AP Statistics Unit 2: Exploring Two-Variable Data — Free Review Games.

Scatterplots show the relationship between two quantitative variables.

r close to 1 indicates a strong positive linear relationship.

r always falls between -1 and 1 inclusive.

y-hat denotes the predicted value from the regression equation.

r^2 is the coefficient of determination, showing the fraction of variance explained.

Residual = actual - predicted (y - y-hat).

Patterns in residuals indicate the linear model does not capture the relationship well.

Even strong correlation does not prove that one variable causes changes in the other.

The slope is the predicted change in y for each one-unit increase in x.

An influential point markedly affects the regression line when it is removed.

r^2 = (-0.85)^2 = 0.7225.

The observed relationship may not continue beyond the range of the data.

A lurking variable influences both variables and can create a misleading association.

A linear relationship in log(y) vs x indicates y is exponential in x.

The least-squares method minimizes the sum of the squared vertical deviations.

The least-squares regression line always passes through \((\bar{x}, \bar{y})\), the means of the explanatory and response variables. This is a fundamental algebraic property of OLS: substituting \(x = \bar{x}\) into \(\hat{y} = b_0 + b_1 x\) always yields \(\hat{y} = \bar{y}\). The line passes through the origin only by coincidence if \(\bar{y} - b_1\bar{x} = 0\).

A correlation of \(r = 0\) means there is no linear association, but a strong nonlinear relationship could still exist. For example, a perfect U-shaped (quadratic) relationship can produce \(r = 0\). Choice A is incorrect because \(r\) only detects linear patterns. Choices C and D are not implied by \(r = 0\).

By convention the explanatory variable goes on the horizontal \(x\)-axis and the response variable on the vertical \(y\)-axis, reflecting that we use \(x\) to predict \(y\). Swapping the axes would produce a different regression equation (though the correlation \(r\) stays the same).

A perfect negative linear relationship means every point falls exactly on a negatively-sloped line, giving \(r = -1\). A value of \(r = 0\) indicates no linear association, \(r = -0.5\) indicates a moderate negative association, and \(r = 1\) indicates a perfect positive association.

The standard framework for describing a scatterplot's association covers form (linear or curved), direction (positive or negative), and strength (how closely points follow the pattern). Shape, center, and spread describe the distribution of a single variable. Slope, intercept, and residual are regression quantities, not description categories.

Correlation is symmetric: \(r_{xy} = r_{yx}\). Swapping which variable is called explanatory and which is response does not change \(r\). The value \(0.64 = r^2\) is a common distractor. No additional information is needed because symmetry is a built-in property of the Pearson correlation formula.

When every point lies exactly on a positively-sloped line there is a perfect positive linear relationship, giving \(r = 1\). The value \(r = 0\) indicates no linear association. Values such as \(0.50\) or \(0.95\) indicate partial, not perfect, linear associations where scatter still exists around the line.

The \(y\)-intercept is the predicted value of the response variable when the explanatory variable equals zero: \(\hat{y} = 2.5(0) - 4 = -4\). Choice A describes the slope (\(2.5\)). The correlation and average residual are separate quantities not read directly from the regression intercept; in fact, the average residual from a least-squares line is always \(0\).

Residual \(= \text{observed} - \text{predicted}\). If \(e = -6\), then \(\text{observed} - \text{predicted} = -6\), meaning the predicted value exceeds the observed by \(6\). Choice B has the relationship backwards. Choices C and D are unrelated to the definition of a residual.

Because \(r\) is computed using means and standard deviations — both of which are non-resistant — a single outlier can dramatically change \(r\). Choice A is wrong: \(r\) measures only linear association. Choice B is wrong: \(r\) is symmetric (\(r_{xy} = r_{yx}\)). Choice C is wrong: \(r\) is unitless because it is based on standardized \(z\)-scores.

First, compute the predicted score: \(\hat{y} = 1.2(10) + 5 = 12 + 5 = 17\). Then the residual is \(e = \text{observed} - \text{predicted} = 22 - 17 = 5\). A positive residual means the student scored higher than the model predicted. Choice A (\(-5\)) would result from subtracting observed from predicted instead of the other way around.

Since \(r^2 = 0.64\), we have \(|r| = \sqrt{0.64} = 0.80\). The slope and \(r\) always share the same sign, so because the slope is negative, \(r = -0.80\). Choice A omits the negative sign. Choice C confuses \(r\) with \(r^2\). Choice D applies the negative sign to \(r^2\) rather than to \(\sqrt{r^2}\).

Correlation is unitless and invariant under linear transformations (multiplying or adding a constant to either variable). This is because \(r\) is computed from \(z\)-scores, and \(z\)-scores are unit-free. Only a transformation that reverses direction (multiplying by a negative number) would change the sign of \(r\); multiplying by \(2.54 > 0\) has no effect.

A fan-shaped residual plot indicates non-constant variance (heteroscedasticity): the spread of \(y\) around the regression line grows with \(x\). This violates a key assumption of the simple linear model. It is a different concern from a curved (U-shaped) residual plot, which signals nonlinearity. Neither pattern implies that the model fits well.

The correlation coefficient describes the strength and direction of a linear association, not causation. \(r = 0.85\) indicates a strong positive linear association. Choice A incorrectly infers causation from correlation. Choice C misinterprets \(r\) as a proportion of cities. Choice D confuses \(r\) with the slope, which depends on the units of both variables and is not equal to \(r\) in general.

The slope formula is \(b_1 = r \cdot \dfrac{s_y}{s_x} = 0.75 \cdot \dfrac{8}{2} = 0.75 \cdot 4 = 3\). Choice B is simply \(r\) itself. Choice C results from computing \(r \cdot s_y = 0.75 \cdot 8 = 6\), forgetting to divide by \(s_x\). Choice D comes from inverting the ratio: \(r \cdot \dfrac{s_x}{s_y} = 0.75 \cdot \dfrac{2}{8} = 0.1875\).

Because \(x\) is measured in thousands of dollars, \(x = 15\). Then \(\hat{y} = 4(15) + 10 = 60 + 10 = 70\) thousand dollars \(= \\)70{,}000$. Choice C (\(\\)60{,}000$) omits the intercept. A common unit error is substituting \(x = 15{,}000\), which yields an unreasonably large value not among the choices.

This is a classic lurking variable scenario. Age is associated with both shoe size (feet grow as children age) and reading ability (older children read better). When the effect of age is accounted for, the correlation between shoe size and reading likely disappears. The flaw is inferring causation from correlation without controlling for confounding variables. Choice A is incorrect because \(0.78\) is actually a fairly strong correlation.

First, $r^2 = 1 - \dfrac{SSE}{SST} = 1 - \dfrac{100}{400} = 0.75$. Since the slope is positive, \(r = \sqrt{0.75} \approx 0.866 \approx 0.87\). Choice C (\(0.75\)) is \(r^2\), not \(r\) — a very common error. Choice A (\(0.25\)) is $\dfrac{SSE}{SST}$, and choice B (\(0.50\)) has no direct connection to these values.

Substitute \(x = 4\) into the transformed equation: \(\widehat{\sqrt{y}} = 0.5(4) + 2 = 2 + 2 = 4\). To reverse the square-root transformation, square both sides: \(\hat{y} = 4^2 = 16\). Choice A (\(4\)) is \(\widehat{\sqrt{y}}\), not \(\hat{y}\) — the most common error of forgetting to undo the transformation. Choice D (\(64\)) results from cubing rather than squaring.

Multiplying all \(y\)-values by \(3\) scales \(\bar{y}\) by \(3\) and \(s_y\) by \(3\). The new slope is \(b_1' = r \cdot \dfrac{3s_y}{s_x} = 3 \times 2 = 6\). The new intercept is \(b_0' = 3\bar{y} - 6\bar{x} = 3(\bar{y} - 2\bar{x}) = 3(1) = 3\). So the new line is \(\hat{y} = 6x + 3\). Choice B scales only the intercept; choice C scales only the slope; choice D incorrectly leaves the line unchanged.

Leverage depends on how extreme \(x\) is relative to \(\bar{x}\). Since \(x = 10 = \bar{x}\), this point has low leverage and exerts little pull on the slope. However, its \(y\)-value of \(80\) is far from the predicted \(\hat{y} = 2(10) + 10 = 30\), making it an outlier in \(y\). A low-leverage, high-residual point primarily affects the intercept rather than the slope. Choice C is incorrect: the point can still influence the intercept even when \(x = \bar{x}\).

Two guaranteed properties of OLS residuals: (1) \(\sum e_i = 0\), because the regression line passes through \((\bar{x}, \bar{y})\), and (2) \(\sum e_i^2\) is minimized — that is the definition of least squares. Choice A is wrong; individual residuals are zero only for points that happen to lie exactly on the line. Choice D confuses leverage (distance from \(\bar{x}\)) with residual magnitude.

Correlation is invariant under linear transformations, and standardizing is a linear transformation. In fact, the Pearson correlation can be written as \(r = \dfrac{1}{n-1}\sum z_x z_y\), so \(r\) is literally the average product of paired \(z\)-scores. The correlation between \(z_x\) and \(z_y\) is therefore exactly \(r\). Choice A would only hold if \(r = 1\) originally. Choice C (\(r^2\)) is the coefficient of determination, a different quantity.

A curved residual plot for the linear model reveals systematic error — the model is wrong in a predictable way, regardless of how high \(r^2\) is. A randomly scattered residual plot for the exponential model signals that model assumptions are satisfied. Residual plot analysis is the primary diagnostic for model appropriateness and takes precedence over \(r^2\) alone. Choice D is incorrect: a well-fitting model with a lower \(r^2\) is preferable to a misspecified model with a higher \(r^2\).

The correlation coefficient \(r\) is bounded by \(-1\) and \(1\), inclusive. A value of \(r = 1\) indicates a perfect positive linear relationship, \(r = -1\) indicates a perfect negative linear relationship, and \(r = 0\) indicates no linear relationship. Choice A is incorrect because it excludes \(-1\) and \(1\), which are achievable when all data points fall exactly on a straight line.

When one variable tends to decrease as the other increases, the variables have a negative association. Positive association occurs when both variables tend to increase together. This describes the direction of the relationship only — the form could still be linear.

The residual is defined as \(\text{residual} = \text{observed} - \text{predicted} = 74 - 79 = -5\). A negative residual means the model overpredicted the student's actual score. Choice B reverses the order of subtraction, computing \(79 - 74 = 5\) instead.

When points show no clear linear pattern, \(r\) is close to \(0\). Values of \(r\) close to \(1\) or \(-1\) indicate strong positive or negative linear relationships, respectively. Values outside the interval \([-1, 1]\) are impossible for a correlation coefficient.

The response variable (also called the dependent variable) is the outcome being predicted or explained. Here, body weight is what is being predicted, so it is the response variable. Daily caloric intake is the explanatory variable because it is used to predict body weight.

\(r = -1\) represents a perfect negative linear relationship — every data point lies exactly on a line with a negative slope. Choice A describes \(r = 0\). Choice C describes \(r = 1\). Choice D is incorrect because \(r = -1\) indicates the strongest possible linear dependence, the opposite of independence.

Correlation measures the strength and direction of a linear association, but it does not establish cause and effect. A high correlation may be explained by a lurking variable, coincidence, or reverse causation. Only controlled, randomized experiments provide evidence for causal conclusions.

The correlation coefficient is not affected by linear transformations (multiplying by a positive constant or adding any constant) applied to either variable. The strength and direction of the linear relationship are preserved, so \(r\) remains \(0.88\). If the multiplying constant were negative, only the sign of \(r\) would flip.

Substitute \(x = 80\) into the regression equation: \(\hat{y} = 15 + 0.7(80) = 15 + 56 = 71\). Choice A results from computing \(0.7 \times 80 = 56\) but forgetting to add the intercept of \(15\).

The slope represents the predicted change in \(\hat{y}\) for each one-unit increase in \(x\). Since \(x\) is in thousands of pounds, each additional thousand pounds is associated with a predicted decrease of \(6.5\) miles per gallon. Choice C confuses the slope with the \(x\)-intercept, and Choice D reverses the roles of the variables.

Extrapolation occurs when a regression equation is used to predict for \(x\)-values outside the range of the data used to build it. The model was based on weeks \(1\) through \(10\), so using it at \(x = 50\) extends far beyond the observed range. The linear pattern that held for weeks \(1\)–\(10\) may not continue to hold at \(x = 50\).

An influential point is one whose removal substantially changes the regression results. Removing this point dramatically changed both the slope (from \(2.8\) to \(0.4\)) and the correlation (from \(0.90\) to \(0.35\)), confirming it is highly influential. A high-leverage point with no influence (Choice C) would not cause such large changes when removed.

\(\text{Residual} = \text{observed} - \text{predicted}\). If the actual price is higher than the predicted price, then observed \(>\) predicted, giving a positive residual. Choice A incorrectly associates a higher-than-predicted observation with a negative residual.

A curved (U-shaped or arch-shaped) pattern indicates a nonlinear relationship, making a linear model inappropriate. Choice A suggests a weak linear association — scatter around a line is expected, and a linear model may still be reasonable. A steep slope (Choice D) says nothing about whether the relationship is linear.

A point that falls exactly on the regression line is perfectly consistent with the linear pattern. Adding it extends the range of \(x\) while maintaining the trend, which typically maintains or slightly increases \(r\). Correlation is weakened only when a point departs from the linear trend. Choice D is incorrect because one additional point on the line does not make all \(26\) points perfectly collinear.

The slope of the LSRL is \(b = r \cdot \dfrac{s_y}{s_x}\). Because standard deviations \(s_y\) and \(s_x\) are always positive, the sign of \(b\) is determined entirely by the sign of \(r\). A negative \(r\) guarantees a negative slope. Choice B is only true when both variables have been standardized to \(z\)-scores.

Larger cities naturally have both more hospitals and more total deaths because of their larger populations. Population size is a lurking variable that drives both quantities, producing an association that does not reflect causation. This is a classic illustration of why correlation does not imply causation.

Using the relationship \(b = r \cdot \dfrac{s_y}{s_x}\), solve for \(r\): \(r = b \cdot \dfrac{s_x}{s_y} = 1.5 \cdot \dfrac{4}{10} = 0.60\). Choice C results from computing \(b \cdot \dfrac{s_y}{s_x}\) instead (the inverted ratio), yielding \(3.75\), which is impossible since \(|r| \leq 1\). Choice A uses \(b \cdot \dfrac{s_x}{s_y}\) with inverted values.

When both variables are standardized, the predicted \(z\)-score for \(y\) is \(\hat{z}_y = r \cdot z_x = 0.70 \times 2.5 = 1.75\). This illustrates regression to the mean: students with extreme scores on one test tend to score closer to the mean on a second test. Choice A would only hold if \(r = 1.0\). Choice D results from confusing \(r \cdot 1\) with \(r \cdot z_x\).

Step 1 — slope: \(b = r \cdot \dfrac{s_y}{s_x} = 0.75 \cdot \dfrac{20}{8} = 1.875\). Step 2 — intercept: \(a = \bar{y} - b\bar{x} = 100 - 1.875(40) = 100 - 75 = 25\). Step 3 — predict: \(\hat{y} = 25 + 1.875(56) = 25 + 105 = 130\). Choice B results from an arithmetic error in Step 1 using \(b = 1.5\); Choice D omits the intercept calculation entirely.

In simple linear regression, leverage depends only on how far the \(x\)-value is from \(\bar{x}\). A point with \(x\) far from \(\bar{x}\) has high leverage because the regression line is sensitive to that point's position. The \(y\)-value determines whether a high-leverage point is also influential, but leverage itself is determined solely by the \(x\)-value. The point at \((\bar{x}, \bar{y})\) actually has the minimum possible leverage.

An influential point is one whose removal causes a substantial change in regression results. The slope changed from \(3.8\) to \(1.2\) — a dramatic shift — confirming the point was highly influential. The modest change in \(r\) (from \(0.92\) to \(0.88\)) does not contradict this conclusion; slope change is the primary indicator of influence here.

The coefficient of determination is \(r^2 = (-0.8)^2 = 0.64\), meaning \(64\%\) of the variation in \(y\) is explained by \(x\). The unexplained (residual) variation is \(1 - r^2 = 1 - 0.64 = 0.36 = 36\%\). Choice B gives the explained proportion, not the unexplained. Choice A incorrectly uses \(|r| = 0.80\) instead of \(r^2\).

The LSRL always passes through the centroid \((\bar{x}, \bar{y})\). This follows from the intercept formula \(a = \bar{y} - b\bar{x}\): substituting \(x = \bar{x}\) gives \(\hat{y} = a + b\bar{x} = (\bar{y} - b\bar{x}) + b\bar{x} = \bar{y}\). Choice A describes the least absolute deviations line, not the least-squares line. Choice D is only true when both variables are standardized to \(z\)-scores.

The LSRL must always pass through the centroid \((\bar{x}, \bar{y})\). With \(\bar{x} = 6\) and \(\bar{y} = 14\), check the student's equation: \(\hat{y} = 3 + 2.5(6) = 3 + 15 = 18 \neq 14\). Since the student's line does not pass through \((6, 14)\), it cannot be the LSRL. The original equation does: \(\hat{y} = 2 + 2(6) = 2 + 12 = 14\) ✓.

The predicted final exam score is \(\hat{y} = 10 + 0.9(60) = 10 + 54 = 64\). The residual is \(y - \hat{y} = 62 - 64 = -2\). A negative residual means the student scored below the regression line's prediction. Choice B reverses the subtraction order (\(\hat{y} - y\)). Choice C gives the predicted value \(\hat{y} = 64\), not the residual. Choice D negates the predicted value, which has no statistical meaning.

\(r^2\) is the coefficient of determination: the proportion of total variation in \(y\) accounted for by the linear relationship with \(x\). Here, \(64\%\) of the variability in \(y\) is explained by the regression. Choice B confuses \(r^2\) with \(r\); the correlation would be \(r = \pm\sqrt{0.64} = \pm 0.8\). Choice C is a common misconception — \(r^2\) does not indicate what fraction of points touch the line. Choice D mischaracterizes \(r^2\) as a per-prediction accuracy guarantee.

The correlation coefficient is invariant under positive linear transformations. Multiplying all \(x\)-values by the positive constant \(2.54\) leaves the standardized \(z\)-scores unchanged, so \(r\) is unchanged. Choice B multiplies by a negative constant, reversing the sign of all \(x\) deviations and therefore changing the sign of \(r\) (e.g., \(0.7\) becomes \(-0.7\)). Choice C adds an influential outlier that can substantially alter \(r\). Choice D applies a nonlinear transformation (\(y \to y^2\)) that fundamentally changes the relationship between the variables.

This is a classic lurking variable scenario. Larger fires require more firefighters *and* cause more property damage. The size of the fire drives both variables simultaneously, creating an apparent positive correlation that does not reflect causation. Choice B is false; there is no restriction on the sign of \(r\) for public safety data. Choice C is incorrect; swapping which variable is explanatory does not resolve the causal fallacy. Choice D is irrelevant; the issue is one of causation, not model selection.

\(r\) measures only the strength of the *linear* component of an association. A symmetric U-shape has equal amounts of upward and downward trend, so the linear components cancel and \(r\) can be near \(0\) despite a very strong nonlinear association. This is why scatterplots must always accompany \(r\). Choice B overstates the case; an asymmetric U-shape can still yield a nonzero \(r\). Choice C is incorrect; a U-shape has both increasing and decreasing portions that offset each other. Choice D is false; \(r\) is always defined for two quantitative variables.

The strength of a linear association is measured by \(|r|\), the absolute value of the correlation. Since \(|0.3| = |-0.3| = 0.3\), the two associations are equally strong. They differ only in direction: Study A is positive (more studying predicts higher scores) and Study B is negative (more TV predicts lower scores). Choices B and C are incorrect because the sign of \(r\) indicates direction, not strength. Choice D incorrectly applies a numerical inequality to judge strength — what matters is \(|r|\).

A well-fitting linear model produces residuals that scatter randomly around zero. A curved pattern reveals systematic error: the linear model is not capturing the true nonlinear relationship, and a transformation or curved model should be considered. Choice B is unrelated; a curved residual pattern does not determine the sign of \(r\). Choice C is incorrect; a curved residual plot reveals model inadequacy but says nothing about the absence of influential points. Choice D would require all residuals to equal exactly zero.

\(r^2 = 0.81\) means \(81\%\) of the variability in salary is explained by the linear regression on years of experience. The slope \(2.5\) (thousands) means each additional year is *associated* with a predicted \(\\)2{,}500$ increase. Choice B incorrectly asserts causation. Choice C misidentifies the \(y\)-intercept — for \(x = 0\), \(\hat{y} = 40\), predicting \(\\)40{,}000$ for someone with no experience. Choice D confuses \(r^2 = 0.81\) with \(r\); the actual correlation is \(r = \sqrt{0.81} = 0.9\).

The slope of the least-squares regression line is \(b_1 = r \cdot \dfrac{s_y}{s_x} = 0.7 \cdot \dfrac{20}{10} = 0.7 \times 2 = 1.4\). Choice B results from inverting the ratio: \(0.7 \cdot \frac{s_x}{s_y} = 0.7 \cdot \frac{10}{20} = 0.35\). Choice C is simply \(r = 0.7\), omitting the \(s_y/s_x\) multiplier. Choice D is the ratio \(s_y/s_x = 2\) without multiplying by \(r\).

The \(y\)-intercept is the predicted value of \(y\) when \(x = 0\). Here it predicts a blood pressure of \(90\) mmHg for age \(= 0\). While mathematically straightforward, this may not be meaningful in context — if the study only included adults, using the model at age \(0\) is extrapolation beyond the observed data. Choice B describes the slope \(0.6\), not the intercept. Choice C confuses the intercept value (\(90\)) with \(r\), which must lie in \([-1, 1]\). Choice D confuses the intercept with \(r^2\).

The correlation coefficient is $r = \dfrac{\sum(x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum(x_i - \bar{x})^2 \cdot \sum(y_i - \bar{y})^2}} = \dfrac{300}{\sqrt{200 \times 800}} = \dfrac{300}{\sqrt{160{,}000}} = \dfrac{300}{400} = 0.75$. Choice B is \(r^2 = 0.75^2 = 0.5625\), the error of stopping after squaring. Choice C exceeds \(1\), which is impossible for a correlation coefficient. Choice D divides \(300\) by \(800\) alone, incorrectly using only one sum of squares.

One fundamental property of the least-squares regression line is that \(\sum(y_i - \hat{y}_i) = 0\): positive and negative residuals cancel exactly. This follows because the line is constrained to pass through \((\bar{x}, \bar{y})\). Choice B would require a perfect fit where all points lie on the line. Choice C is false; \(r^2\) is the ratio of explained to total variation, not the sum of squared residuals \(\sum e_i^2\). Choice D (normality of residuals) is an *assumption* made for inference, but it is not mathematically guaranteed for all datasets.

Substituting \(x = 4\) gives \(\widehat{\ln y} = 1.2 + 0.5(4) = 3.2\). This is the predicted value of \(\ln(y)\). To recover \(\hat{y}\), apply the inverse of \(\ln\): \(\hat{y} = e^{3.2}\). Choice B gives the predicted \(\ln(y) = 3.2\) and stops, forgetting to back-transform. Choice C applies another logarithm instead of exponentiating. Choice D incorrectly exponentiates only the intercept (\(e^{1.2}\)) before adding the slope term, rather than exponentiating the entire predicted value \(3.2\).

The slope of \(y\) on \(x\) is \(b_1 = r \cdot \frac{s_y}{s_x} = 1.8\). Since \(r = 0.6\), we get \(\frac{s_y}{s_x} = \frac{1.8}{0.6} = 3\), so \(\frac{s_x}{s_y} = \frac{1}{3}\). The slope of the regression of \(x\) on \(y\) is \(b^* = r \cdot \frac{s_x}{s_y} = 0.6 \cdot \frac{1}{3} = 0.2\). Choice B assumes the two regression lines share the same slope, which holds only when \(r = \pm 1\). Choice C takes the reciprocal of \(b_1\), valid only when \(r = \pm 1\). Choice D uses \(r\) directly as the slope, ignoring the standard deviation ratio.

The \(t\)-statistic for the slope tests \(H_0: \beta_1 = 0\) (no linear relationship) against \(H_a: \beta_1 \neq 0\). A large \(|t|\) value like \(4.5\) provides strong evidence against the null hypothesis, indicating the slope is significantly nonzero and the linear model is statistically useful. Choice B is unrelated; passing through the origin is a separate constraint determined by the intercept estimate. Choice C is incorrect; \(t = 4.5\) tests only the slope; the intercept requires its own separate \(t\)-test. Choice D is false; \(t\) depends on the estimate divided by its standard error, and a large \(t\) for the slope does not determine \(r^2\).

The removed point raises \(r\) from \(0.55\) to \(0.88\), meaning it *strengthens* the linear association. Such points are typically high-leverage (extreme in \(x\)) and fall close to the regression line, reinforcing the apparent linear pattern — removing them reduces \(r\) substantially. Choice B describes a point that *weakens* the correlation; removing a contradicting point would *increase* \(r\), not decrease it. Choice C is contradicted by the large change in \(r\), which demonstrates high influence. Choice D describes a point near \(\bar{x}\), which has low leverage and would produce only a small change in \(r\).

A high \(r^2\) is desirable but does not by itself confirm that a linear model is appropriate. Model 2's curved residual plot reveals that the linear model is misspecified for \(x_2\) — despite explaining \(85\%\) of the variance, it is systematically wrong in a patterned way. Model 1's random residual plot confirms the linear model is appropriate for \(x_1\). A correctly specified model with \(r^2 = 0.72\) is preferable to a misspecified one with \(r^2 = 0.85\). Choice B is a common and important misconception. Choice C misreads the curved residual pattern — it indicates the model is *missing* nonlinear structure, not capturing it. Choice D is false; there is no universal \(r^2\) threshold for model validity.

The correlation coefficient \(r\) is computed from standardized \(z\)-scores, so it is unitless and unaffected by changes in units (for example, converting inches to centimeters leaves \(r\) unchanged). Choice A is wrong because \(r\) ranges from \(-1\) to \(1\), including negative values for negative associations. Choice C is wrong because \(|r| \leq 1\) always holds. Choice D is wrong because \(r\) only measures the strength of linear association — a perfect U-shaped relationship can yield \(r = 0\).

When one variable increases as the other decreases, the two variables have a negative association. The sign of the association describes direction. Choice A describes the opposite direction. Choice C would mean no discernible trend at all. Choice D addresses form (curved vs. straight), not direction — a downward-sloping straight line is linear and negative, not nonlinear.

\(r^2\) represents the proportion of variability in the response variable that is explained by the linear model. Here, \(81\%\) of the variation in \(y\) is accounted for by its linear relationship with \(x\). Choice A confuses \(r^2\) with \(r\); since \(r^2 = 0.81\), the correlation is \(|r| = 0.9\), not \(0.81\). Choice C is a common misconception — \(r^2\) is about variance explained, not about points sitting on the line. Choice D describes per-observation accuracy, which is an entirely different concept.

A fundamental property of the least-squares regression line is that it passes through the point of means \((\bar{x},\, \bar{y})\). This follows directly from the \(y\)-intercept formula \(b_0 = \bar{y} - b_1 \bar{x}\), which guarantees \(\hat{y} = \bar{y}\) when \(x = \bar{x}\). The other choices describe points that have no guaranteed relationship to the regression line unless special conditions hold (such as \(\bar{x} = 0\) for choices A and D).

The strength of a linear relationship is determined by \(|r|\), not the sign. Computing the absolute values: \(0.85\), \(0.72\), \(0.10\), and \(0.60\). The smallest is \(|-0.10| = 0.10\), indicating the weakest linear association. A common error is to think negative correlations are weaker than positive ones — but \(r = -0.85\) actually indicates a stronger linear relationship than \(r = 0.60\).

The explanatory variable (also called the predictor or independent variable) is used to explain or predict changes in the response variable. By convention it is placed on the \(x\)-axis — not the \(y\)-axis — making Choice A incorrect. Choice C is false: there is no requirement about the relative sizes of the standard deviations. Choice D is also false: in simple linear regression both variables are typically quantitative.

A residual is defined as \(e = y - \hat{y}\) (observed minus predicted). A positive residual means \(y > \hat{y}\), so the actual data point lies above the regression line. Choice D describes a negative residual (where \(\hat{y} > y\)). Choices A and C describe properties of the overall model, not of individual residuals — a positive residual can occur even when the slope or correlation is negative.

\(r \approx -0.95\) indicates a strong negative linear association, meaning points cluster tightly around a line with a negative (downward) slope. Choice A corresponds to \(r \approx 0\). Choice B corresponds to a strong positive correlation near \(+1\). Choice D describes a nonlinear (quadratic) pattern — a symmetric U-shape can yield \(r \approx 0\) even though the association is strong, because \(r\) only measures linear association.

The slope \(b_1 = 0.05\) means that for each one-unit increase in the explanatory variable (volume, in cubic inches), the predicted response (weight, in pounds) increases by \(0.05\) pounds. Choice A describes the \(y\)-intercept \(b_0 = 0.8\). Choice C reverses the roles of \(x\) and \(y\) — the regression of \(y\) on \(x\) does not describe how \(x\) changes per unit of \(y\). Choice D describes \(r^2\), which is a different statistic.

Extrapolation — predicting outside the range of observed \(x\) values — is unreliable because the linear trend established within the data may not continue beyond it. Here \(x = 4.8\) lies outside the observed range of \(2.0\) to \(4.0\). Choice A addresses the magnitude of the slope but has no bearing on whether the prediction is valid. Choice C is a separate (and minor) concern about the intercept. Choice D contradicts the given positive slope.

A lurking variable is an unobserved variable associated with both the explanatory and response variables, producing a misleading correlation. Fire size is the lurking variable: larger fires require more firefighters and also cause more damage. This does not mean firefighters cause damage. Choice A incorrectly infers causation from correlation. Choice C dismisses valid data based on intuition alone. Choice D is doubly wrong — the correlation is unlikely to be exactly \(1.0\), and even a perfect correlation does not establish causation.

The residual is \(e = y - \hat{y}\). The predicted score for \(x = 5\) hours is \(\hat{y} = 40 + 7(5) = 40 + 35 = 75\). The actual score is \(y = 72\). Therefore the residual is \(72 - 75 = -3\). A negative residual means the student scored below what the model predicted. Choice B reverses the sign by computing \(\hat{y} - y\) instead of \(y - \hat{y}\). Choices C and D are not consistent with the correct formula.

An influential observation is one whose removal substantially changes the least-squares regression line — typically a point with high leverage (extreme \(x\)-value). Importantly, influential points often have small residuals, because the regression line is pulled toward them. Choice A describes an outlier in the \(y\)-direction, which may have little influence on the line if it is near \(\bar{x}\). Choices C and D do not define influence.

When a new point falls exactly on the regression line at an extreme \(x\)-value, it has zero residual and high leverage. Because it perfectly supports the linear trend, adding it strengthens the linear association and increases \(|r|\). Choice A incorrectly assumes any \(x\)-outlier weakens correlation. Choice C would only hold if the point were added exactly at \(x = \bar{x}\). Choice D has no basis — a point on the line does not reverse the sign of the association.

\(r^2\) is the coefficient of determination — it represents the proportion of variability in the response variable (plant height) explained by the linear model. Here, \(64\%\) of the variation in plant height is accounted for by fertilizer amount. Choice B confuses \(r^2\) with \(r\); since \(r^2 = 0.64\), the correlation is \(|r| = 0.8\), not \(0.64\). Choices A and D are entirely unrelated to what \(r^2\) measures.

The strength of a linear association is measured by \(|r|\), regardless of sign. Since \(|-0.6| = |0.6| = 0.6\), both datasets have identical linear association strength. The sign of \(r\) indicates only the direction of the relationship (positive or negative slope), not its strength. Choices A and B both incorrectly assign strength based on sign. Choice D introduces variability as a factor, but \(r\) already accounts for the spread of both variables through standardization.

A residual plot with no pattern — random scatter around zero — is the key diagnostic confirming that a linear model is appropriate for the data. If the true relationship were curved, the residuals would show a systematic pattern (such as a bow or arc). Choice A is incorrect: a random residual plot says nothing about the magnitude of \(r^2\); even a moderate \(r^2\) can accompany appropriately random residuals. Choice C describes a perfect fit (\(r^2 = 1\)), which is not implied by random scatter. Choice D relates to the response variable's spread, which is separate from residual patterns.

Substituting into the formula: \(b_1 = 0.5 \cdot \dfrac{8}{4} = 0.5 \cdot 2 = 1.0\). Choice A results from inverting the ratio: \(0.5 \cdot \dfrac{4}{8} = 0.25\). Choice C omits the factor of \(r\) and uses only \(\dfrac{s_y}{s_x} = 2\). Choice D doubles the correct answer. The formula \(b_1 = r \cdot \dfrac{s_y}{s_x}\) shows that the slope depends on both the correlation and the relative spread of the two variables.

This illustrates regression to the mean. In standardized units, the predicted \(z\)-score on \(y\) given a \(z\)-score on \(x\) is \(\hat{z}_y = r \cdot z_x\). Here, \(\hat{z}_y = 0.7 \times 2.0 = 1.4\). Students selected for extreme Exam 1 performance are predicted to score closer to the mean on Exam 2 — not because they decline, but because extreme selection partly reflects chance variation. Choice A assumes no regression to the mean. Choice C uses only \(r\) but ignores \(z_x\). Choice D incorrectly uses \(r^2 = 0.49\) instead of \(r\).

For the regression of \(y\) on \(x\), the slope is \(b_1 = r \cdot \dfrac{s_y}{s_x} = 0.8 \cdot \dfrac{6}{2} = 2.4\). For the regression of \(x\) on \(y\) (roles reversed), the slope is \(b_1^{*} = r \cdot \dfrac{s_x}{s_y} = 0.8 \cdot \dfrac{2}{6} \approx 0.267\). Choice B is the slope of \(y\) on \(x\), not \(x\) on \(y\). Choice C equals \(r\) itself. A key concept is that the two regression lines are not reciprocals of each other — this asymmetry reflects the fact that \(r < 1\).

The linearized model is \(\ln(y) = 1.2 + 2.5\ln(x)\). Substituting \(x = 10\): \(\ln(y) = 1.2 + 2.5\ln(10)\). Exponentiating both sides: \(y = e^{1.2 + 2.5\ln(10)} = e^{1.2} \cdot e^{2.5\ln(10)} = e^{1.2} \cdot 10^{2.5}\). Choice B incorrectly computes \(2.5 \times 10 = 25\) in the exponent instead of \(2.5 \cdot \ln(10)\). Choice C gives \(\ln(y)\), not \(y\). Choice D ignores the \(\ln(x)\) term entirely, treating the model as \(\ln(y) = 1.2 + 2.5\).

First find the slope: \(b_1 = r \cdot \dfrac{s_y}{s_x} = 0.9 \cdot \dfrac{10}{5} = 1.8\). Then use the fact that the line passes through \((\bar{x}, \bar{y})\): \(b_0 = \bar{y} - b_1\bar{x} = 50 - 1.8(20) = 50 - 36 = 14\). Choice B results from using \(b_1 = r = 0.9\) (forgetting \(s_y/s_x\)): \(50 - 0.9(20) = 32\). Choice C results from using \(b_1 = s_y/s_x = 2\) (forgetting \(r\)): \(50 - 2(20) = 10\). Choice D results from inverting the ratio so \(b_1 = 0.9 \cdot (5/10) = 0.45\): \(50 - 0.45(20) = 41\).

Since both models predict the same untransformed response variable \(y\), their \(r^2\) values both measure the proportion of variation in \(y\) explained by the model and are directly comparable. The log-transformed model (\(r^2 = 0.91\)) explains substantially more variation than the linear model (\(r^2 = 0.76\)), making it the better fit. Choice B incorrectly privileges untransformed predictors. Choice C is wrong because different \(r^2\) values indicate different explanatory power. Choice D would apply if the response variable were transformed (making \(r^2\) values measure different things), but only \(x\) is transformed here.

Correlation is invariant under positive linear transformations. Since \(w = 3x + 7\) multiplies \(x\) by a positive constant (\(3 > 0\)) and shifts it, the standardized values of \(w\) are identical to those of \(x\), so \(r_{wy} = r_{xy} = 0.6\). If the multiplier were negative (e.g., \(w = -3x + 7\)), the correlation would become \(-0.6\). Choice B incorrectly multiplies \(r\) by \(3\). Choice C incorrectly divides \(r\) by \(3\). Choice D equals \(r^2 = 0.36\), not the new correlation.

The general slope formula is \(b_1 = r \cdot \dfrac{s_y}{s_x}\). When both variables are standardized, \(s_{z_x} = s_{z_y} = 1\), so the slope becomes \(b_1 = r \cdot \dfrac{1}{1} = r\). The intercept becomes \(b_0 = \bar{z}_y - r \cdot \bar{z}_x = 0 - r \cdot 0 = 0\). The standardized regression equation is therefore \(\hat{z}_y = r \cdot z_x\), revealing that \(r\) is the slope in standardized units. Choice A would require perfect correlation. Choice C uses \(r^2\) instead of \(r\). Choice D omits the factor of \(r\).

A positive association means that larger values of the explanatory variable correspond to larger values of the response variable. This describes only the direction of the trend, not its strength or form. A negative association would mean as \(x\) increases, \(y\) tends to decrease. The association does not need to be perfectly linear to be called positive — \(r = 1\) is a special case of a perfect positive relationship, not a requirement.

By definition, the Pearson correlation coefficient \(r\) is bounded between \(-1\) and \(1\) inclusive. A value of \(r = -1\) indicates a perfect negative linear relationship, \(r = 1\) indicates a perfect positive linear relationship, and \(r = 0\) indicates no linear relationship. Restricting \(r\) to \([0, 1]\) incorrectly excludes negative associations, and allowing \(r > 1\) is mathematically impossible.

A correlation of \(r = 0\) means there is no linear relationship between the variables. It does not rule out a nonlinear relationship — for example, a perfectly curved (quadratic) relationship can yield \(r = 0\). Saying there is no relationship of any kind is too strong a claim. An inverse relationship would correspond to \(r < 0\), not \(r = 0\).

The strength of a linear association is determined by the absolute value \(|r|\), not the sign. Computing the absolute values: \(|-0.92| = 0.92\), \(|0.85| = 0.85\), \(|-0.50| = 0.50\), and \(|0.10| = 0.10\). The largest absolute value is \(0.92\), so \(r = -0.92\) represents the strongest linear association, even though the association is negative.

A residual is defined as \(e = y - \hat{y}\), the observed response value minus the predicted (fitted) value. A positive residual means the actual value is above the regression line; a negative residual means it is below. The order matters — computing \(\hat{y} - y\) gives the negative of the residual and is a common error.

The slope \(b_1\) represents the predicted change in the response variable \(y\) for each one-unit increase in the explanatory variable \(x\). A value of \(2.5\) when \(x = 0\) describes the \(y\)-intercept, not the slope. The correlation coefficient \(r\) cannot exceed \(1\) in absolute value, so \(r = 2.5\) is impossible. The slope of the regression of \(x\) on \(y\) is a different quantity entirely.

Extrapolation means using the regression model to make predictions for \(x\)-values that lie beyond the range of the data used to fit the model. This is considered unreliable because the relationship that holds within the observed range may not extend beyond it. Computing residuals, transforming data, and minimizing squared residuals are all distinct regression procedures.

By convention, the response (or dependent) variable is always plotted on the vertical \(y\)-axis, and the explanatory (or independent) variable is plotted on the horizontal \(x\)-axis. This placement reflects the idea that we are using \(x\) to explain or predict \(y\). The choice is not based on variability or measurement order.

A circular or random scatter with no directional trend is characteristic of little to no linear association, corresponding to a correlation near \(r = 0\). A strong negative association would show a clear downward trend from left to right. A perfect nonlinear relationship would still show a systematic (curved) pattern, not a random cloud. \(r = 1\) would require all points to lie exactly on an upward-sloping line.

The predicted value at \(x = 5\) is \(\hat{y} = 3 + 2(5) = 13\). The residual is \(e = y - \hat{y} = 14 - 13 = 1\). A residual of \(-1\) results from reversing the formula to \(\hat{y} - y\). The values \(13\) and \(14\) are the predicted and observed values respectively, not the residual.

\(r^2\) is the coefficient of determination and represents the proportion of the total variability in the response variable (\(y\)) that is accounted for by the linear relationship with the explanatory variable (\(x\)). If \(r^2 = 0.64\), then \(r = \pm 0.80\), not \(0.64\). The interpretation about individual students describes a causal claim, which cannot be established from correlation alone. The slope describes the per-unit change in \(y\), which is a separate quantity.

When removing a single point causes a large drop in \(r\), that point was exerting strong influence on the correlation. Points that strengthen a linear pattern (lying in the direction of the trend but far from the center) inflate \(r\). A point with a large positive residual lies far above the line and typically weakens, not strengthens, the correlation. A point exactly on the line has zero residual and minimal effect. Points near \(\bar{x}\) have low leverage and little influence on \(r\).

Correlation does not imply causation. In this case, both ice cream sales and drowning rates increase in warm months due to a lurking variable — hot weather drives people to swim and also to buy ice cream. Neither variable causes the other. Attributing causation in either direction ignores the role of this confounding factor. A correlation of \(0.91\) is mathematically possible for any two variables that move together, regardless of any direct causal link.

A residual plot with a systematic curved pattern signals that the linear model is misspecified — the true relationship likely has a nonlinear component that the linear model fails to capture. A good fit produces a residual plot with no pattern (random scatter around zero). A U-shape suggests the opposite of normally distributed residuals. The shape of the residual plot reveals nothing about the sign of the slope or the magnitude of \(r^2\).

The Pearson correlation coefficient \(r\) is dimensionless and scale-invariant. Multiplying every \(x\)-value by a positive constant (such as \(2.54\)) does not change the relative ordering or spread of the data in a way that affects \(r\). This is because \(r\) is computed from standardized values (z-scores), and a linear rescaling cancels out. Changing sign would only occur if the constant were negative.

Substituting \(x = 175\): \(\hat{y} = -110 + 0.8(175) = -110 + 140 = 30\) kg. A common error is to compute only the slope term \(0.8 \times 175 = 140\) without adding the intercept. The value \(-110\) is the \(y\)-intercept (the predicted weight when \(x = 0\), which is outside the realistic data range), and \(0.8\) is the slope — neither is the answer to a prediction question.

\(r\) ranges from \(-1\) to \(1\) and its sign indicates direction; \(r^2\) ranges from \(0\) to \(1\) and quantifies the proportion of variability in \(y\) explained by \(x\). When \(0 < r < 1\), squaring produces \(r^2 < r\), so \(r^2\) is actually smaller, not larger. Since \(r^2 \geq 0\) always, \(r\) and \(r^2\) do not always share the same sign (negative \(r\) gives positive \(r^2\)). When \(r = \pm 1\), \(r^2 = 1\), but \(r^2 \neq r\) when \(r = -1\).

Leverage measures how far a point's \(x\)-value is from \(\bar{x}\) — extreme \(x\)-values produce high leverage. Influence measures how much removing the point would change the regression line. A high-leverage point that happens to fall on (or near) the line extending from the rest of the data will not substantially change the slope when removed, making it high-leverage but not highly influential. A small residual directly contradicts being an outlier in the \(y\)-direction.

The slope of the least-squares regression line is related to the correlation by \(b_1 = r \cdot \dfrac{s_y}{s_x}\). Solving for \(r\): \(r = b_1 \cdot \dfrac{s_x}{s_y} = 0.75 \cdot \dfrac{4}{5} = 0.60\). A common error is confusing \(r\) with \(b_1\) directly, giving \(r = 0.75\). Another error is inverting the ratio: \(r = b_1 \cdot \dfrac{s_y}{s_x} = 0.75 \cdot \dfrac{5}{4} = 0.9375\), which exceeds \(1\) in some cases. The choice \(r = 0.5625 = b_1^2\) incorrectly applies the \(r^2\) relationship.

An influential point is one whose removal substantially changes the regression results. Here, removing the point drops the slope from \(3.2\) to \(1.8\), a large change, confirming high influence. Such a point typically has extreme \(x\)-values (high leverage) and lies above or below the pattern formed by the rest of the data, pulling the slope in its direction. Choices B and C are contradicted by the observed change in slope, and an outlier only in the \(y\)-direction would have low leverage and minimal effect on the slope.

First, substitute \(x = 4\) into the linearized model: \(\widehat{\ln(y)} = 1.2 + 0.5(4) = 1.2 + 2.0 = 3.2\). Then back-transform by exponentiating: \(\hat{y} = e^{3.2}\). Stopping at \(3.2\) without back-transforming gives the prediction on the log scale, not the original scale. The value \(e^{1.7}\) arises from incorrectly computing \(1.2 + 0.5\) (using \(x = 1\)). The value \(e^{2.0}\) comes from using only the slope term \(0.5(4) = 2.0\) and omitting the intercept.

The least-squares regression line is defined as the line that minimizes \(\sum(y_i - \hat{y}_i)^2\) — the sum of the squared vertical residuals. This criterion leads to a unique, computationally tractable solution. Minimizing the sum of absolute residuals (choice C) defines a different method called least absolute deviations (LAD) regression. Minimizing horizontal distances corresponds to a different regression direction (regressing \(x\) on \(y\)). Passing through extreme points is not a property of the least-squares line.

\(r^2 = 0.81\) means \(81\%\) of the variation in \(y\) is explained by the linear relationship with \(x\), but the remaining \(19\%\) is unexplained. Predictions will have scatter around the line — only when \(r^2 = 1\) (all points on the line) are predictions exact. Saying \(81\%\) of predictions match exactly conflates \(r^2\) with a proportion of correct predictions. The threshold \(r^2 > 0.90\) is not a statistical criterion — accuracy depends on context, not an arbitrary cutoff.

The slope of the regression of \(y\) on \(x\) is \(b_1 = r(s_y / s_x)\), while the slope of \(x\) on \(y\) is \(b_1^* = r(s_x / s_y)\). For the two lines to be identical, these slopes (when both expressed as lines in the \(xy\)-plane) must agree, which occurs only when \(r = \pm 1\) — all points lie perfectly on a single line. Equal means (\(\bar{x} = \bar{y}\)) only guarantees both lines pass through the same point on the diagonal, not that they are the same line. Equal standard deviations (\(s_x = s_y\)) gives equal slopes numerically but only ensures the lines coincide when combined with \(|r| = 1\). Sample size has no bearing on this property.

When \(\log y\) vs. \(\log x\) is linear, the original relationship follows a power model \(y = ax^b\), because \(\log y = \log a + b \log x\) is the linearized form. The higher \(r\) after transformation indicates the power model is a better description of the data structure. The 'true' correlation is a fixed property of the joint distribution — no transformation changes it. Transformations should be chosen based on the theoretical or observed data pattern, not to blindly maximize \(r\). An improvement in \(r\) after transformation does not imply that outliers were removed.

The correlation coefficient \(r\) always falls between \(-1\) and \(1\), inclusive. \(r = 1\) indicates a perfect positive linear relationship, \(r = -1\) a perfect negative linear relationship, and \(r = 0\) indicates no linear relationship. Values outside this range are mathematically impossible. Choice B is incorrect because \(r\) can be negative when the association is negative.

When \(y\) tends to decrease as \(x\) increases, the two variables have a negative association, so the correlation coefficient \(r\) will be negative (i.e., \(r < 0\)). A positive \(r\) would indicate that \(y\) tends to increase as \(x\) increases. The value \(r = 0\) would indicate no linear association.

A residual is defined as \(e = y - \hat{y}\), the difference between the observed value and the predicted value. Choice B reverses the subtraction, giving the negative of the residual. Choice D gives the deviation from the mean, which is used in computing variance but is not a residual.

The slope of a regression line represents the predicted change in \(y\) for each one-unit increase in \(x\). Here, for each additional hour of daily exercise, the predicted health score increases by \(2.3\). Choice A describes the intercept (\(5\)), not the slope. The slope is not the correlation coefficient, which must be between \(-1\) and \(1\).

The correlation coefficient \(r\) is dimensionless — it measures the strength and direction of a linear relationship and is unaffected by linear changes in units. Converting temperature from Fahrenheit to Celsius (\(C = \frac{5}{9}(F - 32)\)) and sales to euros are both linear transformations, so \(r\) does not change. Only reversals of scale direction (multiplying by a negative constant) would change the sign of \(r\).

When points in a scatterplot roughly follow a straight-line pattern, the association is described as linear. An exponential association would show a curve that increases or decreases at an accelerating rate, and a quadratic association would show a parabolic (U-shaped or inverted-U) curve. The correlation coefficient \(r\) is an appropriate measure of strength only for linear associations.

The \(y\)-intercept of a regression line is the predicted value of \(y\) when \(x = 0\). Here, \(18\) thousand dollars is the predicted starting salary for a person with zero years of experience. Choice B describes the slope (\(3.5\)), not the intercept. In practice, the intercept may not always have a meaningful real-world interpretation if \(x = 0\) is outside the range of the data.

An outlier in a scatterplot is a data point that does not follow the overall trend or pattern of the remaining data — it falls unusually far from where we would expect it to be. Having the largest \(x\)-value does not make a point an outlier if it still follows the overall pattern. Having a positive residual simply means the observed value is above the regression line, which is true for roughly half of all points.

The residual is \(e = y - \hat{y}\). First, compute the predicted score: \(\hat{y} = 50 + 8(3) = 50 + 24 = 74\). The student's actual score is \(76\), so the residual is \(76 - 74 = 2\). A positive residual means the model underestimated the student's score. Choice A (\(-2\)) results from computing \(\hat{y} - y\) instead of \(y - \hat{y}\).

The coefficient of determination \(r^2\) represents the proportion of variation in \(y\) explained by the linear regression on \(x\). Here, \(r^2 = (0.8)^2 = 0.64\), meaning \(64\%\) of the variation in sleep duration is explained by the regression on screen time. Choice A incorrectly uses \(r\) itself (\(0.80\)) instead of \(r^2\). Choice D is incorrect; \(\sqrt{0.8} \approx 0.894\) has no direct interpretation here.

Correlation does not imply causation. Both ice cream consumption and drowning incidents tend to increase during hot weather — warm weather (or season) is a lurking variable that drives both. Even a very high \(r\) does not establish causation. Choices A and B confuse association with causation, and choice D incorrectly claims that the magnitude of \(r\) can establish a causal link.

Extrapolation occurs when a regression model is applied to \(x\)-values outside the range used to build the model. The linear relationship observed between \(1.0\) and \(4.0\) liters may not continue to \(7.5\) liters. Using the model in this region is unreliable because we have no data to confirm the relationship still holds. The size of \(r\) does not determine whether a prediction constitutes extrapolation.

A fundamental property of the least-squares regression line is that the residuals sum to zero: \(\sum (y_i - \hat{y}_i) = 0\). This follows because the regression line is fitted to minimize the sum of squared residuals, and it always passes through \((\bar{x}, \bar{y})\), which balances positive and negative residuals. Residuals can be positive or negative; no relationship between the largest residual and the slope exists in general.

A point with an extreme \(x\)-value has high leverage — it has the potential to pull the regression line. However, influence requires that the point actually changes the regression results when removed. If the new point falls close to the line, it confirms the existing trend and has little influence on the slope or intercept. High leverage alone does not guarantee high influence; the point's vertical deviation from the line also matters.

Substituting \(x = 160\) into the regression equation: \(\hat{y} = 10 + 0.95(160) = 10 + 152 = 162\) cm. Choice A (\(152\) cm) results from computing \(0.95 \times 160\) without adding the intercept. Choice C (\(160\) cm) would result from ignoring the equation entirely and assuming \(\hat{y} = x\).

Scatterplots are designed to display the relationship between two quantitative (numerical) variables. The direction of association can be positive, negative, or neither. A strong association may be nonlinear — for example, a clearly curved (quadratic or exponential) pattern can represent a very strong association that is not linear. Categorical variables are not plotted on scatterplots.

The strength of a linear association is determined by the absolute value of \(r\), not its sign. Since \(|{-0.78}| = 0.78 > |0.65| = 0.65\), the association with \(r_1 = -0.78\) is stronger. The sign indicates direction only. For choice D, \(r_1^2 = 0.608\) and \(r_2^2 = 0.423\), so \(r_1\) also explains more variation — further confirming that \(r_1\) represents the stronger association.

The correlation coefficient \(r\) measures only the strength of the linear association between two variables. A value of \(r \approx 0\) indicates little or no linear association, but a strong nonlinear relationship — such as a U-shaped or exponential curve — could still be present. In such a case, \(r \approx 0\) is misleading, and concluding 'no association' would be wrong. Sample size affects the significance of \(r\), but a large sample does not make \(r\) detect nonlinear patterns.

Regression to the mean describes the statistical phenomenon where extreme values on one measurement tend to be closer to the mean on a subsequent measurement. Students scoring unusually high or low on the first exam may have been affected by chance factors (luck, illness); on the second exam, their scores tend to move back toward the average. This occurs whenever \(r < 1\) between two measurements. It does not require leniency in grading or a negative correlation — in fact, this pattern occurs even when the two exams are positively correlated.

The slope represents \(\Delta y / \Delta x\). When \(y\) is scaled by \(0.454\) (lb to kg) and \(x\) is scaled by \(2.54\) (inches to cm), the new slope is \(b_{\text{new}} = b \times \dfrac{0.454}{2.54} = \dfrac{4.5 \times 0.454}{2.54}\). In general, if \(y_{\text{new}} = c_1 y\) and \(x_{\text{new}} = c_2 x\), then \(b_{\text{new}} = b \cdot \dfrac{c_1}{c_2}\). Note that while the slope changes with unit conversions, the correlation coefficient \(r\) does not change — it is unitless.

Even a strong association does not imply causation when a lurking variable is present. Socioeconomic status — including household income and parental education — plausibly causes both higher book ownership and better reading scores. Families with more resources tend to own more books AND provide environments that support literacy. Without controlling for such variables (as in a randomized experiment), the causal claim is unwarranted. The absence of a reported \(r\) value does not prevent identifying direction, and a positive slope is expected, not negative.

\(r^2 = 0.49\) means that \(49\%\) of the total variation in \(y\) is explained by its linear relationship with \(x\); the remaining \(51\%\) is attributed to other sources. Choice A confuses \(r^2\) with \(r\): since \(r^2 = 0.49\), the correlation is \(r = \pm 0.7\), not \(0.49\). Choice C describes the slope, not \(r^2\). Choice B misinterprets \(r^2\) as referring to data points near the line, which is not its definition.

A well-fitting linear model should produce residuals that are randomly scattered around \(e = 0\) with no discernible pattern. A U-shaped curve in the residual plot reveals that the linear model is systematically underestimating in the middle range of \(x\) and overestimating at the extremes (or vice versa). This indicates that the true relationship is nonlinear, and a more appropriate model — such as a quadratic or exponential function — should be considered. The absence of outliers does not validate a misspecified model.

A high-leverage point has an \(x\)-value far from \(\bar{x}\), which gives it the potential to exert strong influence on the regression line. An influential point is operationally defined as one whose removal substantially changes key regression results (slope, intercept, or \(r\)). A high-leverage point that happens to fall near the regression line will have little actual influence. Conversely, points near \(\bar{x}\) can occasionally be influential if they deviate sharply from the pattern. The sign or size of the residual does not define either term.

When \(\ln(y)\) vs. \(x\) is linear — say \(\ln(y) = a + bx\) — this is equivalent to \(y = e^a \cdot e^{bx}\), an exponential model of the form \(y = Ab^x\). The high correlation (\(r = 0.97\)) for the transformed data confirms that the exponential model fits the original data well, much better than a straight line through the original curved scatterplot. The correlation \(r = 0.97\) applies to the linearized version (\(\ln(y)\) vs. \(x\)), not the original variables, so choice A is incorrect. Choice D is wrong because the original \(r\) between \(x\) and \(y\) is simply lower than \(0.97\), not necessarily zero.

By definition, \(r\) is bounded between \(-1\) and \(1\) inclusive. \(r = 1\) indicates a perfect positive linear relationship, \(r = -1\) indicates a perfect negative linear relationship, and \(r = 0\) indicates no linear association. Choice A is incorrect because \(r\) can be negative whenever there is a negative association between the two variables.

A residual is \(e = y - \hat{y}\), that is, observed minus predicted. A positive residual means the observed value falls above the regression line; a negative residual means it falls below. Choice C reverses the subtraction and gives \(\hat{y} - y\), the negative of the residual. Residuals are used to assess how well the model fits individual data points.

When one variable increases as the other decreases, the two variables have a negative association, and the correlation \(r\) will be negative. A positive association (choice A) describes variables that increase together. The description says nothing about whether the pattern is linear or curved, so choice D is not supported — a negative association can be perfectly linear.

A fundamental property of least-squares regression is that the line always passes through \((\bar{x}, \bar{y})\), the point of the sample means. This follows from how the intercept is computed: \(a = \bar{y} - b\bar{x}\), which guarantees \(\hat{y} = \bar{y}\) when \(x = \bar{x}\). The line does not necessarily pass through the origin (choice A) unless the intercept \(a\) happens to equal \(0\).

\(r = 1\) signifies a perfect positive linear relationship — all points fall exactly on a line with positive slope. It does NOT mean the slope equals \(1\); the slope is \(b = r \cdot \frac{s_y}{s_x} = \frac{s_y}{s_x}\), which equals \(1\) only when \(s_y = s_x\). Choice C commits the error of inferring causation from correlation, which is never justified by \(r\) alone.

\(r^2\) represents the fraction of total variability in the response variable \(y\) that is accounted for by the linear regression on \(x\). For example, \(r^2 = 0.64\) means \(64\%\) of the variation in \(y\) is explained by the linear model. It is not a probability (choice D cannot exceed \(1\) and probabilities have a specific interpretation). Note that \(r^2\) is the same whether you regress \(y\) on \(x\) or \(x\) on \(y\), so the asymmetry in choice C is misleading.

The correlation coefficient \(r\) is computed from standardized values of \(x\) and \(y\) — each variable is divided by its own standard deviation — so the units cancel and \(r\) is dimensionless. This makes \(r\) a useful measure for comparing the strength of linear relationships across different types of data. The slope \(b\), by contrast, does carry units: units of \(y\) per unit of \(x\).

A strong positive linear association combines two features: (1) positive direction — \(y\) tends to increase as \(x\) increases, so the line goes upward from left to right; and (2) strong — the points are tightly clustered around the line with little scatter. Choice A describes a strong negative association. Choice D describes a nonlinear (curved) increasing pattern, which would not be called a linear association regardless of strength.

Extrapolation is the practice of using a regression model to predict beyond the range of observed \(x\) values. It is generally unreliable because there is no guarantee the linear pattern continues outside the observed range. Interpolation (choice A) refers to predicting within the range of the data, which is generally safer. Neither confounding nor regression to the mean (choices C and D) describes this concept.

The slope \(b = 4.5\) represents the predicted change in \(y\) for each one-unit increase in \(x\). Here, each additional hour of study per week is associated with a \(4.5\)-point increase in predicted exam score. Choice A describes the \(y\)-intercept (\(52\)), not the slope. Choice C reverses the roles of \(x\) and \(y\), describing the slope of a different regression (predicting hours from score). Choice D confuses the slope with the correlation coefficient \(r\), which must satisfy \(-1 \leq r \leq 1\).

The coefficient of determination is \(r^2 = (-0.8)^2 = 0.64\), so \(64\%\) of the variation in \(y\) is explained by the linear regression on \(x\). Note that \(r^2\) is always non-negative — squaring eliminates the negative sign. Choices A and B are impossible because a proportion of variance cannot be negative. Choice D (\(80\%\)) confuses \(|r|\) with \(r^2\).

In a well-fitting regression model, residuals should be scattered randomly around \(0\) with no discernible pattern. A U-shaped (curved) pattern in the residual plot indicates that the linear model is systematically over- or under-predicting in different regions — a sign that the true relationship is curved, not linear. A nonlinear model (e.g., quadratic) may fit better. Choice B describes a fan-shaped pattern, which indicates non-constant variance (heteroscedasticity), not a U-shape.

The correlation coefficient \(r\) is invariant under linear transformations of the variables. Since \(F = 1.8C + 32\) is a linear transformation with a positive multiplier, it does not change \(r\). This is because \(r\) is computed using standardized values: \(\frac{x_i - \bar{x}}{s_x}\), and linear transformations do not affect standardized scores. A linear transformation with a negative multiplier would only flip the sign of \(r\), not its magnitude. Choice A incorrectly assumes that \(r\) depends on the raw magnitude of the values.

The slope of the least-squares regression line is \(b = r \cdot \frac{s_y}{s_x} = 0.75 \cdot \frac{6}{2} = 0.75 \cdot 3 = 2.25\). Choice C (\(0.75\)) confuses the slope with \(r\) directly. Choice A (\(0.25\)) results from inverting the ratio, using \(\frac{s_x}{s_y} = \frac{2}{6}\) instead of \(\frac{s_y}{s_x}\). Choice D (\(3.00\)) omits multiplication by \(r\) and uses only \(\frac{s_y}{s_x}\).

The correlation \(r\) between \(x\) and \(y\) equals the correlation between \(y\) and \(x\), so \(r\) and \(r^2\) are unchanged (choices A and B are incorrect). Both regression lines pass through \((\bar{x}, \bar{y})\), so choice D is also incorrect. The slope, however, does change: the slope of \(y\) on \(x\) is \(b_1 = r \cdot \frac{s_y}{s_x}\), while the slope of \(x\) on \(y\) is \(b_2 = r \cdot \frac{s_x}{s_y}\). These are equal only in the special case where \(s_x = s_y\).

The \(y\)-intercept of \(-15\) would predict that a person exercising \(0\) minutes burns \(-15\) calories — a physically impossible value. In many regression contexts, the intercept serves a mathematical role but has no meaningful real-world interpretation, especially when \(x = 0\) is outside the range of the observed data. This is a common caution about overinterpreting regression coefficients. Choice B incorrectly describes the intercept as if it were the slope. Choices C and D are contradicted by the positive slope of \(9\).

The predicted value is \(\hat{y} = 10 + 3(4) = 10 + 12 = 22\). The residual is \(e = y - \hat{y} = 25 - 22 = 3\). A positive residual means the observed value (\(25\)) lies above the regression line. Choice B (\(-3\)) reverses the subtraction, computing \(\hat{y} - y\) instead. Choice C (\(22\)) is the predicted value \(\hat{y}\), not the residual.

This is a classic example of a lurking (confounding) variable. Age is associated with both shoe size (older children have larger feet) and reading ability (older children read better). The observed correlation is real but spurious — it is driven by the common cause of age, not a direct link between feet and reading. Choices A and B both commit the error of inferring causation from correlation. Choice D is incorrect; the correlation is genuine in the data even though there is no causal relationship.

First, compute the slope: \(b = r \cdot \frac{s_y}{s_x} = 0.6 \cdot \frac{15}{5} = 0.6 \cdot 3 = 1.8\). Then compute the intercept: \(a = \bar{y} - b\bar{x} = 50 - 1.8(8) = 50 - 14.4 = 35.6\). So the equation is \(\hat{y} = 35.6 + 1.8x\). Choice A uses the correct slope but forgets to subtract \(b\bar{x}\) from \(\bar{y}\). Choice C results from inverting the ratio to \(\frac{s_x}{s_y} = \frac{5}{15}\), giving \(b = 0.2\). Choice D uses \(r\) itself as the slope instead of computing \(r \cdot \frac{s_y}{s_x}\).

A fan-shaped residual plot is the signature of heteroscedasticity — the variance of the residuals is not constant across all fitted values. Standard regression inference (confidence intervals, hypothesis tests for slope) assumes constant variance (homoscedasticity), so violating this condition can make those procedures unreliable. A fan shape does not indicate a curved relationship; that would appear as a U-shape or arc in residuals (choice A). Choice D is unrelated: a strong linear relationship (large \(|r|\)) can still exhibit heteroscedasticity.

A data point with a very large \(x\) value has high leverage — it exerts strong influence on the regression line. Because its \(y\) value is far below the predicted value, it pulls the right end of the regression line downward, decreasing the slope. Additionally, since this point deviates sharply from the strong linear pattern of the other data, it weakens the overall linear relationship, causing \(r\) to decrease. Choice A is incorrect: an inconsistent outlier reduces, not increases, correlation. Choice D is incorrect because a high-leverage point that falls off the trend line changes both slope and intercept.

Use the relationship $r^2 = 1 - \frac{SSE}{SST}$. Substituting: $0.64 = 1 - \frac{900}{SST}$, so $\frac{900}{SST} = 0.36$, giving $SST = \frac{900}{0.36} = 2500$. Choice B (\(1406.25\)) results from incorrectly dividing by \(r^2\) rather than \(1 - r^2\): \(\frac{900}{0.64}\). Choice A (\(576\)) comes from multiplying $SSE$ by \(r^2\): \(900 \times 0.64\). Choice D (\(324\)) multiplies $SSE$ by \(1 - r^2\): \(900 \times 0.36\).

If \(r = 1\), every data point lies exactly on the regression line, meaning the predicted value equals the observed value for every point: \(\hat{y}_i = y_i\). Therefore every residual \(e_i = y_i - \hat{y}_i = 0\). The slope is \(b = r \cdot \frac{s_y}{s_x} = \frac{s_y}{s_x}\), which equals \(1\) only if \(s_y = s_x\) — not a requirement for \(r = 1\), so choices A and D are not necessarily true. Choice C is also not required: \(\bar{x}\) and \(\bar{y}\) can differ even when all points lie on a line.

When a strong linear trend dominates a scatterplot, a mild curvature or systematic pattern can be visually obscured. By computing residuals — the vertical distances from each point to the fitted line — the linear component is subtracted out. Any remaining pattern (curvature, fan shape, cycles) is then displayed on a much smaller scale and becomes far easier to detect. A well-fitting linear model should produce a residual plot with no pattern at all. Choice C is incorrect: residuals are signed values (\(y - \hat{y}\)) and are not converted to absolute values in standard residual plots.

The predicted value is \(\hat{y} = 500 - 20(7) = 500 - 140 = 360\) ms. The residual standard deviation \(s\) measures the typical size of prediction errors — the typical distance between observed and predicted values. A value of \(s = 45\) ms means that this model's predictions are typically off by about \(45\) ms, giving a sense of the model's precision. Choice B misinterprets \(s\) as a count of people. Choice C incorrectly states the predicted value and dismisses \(s\), which is the primary measure of model accuracy around individual predictions.