Correlation & Regression - IB Mathematics Analysis and Approaches

The correlation coefficient between the height (in cm) and weight (in kg) of a group of students is calculated to be $r = 0.78$ . If the heights are converted to metres (by dividing by 100) and the weights are converted to grams (by multiplying by 1000), what will be the new correlation coefficient?
Answer: 0.78 Pearson's product-moment correlation coefficient (r) is a dimensionless quantity and is unaffected by linear transformations (such as changing units) of the variables. Therefore, converting height from cm to m and weight from kg to g will not change the value of $r$ .
The correlation coefficient between hours spent exercising per week ( $h$ ) and body fat percentage ( $p$ ) is found to be $r = -0.8$ . What percentage of the variation in body fat percentage can be explained by the linear relationship with hours spent exercising?
Answer: 64% When you encounter correlation coefficient problems in IB Mathematics, remember that the correlation coefficient $r$ tells you about the strength and direction of a linear relationship, but to find the percentage of variation explained, you need the coefficient of determination, which is $r^2$ . The coefficient of determination $r^2$ represents what percentage of the variation in the dependent variable (body fat percentage) can be explained by the linear relationship with the independent variable (hours exercising). Here, $r = -0.8$ , so $r^2 = (-0.8)^2 = 0.64 = 64\%$ . Notice that we square the correlation coefficient, which always gives a positive result regardless of whether the original correlation was positive or negative. Looking at the wrong answers: Choice A (-80%) is impossible because percentages of variation explained cannot be negative - $r^2$ is always positive since it's a squared value. Choice B (36%) likely comes from incorrectly calculating $(0.8 - 0.44)$ or some other computational error. Choice C (80%) is the trap of using the absolute value of $r$ directly instead of squaring it - this is a very common mistake. The key insight is that correlation and explained variation are different concepts. A correlation of -0.8 indicates a strong negative relationship, but to find explained variation, you must square this value. Study tip: Always remember $r^2$ for explained variation questions. The sign of $r$ tells you direction, but $r^2$ tells you explanatory power. This distinction appears frequently on IB exams.
A scatter plot of variables $x$ and $y$ shows a perfect parabolic relationship, defined by $y = x^2$ for $x$ values symmetrically distributed around 0 (e.g., from -5 to 5). The Pearson's product-moment correlation coefficient $r$ is calculated for this data. Which value is the most likely for $r$ ?
Answer: $r = 0$ Pearson's correlation coefficient $r$ measures the strength and direction of a linear relationship. For the function $y = x^2$ over a symmetric interval like [-5, 5], there is a perfect non-linear relationship. For every positive value of $x$ that suggests a positive trend, there is a corresponding negative value of $x$ that suggests a negative trend. These effects cancel each other out, resulting in a correlation coefficient of exactly 0. This highlights that $r=0$ means no linear correlation, not necessarily no relationship at all.
A biologist models the relationship between the length of a fish ( $L$ cm) and its weight ( $W$ g). The data was collected for fish with lengths between 10 cm and 30 cm. The resulting regression line is $W = 15.2L - 80$ and the correlation coefficient is $r = 0.92$ . The biologist wants to estimate the weight of a fish that is 5 cm long. Which statement is the most appropriate?
Answer: The estimate is unreliable because it involves extrapolation. The data used to create the model was for fish with lengths between 10 cm and 30 cm. Using the model to predict the weight for a fish of 5 cm length is an example of extrapolation, as it is outside the range of the original data. Predictions made by extrapolation are considered unreliable, regardless of how strong the correlation coefficient is. While correlation doesn't imply causation (D), the primary reason for unreliability here is extrapolation.
The regression line for the number of ice creams sold ( $N$ ) versus the daily temperature ( $T$ in degrees Celsius) is $N = 12T - 50$ . The data was collected for temperatures between $15^\circ\text{C}$ and $35^\circ\text{C}$ . What is the most appropriate interpretation of the $N$ -intercept?
Answer: The model is likely not valid for a temperature of $0^\circ\text{C}$ as this represents a significant extrapolation from the data range. The $N$ -intercept is the value of $N$ when $T=0$ . In this model, the intercept is -50. However, the data was collected for temperatures between $15^\circ\text{C}$ and $35^\circ\text{C}$ . A temperature of $0^\circ\text{C}$ is far outside this range. Interpreting the intercept is an act of extrapolation, and therefore the model is not reliable at this value. Choice A is a literal interpretation but B provides the better statistical reason why that interpretation is flawed.
Consider three distinct bivariate datasets. Dataset A has a correlation coefficient of $r = 0.8$ . Dataset B has a correlation coefficient of $r = -0.9$ . Dataset C has a correlation coefficient of $r = 0.7$ . Which statement correctly compares the strength of the linear relationships?
Answer: Dataset B has the strongest relationship. The strength of a linear correlation is determined by the absolute value of the correlation coefficient, $|r|$ . A value closer to 1 or -1 indicates a stronger linear relationship. We compare the absolute values: $|r_A| = |0.8| = 0.8$ , $|r_B| = |-0.9| = 0.9$ , and $|r_C| = |0.7| = 0.7$ . Since 0.9 is the largest value, Dataset B has the strongest linear relationship. Dataset C has the weakest.
A researcher finds the Pearson's product-moment correlation coefficient between two variables, $x$ and $y$ , is $r = -0.95$ . Which of the following statements is the most accurate interpretation of this result?
Answer: There is a strong negative linear correlation between $x$ and $y$ , and as $x$ increases, $y$ tends to decrease. The value of $r$ indicates the strength and direction of a linear correlation. A value close to -1, such as -0.95, indicates a very strong negative linear correlation. This means that as one variable increases, the other variable tends to decrease. Correlation does not imply causation, so a causal link cannot be established from the $r$ value alone.
The equation of the regression line of $y$ on $x$ for a dataset is given by $y = 2.5x - 4$ . The mean of the $x$ -values is $\bar{x} = 6$ . What is the mean of the $y$ -values, $\bar{y}$ ?
Answer: 11 A key property of the least squares regression line is that it always passes through the point of means, $(\bar{x}, \bar{y})$ . Therefore, the coordinates of the mean point must satisfy the equation of the line. We can substitute $\bar{x} = 6$ into the equation to find $\bar{y}$ : $\bar{y} = 2.5(6) - 4 = 15 - 4 = 11$ .
For a dataset, the regression line of $y$ on $x$ is $y = -0.7x + 15.4$ . The mean of the $x$ -values is $\bar{x} = 8$ . Use the regression line to predict the value of $y$ when $x = 10$ .
Answer: 8.4 To predict the value of $y$ for a given $x$ , substitute the value of $x$ into the regression equation. Here, we substitute $x=10$ : $y = -0.7(10) + 15.4 = -7 + 15.4 = 8.4$ . The information about the mean $\bar{x}$ is extra information not needed for this specific calculation, though it can be used to find $\bar{y}$ (which is 9.8, a distractor).
For a set of data, the regression line of the price of a used car ( $P$ , in thousands of dollars) on its age ( $A$ , in years) is given by $P = -1.8A + 25$ . What is the best interpretation of the gradient of this line?
Answer: For every year older a car is, its price is expected to decrease by $1800. The gradient (slope) of the regression line represents the change in the dependent variable (Price) for a one-unit increase in the independent variable (Age). Here, the gradient is -1.8. Since the price $P$ is in thousands of dollars, a gradient of -1.8 means that for each additional year of age, the price is expected to decrease by 1.8 thousand dollars, which is $1800.
The least squares regression line of $y$ on $x$ is determined for a set of data points. Which of the following quantities does this line minimize?
Answer: The sum of the squares of the vertical distances from the points to the line. When you encounter questions about least squares regression, remember that this method gets its name from literally minimizing the sum of squared residuals. Understanding what "residual" means is key to answering this correctly. The least squares regression line is positioned to minimize the sum of the squares of the vertical distances from each data point to the line. These vertical distances are called residuals - they represent the difference between the actual y-value and the predicted y-value for each x. The line is called "least squares" because it finds the position where $\sum(y_i - \hat{y_i})^2$ is minimized, where $y_i$ is the actual value and $\hat{y_i}$ is the predicted value. Answer D correctly identifies this fundamental principle of least squares regression. Answer A is incorrect because regression minimizes vertical distances, not horizontal ones. Horizontal distances would be relevant if we were regressing x on y instead. Answer B describes minimizing the maximum distance (a "minimax" approach), which is a completely different optimization criterion not used in least squares regression. Answer C refers to minimizing the sum of absolute values of vertical distances, which describes a different type of regression called "least absolute deviations" or "robust regression." This method is less sensitive to outliers but is not what least squares does. Remember this connection: "least squares" = minimizing the sum of squared vertical distances. The squaring serves two purposes: it eliminates negative values (since we want total deviation) and it penalizes larger deviations more heavily than smaller ones.
A dataset on study hours ( $x$ ) and exam scores ( $y$ ) shows a strong positive correlation. A data point $(100, 95)$ is found to be a data entry error and is corrected to $(10, 95)$ . Assuming this corrected point better fits the linear trend of the other data points, what is the likely effect of this correction on the correlation coefficient $r$ and the gradient of the regression line?
Answer: $r$ will increase and the gradient will increase. The original point $(100, 95)$ is an influential outlier in the $x$ -direction. It has high leverage and would pull the regression line towards it, making the line less steep (decreasing the gradient) than it otherwise would be. It also lies far from the general linear trend, weakening the correlation (decreasing $r$ ). Correcting this point to $(10, 95)$ , which is assumed to fit the trend, removes the outlier's influence. As a result, the line will become steeper (the gradient increases) and the points will be more tightly clustered around the line (the correlation becomes stronger, so $r$ increases).
For a bivariate dataset, the following summary statistics are given: $n=10$ , $\sum x = 50$ , $\sum y = 80$ , $S_{xx} = 40$ , and $S_{xy} = 60$ . Find the equation of the regression line of $y$ on $x$ , in the form $y=ax+b$ .
Answer: $y = 1.5x + 0.5$ The gradient of the regression line $y$ on $x$ is given by $a = \frac{S_{xy}}{S_{xx}}$ . So, $a = \frac{60}{40} = 1.5$ . The line passes through the mean point $(\bar{x}, \bar{y})$ . We calculate the means: $\bar{x} = \frac{\sum x}{n} = \frac{50}{10} = 5$ and $\bar{y} = \frac{\sum y}{n} = \frac{80}{10} = 8$ . Now we find the y-intercept $b$ using $\bar{y} = a\bar{x} + b$ . Substituting the values: $8 = 1.5(5) + b \Rightarrow 8 = 7.5 + b \Rightarrow b = 0.5$ . Thus, the equation is $y = 1.5x + 0.5$ .
The equation of the regression line of $y$ on $x$ is $y = -3x + c$ . The line passes through the point $(5, 10)$ and the mean point $(\bar{x}, 16)$ . Find the value of $\bar{x}$ .
Answer: 3 This is a two-step problem. First, use the fact that the line passes through $(5, 10)$ to find the constant $c$ . Substitute $x=5$ and $y=10$ into the equation: $10 = -3(5) + c \Rightarrow 10 = -15 + c \Rightarrow c = 25$ . So, the full equation of the regression line is $y = -3x + 25$ . Second, use the property that the regression line must pass through the mean point $(\bar{x}, \bar{y})$ . Substitute $(\bar{x}, 16)$ into the equation: $16 = -3\bar{x} + 25 \Rightarrow -9 = -3\bar{x} \Rightarrow \bar{x} = 3$ .
For two variables $x$ and $y$ , the regression line of $y$ on $x$ has a negative gradient. Which of the following statements about the Pearson's product-moment correlation coefficient $r$ must be true?
Answer: $r < 0$ The gradient $a$ of the regression line $y=ax+b$ is calculated as $a = \frac{S_{xy}}{S_{xx}}$ . The correlation coefficient is $r = \frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}$ . Since $S_{xx}$ and $\sqrt{S_{xx}S_{yy}}$ are always non-negative, the sign of both $a$ and $r$ is determined by the sign of $S_{xy}$ . Therefore, if the gradient is negative, $S_{xy}$ must be negative, and thus $r$ must also be negative.

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