AP Statistics Exploring Two-Variable Data — Worked Answer Explanations

Unit 2 · 12 questions explained

Below is a complete answer key for our AP Statistics Exploring Two-Variable Data practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Exploring Two-Variable Data practice test and come back here to review, or head back to the Exploring Two-Variable Data unit overview.

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Question 1 · Easy
A scatterplot of hours studied ( $x$ ) versus exam score ( $y$ ) for 20 students shows a positive linear association with no obvious outliers. The correlation coefficient is $r = 0.83$ . What does this value tell us?
- A
  83% of the variation in exam scores is explained by hours studied.
  Why not A: This describes $r^{2} = 0.69$ , not $r = 0.83$ .
- B
  There is a strong positive linear association between hours studied and exam score.Correct
- C
  Studying more hours causes higher exam scores.
  Why not C: Correlation does not imply causation; association is not the same as a causal relationship.
- D
  Each additional hour of studying increases the score by 0.83 points.
  Why not D: This describes the slope, not the correlation coefficient.
Explanation
$r = 0.83$ describes the strength and direction of a linear association. Values of $∣ r ∣$ near 1 indicate a strong linear relationship; $r > 0$ indicates a positive direction. $r$ does not measure causation or slope.

Key takeaway
$r$ measures the strength and direction of a linear association; it does not imply causation or represent the slope.
Question 2 · Easy
Computer output for a regression of weight ( $y$ , in lbs) on height ( $x$ , in inches) gives: $\overset{y}{^} = - 130 + 4.5 x$ , $r^{2} = 0.71$ . Interpret the slope.
- A
  For each additional pound, height is predicted to increase by 4.5 inches.
  Why not A: This reverses the roles of $x$ and $y$ .
- B
  For each additional inch of height, weight is predicted to increase by 4.5 lbs.Correct
- C
  The average weight is 4.5 lbs.
  Why not C: 4.5 is the slope (rate of change), not an average value of $y$ .
- D
  71% of the increase in weight is caused by height.
  Why not D: This misinterprets $r^{2}$ and incorrectly claims causation.
Explanation
The slope $b = 4.5$ is interpreted as: for each 1-unit increase in $x$ (height), the predicted value of $y$ (weight) increases by 4.5 units (lbs). The slope is a rate of change of the predicted response per unit of the explanatory variable.

Key takeaway
Slope interpretation: for each 1-unit increase in $x$, the predicted $y$ changes by $b$ units.
Question 3 · Easy
A least-squares regression line is fit to data on advertising spending ( $x$ , thousands of dollars) and sales ( $y$ , thousands of units). The residual for one observation is $+ 12$ . What does this mean?
- A
  The actual sales were 12 thousand units less than predicted.
  Why not A: A positive residual means actual exceeds predicted, not the reverse.
- B
  The actual sales were 12 thousand units more than predicted.Correct
- C
  Advertising spending was overestimated by 12 thousand dollars.
  Why not C: Residuals refer to the response variable $y$ , not the explanatory variable $x$ .
- D
  The regression line passes 12 units above the data point.
  Why not D: A positive residual means the data point is above the line, not below.
Explanation
Residual $= actual - \overset{y}{^}$ . A residual of $+ 12$ means the actual sales exceeded the predicted sales by 12 thousand units. The point lies above the regression line.

Key takeaway
Residual $= y - \hat{y}$. Positive residual: point above the line; negative residual: point below the line.
Question 4 · Easy
A residual plot for a linear regression model shows a clear curved (U-shaped) pattern. What does this indicate?
- A
  The regression line fits the data well.
  Why not A: A well-fitting linear model produces a residual plot with no pattern.
- B
  There is likely an outlier in the data.
  Why not B: A U-shaped pattern across the whole plot indicates a systematic non-linear trend, not just an outlier.
- C
  A linear model is not appropriate; the relationship is non-linear.Correct
- D
  The standard deviation of the residuals is too large.
  Why not D: The size of residuals relates to variability, not the pattern shape, which indicates non-linearity.
Explanation
A residual plot for a well-fitting linear model should show random scatter with no pattern. A systematic curve in the residual plot reveals that the linear model does not capture the true relationship — a non-linear model would be more appropriate.

Key takeaway
Patterns in a residual plot (curves, fans) signal that the linear model is inadequate.
Question 5 · Easy
For a regression of final exam score on midterm score, the regression equation is $\overset{y}{^} = 10 + 0.80 x$ , $r = 0.90$ . A student scored 70 on the midterm. What is the student's predicted final exam score?
- A
  63
  Why not A: Using $r$ as the coefficient instead of the slope: $0.90 \times 70 = 63$ .
- B
  66Correct
- C
  80
  Why not C: Using the slope as the predicted value without computing $10 + 0.80 (70)$ .
- D
  70
  Why not D: Using the midterm score directly without applying the regression equation.
Explanation
Substitute $x = 70$ into the regression equation: $\overset{y}{^} = 10 + 0.80 (70) = 10 + 56 = 66$ . The correct answer is 66.

Key takeaway
To predict, substitute the given $x$-value into $\hat{y} = a + bx$ and compute.
Question 6 · Easy
A regression of salary ( $y$ , thousands of dollars) on years of experience ( $x$ ) yields $r^{2} = 0.64$ . Which interpretation is correct?
- A
  $r = 0.64$ , indicating a moderately positive linear association.
  Why not A: The value given is $r^{2}$ , not $r$ ; $r = 0.64 = 0.80$ .
- B
  64% of the variation in salary is explained by the linear relationship with years of experience.Correct
- C
  64% of employees have salaries predicted correctly by the model.
  Why not C: $r^{2}$ measures explained variation, not prediction accuracy rate for individuals.
- D
  Years of experience causes 64% of the increase in salary.
  Why not D: $r^{2}$ describes association strength, not causal attribution.
Explanation
$r^{2} = 0.64$ means 64% of the variability in salary is accounted for by the least-squares regression on years of experience. Note that $r = 0.64 = 0.80$ , not 0.64.

Key takeaway
$r^2$ = proportion of variability in $y$ explained by the linear model; it is always between 0 and 1.
Question 7 · Easy
A researcher removes one influential point from a scatterplot. After removing it, $r$ increases from 0.60 to 0.91. What type of point was removed?
- A
  A point with a large positive residual
  Why not A: Large residuals indicate poor fit but don't necessarily reduce correlation if the point follows the linear trend direction.
- B
  An outlier in the $y$ -direction only (high-leverage point)
  Why not B: High leverage is about extreme $x$ -values, not $y$ -values, and pulling $r$ down suggests the point did not fit the linear pattern.
- C
  An influential outlier that did not follow the linear pattern of the other pointsCorrect
- D
  A data point with $x$ -value equal to the mean of $x$
  Why not D: Points at $\overset{x}{ˉ}$ have zero leverage and minimal influence on the regression line.
Explanation
When removing a point dramatically increases $r$ (0.60 → 0.91), that point weakened the linear association. It was an influential outlier — a point that did not fit the overall linear pattern and pulled $r$ downward. Influential points are often far from the bulk of the data in the $x$ -direction.

Key takeaway
An influential point can dramatically change the regression line and $r$ when removed; identify them with leverage and residual analysis.
Question 8 · Easy
The regression equation for predicting a city's high temperature in July ( $y$ , °F) from its latitude ( $x$ , degrees north) is $\overset{y}{^} = 115 - 1.1 x$ . Interpret the $y$ -intercept in context.
- A
  At 0° latitude (equator), the predicted July high is 115°F.Correct
- B
  For each degree increase in latitude, temperature decreases by 115°F.
  Why not B: This describes the slope, not the $y$ -intercept.
- C
  The average July high temperature across all cities is 115°F.
  Why not C: The $y$ -intercept is the predicted $y$ when $x = 0$ , not the average $y$ .
- D
  The $y$ -intercept has no meaningful interpretation because latitude cannot equal zero.
  Why not D: Latitude can equal 0 (the equator), so the intercept has a technically valid interpretation even if extrapolation beyond the data is risky.
Explanation
The $y$ -intercept $a = 115$ is the predicted value of $y$ when $x = 0$ . In context, when $x = 0$ (equator, 0° latitude), the model predicts a July high of 115°F. Whether this is a practically meaningful prediction depends on whether 0° is within the range of the data.

Key takeaway
$y$-intercept is the predicted $y$ when $x = 0$; always interpret it in the context of the data.
Question 9 · Medium
A scatterplot of two variables shows a strong curved relationship. A student computes $r = 0.45$ and concludes the association is weak. What is the error in the student's reasoning?
- A
  The student should use $r^{2}$ instead of $r$ to assess strength.
  Why not A: $r^{2}$ also measures strength of linear fit; it would similarly underestimate a curved relationship.
- B
  $r$ measures only the strength of the linear association; a low $r$ with a curved pattern means the relationship is strong but non-linear.Correct
- C
  The student should use Spearman's rank correlation for curved data.
  Why not C: While rank correlation is more robust, the AP Stats curriculum focuses on $r$ and its limitations with non-linear data.
- D
  A value of $r = 0.45$ can still indicate a strong relationship.
  Why not D: Without additional context, 0.45 is not considered a strong linear association; the issue is that $r$ is the wrong tool here.
Explanation
The correlation coefficient $r$ measures only the strength of a linear association. If the scatterplot shows a curved relationship, $r$ may be low even though the association is actually very strong. The student should use the scatterplot itself, not $r$ alone, to characterize the association.

Key takeaway
$r$ only measures linear association; always look at the scatterplot — a low $r$ with a clear curve means a strong non-linear relationship.
Question 10 · Medium
The regression of $ln (y)$ on $x$ produces a better fit than the regression of $y$ on $x$ . What does this suggest about the relationship between $x$ and $y$ ?
- A
  The relationship between $x$ and $y$ is exponential.Correct
- B
  The relationship between $x$ and $y$ is linear.
  Why not B: If the relationship were linear, the untransformed regression would fit as well as the transformed one.
- C
  The relationship between $x$ and $y$ is a power function.
  Why not C: A power function would be linearized by taking $ln$ of both $x$ and $y$ , not just $y$ .
- D
  $x$ must be transformed, not $y$ .
  Why not D: Taking $ln (y)$ (not $ln (x)$ ) linearizes exponential growth in $y$ ; the choice of transformation depends on the pattern.
Explanation
An exponential model $y = a b^{x}$ can be linearized by taking $ln (y) = ln (a) + x ln (b)$ . If the regression of $ln (y)$ on $x$ is linear, the original relationship is exponential. A power model $y = a x^{b}$ is linearized by taking $ln$ of both variables.

Key takeaway
Regressing $\ln(y)$ on $x$ linearizes exponential relationships; regressing $\ln(y)$ on $\ln(x)$ linearizes power relationships.
Question 11 · Hard
A statistician notes that cities with more churches tend to have more crime. Before concluding that religion causes crime, what is the most important alternative explanation?
- A
  The correlation between churches and crime is not strong enough to draw any conclusion.
  Why not A: The issue is not the strength of the correlation but its causal interpretation.
- B
  Population size is a confounding variable — larger cities have both more churches and more crime.Correct
- C
  The data should be analyzed with a $χ^{2}$ test, not correlation.
  Why not C: The type of test is not the issue; the issue is the causal interpretation of correlation.
- D
  Crime causes people to build more churches for moral guidance.
  Why not D: This still claims a causal relationship without evidence and in the wrong direction.
Explanation
This is a classic example of a confounding variable (lurking variable). Population is the confound: larger cities have more of everything, including churches and crime. Both variables are associated with population, creating a spurious correlation. Correlation alone cannot establish causation.

Key takeaway
A confounding variable is associated with both the explanatory and response variables, creating a spurious association; correlation does not imply causation.
Question 12 · Hard
Regression output for predicting plant height ( $y$ , cm) from weekly fertilizer dose ( $x$ , grams) is shown below:

Predictor Coef SE Coef T P
Constant 5.2 1.1 4.73 0.000
Dose 3.8 0.6 6.33 0.000

$r^{2} = 0.77$

A plant received 4 grams of fertilizer per week. What is its predicted height, and what does $r^{2}$ tell us?
- A
  Predicted height = 20.4 cm; 77% of the variation in height is explained by fertilizer dose.Correct
- B
  Predicted height = 20.4 cm; $r = 0.77$ indicates a moderately strong positive association.
  Why not B: $r^{2} = 0.77$ means $r = 0.77 \approx 0.88$ ; calling 0.77 itself a correlation confuses $r$ and $r^{2}$ .
- C
  Predicted height = 20.4 cm; 23% of the variation in height is unexplained.
  Why not C: While 23% is unexplained, the answer misses the main interpretation of $r^{2}$ as the explained portion.
- D
  Predicted height = 17.4 cm; 77% of plants are predicted correctly.
  Why not D: Omits the constant (5.2 + 3.8×4 = 20.4, not 3.8×4 = 15.2) and misinterprets $r^{2}$ .
Explanation
$\overset{y}{^} = 5.2 + 3.8 (4) = 5.2 + 15.2 = 20.4$ cm. The $r^{2} = 0.77$ means 77% of the variation in plant height is explained by the linear regression on fertilizer dose. The remaining 23% is due to other factors.

Key takeaway
Predicted value: substitute $x$ into $\hat{y} = a + bx$. $r^2$: proportion of $y$-variation explained by the model.

Predictor	Coef	SE Coef	T	P
Constant	5.2	1.1	4.73	0.000
Dose	3.8	0.6	6.33	0.000