Given fitted y = 2 + 1.5x1 - 0.5x2, find fitted y at x1 = 4 and x2 = 2.
Answer: ______________
Two predictors are nearly, but not exactly, linearly dependent. What problem can this cause?
- Imprecise coefficients sensitive to small data changes
- A necessary drop of training R² to zero in every fitted sample
- A necessary disappearance of every residual in the fitted model
In fitted y = 2 + 1.5x1 - 0.5x2, what does the coefficient 1.5 describe?
- A guaranteed causal increase of 1.5 in observed y whenever x1 rises
- An increase of 1.5 in fitted y regardless of how much x2 changes
- An increase of 1.5 in fitted y per unit of x1, holding x2 fixed
In fitted y = 8 + 4x1 - 2x2, x1 increases by 2 and x2 increases by 3. Find the change in fitted y.
Answer: ______________
An ordinary least-squares model includes an intercept and x1, x2 and x3. Every row satisfies x3 = 2x1 + x2. What follows?
- All three predictor coefficients must equal zero in every least-squares fit
- The design columns must be independent because three predictors are included
- The coefficients are not uniquely determined
- No fitted response can be calculated because no least-squares solution exists
An observation has unusual predictor values, but removing it changes the fitted model very little. Which conclusion is supported?
- Unusual predictor values alone do not establish influence
- Every unusual predictor combination must be deleted before fitting a model
- The small change proves that the observation has the largest residual
- The small change proves that all predictor coefficients are exactly zero
Removing one observation substantially changes several fitted coefficients. What does this show about that observation?
- It must have the largest residual among all observations
- It must make two predictors exactly linearly dependent
- It is influential for this fitted model
An ordinary least-squares model has an intercept and residuals summing to zero. That fact alone establishes that the model is appropriate.
Circle one: True False
An additive regression fitted to observational data has a positive x1 coefficient. Does this alone establish that intervening to raise x1 will raise y?
- No; the coefficient describes a conditional association
- Yes; holding measured predictors fixed removes every possible source of bias
- Yes; the sign of a fitted coefficient establishes the direction of a causal effect
A team repeatedly adds predictors and selects its model using training R² alone. Which risk has it not checked?
- Every added predictor must reduce training fit and increase prediction error
- The model may fit sample noise and predict poorly on new data
- Predictor collinearity must vanish once training R² becomes sufficiently high