Scatter diagrams and linear regressionEdexcel International A Level Maths: Revision notes

A scatter diagram plots paired data $(x,y)$ and shows whether there is a relationship between two variables. Look for the direction (positive or negative), the strength (how closely points follow a line) and whether the pattern is roughly linear. The explanatory (independent) variable $x$ is the one that is controlled or used to explain; the response (dependent) variable $y$ is the one being predicted. Example: in a study of fertiliser and plant height, the fertiliser $x$ is explanatory and the height $y$ is the response. Other letters may be used, such as $t$ for temperature or $s$ for sales.

The regression line of $y$ on $x$ is $y=a+bx$. It is the least squares line, the one that makes the sum of the squared vertical distances from the points to the line as small as possible. Derivations are not required. Using $S_{xx}=\sum x^2-\frac{(\sum x)^2}{n}$ and $S_{xy}=\sum xy-\frac{\sum x\sum y}{n}$: $b=\frac{S_{xy}}{S_{xx}},\qquad a=\bar{y}-b\bar{x}.$ The line always passes through $(\bar{x},\bar{y})$, so you can draw it by plotting that point and one other. Example: $\sum x=30$, $\sum y=55$, $\sum x^2=220$, $\sum xy=386$, $n=5$ gives $S_{xx}=40$, $S_{xy}=56$, $b=1.4$, $\bar{x}=6$, $\bar{y}=11$, so $a=11-1.4\times6=2.6$ and $y=2.6+1.4x$.

Always interpret in the context of the question.

• The gradient $b$ is the change in the response for each one-unit increase in the explanatory variable. For $s=120+8.5t$: each 1 °C rise in temperature increases predicted sales by 8.50 pounds.
• The intercept $a$ is the predicted response when $x=0$. It only has meaning if $x=0$ is sensible and within the data. Regression of $y$ on $x$ is for predicting $y$ from $x$. It should not be rearranged to predict $x$ from $y$.

Substitute a value of $x$ into the regression line to predict $y$.

• Interpolation: $x$ lies inside the range of the data. The prediction is usually reliable.
• Extrapolation: $x$ lies outside the range of the data. The prediction may be unreliable, because the linear relationship may not continue. Example: a model fitted for temperatures 12 °C to 28 °C should not be used at 5 °C. A good answer says the value is outside the range, so it is extrapolation, and the relationship may not continue.

Data are sometimes coded, for example $u=\frac{x-40}{5}$, to make the numbers smaller. A regression line of $y$ on $u$ can be found as usual, then converted back by substituting the coding. Example: $y=89.4-7.2u$ with $u=\frac{x-40}{5}$ gives $y=89.4-1.44(x-40)=147-1.44x$. Notice that the gradient is divided by 5, the coding factor, and that the original units apply to the final line. Remember to interpret the gradient of the line in the original variables.