11.4 Scatter diagrams, regression lines and correlation — Practise

Work through these on your own. For each prediction, say whether it is an interpolation or an extrapolation, and whether causation is justified.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A line of best fit passes through the points $(2, 10)$ and $(6, 22)$ . Find its gradient.
A regression line is $y = 3x + 4$ . Use it to predict $y$ when $x = 10$ .
A line of best fit passes through $(1, 20)$ and $(5, 4)$ . Find its gradient and state whether the correlation is positive or negative.
A line of best fit passes through $(0, 25)$ and $(5, 5)$ . Find its gradient and its intercept.
A regression line is $y = 2x + 5$ . Predict $y$ when $x = 4$ .
A scatter diagram has two distinct sections. The first section has line $y = 2x + 1$ and the second has line $y = 2x + 6$ . Find the difference between the two intercepts.
A regression line from data with $x$ between 1 and 12 is $y = 3x + 4$ . Predict $y$ when $x = 100$ , and state whether the prediction is an interpolation or an extrapolation.
The line $y = 2x + 1$ is drawn on a scatter diagram. The points are $(1, 2)$ , $(2, 3)$ , $(3, 9)$ and $(4, 8)$ . Find how many points lie above the line.
The points $(1, 2)$ , $(2, 4)$ , $(3, 6)$ , $(4, 8)$ and $(5, 10)$ all lie on the line $y = 2x$ . Find the mean of the $x$ -values and the mean of the $y$ -values.
A line of best fit is drawn through $(10, 30)$ and $(30, 50)$ . Find its gradient.
Two lines of best fit are $y = 2x + 5$ and $y = 2x + 1$ for two distinct sections of a population. Find the difference between their predictions at $x = 4$ .
The line of best fit for a scatter diagram is $y = 0.8x + 12$ . Predict $y$ when $x = 15$ . Say which variable is the explanatory variable if $x$ is the number of hours of sunshine and $y$ is the temperature.

Answers and concise solutions

1. The gradient

Answer: Gradient $= \dfrac{\text{change in } y}{\text{change in } x} = \dfrac{22 - 10}{6 - 2} = 3$ . The line rises by 3 for each extra unit of $x$ .

2. The predicted value

Answer: Substitute $x = 10$ : $y = 3(10) + 4 = 34$ . The prediction lies inside the range of the data.

3. The gradient

Answer: The gradient is $\dfrac{4 - 20}{5 - 1} = -4$ . It is negative, so the correlation is negative: as $x$ increases, $y$ tends to decrease.

4. The gradient and intercept

Answer: The gradient is $\dfrac{5 - 25}{5 - 0} = -4$ , and the line crosses the $y$ -axis at $(0, 25)$ , so the intercept is 25. The equation is $y = 25 - 4x$ .

5. The predicted value

Answer: Substitute $x = 4$ : $y = 2(4) + 5 = 13$ . The gradient 2 means each extra unit of $x$ adds about 2 to $y$ .

6. The difference

Answer: The intercepts are 1 and 6, so the difference is 5. The two sections sit at different levels, so one line for both would fit neither well.

7. The predicted value

Answer: The value is $y = 3(100) + 4 = 304$ . It is an extrapolation, because $x = 100$ lies far outside the range of the data. The trend may not continue, so the prediction is unreliable.

8. The number of points above the line

Answer: On the line, $y = 2x + 1$ gives 3, 5, 7 and 9 at $x = 1, 2, 3, 4$ . Only the point $(3, 9)$ has a larger $y$ -value than the line, so one point lies above the line.

9. The two means

Answer: The means are $\bar{x} = 3$ and $\bar{y} = 6$ . The point $(3, 6)$ lies on the line, which is why every line of best fit from these data passes through $(\bar{x}, \bar{y})$ .

10. The gradient

Answer: The gradient is $\dfrac{50 - 30}{30 - 10} = 1$ . Each extra unit of $x$ is associated with about one extra unit of $y$ .

11. The difference

Answer: The predictions are 13 and 9, so the difference is 4 in the predicted $y$ . The two sections need separate lines, because their levels differ.

12. The predicted value

Answer: Substitute $x = 15$ : $y = 0.8(15) + 12 = 24$ . The explanatory variable is the number of hours of sunshine, because it is used to explain or predict the temperature. A correlation between sunshine and temperature does not prove that sunshine causes a rise in temperature, although a mechanism is plausible.