Correlation and regression flashcards

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Bivariate data
Pairs of values of two variables measured on each member of a sample, e.g. the height and arm span of each student.
Positive correlation
As one variable increases, the other tends to increase: the points on a scatter diagram rise from left to right.
Product moment correlation coefficient r
A measure of the strength and direction of the linear relationship between two variables; −1 ≤ r ≤ 1.
Least-squares regression line of y on x
The straight line that makes the sum of the squares of the vertical distances (residuals) from the points to the line as small as possible.
Explanatory and response variables
Explanatory (independent): the variable that is set or changes on its own, plotted as x. Response (dependent): the variable that depends on it, plotted as y.
Interpolation and extrapolation
Interpolation: estimating y for an x inside the range of the data. Extrapolation: for an x outside it — unreliable.
Residual
Observed y minus the value of y on the regression line, for the same x: the vertical distance from the point to the line.
Sxx and Sxy from summary sums
Sxx = Σx² − (Σx)²/n; Sxy = Σxy − (Σx)(Σy)/n. (Yes)
Product moment correlation coefficient
r = Sxy / √(Sxx Syy). (Yes)
Gradient and intercept of the regression line of y on x
b = Sxy / Sxx and a = ȳ − b x̄, giving y = a + bx. (Yes)
Point that every regression line of y on x passes through
The mean point (x̄, ȳ), because a = ȳ − b x̄ rearranges to ȳ = a + b x̄. (No — derive it)
S values after coding u = (x − p)/q, v = (y − s)/t
Suu = Sxx/q², Svv = Syy/t², Suv = Sxy/(qt). The constants p and s have no effect. (No — learn it)
Effect of coding on r
r is unchanged by adding/subtracting constants and by multiplying/dividing by positive constants; a negative multiplier reverses its sign. (No — learn it)
Scatter diagram with r close to −1
Points lie close to a straight line falling from left to right: strong negative linear correlation.
Points on a U-shaped curve, symmetric about the middle x-value
A strong (perfect) relationship but r = 0: the PMCC measures only straight-line association.
One point far below a tight upward line of points
An outlier (often a recording error): it lowers r a lot and pulls the regression line towards it.
Regression line extended beyond the data
The dashed continuation is extrapolation: e.g. a battery-level line eventually predicts negative percentages.
Coded scales drawn on the same scatter diagram
Coding only relabels the axes; the points and the pattern are unchanged, so r is unchanged (for positive scale factors).
Does a strong correlation show that x causes y?
No. Correlation is not causation: both may depend on a third variable (e.g. ice-cream sales and sunburn both rise in sunny weather).
Does r = 0 mean the variables are unrelated?
No — only that there is no linear correlation. There may be a strong curved relationship; look at the scatter diagram.
Why measure residuals vertically for the line of y on x?
y is the variable being predicted, so the errors that matter are in y, measured parallel to the y-axis.
Why not rearrange y = a + bx to estimate x from y?
The line of y on x minimises vertical distances only; estimating x needs the line of x on y, which is different unless r = ±1.
When does the intercept a have no practical meaning?
When x = 0 is impossible or far outside the data (e.g. hardness at 0 °C, fuel economy at 0 km/h): using it is extrapolation.
What makes an estimate from a regression line reliable?
It is interpolation (inside the data range) and r is close to ±1 (strong linear correlation).
n = 10, Σx = 308, Σx² = 9668, Σy = 2980, Σxy = 95270. Sxx and Sxy?
Sxx = 9668 − 308²/10 = 181.6; Sxy = 95270 − 308 × 2980/10 = 3486.
Sxx = 40, Syy = 90, Sxy = −48. Find r.
r = −48/√3600 = −48/60 = −0.8: fairly strong negative linear correlation.
Sxy = −252, Sxx = 23, x̄ = 3.5, ȳ = 61. Regression line of y on x?
b = −252/23 = −10.957…; a = 61 − (−10.957…)(3.5) = 99.35 (keep b unrounded); y = 99.3 − 11.0x.
Line y = 12.4 + 0.75x and x̄ = 20. Find ȳ.
The line passes through (x̄, ȳ): ȳ = 12.4 + 0.75 × 20 = 27.4.
u = (x − 40)/5, v = (y − 120)/3, Suu = 52, Suv = 61.5. Sxx and Sxy?
Sxx = 5² × 52 = 1300; Sxy = 5 × 3 × 61.5 = 922.5.
v = 8.4547 + 1.8104u, where u = (x − 800)/5 and v = y − 200. Line of y on x?
y − 200 = 8.4547 + 1.8104(x − 800)/5 → y = 208.4547 + 0.36208x − 289.664 → y = −81.2 + 0.362x.
Remove the pair (172, 117) from Σx² = 292876 and Σxy = 283512.
Σx² = 292876 − 172² = 263292; Σxy = 283512 − 172 × 117 = 263388 (and reduce n by 1).
r = 0.6, Sxx = 1600, Syy = 4.84. Find Sxy.
Sxy = r√(SxxSyy) = 0.6 × √7744 = 0.6 × 88 = 52.8.
Interpreting r in an answer
Give direction, strength and context: “strong positive correlation: days with more sunshine tend to generate more energy”.
Rounding b before finding a
Keep the full value of b in the calculator: rounding it first can change the third significant figure of a.
Interpreting b in context
“For each extra (unit of x), (y) increases/decreases by about b (units of y)” — name both variables and units.
Correcting one wrong data value
Change every sum that contains it: Σy, Σy² and Σxy (the Σxy change is x × the change in y). Σx and Σx² stay the same.
Checking your answers
r, Sxy and b share the sign of the trend; −1 ≤ r ≤ 1; the line passes through (x̄, ȳ).

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