Drawing the line that misses least

Computing · Machine Learning · ages 19-20

Name ______________________ Date ____________

Which calculation defines a residual?
- Predicted y minus observed y
- Observed y minus predicted y
- The unsigned distance between y and its prediction
For ordinary unweighted least squares with an intercept and varying x values, the fitted line passes through (mean x, mean y).
Circle one: True False
A fitted model is s hat = 52 + 6.4h, with h in study hours and s in exam points. What does 6.4 mean?
- Predicted score rises 6.4 points per additional hour
- The predicted score at zero study hours is 6.4 points
- Every student's observed score rises exactly 6.4 points
Two candidate lines fit the same two observations. Fit A has residuals -2 and 2. Fit B has residuals -1 and 3. Which has the smaller least-squares objective?
- Fit B, because its signed residual total is 2
- Fit A, because its squared residual total is 8
- Both fits, because each has two residuals to compare
- Neither fit, because negative residuals cannot be squared
A two-feature model is y hat = 10 + 2*x1 - 3*x2. A row has x1 = 4, x2 = 2 and observed y = 15. What are its prediction and residual?
- Prediction 18; residual -3
- Prediction 24; residual -9
- Prediction 12; residual 3
- Prediction 12; residual -3
Four practice sessions have (hours, points) pairs (1,1), (2,2), (3,2), (4,5). Find the slope of the ordinary unweighted least-squares line with an intercept. Give a decimal.
Answer: ______________
A house-price model uses size (m^2) and age (years). Their coefficients are 180 and -1200. Which gives the predicted changes when each feature rises by one, holding the other fixed?
- Size: +$180 per m^2; age: -$1200 per year
- Size: -$1200 per m^2; age: +$180 per year
- Size: +$180 per year; age: -$1200 per m^2
A two-feature least-squares fit has equations 3b0+b1+b2=8, b0+b1=3 and b0+b2=4, with a unique solution. A report claims b0=2, b1=1 and b2=2 are the fitted coefficients. Is the report correct?
Circle one: True False
Residuals stay centered near zero but form a wider fan as fitted values grow. What should you do before relying on uncertainty estimates that assume constant error spread?
- Set the fitted slope to zero so every prediction has exactly the same value
- Keep constant-spread estimates because a zero center proves equal spread
- Investigate the changing spread and use an uncertainty method that allows it
An observational study fits earnings against hours worked, with slope $12 per hour. A friend says this proves that working another hour causes earnings to rise by $12. What is the problem?
- Dollars per hour cannot be a valid slope unit for an earnings model
- It treats an association as proof of cause
- The intercept always measures the earnings caused by another hour

Answer key

For grown-ups. Fold this page away before handing over the rest.

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Observed y minus predicted y · A residual is observed minus predicted. Its sign distinguishes an observation above the prediction from one below it.
True · The intercept is a = mean y - b*mean x. Substituting mean x into a + bx therefore gives mean y.
Predicted score rises 6.4 points per additional hour · 6.4 multiplies hours, so it is the slope: 6.4 predicted points per additional hour. The intercept is 52 points.
Fit A, because its squared residual total is 8 · Fit A has total (-2)^2 + 2^2 = 8. Fit B has total (-1)^2 + 3^2 = 10. Least squares prefers the smaller squared total.
Prediction 12; residual 3 · The prediction is 10 + 2(4) - 3(2) = 12. Residual means observed minus predicted, so 15 - 12 = 3.
1.2 · Both means are 2.5. The products of paired deviations sum to 6, and the squared deviations in hours sum to 5. The slope is 6/5 = 1.2 points per hour.
Size: +$180 per m^2; age: -$1200 per year · At fixed age, one more m^2 adds $180 to predicted price. At fixed size, one more year subtracts $1200. These are model predictions, not causal guarantees.
False · The claimed coefficients fail the first equation: 3(2)+1+2=9, not 8. Solving the equations gives b0=1, b1=2 and b2=3.
Investigate the changing spread and use an uncertainty method that allows it · The fan suggests changing error variability. It does not by itself show that the fitted mean relationship is curved.
It treats an association as proof of cause · The fitted slope describes predicted change in the observational data. The coefficient alone does not establish what changing hours would cause.