1
The scatter plot comes first.
Sign of r
+ or −
the direction of the slope
Size of |r|
0 to 1
how tightly the points hug a line
r = ±1
a perfect line
every point exactly on it
r measures linear association only. A perfect U-shape can give r near 0 while the
two variables are tied together completely — which is why the plot is drawn before r
is trusted.
2
Reading the strength.
| |r| | Usually described as |
|---|
| 0.00 – 0.29 | weak or none |
| 0.30 – 0.69 | moderate |
| 0.70 – 1.00 | strong |
Correlation is not causation, and the wording matters
“Hours of revision and exam marks are strongly positively correlated” is
safe. “Revising more makes marks go up” is a causal claim that a
correlation cannot support on its own. Three things can produce a correlation: a real
cause, a lurking variable driving both, or plain coincidence in a small
sample.
3
The regression line, and what its numbers mean.
ŷ = a + bx
ŷ = 38 + 4.2x
mark predicted from hours revised
b = 4.2
each extra hour adds 4.2 marks
a = 38
predicted mark at zero hours
x = 6 → ŷ = 38 + 25.2 = 63.2
a prediction, not a promise
“Least squares” names what the line does: of all the lines you could draw, it
is the one making the total of the squared vertical gaps as small as possible. The hat on
ŷ marks a predicted value rather than an observed one.
4
Two things the line must not be asked to do.
Extrapolate
predict outside the data range
the pattern may not continue
Swap the axes
predict x from y
a different line entirely
If nobody in the sample revised more than 10 hours, the line says nothing about 30
hours. Push it far enough and ŷ = 38 + 4.2(30) = 164 on a test marked out of 100
— an answer the arithmetic gives and the data never supported.
The intercept is often meaningless for the same reason. If nobody revised zero hours, the
38 is where the line happens to cross, not a measurement of anybody.
5
Reporting a result properly.
1. Plot the data
is a line even the right shape?
2. Quote r with a word
“strong positive”
3. Give the line and read b
in the units of the question
4. Predict inside the range only
and call it a prediction
5. Say nothing about cause
unless the data were experimental
6
Variations: a negative slope, x as the unknown, a residual.
The same ŷ = a + bx, read downhill, backwards, and against a real point.
Grade from absences: ŷ = 92 − 3.5x. Read the slope, then predict for 4 absences.
b = −3.5
each extra absence lowers the predicted grade by 3.5
ŷ = 92 − 3.5(4) = 92 − 14 = 78
substitute x = 4
A negative slope goes with a negative r; if the signs disagree, one of them is wrong.
Seedling height in cm after x weeks: ŷ = 45 + 6x. After how many weeks is 87 cm predicted?
87 = 45 + 6x
the y-value is given
6x = 42, so x = 7 weeks
check: 45 + 42 = 87 ✓
Solving the line for x is fine for a rough reading; a study built to predict x from y would fit its own line.
Same line: one seedling measured 71 cm at 5 weeks. Find its residual.
ŷ = 45 + 6(5) = 75
what the line predicts
residual = y − ŷ = 71 − 75 = −4
observed minus predicted
A negative residual means the point sits below the line; least squares makes the squares of these gaps as small as possible.
On the exam
Items ask you to describe r in words, interpret the slope and intercept of a given line in the question’s units, make a prediction, and spot causation or extrapolation claims. Read b as “per one more x”, substitute for predictions, and stay inside the data range. The 10-second check: the sign of b matches the sign of r, and a prediction lands on a sensible scale.
