IB Mathematics: Applications and Interpretation Quiz: Correlation And Regression

20 questions · exam conditions

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Correlation And RegressionQuestion 1 of 20

Two different sales teams, Team A and Team B, sell a product. Their performance is modelled by relating the number of sales calls made in a week ( $c$ ) to the total value of sales ( $V$ ).

Team A's model: $V_A = 150c + 500$
Team B's model: $V_B = 120c + 800$ Both models are valid for $c$ between 10 and 40 calls.

Based on these models, which statement is the most accurate comparison of the teams' performance?

Team A is always more effective as their sales value increases more per call.

Team B starts with a higher base value, but Team A becomes more effective as the number of calls increases.

Team B is always more effective because their base sales value (intercept) is higher.

The models show that making more than 40 calls would be beneficial for both teams.

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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Correlation And Regression

Practice Correlation And Regression in IB Mathematics: Applications and Interpretation 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: Applications and Interpretation.

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

Two different sales teams, Team A and Team B, sell a product. Their performance is modelled by relating the number of sales calls made in a week ( $c$ ) to the total value of sales ( $V$ ).

Team A's model: $V_A = 150c + 500$
Team B's model: $V_B = 120c + 800$ Both models are valid for $c$ between 10 and 40 calls.

Based on these models, which statement is the most accurate comparison of the teams' performance?

Team A is always more effective as their sales value increases more per call.
Team B starts with a higher base value, but Team A becomes more effective as the number of calls increases. (correct answer)
Team B is always more effective because their base sales value (intercept) is higher.
The models show that making more than 40 calls would be beneficial for both teams.

Explanation: To compare the models, we find their intersection:

150c + 500 = 120c + 800

, which gives

c = 10

. At

c = 10

, both teams have the same sales value of $2000. Team B has a higher y-intercept (800 vs 500), representing better performance with fewer calls, but Team A has a higher slope (150 vs 120), meaning they gain more value per additional call. Within the valid domain [10, 40], Team A performs better for

c > 10

due to their higher rate of improvement.

A is incorrect because at $c = 10$ , they are equal.
C is incorrect because Team A outperforms Team B for most of the valid domain.
D discusses extrapolation beyond the model's validity.

Question 2

A study of the relationship between the age of a used car, $x$ in years, and its value, $y$ in euros, finds a coefficient of determination $r^2 = 0.7225$ .

Given the context, find the Pearson's correlation coefficient, $r$ .

-0.85 (correct answer)
0.7225
0.85
0.95

Explanation: The Pearson's correlation coefficient,

r

, is the square root of the coefficient of determination,

r^2

. The sign of

r

must match the sign of the slope of the regression line. In the context of a used car's age and value, as age (

x

) increases, value (

y

) decreases. This implies a negative relationship, so

r

must be negative.

r = -\sqrt{0.7225} = -0.85

C is incorrect because it uses the positive root, which would imply that older cars are more valuable.
B is incorrect as it is the value of $r^2$ , not $r$ .
D is an incorrect calculation of the square root.

Question 3

A café manager models the number of hot chocolates sold per day, $N$ , based on the average daily temperature, $T$ , in degrees Celsius. The regression equation is $N = -3.5T + 80$ . What is the interpretation of the value -3.5 in this context?

For every 3.5 degrees the temperature drops, one extra hot chocolate is sold.
The number of hot chocolates sold is expected to decrease by 3.5 for each one-degree Celsius increase in temperature. (correct answer)
The minimum number of hot chocolates sold on any day is 3.5.
When the temperature is 0°C, the number of hot chocolates sold is 3.5 less than the maximum.

Explanation: The slope of the regression line, -3.5, represents the change in the dependent variable (

N

) for a one-unit increase in the independent variable (

T

). Therefore, for each 1°C increase in temperature, the number of hot chocolates sold is predicted to change by -3.5, which means it decreases by 3.5.

A is incorrect because it misinterprets the ratio of the change.
C is incorrect because the slope is a rate of change, not a minimum value.
D is an incorrect interpretation that confuses the slope with a specific prediction at T=0.

Question 4

An analyst is investigating different models for a dataset relating variables $x$ and $y$ . They find the following Pearson's correlation coefficients ( $r$ ) for different transformations:

Model 1: $y$ vs $x$ , $r = 0.82$
Model 2: $y$ vs $\log(x)$ , $r = 0.91$
Model 3: $\log(y)$ vs $x$ , $r = -0.95$
Model 4: $\log(y)$ vs $\log(x)$ , $r = 0.88$

Based on this information, which model provides the best fit for the data?

Model 1, because it uses the original untransformed data.
Model 2, because it has the highest positive correlation coefficient.
Model 3, because its correlation coefficient has the greatest magnitude. (correct answer)
Model 4, because it transforms both variables, making it more robust.

Explanation: The best linear fit is indicated by the Pearson's correlation coefficient,

r

, whose absolute value is closest to 1. We compare the magnitudes:

|0.82| = 0.82

|0.91| = 0.91

|-0.95| = 0.95

, and

|0.88| = 0.88

. The value with the greatest magnitude is -0.95, which corresponds to Model 3. This indicates the strongest linear relationship is between

\log(y)

and

x

A is incorrect because the original data may not have the most linear relationship.
B is incorrect because the sign of $r$ indicates direction, not strength of fit. A value of -0.95 indicates a stronger fit than 0.91.
D is incorrect; transforming both variables does not automatically make a model better.

Question 5

A real estate agent uses the linear regression equation $P = 2150A + 45000$ to model the price, $P$ , of a house in US dollars based on its floor area, $A$ , in square metres. Which statement is the correct interpretation of the slope of this regression line?

The minimum price of any house in this model is ($2150).
For each additional US dollar in price, the floor area is expected to increase by 2150 square metres.
For each additional square metre of floor area, the price of the house is expected to increase by ($2150). (correct answer)
A house with no floor area would have a price of ($45000), and the price increases from there.

Explanation: The slope of a linear regression equation

y = mx + c

represents the change in the dependent variable (

y

) for a one-unit increase in the independent variable (

x

). In this model,

P = 2150A + 45000

, the slope is 2150. This means for each one-unit increase in area

A

(i.e., 1 square metre), the price

P

is predicted to increase by ($2150).

A is incorrect because it confuses the slope with a minimum price.
B is incorrect because it swaps the dependent and independent variables.
D is incorrect because it describes the y-intercept, not the slope.

Question 6

The regression line for a dataset is $y = 3.1x - 11.4$ . The data used to generate the model had $x$ values ranging from 5 to 20. Two predictions are made using the model: one for $x=15$ and one for $x=30$ .

Which statement best describes the reliability of these predictions?

The prediction at $x=30$ is an interpolation and is therefore reliable.
Both predictions are extrapolations and are therefore unreliable.
The prediction at $x=30$ is considered more reliable because it reflects a longer-term trend.
The prediction at $x=15$ is an interpolation and is considered more reliable than the prediction at $x=30$ . (correct answer)

Explanation: When you encounter regression problems involving predictions, the key distinction is between interpolation (predicting within the data range) and extrapolation (predicting outside the data range). Since your original data had x-values from 5 to 20, any prediction within this range is interpolation, while predictions outside this range are extrapolation. For the prediction at

x = 15

: Since 15 falls between 5 and 20, this is interpolation. The model has actual data points around this value, making the prediction reliable because you're working within the established pattern. For the prediction at

x = 30

: Since 30 is well beyond the upper limit of 20, this is extrapolation. You're assuming the linear relationship continues far beyond your data, which may not be valid. Real-world relationships often change or break down outside the observed range. Answer A incorrectly calls the

x = 30

prediction an interpolation when it's actually extrapolation. Answer B wrongly claims both predictions are extrapolations—the

x = 15

prediction is clearly within the data range. Answer C suggests extrapolation is more reliable for long-term trends, which contradicts statistical principles that favor predictions within the data range. Answer D correctly identifies that

x = 15

is interpolation and is more reliable than the extrapolation at

x = 30

. Study tip: Always check whether prediction values fall within or outside your original data range. Interpolation (within range) is generally reliable, while extrapolation (outside range) becomes increasingly unreliable the further you move from your data.

Question 7

A plot of the residuals versus the independent variable ( $x$ ) for a linear regression shows a distinct curve, like a parabola.

What is the most likely conclusion an analyst should draw from this residual plot?

The linear model is a good fit for the data.
There is a strong positive correlation between the variables.
The variance of the residuals is increasing as x increases.
The relationship between the variables is non-linear. (correct answer)

Explanation: When you encounter residual plots in regression analysis, you're examining whether your model captures the true relationship in the data. Residuals are the differences between observed and predicted values, and their pattern reveals crucial information about model adequacy. A parabolic curve in the residual plot indicates that your linear model is systematically missing something important. When residuals follow a curved pattern, it means the relationship between your variables isn't actually linear—it has curvature that a straight line cannot capture. This systematic pattern suggests you need a quadratic or other non-linear model instead. Option D correctly identifies this non-linear relationship. The curved residual pattern is a classic diagnostic sign that the true relationship has curvature. Option A is wrong because a good linear fit would produce residuals scattered randomly around zero with no discernible pattern. A parabolic curve indicates poor fit. Option B misses the point entirely. Correlation strength isn't what residual plots assess—they evaluate whether your chosen model type (linear vs. non-linear) is appropriate. You could have strong correlation but still need a non-linear model. Option C describes heteroscedasticity (changing variance), which would appear as a fan or funnel shape in residuals, not a parabolic curve. The curve pattern specifically indicates wrong model form, not changing variance. Remember this key principle: random scatter in residuals = good model fit, while systematic patterns (curves, funnels, waves) = model problems. Curved residuals almost always mean you need a non-linear approach.

Question 8

A business analyst creates a linear model for monthly sales ( $S$ ) based on advertising spending ( $A$ ). The coefficient of determination is $r^2 = 0.55$ .

Which statement provides the best interpretation of this $r^2$ value?

55% of the data points lie on the least squares regression line.
There is a 55% probability that an increase in advertising will lead to an increase in sales.
55% of the variation in monthly sales can be explained by the variation in advertising spending. (correct answer)
The correlation between advertising spending and sales is 0.55, indicating a moderate relationship.

Explanation: The coefficient of determination,

r^2

, measures the proportion of the variance in the dependent variable that is predictable from the independent variable(s). It is expressed as a percentage.

C gives the correct statistical interpretation.
A is incorrect. $r^2$ does not describe how many points are on the line itself.
B is incorrect because $r^2$ is not a probability of a specific outcome.
D is incorrect because it confuses the coefficient of determination ( $r^2$ ) with the correlation coefficient ( $r$ ).

Question 9

A researcher is studying the relationship between two variables, $x$ and $y$ . They perform a linear regression of $y$ on $x$ and obtain the equation $y = 2x + 3$ . They then decide to swap the variables and perform a linear regression of $x$ on $y$ . The Pearson's correlation coefficient $r$ between $x$ and $y$ is 0.8.

Which of the following statements about the new regression line, $x$ on $y$ , is true?

The slope of the new regression line will be 0.5.
The correlation coefficient for the new regression will be -0.8.
The new regression line is mathematically distinct from the first line. (correct answer)
The y-intercept of the new line will be -1.5.

Explanation: The regression line of

y

x

is different from the regression line of

x

y

, unless the data lies perfectly on a line (

r = \pm 1

). The first line minimizes vertical distances (residuals in

y

), while the second line minimizes horizontal distances (residuals in

x

). Therefore, they are distinct lines.

A is incorrect. The slope of the $x$ on $y$ line is not simply the reciprocal of the $y$ on $x$ line slope. The new slope is given by $r \frac{\sigma_x}{\sigma_y}$ , while the old slope was $r \frac{\sigma_y}{\sigma_x}$ . Their product is $r^2$ , not 1.
B is incorrect. The correlation coefficient $r$ is symmetric; it remains the same regardless of which variable is treated as independent.
D is incorrect. There's no simple relationship that guarantees the new intercept is -1.5. One cannot determine the new line without more information.

Question 10

A researcher finds a strong correlation of $r=0.95$ between a country's annual chocolate consumption per capita and the number of Nobel laureates it has per 10 million people. Which of the following is the most sound conclusion?

Eating more chocolate is a highly effective way to increase a person's intelligence and chances of winning a Nobel prize.
The relationship is statistically significant, meaning there is a less than 5% chance it is due to random variation.
Countries with many Nobel laureates can afford to import more chocolate, so winning Nobel prizes causes chocolate consumption.
This high correlation is likely coincidental or due to a lurking variable, such as national wealth, affecting both. (correct answer)

Explanation: When you encounter correlation problems on the IB exam, remember that correlation never implies causation – even when the correlation coefficient is very high like

r = 0.95

. The correct answer is D because a strong correlation between two variables often reflects the influence of a third, unobserved variable called a lurking variable. In this case, national wealth likely affects both chocolate consumption (wealthier countries can afford luxury foods) and Nobel laureates (wealthier countries typically have better educational systems, research funding, and universities). This creates an indirect relationship that appears as correlation without any direct causal link. Option A commits the classic correlation-causation fallacy. Just because two variables move together doesn't mean one causes the other – eating chocolate doesn't make individuals smarter or more likely to win Nobel prizes. Option B confuses correlation strength with statistical significance. The value

r = 0.95

tells us about the strength of the linear relationship, not about statistical significance, which requires additional information like sample size and hypothesis testing. Option C attempts reverse causation but still assumes a direct causal relationship exists. While this reverses the direction, it's still incorrectly assuming that Nobel prizes directly influence chocolate consumption rather than both being influenced by a third factor. Remember this key strategy: when you see high correlations between seemingly unrelated variables, always consider lurking variables. Look for underlying factors that could logically influence both variables simultaneously – this approach will help you avoid correlation-causation traps throughout the statistics portion of your exam.

Question 11

An environmental scientist models the concentration of a certain chemical in a lake, $C$ , in parts per million (ppm), against the depth, $d$ , in metres. The regression line is $C = -0.12d + 5.4$ . For a measurement taken at a depth of 10 metres, the residual was $+0.5$ ppm.

What was the actual measured concentration of the chemical at this depth?

3.7 ppm
4.2 ppm
4.7 ppm (correct answer)
5.9 ppm

Explanation: The residual is defined as

e = y_{actual} - y_{predicted}

. We can rearrange this to find the actual value:

y_{actual} = y_{predicted} + e

First, calculate the predicted concentration at $d=10$ metres: $C_{predicted} = -0.12(10) + 5.4 = -1.2 + 5.4 = 4.2$ ppm.
The residual is given as $e = +0.5$ ppm.
Calculate the actual concentration: $C_{actual} = 4.2 + 0.5 = 4.7$ ppm.

A is incorrect; it results from subtracting the residual from the predicted value and also making a calculation error.
B is incorrect as this is the predicted value, not the actual measured value.
D is incorrect as it results from adding the residual to the y-intercept.

Question 12

A teacher uses the regression line $y = 0.85x + 12$ to predict a student's score on a final exam, $y$ , based on their score in a midterm exam, $x$ . One student scored 80 on the midterm and 79 on the final exam.

Calculate the residual for this student.

-1.0 (correct answer)
1.0
12.0
80.0

Explanation: The residual is the difference between the observed value and the predicted value (residual = observed

y

- predicted

y

First, find the predicted score using the regression equation: $y_{predicted} = 0.85(80) + 12 = 68 + 12 = 80$ .
The observed score is $y_{observed} = 79$ .
Calculate the residual: $79 - 80 = -1.0$ .

B is incorrect due to a sign error (predicted - observed).
C is incorrect as it is the y-intercept of the model.
D is incorrect as it is the predicted value, not the residual.

Question 13

A study of the relationship between the age of a used car, $x$ in years, and its value, $y$ in euros, finds a coefficient of determination $r^2 = 0.7225$ .

Given the context, find the Pearson's correlation coefficient, $r$ .

-0.85 (correct answer)
0.7225
0.85
0.95

Explanation: The Pearson's correlation coefficient,

r

, is the square root of the coefficient of determination,

r^2

. The sign of

r

must match the sign of the slope of the regression line. In the context of a used car's age and value, as age (

x

) increases, value (

y

) decreases. This implies a negative relationship, so

r

must be negative.

r = -\sqrt{0.7225} = -0.85

C is incorrect because it uses the positive root, which would imply that older cars are more valuable.
B is incorrect as it is the value of $r^2$ , not $r$ .
D is an incorrect calculation of the square root.

Question 14

What was the actual measured concentration of the chemical at this depth?

3.7 ppm
4.2 ppm
4.7 ppm (correct answer)
5.9 ppm

Explanation: The residual is defined as

e = y_{actual} - y_{predicted}

. We can rearrange this to find the actual value:

y_{actual} = y_{predicted} + e

First, calculate the predicted concentration at $d=10$ metres: $C_{predicted} = -0.12(10) + 5.4 = -1.2 + 5.4 = 4.2$ ppm.
The residual is given as $e = +0.5$ ppm.
Calculate the actual concentration: $C_{actual} = 4.2 + 0.5 = 4.7$ ppm.

A is incorrect; it results from subtracting the residual from the predicted value and also making a calculation error.
B is incorrect as this is the predicted value, not the actual measured value.
D is incorrect as it results from adding the residual to the y-intercept.

Question 15

For every 3.5 degrees the temperature drops, one extra hot chocolate is sold.
The number of hot chocolates sold is expected to decrease by 3.5 for each one-degree Celsius increase in temperature. (correct answer)
The minimum number of hot chocolates sold on any day is 3.5.
When the temperature is 0°C, the number of hot chocolates sold is 3.5 less than the maximum.

Explanation: The slope of the regression line, -3.5, represents the change in the dependent variable (

N

) for a one-unit increase in the independent variable (

T

). Therefore, for each 1°C increase in temperature, the number of hot chocolates sold is predicted to change by -3.5, which means it decreases by 3.5.

A is incorrect because it misinterprets the ratio of the change.
C is incorrect because the slope is a rate of change, not a minimum value.
D is an incorrect interpretation that confuses the slope with a specific prediction at T=0.

Question 16

Two different sales teams, Team A and Team B, sell a product. Their performance is modelled by relating the number of sales calls made in a week ( $c$ ) to the total value of sales ( $V$ ).

Team A's model: $V_A = 150c + 500$
Team B's model: $V_B = 120c + 800$ Both models are valid for $c$ between 10 and 40 calls.

Based on these models, which statement is the most accurate comparison of the teams' performance?

Team A is always more effective as their sales value increases more per call.
Team B starts with a higher base value, but Team A becomes more effective as the number of calls increases. (correct answer)
Team B is always more effective because their base sales value (intercept) is higher.
The models show that making more than 40 calls would be beneficial for both teams.

Explanation: To compare the models, we find their intersection:

150c + 500 = 120c + 800

, which gives

c = 10

. At

c = 10

c > 10

due to their higher rate of improvement.

A is incorrect because at $c = 10$ , they are equal.
C is incorrect because Team A outperforms Team B for most of the valid domain.
D discusses extrapolation beyond the model's validity.

Question 17

Eating more chocolate is a highly effective way to increase a person's intelligence and chances of winning a Nobel prize.
The relationship is statistically significant, meaning there is a less than 5% chance it is due to random variation.
Countries with many Nobel laureates can afford to import more chocolate, so winning Nobel prizes causes chocolate consumption.
This high correlation is likely coincidental or due to a lurking variable, such as national wealth, affecting both. (correct answer)

Explanation: When you encounter correlation problems on the IB exam, remember that correlation never implies causation – even when the correlation coefficient is very high like

r = 0.95

r = 0.95

Question 18

A study finds a strong negative correlation ( $r = -0.92$ ) between the number of hours a person spends watching television per week, $h$ , and their score on a physical fitness test, $s$ .

Which of the following is a valid conclusion that can be drawn from this information?

Watching more television causes a person's physical fitness score to decrease.
A high score on the physical fitness test causes a person to watch less television.
There is a strong linear association where higher hours of television watching correspond to lower fitness scores. (correct answer)
If a person stops watching television, their score on the physical fitness test is guaranteed to increase.

Explanation: Correlation measures the strength and direction of a linear association between two variables, but it does not imply causation. A correlation coefficient of

r = -0.92

indicates a strong, negative, linear relationship.

C correctly states this association without implying cause and effect.
A and B are incorrect because they infer a causal relationship, which correlation alone cannot establish.
D is incorrect because it makes a definitive prediction of an increase, which is not guaranteed and also implies causation.

Question 19

The height of a growing plant, $H$ in cm, is modelled by the linear regression equation $H = 1.5t + 8$ , where $t$ is the number of days after the seedling was planted. The data used to create this model was collected for $t$ values between 5 and 30 days.

What is the correct interpretation of the $y$ -intercept of this equation?

The plant is predicted to be 8 cm tall at the moment it is planted ( $t=0$ ), which is a valid estimate.
The model predicts the plant was 8 cm tall at $t=0$ , but this is an unreliable extrapolation as it is outside the data range. (correct answer)
The minimum height of the plant during the experiment was 8 cm.
For every day that passes, the height of the plant increases by a minimum of 8 cm.

Explanation: The

y

-intercept of a regression line is the predicted value of the dependent variable when the independent variable is zero. Here, it is the height at

t=0

. However, the model was built using data from

t=5

t=30

. Predicting a value for

t=0

is an extrapolation, meaning we are making a prediction outside the range of our data. Such predictions are often unreliable.

A is incorrect because it fails to recognize the unreliability of the extrapolation.
C is incorrect because the intercept is a predicted starting value, not an observed minimum.
D is incorrect because it misinterprets the intercept as the slope (rate of change).

Question 20

Calculate the residual for this student.

-1.0 (correct answer)
1.0
12.0
80.0

Explanation: The residual is the difference between the observed value and the predicted value (residual = observed

y

- predicted

y

First, find the predicted score using the regression equation: $y_{predicted} = 0.85(80) + 12 = 68 + 12 = 80$ .
The observed score is $y_{observed} = 79$ .
Calculate the residual: $79 - 80 = -1.0$ .

B is incorrect due to a sign error (predicted - observed).
C is incorrect as it is the y-intercept of the model.
D is incorrect as it is the predicted value, not the residual.