A-Level Correlation and regression Practice Questions

Free A-Level Correlation and regression practice questions with full step-by-step worked solutions. Covers regression line, prediction, substitution, inverse prediction. Practise exam-style problems and check your method.

regression linepredictionsubstitutioninverse predictionrearrangegradient
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A cafe owner records data over several days. They record the daily maximum temperature (°C) as xx and the number of iced drinks sold as yy. The regression line of yy on xx is y=10+2xy = 10 + 2x, valid for 0≤x≤200 \le x \le 20. Use the model to predict yy when x=4x = 4.
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Worked solution

  1. Write down the regression line of y on x.

    y=10+2xy = 10 + 2x

    Begin from the given equation of the line of y on x.

  2. Substitute x = 4 into the equation.

    y=10+2×4y = 10 + 2 \times 4

    Replace x with the required value.

  3. Evaluate to obtain the predicted value.

    y=18y = 18

    This is the value of y predicted by the model.

Answer
y=18y = 18
Question 2
2 markseasy
A gym records member data. A strong correlation is found between the weekly training hours and the distance run in a test (km). Can we conclude that changes in xx cause the changes in yy?
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Worked solution

  1. Recall what correlation measures.

    correlation: linear association\text{correlation: linear association}

    Correlation only measures how points cluster about a line.

  2. Recall the causation principle.

    association≠cause\text{association} \ne \text{cause}

    Association does not establish a causal mechanism.

  3. Apply the principle to this context.

    possible lurking variable\text{possible lurking variable}

    A third factor could drive both variables.

Answer
No - a strong correlation does not by itself show that one variable causes the other
Question 3
3 marksintermediate
An economist studies towns. The regression line is y=8+5xy = 8 + 5x, with yy representing the savings rate (%) and xx representing the average income (£1000s). What does the intercept represent in this context?
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Worked solution

  1. Identify the intercept a of the line.

    a=8a = 8

    The intercept is the constant term.

  2. Interpret the intercept as the value at x = 0.

    x=0⇒y=ax = 0 \Rightarrow y = a

    It is the predicted y when x is zero.

  3. Express the intercept in the context.

    predicted y when x=0\text{predicted } y \text{ when } x = 0

    State its meaning using the real-world variables.

  4. Interpret the gradient in context.

    b=5b = 5

    For each extra unit of the average income (£1000s), the model changes the savings rate (%) by 5 percentage points.

  5. Interpret the intercept in context.

    a=8a = 8

    When the average income (£1000s) is 0 the model predicts the savings rate (%) = 8 percentage points.

  6. State the valid range of the explanatory variable.

    0≤x≤180 \le x \le 18

    Predictions are most trustworthy inside this range.

Answer
The predicted the savings rate (%) when the average income (£1000s) is 0, namely 8 percentage points
Question 4
5 markshard
A marketing team reviews weekly figures. They investigate how the units sold depends on the advertising spend (£100s). Plotting xx on the horizontal axis and yy on the vertical axis, which statement correctly identifies the variables?
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Worked solution

  1. Decide which variable is chosen or controlled.

    x=explanatoryx = \text{explanatory}

    The explanatory (independent) variable is placed on the x-axis.

  2. Decide which variable responds.

    y=responsey = \text{response}

    The response (dependent) variable is placed on the y-axis.

  3. Match the variables to this context.

    context→x,y\text{context} \to x, y

    Assign the roles using the wording of the problem.

  4. Interpret the gradient in context.

    b=1.5b = 1.5

    For each extra unit of the advertising spend (£100s), the model changes the units sold by 1.5 units.

  5. Interpret the intercept in context.

    a=5a = 5

    When the advertising spend (£100s) is 0 the model predicts the units sold = 5 units.

  6. State the valid range of the explanatory variable.

    5≤x≤205 \le x \le 20

    Predictions are most trustworthy inside this range.

  7. Distinguish interpolation from extrapolation.

    inside range→interpolation\text{inside range} \to \text{interpolation}

    Interpolation is reliable; extrapolation beyond the data is not.

  8. Identify the response variable.

    y is the response variabley \text{ is the response variable}

    The units sold responds to changes in the explanatory variable.

  9. Identify the explanatory variable.

    x is the explanatory variablex \text{ is the explanatory variable}

    The advertising spend (£100s) is the controlled/chosen variable.

  10. Note that correlation does not imply causation.

    correlation≠causation\text{correlation} \ne \text{causation}

    A linear relationship alone does not prove one variable causes the other.

Answer
The advertising spend (£100s) is the explanatory variable and the units sold is the response variable
Question 5
8 markschallenging
A shop tracks heaters in winter. They investigate how the number of heaters sold depends on the outside temperature (°C). Plotting xx on the horizontal axis and yy on the vertical axis, which statement correctly identifies the variables?
Show worked solution

Worked solution

  1. Decide which variable is chosen or controlled.

    x=explanatoryx = \text{explanatory}

    The explanatory (independent) variable is placed on the x-axis.

  2. Decide which variable responds.

    y=responsey = \text{response}

    The response (dependent) variable is placed on the y-axis.

  3. Match the variables to this context.

    context→x,y\text{context} \to x, y

    Assign the roles using the wording of the problem.

  4. Interpret the gradient in context.

    b=−2.2b = -2.2

    For each extra unit of the outside temperature (°C), the model changes the number of heaters sold by -2.2 heaters.

  5. Interpret the intercept in context.

    a=12a = 12

    When the outside temperature (°C) is 0 the model predicts the number of heaters sold = 12 heaters.

  6. State the valid range of the explanatory variable.

    6≤x≤346 \le x \le 34

    Predictions are most trustworthy inside this range.

  7. Distinguish interpolation from extrapolation.

    inside range→interpolation\text{inside range} \to \text{interpolation}

    Interpolation is reliable; extrapolation beyond the data is not.

  8. Identify the response variable.

    y is the response variabley \text{ is the response variable}

    The number of heaters sold responds to changes in the explanatory variable.

  9. Identify the explanatory variable.

    x is the explanatory variablex \text{ is the explanatory variable}

    The outside temperature (°c) is the controlled/chosen variable.

  10. Note that correlation does not imply causation.

    correlation≠causation\text{correlation} \ne \text{causation}

    A linear relationship alone does not prove one variable causes the other.

  11. Comment on the correlation coefficient.

    r=−0.75r = -0.75

    Values close to +1 or -1 indicate strong linear correlation.

  12. Give the answer to a sensible degree of accuracy.

    round to match the data\text{round to match the data}

    Do not quote more precision than the data supports.

  13. Restate the regression equation used.

    y=12−2.2xy = 12 - 2.2x

    This is the fitted line of y on x.

  14. Check the sign of the gradient.

    b<0b < 0

    The sign of the gradient agrees with the direction of the correlation.

  15. Reflect on the limitations of a linear model.

    relationship may be non-linear\text{relationship may be non-linear}

    The straight-line model may not hold outside the observed data.

Answer
The outside temperature (°c) is the explanatory variable and the number of heaters sold is the response variable

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