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İKT379 / Exam graphs
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
X (income)Y (consumption)Y = β₀ + β₁Xuobserved (X, Y)
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
8012016020024060100140180weekly income X (dollars)weekly consumption Y (dollars)E(Y | X = 140) = 101PRF
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 uiu_i and eie_i 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
XYYᵢE(Y|Xᵢ)ŶᵢPRFSRFXᵢAuᵢeᵢ
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
8012016020024060100140180weekly income X (dollars)weekly consumption Y (dollars)SRF₁SRF₂
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
801201602002406090120150weekly income X (dollars)weekly consumption Y (dollars)SRFe₂ = −10.36e₉ = +8.36
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 YY graphically” (the slides’ decomposition-of-variance figure).

Diagram · Ordinary least squaresŶ = 24.45 + 0.509X, n = 10
XYȲSRFXᵢYᵢYᵢ − Ȳ (total)Ŷᵢ − Ȳ(explained)eᵢΣ(Yᵢ − Ȳ)² = Σ(Ŷᵢ − Ȳ)² + Σeᵢ²TSS = ESS + RSS
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
XYPRFX₁X₂X₃same spread σ² at every X
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
0uₜtime trun of u > 0run of u < 0
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
value of the estimateβ₂ (true)E(β̃₂)biasOLS: E(b₂) = β₂biased
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 r2r^2 with a Venn diagram.”

Diagram · The Ballentine: r² as overlap0 ≤ r² ≤ 1
YXESSRSSr² = ESS/TSS
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=0H_0: \beta_2 = 0 at 5% and show it on the t distribution.”

Diagram · t test and confidence intervaldf = 8, consumption example
t0−2.3062.306α/2 = 0.025α/2 = 0.0250.95do not reject H₀rejectreject
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\beta_2 and use it to test H0:β2=0.60H_0: \beta_2 = 0.60.”

Diagram · t test and confidence intervaldf = 8, consumption example
β₂0.400.450.500.550.600.650.4280.590b₂ = 0.5090.509 ± 2.306 × 0.035c = 0.55 insidedo not reject H₀c = 0.60 outsidereject H₀c = 0 is far to the left: reject H₀ (slope significant)
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
801001401702002402606080100120140160Weekly income X ($)Weekly consumption Y ($)(X̄=170, Ȳ=111)Narrowest at X̄CI width: ±4.7395% CI: [67.90, 82.84]Width: ±7.46
SRF: Ŷ = 24.45 + 0.5091X95% Confidence bandNarrowest at X̄ = 170

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\beta_1 as an asymptote. Draw the Phillips curve.”

Diagram · Reciprocal regression modelsY = β₁ + β₂(1/X)
200050001000015000200005082150250350Per capita GNP ($)Child mortality (deaths/1000)Asymptote β₁ ≈ 81.79Slope = −β₂/X² < 0Drop tapers off as PGNP rises
Reciprocal curve YAsymptote (limit β₁)

Reciprocal model CM = 81.79 + 27,273(1/PGNP). As per capita GNP increases indefinitely, 1/PGNP → 0 and child mortality falls toward horizontal asymptote β₁ ≈ 82 deaths per 1000. Slope = −β₂/X² < 0.

15 · Log-reciprocal S-shaped production function

Asked as: “Draw the log-reciprocal (log hyperbola) model. Show the inflection point and explain its economic meaning.”

Diagram · Log hyperbola / Logarithmic reciprocalln Y = β₁ − β₂(1/X)
124681020406080100Labor input X (or L)Output Y (or Q)Asymptote e^β₁ = 100Inflection point (X = β₂/2 = 2)Increasing returns(d²Y/dX² > 0)Diminishing returns(d²Y/dX² < 0)
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
04812162024240025002600270028002900Quarterly time t (1993:1 to 1998:3)Expenditure on services ($ billions)Log-lin (+0.743%/qtr)Linear trend (+$19.69B/qtr)t=0: $2413.9B
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 eie_i 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=1H_0: \beta_2 = 1, H1:β2<1H_1: \beta_2 \lt 1, α=0.05\alpha = 0.05, df = 8. Where do you shade, and what is the critical value?