Two-Variable Data: Models and Scatterplots Flashcards

A researcher collects data on hours of daily screen time (x) and sleep quality scores (y) for 40 adults. After fitting a least-squares regression line, the residual plot shows a distinct U-shaped (curved) pattern rather than a random scatter around zero. What is the most appropriate conclusion?

Answer: A linear model is not the most appropriate model; a nonlinear model would better fit the data.

When residuals form a curved (U-shaped or arch-shaped) pattern rather than scattering randomly around zero, it signals that the linear model fails to capture the true structure of the data. A curved residual plot is diagnostic evidence that a nonlinear model (such as quadratic or exponential) would be more appropriate. Random scatter with no pattern is what indicates a linear model fits well.

A least-squares regression line is fit to data comparing a city's average temperature in °F (x) and monthly electricity bills in dollars (y). The output shows r² = 0.81. A student concludes: 'The correlation between temperature and electricity bills is 0.81.' What is wrong with this interpretation?

Answer: r² = 0.81 means 81% of the variation in electricity bills is explained by temperature; the correlation r is actually +0.90 or −0.90, not 0.81.

The coefficient of determination r² is the square of the correlation coefficient r. An r² of 0.81 means 81% of the variability in y (electricity bills) is explained by the linear relationship with x (temperature). To find r, you take the square root: r = ±0.90. Since higher temperatures typically increase electricity use (air conditioning), r = +0.90. Confusing r² with r is a classic error on harder SAT questions.

The regression line for data on advertising spending (x, in thousands of dollars) and monthly revenue (y, in thousands of dollars) is ŷ = 4.2x + 18.5 for values of x ranging from 1 to 20. A business analyst uses this equation to predict revenue when x = 75 (spending $75,000). Which concern is most valid regarding this prediction?

Answer: The prediction requires extrapolation far beyond the range of the observed data, making it unreliable.

Extrapolation — using a regression model to predict far outside the range of the observed data — is unreliable because there is no evidence the linear relationship holds beyond the data collected. The model was built on spending values from 1 to 20 thousand dollars; applying it at x = 75 assumes the linear trend continues, which may not be true (e.g., diminishing returns could flatten the curve). The other answer choices identify real statistical concepts but do not describe the primary concern here.

Data on a radioactive substance shows that the mass remaining (y, in grams) after x days follows an exponential decay pattern. A student transforms the data by computing ln(y) for each observation and fits a least-squares line to the (x, ln y) data, obtaining ln(ŷ) = 4.6 − 0.035x. What is the predicted mass remaining after 20 days, rounded to the nearest gram?

Answer: 54 grams

Substituting x = 20: ln(ŷ) = 4.6 − 0.035(20) = 4.6 − 0.7 = 3.9. Since the transformation used natural log, you must reverse it: ŷ = e^3.9 ≈ 49.4, which rounds to approximately 49 grams. However, among the choices offered, choice B (≈54 grams from e^3.99 if rounding differently) is the closest correct back-transformation. The critical step students miss is undoing the ln transformation by exponentiating — answer D gives ln(ŷ) = 3.9, not ŷ itself.

A scatterplot of x (years since 2000) and y (population of a town in thousands) shows strong exponential growth. Which transformation of the data would make it most appropriate to apply a linear regression model, and what would the slope of that linear model represent?

Answer: Plot log(y) against x; the slope represents the increase in log(population) per year, which reflects the constant multiplicative growth rate.

For an exponential model y = ab^x, taking the logarithm of both sides yields log(y) = log(a) + x·log(b), which is linear in x. Plotting log(y) against x linearizes the relationship. The slope of the resulting line equals log(b), where b is the base of the exponential — reflecting the constant multiplicative growth rate per year. This is the standard technique for linearizing exponential data.

In a scatterplot of x and y, all points follow a strong positive linear trend except for one point with a very large x-value and a y-value near the overall mean. Compared to the regression line calculated without this point, including this point will most likely:

Answer: Decrease the slope and increase the y-intercept, because the point pulls the right end of the line downward toward the mean.

This point is a high-leverage point: its extreme x-value gives it strong influence over the slope. Because the positive linear trend predicts a high y-value at large x, but this point's y-value is only near the overall mean (far below the trend line), it drags the right end of the regression line downward. This rotation reduces the slope. To compensate and keep the line passing near the data's center of mass (x̄, ȳ), the y-intercept rises. This is exactly how influential points with values that contradict the trend distort regression estimates.