Bivariant Data
Bivariate Data in NSW Year 10 Core maths looks more closely at relationships between two variables (NESA outcome MA5‑DAT‑C‑02). It separates the independent variable from the dependent variable, and an association between two variables from one of them actually causing the other.
You will decide which situations involve bivariate data, identify independent and dependent variables, and explain why two quantities that rise together, such as umbrella sales and traffic delays, need not be cause and effect. You will also use the equation of a line of best fit to make predictions, explain what its gradient and intercept mean in context, and decide whether a prediction is interpolation or extrapolation. It is part of Year 10 Core Statistics.
Every question with a fully worked solution.
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Theory
Bivariate data pairs two numerical variables, such as temperature and drinks sold. This Year 10 maths page shows how to choose the independent and dependent variables, why an association is not the same as cause and effect, and how to use and interpret the equation of a line of best fit, including when its predictions can be trusted.
One-variable (univariate) data records one value for each item, such as the heights of a class. Bivariate data records two values for each item, such as the height and shoe size of each student.
The independent variable is the one that may affect the other; it goes on the horizontal axis. The dependent variable is the one that may respond; it goes on the vertical axis.
Two variables are associated if they change together. A causal relationship means a change in one actually produces a change in the other. An association can happen by chance, or because a third (lurking) variable affects both.
A line of best fit can be written as an equation \(y = c + mx\) and used as a model to make predictions.
A line of best fit and its gradient:
How to use a line of best fit equation
- To predict \(y\): substitute the \(x\)-value and calculate.
- To find \(x\): substitute the \(y\)-value and solve the equation.
- Check the range: inside the data is interpolation; outside is extrapolation and may be unreliable. Say why in context.
How to find the equation from two points on the line
- Gradient: \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\).
- Intercept: substitute one point into \(y = c + mx\) and solve for \(c\).
- Write the equation using the variables in the question, such as \(D = -12 + 3T\).
How to compare two models
- Substitute the same value into both.
- To find when they agree, make the two expressions equal and solve.
(i) Is this one-variable or bivariate data?
(ii) Which is the independent variable?
(iii) Can the owner say that cold weather causes more hot-drink sales?
(i) Two numerical values are recorded for each day, so it is bivariate data.
(ii) The temperature may affect the sales, not the other way around. The independent variable is the temperature; the number of drinks sold is the dependent variable. The temperature goes on the horizontal axis.
(iii) A scatter plot can only show an association. Cold weather is a likely cause, but other factors, such as school holidays or a nearby event, could also change the sales. Proving cause needs a controlled experiment.
(i) Predict the height of a \(10\)-year-old tree.
(ii) Predict the age of a tree that is \(7.9\) m tall.
(iii) What do \(0.8\) and \(1.5\) mean?
| \(\text{(i) } H\) | \(=\) | \(1.5 + 0.8 \times 10 = \mathbf{9.5}\textbf{ m}\) |
| \(\text{(ii) } 7.9\) | \(=\) | \(1.5 + 0.8A\) |
| \(6.4\) | \(=\) | \(0.8A\) |
| \(A\) | \(=\) | \(\mathbf{8}\textbf{ years}\) |
(iii) The gradient \(0.8\): a tree grows about \(0.8\) m taller each year. The intercept \(1.5\) is the height the model predicts at age \(0\). Age \(0\) is outside the data (\(2\) to \(15\) years), so \(1.5\) m is only a model value, not a real height.
(i) Find its equation.
(ii) On a \(26\)°C day \(70\) drinks were sold. By how much is the prediction out?
(i) Gradient first, then substitute a point to find the intercept.
| \(m\) | \(=\) | \(\dfrac{78 - 48}{30 - 20} = 3\) |
| \(48\) | \(=\) | \(c + 3 \times 20\) |
| \(c\) | \(=\) | \(-12\) |
| \(D\) | \(=\) | \(\boldsymbol{-12 + 3T}\) |
(ii) Predict, then compare with the real value.
| \(D\) | \(=\) | \(-12 + 3 \times 26 = 66\) |
| \(70 - 66\) | \(=\) | \(\mathbf{4}\) |
The prediction is \(\mathbf{4}\) drinks too low.
(i) Find each height after \(3\) weeks.
(ii) When do the models give the same height?
(iii) Comment on using B to predict the height after \(52\) weeks.
| \(\text{(i) A}\) | \(:\) | \(20 + 4 \times 3 = \mathbf{32}\textbf{ cm}\) |
| \(\text{B}\) | \(:\) | \(8 + 6 \times 3 = \mathbf{26}\textbf{ cm}\) |
(ii) Make the two expressions equal.
| \(20 + 4w\) | \(=\) | \(8 + 6w\) |
| \(12\) | \(=\) | \(2w\) |
| \(w\) | \(=\) | \(6\) |
| \(H\) | \(=\) | \(20 + 4 \times 6 = 44\) |
Both give \(\mathbf{44}\) cm after \(\mathbf{6}\) weeks.
(iii) B gives \(8 + 6 \times 52 = 320\) cm. That is extrapolation, far beyond \(10\) weeks; plants slow down, so the prediction is not reliable.
Common pitfalls
Frequently asked questions
What is bivariate data?
Data with two numerical values recorded for each person or item, such as height and weight. It is shown on a scatter plot.
How do you choose the independent variable?
Ask which variable might affect the other. That one is the independent variable and goes on the horizontal axis. The variable that responds is the dependent variable.
What is the difference between association and causation?
Association means two variables change together. Causation means one actually produces the change in the other. A strong association alone does not prove causation.
What does the gradient of a line of best fit mean?
It is the average change in the dependent variable for each increase of 1 in the independent variable, in the units of the question.
When is a prediction from a line of best fit unreliable?
When it is an extrapolation far outside the range of the data, or when the relationship is weak. The pattern may not continue.