IB Mathematics: Analysis and Approaches Quiz: Correlation And Regression

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Correlation And RegressionQuestion 1 of 20

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?

r

will increase and the gradient will increase.

r

will increase and the gradient will decrease.

r

will decrease and the gradient will increase.

r

will decrease and the gradient will decrease.

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Correlation And Regression

Practice Correlation And Regression in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Correlation And Regression, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$r$ will increase and the gradient will increase. (correct answer)
$r$ will increase and the gradient will decrease.
$r$ will decrease and the gradient will increase.
$r$ will decrease and the gradient will decrease.

Explanation: 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).

Question 2

$r$ will increase and the gradient will increase. (correct answer)
$r$ will increase and the gradient will decrease.
$r$ will decrease and the gradient will increase.
$r$ will decrease and the gradient will decrease.

Explanation: The original point

(100, 95)

is an influential outlier in the

x

r

). Correcting this point to

(10, 95)

r

increases).

Question 3

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?

For every year older a car is, its price is expected to increase by $1800.
For every year older a car is, its price is expected to decrease by $1800. (correct answer)
A brand new car ( $A=0$ ) is expected to cost $1800.
The average price of a car in the dataset is $25,000.

Explanation: 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.

Question 4

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?

Dataset A has the strongest relationship because its $r$ value is the highest positive number.
Dataset C has the weakest relationship.
Dataset B has the strongest relationship. (correct answer)
The relationships in Datasets A and B cannot be compared because their signs are different.

Explanation: 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.

Question 5

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?

On a day with temperature $0^\circ\text{C}$ , the model predicts -50 ice creams will be sold, which is physically impossible.
The model is likely not valid for a temperature of $0^\circ\text{C}$ as this represents a significant extrapolation from the data range. (correct answer)
For every degree increase in temperature, the number of ice creams sold decreases by 50.
The minimum number of ice creams ever sold on a day was 50.

Explanation: 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.

Question 6

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?

-80%
36%
80%
64% (correct answer)

Explanation: 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.

Question 7

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?

0.0078
0.78 (correct answer)
7.8
It cannot be determined without the original data.

Explanation: 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

Question 8

For every year older a car is, its price is expected to increase by $1800.
For every year older a car is, its price is expected to decrease by $1800. (correct answer)
A brand new car ( $A=0$ ) is expected to cost $1800.
The average price of a car in the dataset is $25,000.

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.

Question 9

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 $r$ ?

$r \approx 1$
$r \approx -1$
$r = 0$ (correct answer)
$r \approx 0.5$

Explanation: 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.

Question 10

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?

$r = -1$
$r = 0$
$r > 0$
$r < 0$ (correct answer)

Explanation: 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.

Question 11

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?

There is a strong positive linear correlation between $x$ and $y$ , and as $x$ increases, $y$ tends to increase.
There is a strong negative linear correlation between $x$ and $y$ , and as $x$ increases, $y$ tends to decrease. (correct answer)
There is a weak negative linear correlation between $x$ and $y$ , and as $x$ increases, $y$ tends to decrease.
An increase in $x$ causes a decrease in $y$ .

Explanation: 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.

Question 12

0.0078
0.78 (correct answer)
7.8
It cannot be determined without the original data.

r

Question 13

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?

The estimate is reliable because the correlation is strong.
The estimate is unreliable because it involves extrapolation. (correct answer)
The estimate is reliable and the weight is -3 g.
The estimate is unreliable because correlation does not imply causation.

Explanation: 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.

Question 14

$r \approx 1$
$r \approx -1$
$r = 0$ (correct answer)
$r \approx 0.5$

Explanation: 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.

Question 15

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 $y=ax+b$ .

$y = 1.5x + 0.5$ (correct answer)
$y = 1.5x + 5$
$y = 0.67x + 4.67$
$y = 0.67x + 8$

Explanation: The gradient of the regression line

y

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

Question 16

$r = -1$
$r = 0$
$r > 0$
$r < 0$ (correct answer)