Exam graphs · Lessons 1–6 · every graph they can ask you to draw
Exam Graph Drill
Each econometrics graph the midterm can ask for, with a numbered drawing recipe, the mistakes that lose marks, and the live diagram to check your drawing against. Draw each one on paper first, then compare.
By the end you can
Draw every Lesson 1–6 graph from memory with full labels
Mark u and e, the conditional means, the rejection regions and the critical values in the right places
Write the one-sentence conclusion that goes under each graph
Rules for every graph
Practise like this: read the recipe, draw on paper, then open the diagram and compare.
1 · Deterministic vs stochastic model
Asked as: “Explain, with a graph, the difference between the mathematical and the econometric model of consumption.”
Diagram · Deterministic vs stochastic modelKeynesian consumption function
Y = β₀ + β₁Xdisturbance u
Econometric model Y = β₀ + β₁X + u: households with the same income consume different amounts, so the points scatter around the line. The vertical distance from a point to the line is the disturbance u (above the line u > 0, below u < 0).
2 · Conditional distributions and the PRF
Asked as: “What is the population regression function? Show it on a graph.”
Diagram · Population and sample regression functionsGujarati Tables 2.1, 2.4, 2.5
PRF E(Y | Xᵢ) = 17 + 0.6Xᵢconditional mean
Gujarati's 60 families (Table 2.1). At each income X there is a whole distribution of consumption Y. Its average is the conditional mean E(Y | Xᵢ) (big dots: 65, 77, …, 173). The line joining the conditional means is the population regression function, PRF: E(Y | Xᵢ) = 17 + 0.6Xᵢ.
3 · PRF vs SRF with uᵢ and eᵢ
Asked as: “Show ui and ei for one observation on a graph” (Gujarati’s figure 2.5). A favourite.
Diagram · Population and sample regression functionsGujarati Tables 2.1, 2.4, 2.5
PRFSRFuᵢ (to the PRF)
Gujarati's figure 2.5. For one observation A at Xᵢ: in terms of the PRF, Yᵢ = E(Y | Xᵢ) + uᵢ (uᵢ = vertical gap to the PRF); in terms of the SRF, Yᵢ = Ŷᵢ + eᵢ (eᵢ = vertical gap to the SRF). Here the SRF underestimates the true E(Y | Xᵢ), so eᵢ > uᵢ. Data example: A = (240, 189) gives u = 189 − 161 = 28 and, with Ŷ = 24.45 + 0.509X, e = 189 − 146.65 = 42.35.
4 · Two samples, two SRFs
Asked as: “Can we estimate the PRF exactly from a sample? Explain with a graph.”
Diagram · Population and sample regression functionsGujarati Tables 2.1, 2.4, 2.5
sample 1 · SRF₁sample 2 · SRF₂PRF (unknown)
Two random samples of 10 families from the same population (Tables 2.4 and 2.5) give two different SRFs: Ŷ = 24.45 + 0.509X and Ŷ = 17.17 + 0.576X. Neither is the true PRF; we can never be sure which is closer. That is sampling fluctuation.
5 · OLS residuals and the line through the means
Asked as: “Explain the least-squares principle with a graph” or “Show the properties of the OLS line.”
Diagram · Ordinary least squaresŶ = 24.45 + 0.509X, n = 10
OLS line (SRF)e > 0e < 0
Sample 1 (Table 2.4) with the OLS line Ŷ = 24.45 + 0.509X. Each residual eᵢ = Yᵢ − Ŷᵢ is the vertical distance from the point to the line: green above the line (e > 0), red below (e < 0). Positive and negative residuals cancel: Σeᵢ = 0.
6 · TSS = ESS + RSS for one point
Asked as: “Decompose the variation of Y graphically” (the slides’ decomposition-of-variance figure).
Diagram · Ordinary least squaresŶ = 24.45 + 0.509X, n = 10
SRFtotalexplainedresidual
For each observation the total deviation from the mean, Yᵢ − Ȳ, splits into the part the line explains, Ŷᵢ − Ȳ, plus the residual eᵢ. Square and sum over all i: TSS = ESS + RSS. Data example: Yᵢ = 155 at Xᵢ = 240 gives 44 = 35.65 + 8.36; for the whole sample 8890 = 8552.73 + 337.27.
7 · Homoscedasticity vs heteroscedasticity
Asked as: “Explain the assumption of homoscedasticity with a graph” (Gujarati’s figures 3.4 and 3.5).
Diagram · What the CLRM assumes about uᵢdraw these on the exam
PRF E(Y | Xᵢ) = β₁ + β₂Xᵢf(u | Xᵢ)
Homoscedasticity (equal spread): Var(uᵢ | Xᵢ) = σ². The conditional distributions have the SAME width at every X, so the points form a band of constant thickness around the PRF.
8 · Autocorrelation
Asked as: “What does no autocorrelation mean? Show it.”
Diagram · What the CLRM assumes about uᵢdraw these on the exam
uₜ over time
Positive autocorrelation (violation): a positive u tends to be followed by another positive u, so the series moves in long runs above and below zero. (Negative autocorrelation would flip sign almost every period.)
9 · Unbiased and efficient (Gauss–Markov)
Asked as: “Explain BLUE with the sampling distributions of the estimators.”
Diagram · Sampling distribution of b₂Gauss–Markov: OLS is BLUE
OLS b₂biased β̃₂
Repeat the sampling many times and b₂ takes a different value each time: it has a sampling distribution. OLS is unbiased: its distribution is centred on the true β₂, E(b₂) = β₂. A biased estimator β̃₂ is centred somewhere else; the gap is the bias.
10 · The Ballentine (r²)
Asked as: “Show r2 with a Venn diagram.”
Diagram · The Ballentine: r² as overlap0 ≤ r² ≤ 1
Y (TSS)Xexplained (ESS)
Overlap (shaded) = the part of Y's variation explained by X (ESS). The rest of the Y circle is unexplained (RSS). r² = shaded area / whole Y circle = ESS/TSS.
11 · The t test: rejection regions
Asked as: “Test H0:β2=0 at 5% and show it on the t distribution.”
Diagram · t test and confidence intervaldf = 8, consumption example
t distribution, 8 dfrejection regiondo-not-reject region
H₀: β₂ = c vs H₁: β₂ ≠ c at α = 5%, df = n − 2 = 8. Put α/2 = 0.025 in EACH tail: critical values ±2.306. If the computed t falls in a red tail (|t| > 2.306), reject H₀; in the middle 95% region, do not reject.
12 · The confidence interval
Asked as: “Construct the 95% interval for β2 and use it to test H0:β2=0.60.”
Diagram · t test and confidence intervaldf = 8, consumption example
95% intervalinside: do not rejectoutside: reject
The 95% interval for β₂ is 0.509 ± 2.306 × 0.035 = [0.428, 0.590]. The confidence-interval approach (two-sided only): c = 0.55 lies inside → do not reject; c = 0.60 lies outside → reject. 0 is far outside → slope significant.
13 · Prediction confidence band around the SRF
Asked as: “Draw the sample regression line and its 95% confidence band for mean prediction. Explain why the band is shaped the way it is.”
Diagram · Mean prediction & confidence bandTable 2.4 consumption regression
95% confidence band for E(Y|X₀): Ŷ₀ ± t₀.₀₂₅ × se(Ŷ₀). Because se(Ŷ₀) has (X₀ − X̄)², the band is narrowest at the mean income X̄ = 170 (se = 2.05) and widens hyperbolically as X departs from X̄.
14 · Reciprocal models and the Phillips curve
Asked as: “Sketch the reciprocal model and explain the role of β1 as an asymptote. Draw the Phillips curve.”
S-shaped output curve YAsymptote exp(β₁) = 100Inflection point X = β₂/2
Logarithmic reciprocal model ln Y = β₁ − β₂(1/X), or Y = exp(β₁ − β₂/X). At first Y increases at an increasing rate, reaches an inflection point at X = β₂/2, then increases at a decreasing rate toward the asymptote exp(β₁). Models short-run production.
16 · Log-lin growth vs linear trend
Asked as: “Compare a linear trend model with a constant growth (log-lin) model on a diagram.”
Diagram · Growth model vs linear trendServices expenditure 1993:1–1998:3
Linear trend (+$19.69B/qtr)Log-lin model (+0.743%/qtr)
Expenditure on services 1993:1-1998:3. Linear trend: Ŷ = 2405.85 + 19.69t (constant $19.69B per quarter). Log-lin model: Ŷ = 2413.90 × e^(0.00743t) (constant quarterly growth 0.743%, compound 0.746%).
Final self-test
Check your understanding
On the PRF–SRF graph, the residual ei of point A is the vertical distance from A to which line?
Check your understanding
You draw three sideways bell curves along the PRF that get wider as X rises. Which assumption is violated?
Check your understanding
H0:β2=1, H1:β2<1, α=0.05, df = 8. Where do you shade, and what is the critical value?