model for the same data yields R² = 0.74. Which statement is most statistically sound?

Accepted Answer

The quadratic model explains more variation in cost, but the gain in R² should be weighed against added complexity and checked with a residual analysis

Answer

R² measures the proportion of variability in y explained by the model—not the percentage of predictions that are 'correct.' A higher R² does not automatically justify a more complex model; adding parameters always increases R², even if the improvement is trivial or due to overfitting. The right choice depends on whether the scatterplot and residual plot support the quadratic form and whether the 16-point gain in R² is meaningful in context.

Question 5

A regression line is fit to 20 data points with x-values ranging from 1 to 8, yielding ŷ = 4x + 10 and r = 0.87. A 21st point at (x = 30, y = 130) is added. The new line is ŷ = 3.96x + 10.4 with r = 0.91. This new point is best described as:

Accepted Answer

A high-leverage point that had little influence on the line because it falls near the predicted trend

Answer

The new point has extremely high leverage because its x-value (30) is far from the existing x-range (1–8). However, plugging x = 30 into the original equation gives ŷ = 4(30) + 10 = 130—exactly where the new point lands. Because the point lies on (or very near) the existing trend line, it does not change the slope or intercept much. High-leverage points are only influential when they deviate from the predicted trend; this one does not.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Two-Variable Data: Models and Scatterplots 5