Module M02 Quiz: Simple Linear Regression
18 questions · Graduate (ECMT5001) · MCQ, calculation, short answer
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Question 1
Lesson L02 · OLS Objective
OLS estimation minimises which quantity?
Type: Multiple Choice
- A) \(\sum_{i=1}^n \hat{u}_i\)
- B) \(\sum_{i=1}^n |\hat{u}_i|\)
- C) \(\sum_{i=1}^n \hat{u}_i^2\)
- D) \(\max_i |\hat{u}_i|\)
Show Answer
Answer: C
OLS minimises the sum of squared residuals: \(\text{SSR} = \sum_{i=1}^n \hat{u}_i^2 = \sum_{i=1}^n (y_i - \hat{\beta}_0 - \hat{\beta}_1 x_i)^2\). Squaring penalises large deviations more heavily than small ones, and makes the first-order conditions linear — yielding closed-form estimators.
Question 2
Lesson L02 · OLS Computation
[Calculation] Given \(x = \{1,2,3,4,5\}\) and \(y = \{2,4,5,4,6\}\), compute \(\hat{\beta}_1\) using: $\(\hat{\beta}_1 = \frac{\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^n (x_i - \bar{x})^2}\)$
Type: Calculation
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\(\bar{x} = 3\), \(\bar{y} = 4.2\)
| \(i\) | \(x_i - \bar{x}\) | \(y_i - \bar{y}\) | product | \((x_i-\bar{x})^2\) |
|---|---|---|---|---|
| 1 | \(-2\) | \(-2.2\) | \(4.4\) | \(4\) |
| 2 | \(-1\) | \(-0.2\) | \(0.2\) | \(1\) |
| 3 | \(0\) | \(0.8\) | \(0\) | \(0\) |
| 4 | \(1\) | \(-0.2\) | \(-0.2\) | \(1\) |
| 5 | \(2\) | \(1.8\) | \(3.6\) | \(4\) |
| Sum | 8.0 | 10 |
Question 3
Lesson L02 · OLS Intercept
[Calculation] Using the result from Q2 (\(\hat{\beta}_1 = 0.8\), \(\bar{x}=3\), \(\bar{y}=4.2\)), compute \(\hat{\beta}_0\).
Type: Calculation
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Fitted line: \(\hat{y} = 1.8 + 0.8x\). This passes through the sample means \((\bar{x}, \bar{y}) = (3, 4.2)\) — a guaranteed property of OLS.
Question 4
Lesson L03 · Residual Properties
Which of the following is NOT a guaranteed algebraic property of OLS residuals?
Type: Multiple Choice
- A) \(\sum_{i=1}^n \hat{u}_i = 0\)
- B) \(\sum_{i=1}^n x_i \hat{u}_i = 0\)
- C) \(\sum_{i=1}^n \hat{u}_i^2 = 0\)
- D) \(\bar{y} = \hat{\bar{y}}\) (mean of fitted values equals mean of \(y\))
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Answer: C
\(\sum \hat{u}_i^2 = \text{SSR} > 0\) unless the fit is perfect. The true OLS properties are:
- \(\sum \hat{u}_i = 0\) (from intercept FOC)
- \(\sum x_i \hat{u}_i = 0\) (from slope FOC — residuals orthogonal to \(x\))
- \(\bar{\hat{y}} = \bar{y}\) (consequence of the first property)
Question 5
Lesson L04 · R-squared
A regression of hourly wage on years of education gives \(R^2 = 0.35\). The best interpretation is:
Type: Multiple Choice
- A) Education causes 35% of the variation in wages
- B) 35% of the variation in wages is explained by the linear relationship with education
- C) The regression fits the data well
- D) Education is statistically significant at the 35% level
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Answer: B
\(R^2 = \text{SSE}/\text{SST}\) measures the proportion of total variation in \(y\) explained (in a statistical, not causal, sense) by the regression. 35% is fairly typical for simple wage regressions. Option A is wrong — \(R^2\) says nothing about causality. Option C is misleading — whether 35% is "good" depends on the application.
Question 6
Lesson L04 · SST decomposition
[Calculation] \(\text{SST} = 200\), \(\text{SSE} = 150\). Compute \(\text{SSR}\) and \(R^2\).
Type: Calculation
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Equivalently: \(R^2 = 1 - \text{SSR}/\text{SST} = 1 - 50/200 = 0.75\). The regression explains 75% of the variation in \(y\).
Question 7
Lesson L05 · SLR Assumptions
SLR Assumption 1 (linearity in parameters) requires:
Type: Short Answer
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The population model is linear in the parameters: \(y = \beta_0 + \beta_1 x + u\). The coefficients \(\beta_0, \beta_1\) enter linearly. This does not require \(y\) and \(x\) to have a linear relationship — we can use \(\ln(y)\) or \(x^2\) as variables and still satisfy SLR.1. What cannot be linear in \(\beta\) is something like \(y = e^{\beta_1 x} + u\) (a nonlinear model that requires nonlinear least squares).
Question 8
Lesson L05 · SLR Assumptions
Why is SLR Assumption 4 — the zero conditional mean \(E(u \mid x) = 0\) — considered the key identification assumption?
Type: Short Answer
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\(E(u \mid x) = 0\) means the error term is mean-independent of \(x\): everything in \(u\) that affects \(y\) is uncorrelated with \(x\) on average. This is needed for unbiasedness — if violated (e.g. \(x\) is correlated with an omitted variable in \(u\)), OLS is biased. In practice, SLR.4 is the hardest assumption to justify with observational data. Randomised experiments guarantee it by design (random assignment makes \(x\) independent of \(u\)). With observational data, credible strategies (IV, DiD, controls) are needed.
Question 9
Lesson L05 · Unbiasedness
\(E(\hat{\beta}_1) = \beta_1\) (OLS is unbiased for the slope) requires which SLR assumptions?
Type: Multiple Choice
- A) SLR.1 only
- B) SLR.1, SLR.2, SLR.3, and SLR.4 (not SLR.5)
- C) All five assumptions SLR.1–SLR.5
- D) SLR.4 and SLR.5 only
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Answer: B
Unbiasedness requires SLR.1 (linearity), SLR.2 (random sampling), SLR.3 (variation in \(x\)), and SLR.4 (zero conditional mean). SLR.5 (homoskedasticity) is not needed for unbiasedness — only for the Gauss-Markov result (OLS is BLUE) and for standard error formulas to be correct.
Question 10
Lesson L05 · Variance of OLS
\(\text{Var}(\hat{\beta}_1) = \sigma^2 / \sum(x_i - \bar{x})^2\). Which statements are correct?
Type: Multiple Choice
- A) Variance decreases as \(n\) increases; variance decreases as spread of \(x\) increases
- B) Variance increases as \(n\) increases; variance increases as spread of \(x\) increases
- C) Variance is unaffected by sample size
- D) Variance decreases as \(\sigma^2\) increases
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Answer: A
As \(n \to \infty\), \(\sum(x_i - \bar{x})^2 \to \infty \Rightarrow \text{Var}(\hat{\beta}_1) \to 0\) (consistent estimator). As the spread of \(x\) increases (more variation in the regressor), \(\sum(x_i-\bar{x})^2\) increases \(\Rightarrow\) lower variance. Intuitively: more variation in \(x\) gives more "information" to pin down the slope. This is why experimenters often spread \(x\) values across the design space.
Question 11
Lesson L05 · Error Variance
Why does the unbiased estimator of \(\sigma^2\) use \(n-2\) as the divisor: \(\hat{\sigma}^2 = \text{SSR}/(n-2)\)?
Type: Multiple Choice
- A) Because we have 2 regressors
- B) Because estimating \(\beta_0\) and \(\beta_1\) consumes 2 degrees of freedom, leaving \(n-2\) free residuals
- C) Convention — \(n-1\) would work equally well
- D) Because the formula applies only to samples of size \(\geq 2\)
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Answer: B
Each estimated coefficient "uses up" one degree of freedom. With 2 parameters estimated (\(\hat{\beta}_0, \hat{\beta}_1\)), only \(n-2\) residuals are "free". Using SSR/\((n-2)\) makes \(E(\hat{\sigma}^2) = \sigma^2\) (unbiased). In multiple regression with \(k\) slope coefficients plus intercept: \(\hat{\sigma}^2 = \text{SSR}/(n-k-1)\).
Question 12
Lesson L01 · Log Interpretation
The estimated regression is \(\ln(\text{wage}) = 1.2 + 0.08 \cdot \text{educ}\). The coefficient 0.08 means:
Type: Multiple Choice
- A) An additional year of education raises wages by $0.08
- B) An additional year of education raises wages by approximately 8%
- C) An additional year of education raises log-wages by 8 units
- D) Wages increase by 8 times for each year of education
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Answer: B
In a log-level model, the coefficient approximates the percentage change: \(100 \times 0.08 = 8\%\) wage increase per additional year of education. More precisely: \(e^{0.08} - 1 \approx 8.33\%\). The approximation \(\approx \beta_1 \times 100\%\) is accurate for small \(\beta_1\) (say, \(|\beta_1| < 0.2\)).
Question 13
Lesson L04 · Scale Effects
If we replace \(y\) with \(100y\) in the regression (e.g., convert from dollars to cents), what happens to \(\hat{\beta}_1\) and \(R^2\)?
Type: Multiple Choice
- A) \(\hat{\beta}_1\) is multiplied by 100; \(R^2\) is unchanged
- B) Both \(\hat{\beta}_1\) and \(R^2\) are multiplied by 100
- C) \(\hat{\beta}_1\) is divided by 100; \(R^2\) is unchanged
- D) \(R^2\) is multiplied by 100; \(\hat{\beta}_1\) is unchanged
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Answer: A
Scaling \(y\) by \(c\) scales \(\hat{\beta}_1\) by \(c\) (since \(\hat{\beta}_1 = \hat{\text{Cov}}(x,y)/\hat{\text{Var}}(x)\) and scaling \(y\) scales the numerator). \(R^2\) is scale-invariant — it measures the proportion of variance explained, which is unchanged by a constant multiplicative transformation of either variable.
Question 14
Lesson L05 · Gauss-Markov
OLS is BLUE under SLR.1–SLR.5. What does BLUE stand for and what assumption is most critical?
Type: Short Answer
Show Answer
Best Linear Unbiased Estimator. Among all linear unbiased estimators of \(\beta_1\), OLS has the smallest variance.
The critical assumption is SLR.5 (homoskedasticity): \(\text{Var}(u \mid x) = \sigma^2\) (constant). If violated (heteroskedasticity), OLS is still unbiased but no longer efficient — weighted least squares (WLS) would beat it. Gauss-Markov requires SLR.1–SLR.5; it does not require normality (SLR.6). For valid t/F tests with finite samples, normality is additionally required.
Question 15
Lesson L02 · CEF with Dummy Regressor
[Short Answer] If \(x \in \{0,1\}\) (binary), show that the OLS estimate gives \(\hat{\beta}_1 = \bar{y}_1 - \bar{y}_0\) (difference in group means).
Type: Short Answer
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Let \(n_1\) = number with \(x=1\), \(n_0\) with \(x=0\), \(n = n_0 + n_1\).
\(\bar{x} = n_1/n\). For \(x_i \in \{0,1\}\): \(\sum(x_i - \bar{x})^2 = n_1(1-\bar{x})^2 + n_0(0-\bar{x})^2 = n_1 n_0^2/n^2 + n_0 n_1^2/n^2 = n_1 n_0/n\).
\(\sum(x_i-\bar{x})(y_i-\bar{y}) = \sum_{x_i=1}(1-\bar{x})(y_i-\bar{y}) = (1-\bar{x})\sum_{x_i=1}y_i - (1-\bar{x})n_1\bar{y}\)
After algebra: \(= (n_0/n)(n_1\bar{y}_1) - (n_0/n)n_1\bar{y} = (n_1 n_0/n)(\bar{y}_1 - \bar{y})\).
Since \(\bar{y} = (n_1\bar{y}_1 + n_0\bar{y}_0)/n\), we get \(\bar{y}_1 - \bar{y} = n_0(\bar{y}_1-\bar{y}_0)/n\).
Therefore: \(\hat{\beta}_1 = \frac{(n_1n_0/n)(\bar{y}_1-\bar{y})}{n_1n_0/n} = \bar{y}_1 - \bar{y}_0\). \(\square\)
Question 16
Lesson L02 · t-Test — MIT 14.32 style
[Calculation] A wine price regression: \(\hat{\beta}_1 = 0.032\) (effect of vintage year on \(\ln(\text{price})\)), \(\text{se}(\hat{\beta}_1) = 0.012\), \(n = 25\). Test \(H_0: \beta_1 = 0\) at the 5% level. Critical value \(t_{0.025, 23} = 2.069\).
Type: Calculation
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Since \(|t| = 2.67 > t_{0.025,23} = 2.069\), reject \(H_0\) at 5%. Wine prices are significantly related to vintage year. Each additional year is associated with approximately \(100 \times 0.032 = 3.2\%\) higher price.
Question 17
Lesson L04 · Confidence Interval
[Calculation] \(\hat{\beta}_1 = 1.5\), \(\text{se}(\hat{\beta}_1) = 0.3\), \(n = 50\), \(k = 1\) slope. Construct a 95% CI for \(\beta_1\). Use \(t_{0.025,48} \approx 2.011\).
Type: Calculation
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Interpretation: If we repeated this sampling procedure many times and constructed a CI each time, 95% of such intervals would contain the true \(\beta_1\).
Question 18
Lesson L02 · Prediction
[Calculation] Using \(\hat{y} = 1.8 + 0.8x\) from Q3: for observation \(i=3\) with \(x_3 = 3\) and \(y_3 = 5\), find the fitted value \(\hat{y}_3\) and residual \(\hat{u}_3\).
Type: Calculation
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The positive residual means the actual wage is 0.8 units above the predicted value for this observation.