Assignment 3

STAT 8561 | Fall 2026
Due date: Friday, October 30, 2026
Total: 40 points
Coverage: Chapter 4 and the introductory portions of Chapter 5.

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Questions 1-3 review Chapter 4. Question 4 introduces indicator coding and additive models from Chapter 5; the model and a reading guide are supplied. Complete Question 4 after those topics have been covered in class.

Question Topic Points
1 Model-comparison concepts 10
2 ANOVA and extra sums of squares 10
3 Reading a nested-model comparison 8
4 Guided introduction: two groups and one predictor 12
Total 40
NoteInstructions

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The PDF must be generated from the submitted source. HTML alone does not satisfy the submission requirement.

  • Put your name and course number at the beginning; label each question and subpart.
  • For multiple-choice items, enter exactly one letter. For True/False items, enter only T or F.
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Notation and assumptions. \(p\) counts all independent design columns, including the intercept. Use \(\alpha=0.05\). Assume fixed full-rank designs and independent \(N(0,\sigma^2)\) errors under the full model. Nested comparisons use the same response and observations, except the invalid comparison described in Question 1(e).

Question 1: What Model Comparisons Establish [10]

Select exactly one answer for each part. No written explanation is required.

(a) Adding a predictor [2]

A full OLS model adds an independent predictor column to a reduced model. Which statement must hold for the observed data?

A. \(\mathrm{SSE}_F<\mathrm{SSE}_R\) strictly.
B. \(\mathrm{SSE}_F\leq\mathrm{SSE}_R\), with equality possible.
C. \(\mathrm{SSE}_F>\mathrm{SSE}_R\) because one degree of freedom is lost.
D. The added predictor has a statistically significant coefficient.

B. The reduced-model space is contained in the full-model space. OLS minimizes SSE over that space, so \(\mathrm{SSE}_F\leq\mathrm{SSE}_R\). Equality is possible when the added direction does not improve the fitted response.

(b) The overall regression sum of squares [2]

The full model includes an intercept. Which reduced model gives \(\mathrm{SSE}_R-\mathrm{SSE}_F=\mathrm{SSR}_F\)?

A. Any model obtained by deleting one predictor.
B. The same full model written with different coefficient names.
C. The intercept-only model.
D. The model that fixes every fitted mean at zero.

C. The intercept-only fit is \(\bar Y\), so its SSE is \(\mathrm{SST}\). Hence \(\mathrm{SSE}_R-\mathrm{SSE}_F=\mathrm{SST}-\mathrm{SSE}_F=\mathrm{SSR}_F\).

(c) An overall rejection [2]

An overall F test rejects the null that all slopes are zero. What is the appropriate conclusion under the stated model assumptions?

A. There is evidence that at least one slope is nonzero.
B. Every slope is individually significant at the 5% level.
C. Every predictor has a causal effect on the response.
D. The fitted model has the lowest possible future prediction error.

A. Rejecting the joint null provides evidence that at least one slope is nonzero. It does not establish that every slope is significant, that the associations are causal, or that the model predicts best.

(d) Training fit [2]

Both nested models include an intercept and \(\mathrm{SST}>0\). What can happen when a predictor is added?

A. Training \(R^2\) must increase strictly.
B. Both \(R^2\) and adjusted \(R^2\) must increase.
C. Training \(R^2\) can decrease because the residual degrees of freedom decrease.
D. Training \(R^2\) cannot decrease, but adjusted \(R^2\) may decrease.

D. Since SSE cannot increase and SST is unchanged, training \(R^2=1-\mathrm{SSE}/\mathrm{SST}\) cannot decrease. Adjusted \(R^2\) also accounts for the loss of a residual degree of freedom, so it may decrease.

(e) A missing-data trap [2]

lm(y ~ x) uses 30 observations; lm(y ~ x + z) uses 26 because z is missing for four observations. What should be done before using the nested-model partial F formula?

A. Use both SSE values as printed; nested formulas are sufficient.
B. Refit both models to the same complete-case observations for y, x, and z.
C. Replace the four missing z values by zero without further justification.
D. Divide each SSE by its own sample size before subtracting them.

B. Refit both models using the same complete cases for y, x, and z. Otherwise, the two SSE values describe different response vectors; their difference does not isolate the contribution of z.

Question 2: ANOVA and Extra Sums of Squares [10]

Two models use the same \(n=24\) observations:

\[ \begin{aligned} M_R:\quad \mu_R(x)&=\alpha_0+\alpha_1x,\\ M_F:\quad \mu_F(x,z)&=\beta_0+\beta_1x+\beta_2z. \end{aligned} \]

The design ranks are \(p_R=2\) and \(p_F=3\). You are given

\[ \mathrm{SST}=120,\qquad \mathrm{SSE}_R=54,\qquad \mathrm{SSE}_F=42. \]

(a)-(h) Numerical answers [8]

Give one number per row. Each row is worth 1 point. Here \(\mathrm{SSR}_F\) uses the intercept-only model as its baseline.

Item Quantity Your answer
(a) Full-model \(\mathrm{SSR}_F\) _____
(b) Full-model error degrees of freedom _____
(c) Full-model \(\mathrm{MSE}_F\) _____
(d) Overall F statistic for \(H_0:\beta_1=\beta_2=0\) _____
(e) Full-model \(R^2\) _____
(f) Full-model adjusted \(R^2\) _____
(g) Extra sum of squares for adding \(z\) after \(x\) _____
(h) Partial F statistic for \(H_0:\beta_2=0\) _____

Formula reminders:

\[ \bar R_F^2=1-\frac{\mathrm{SSE}_F/(n-p_F)}{\mathrm{SST}/(n-1)},\qquad F_{\mathrm{partial}}=\frac{(\mathrm{SSE}_R-\mathrm{SSE}_F)/(p_F-p_R)}{\mathrm{SSE}_F/(n-p_F)}. \]

Item Answer Calculation
(a) 78 \(120-42\)
(b) 21 \(24-3\)
(c) 2 \(42/21\)
(d) 19.5 \((78/2)/2\)
(e) 0.65 \(1-42/120\)
(f) 0.617 \(1-(42/21)/(120/23)\)
(g) 12 \(54-42\)
(h) 6 \([(54-42)/(3-2)]/(42/21)\)

The overall test uses the two slopes jointly (\(2\) numerator degrees of freedom); the partial test adds only \(z\) (\(1\) numerator degree of freedom). Both use the full-model error mean square, \(\mathrm{MSE}_F=2\).

(i) Decision [2]

For part (h), the null reference distribution is \(F_{1,21}\), with 5% upper-tail critical value 4.325. Select one conclusion.

A. Fail to reject: the partial F statistic does not exceed 4.325.
B. Reject: there is evidence that \(z\) contributes to the mean after accounting for \(x\).
C. Reject: this establishes that every slope is individually significant.
D. The test cannot be performed without fitting the models in R.

B. The partial F statistic is \(6>4.325\), so reject \(H_0:\beta_2=0\) at the 5% level. There is evidence that \(z\) contributes to the mean after accounting for \(x\).

Question 3: Does Horsepower Add Information Beyond Weight? [8]

R’s mtcars data contain 32 automobile models. The response is fuel economy (mpg, miles per US gallon); predictors are weight (wt, in 1,000 lb) and gross horsepower (hp). The data are observational.

Figure 1 motivates the question: does horsepower contribute to the mean after accounting for weight?

Figure 1: Fuel economy and weight for the 32 automobile models in R’s mtcars data. Point colors identify horsepower bands (legend); horsepower is used as a continuous predictor in the models, not as these bands. The plot shows raw observations.

Model summaries (no R required).

\[ \begin{aligned} M_R:\quad \mu_R(w)&=\alpha_0+\alpha_w w,\\ M_F:\quad \mu_F(w,h)&=\beta_0+\beta_w w+\beta_h h. \end{aligned} \]

Both fits use the same \(n=32\) observations, with \(p_R=2\) and \(p_F=3\). In the full model, the test of \(\beta_h=0\) gives \(t_h=-3.518712\) and a two-sided p-value of 0.001451.

(a) The null hypothesis [2]

Which null tests whether horsepower adds information after accounting for weight?

A. \(\beta_w=0\); B. \(\beta_0=0\); C. \(\beta_h=0\); D. \(\beta_w=\beta_h=0\).

C. Setting \(\beta_h=0\) removes horsepower while retaining weight. Thus the null tests the additional contribution of horsepower after accounting for weight.

(b) Reference distribution [2]

Which is the null reference distribution for the partial F test comparing these models?

A. \(F_{2,29}\); B. \(F_{1,30}\); C. \(F_{2,30}\); D. \(F_{1,29}\).

D. There is one added parameter and \(32-3=29\) full-model residual degrees of freedom, giving \(F_{1,29}\) under the null.

(c) Connect the tests [2]

For this single restriction, \(F=t_h^2\). Report the partial F statistic to three decimal places.

12.381. For one restriction, the partial F statistic equals the square of the corresponding \(t\) statistic:

\[ F=t_h^2=(-3.518712)^2\approx12.381. \]

(d) What does the result establish? [2]

Enter T or F for each statement (1 point each).

Item Statement T/F
(i) At the 5% level, there is evidence that horsepower contributes to the mean after accounting for weight. _____
(ii) This result proves that changing horsepower causes a change in fuel economy. _____

(i) T; (ii) F. The p-value \(0.001451<0.05\) supports a horsepower contribution after accounting for weight. These are observational data, so the result alone does not establish a causal effect.

Question 4: Guided Introduction to Chapter 5 [12]

In a hypothetical plant-growth study, \(Y\) is plant height (cm), \(x\) is daily water volume (mL/day, observed from 8 to 12), and fertilizer is A or B. Define

\[ x_c=x-10,\qquad z=\begin{cases}0,&\text{fertilizer A},\\1,&\text{fertilizer B}.\end{cases} \]

The additive mean model and its fitted equation are

\[ \begin{aligned} \mu(x,z)&=\beta_0+\beta_x(x-10)+\beta_gz,\\ \widehat\mu(x,z)&=18+1.5(x-10)+4z. \end{aligned} \]

Reading guide. Substitute \(z=0\) for A and \(z=1\) for B. The additive model has a common slope because it contains no product term \((x-10)z\). A design row uses column order \((1,x_c,z)\). See Chapter 5: indicator coding and additive models.

(a) Group B’s population mean [2]

Which expression is \(\mu_B(x)\)?

A. \(\beta_0+\beta_x(x-10)\); B. \(\beta_0+\beta_g+\beta_x(x-10)\).
C. \(\beta_0+(\beta_x+\beta_g)(x-10)\); D. \(\beta_g+\beta_x x\).

B. Substituting \(z=1\) gives \(\mu_B(x)=\beta_0+\beta_g+\beta_x(x-10)\). The group coefficient shifts the intercept; the water slope remains \(\beta_x\).

(b) Design rows [2]

Which pair gives the rows for A at \(x=8\) and B at \(x=12\), in that order?

A. \((1,8,0)\) and \((1,12,1)\).
B. \((1,-2,1)\) and \((1,2,0)\).
C. \((1,-2,0)\) and \((1,2,1)\).
D. \((0,-2,0)\) and \((1,2,1)\).

C. The columns are the intercept, centered water volume \(x_c=x-10\), and the indicator for B. At \(x=8\), \(x_c=-2\) and A has \(z=0\); at \(x=12\), \(x_c=2\) and B has \(z=1\).

(c) Fitted means [2]

Give the fitted mean height at \(x=12\) for A and for B, in cm (1 point each).

A: 21 cm; B: 25 cm. At \(x=12\), the centered water volume is \(2\), so

\[ \widehat\mu_A(12)=18+1.5(2)=21,\qquad \widehat\mu_B(12)=18+1.5(2)+4=25. \]

(d) Adjusted group difference [2]

At the same water volume, what is the fitted B-minus-A mean difference, in cm? Give one number.

4 cm. At the same water volume, the common intercept and water terms cancel:

\[ \widehat\mu_B(x)-\widehat\mu_A(x) =[18+1.5(x-10)+4]-[18+1.5(x-10)]=4. \]

This is a difference in fitted means, not an exact difference between two individual plants.

(e) Interpret the slope [2]

Which statement correctly interprets the fitted coefficient 1.5?

A. Within either fertilizer group, one additional mL/day of water is associated with a 1.5-cm increase in fitted mean height.
B. Every individual plant grows exactly 1.5 cm when water increases by one mL/day.
C. At the same water volume, B has a fitted mean height 1.5 cm higher than A.
D. The water slope is 1.5 for A and \(1.5+4=5.5\) for B.

A. Holding fertilizer group fixed, one additional mL/day of water is associated with a 1.5-cm increase in fitted mean height. The additive model gives both groups the same slope; it does not prescribe every individual plant’s response.

(f) Connect to Chapter 4 [2]

Which null hypothesis and reduced model test the fertilizer contribution after accounting for water, within this additive model?

A. \(H_0:\beta_x=0\); reduced mean \(\alpha_0+\alpha_gz\).
B. \(H_0:\beta_0=0\); reduced mean \(\alpha_x(x-10)+\alpha_gz\).
C. \(H_0:\beta_x=\beta_g=0\); reduced mean \(\alpha_0\).
D. \(H_0:\beta_g=0\); reduced mean \(\alpha_0+\alpha_x(x-10)\).

D. Under \(H_0:\beta_g=0\), the fertilizer indicator is removed while centered water remains. The reduced mean is \(\alpha_0+\alpha_x(x-10)\); its coefficients are estimated again when the reduced model is fitted.