Reading a least-squares line
Finding the equation is the calculator's job. Interpreting it is yours, and that is where the marks are. These examples all use one practice dataset.
The data
Twelve students measured their arm span and height, in centimetres. This is a practice dataset made up for these examples.
| Arm span (cm) | 152 | 158 | 161 | 163 | 165 | 167 | 170 | 172 | 174 | 177 | 180 | 186 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Height (cm) | 156 | 156 | 162 | 160 | 166 | 166 | 168 | 173 | 171 | 176 | 177 | 183 |
A calculator gives the least-squares line, the correlation coefficient \(r\) and the coefficient of determination \(r^2\):
\[ \text{height} = 22.9 + 0.859 \times \text{arm span} \] \[ r = 0.978, \qquad r^2 = 0.957 \]Worked examples
Example 1: Which variable is which?
We are using arm span to predict height, so arm span is the explanatory variable and height is the response variable.
Explanatory: arm span. Response: height.
Example 2: Interpret the slope
Use the template: on average, [response] increases (or decreases) by [slope] [units] for each one [unit] increase in [explanatory].
On average, height increases by 0.859 cm for each 1 cm increase in arm span.
The words "on average" matter. The line describes the trend, not every student.
Example 3: Interpret the intercept
The intercept says a student with an arm span of 0 cm would have a height of 22.9 cm.
The intercept has no practical meaning here: an arm span of 0 cm is far outside the data and impossible.
Example 4: Describe the association
\(r = 0.978\) is positive and close to 1.
There is a strong, positive, linear association between arm span and height.
Strength, direction and form, in that order. Check the scatterplot to confirm the form really is linear.
Example 5: Interpret \(r^2\)
\(r^2 = 0.957\). Write it as a percentage and use the template.
About 95.7% of the variation in height can be explained by the variation in arm span.
Example 6: Interpolation or extrapolation?
Predict height for an arm span of 170 cm, and for 120 cm.
\[ 22.9 + 0.859 \times 170 = 168.93 \approx 168.9 \text{ cm} \] \[ 22.9 + 0.859 \times 120 = 126.0 \text{ cm} \]170 cm is inside the data (152 to 186 cm), so 168.9 cm is an interpolation and reasonably reliable. 120 cm is outside it, so 126.0 cm is an extrapolation and may not be reliable.