Inference for Regression Coefficients - Statistics Graduate Level

1. A regression model is written as $Y=\beta_0\mathbf 1+\beta_1x_1+\beta_2x_2+\varepsilon,$ but the observed design columns satisfy $x_2=2x_1$ exactly. Which slope functional can be tested using a generalized inverse in a way that is invariant to the particular least-squares solution selected?
   Answer: $\beta_1+2\beta_2$ When a design matrix is rank-deficient, individual parameters like $\beta_1$ and $\beta_2$ are not estimable — their values shift depending on which generalized inverse you use. The key concept here is estimability: a linear combination $c^\top\beta$ is estimable if and only if $c^\top$ lies in the row space of the design matrix $X$. Here, $x_2 = 2x_1$ exactly, so the column space of $X = [\mathbf{1},\ x_1,\ x_2]$ is spanned by only $\mathbf{1}$ and $x_1$. The row space of $X$ is generated by rows of the form $(1, x_{i1}, 2x_{i1})$, meaning any estimable contrast in $(\beta_0, \beta_1, \beta_2)$ must satisfy $c = (c_0, c_1, c_2)$ where $c_2 = 2c_1$. For slope-only contrasts ($c_0 = 0$), estimability requires $c_2 = 2c_1$. Choice A, $\beta_1 + 2\beta_2$, corresponds to $c = (0, 1, 2)$, satisfying $c_2 = 2c_1$. This lies in the row space, so it is estimable and invariant across all generalized inverse solutions — confirming A is correct. Choice B, $2\beta_1 + \beta_2$, gives $c = (0,2,1)$; here $c_2 = 1 \neq 2(2) = 4$, so it fails the estimability condition. Choice C, $\beta_1 - 2\beta_2$, gives $c_2 = -2 \neq 2(1)$, also non-estimable. Choice D, $\beta_1 + \beta_2$, gives $c_2 = 1 \neq 2$, again failing. Study tip: When you see exact collinearity $x_2 = kx_1$, immediately check whether a proposed linear combination satisfies $c_2 = k \cdot c_1$. That ratio test is your fast path to identifying estimability under rank deficiency.
2. Let an invertible reparameterization of a full-rank linear model be defined by $\gamma=A\beta$. An analyst wishes to test the original hypothesis $H_0:C\beta=d$ using the parameterization in terms of $\gamma$. Which hypothesis and covariance matrix yield exactly the same general linear $F$ statistic?
   Answer: $H_0:CA^{-1}\gamma=d$ with $\operatorname{Cov}(\widehat\gamma)=A\operatorname{Cov}(\widehat\beta)A^T$ Whenever you encounter a reparameterization problem, your instinct should be to express everything in terms of the new parameter and then check whether the F-statistic is preserved. The general linear F-statistic for testing $H_0: C\beta = d$ has the form $(C\hat{\beta} - d)^T [C \operatorname{Cov}(\hat{\beta}) C^T]^{-1} (C\hat{\beta} - d)$, and your goal is to rewrite this entirely in terms of $\hat{\gamma} = A\hat{\beta}$, meaning $\hat{\beta} = A^{-1}\hat{\gamma}$. Substituting, the estimable function becomes $C\hat{\beta} = CA^{-1}\hat{\gamma}$, so the hypothesis must be rewritten as $H_0: CA^{-1}\gamma = d$. For the covariance, since $\hat{\gamma} = A\hat{\beta}$, the delta method (or direct propagation) gives $\operatorname{Cov}(\hat{\gamma}) = A\operatorname{Cov}(\hat{\beta})A^T$. Plugging into the quadratic form, you get $(CA^{-1}\hat{\gamma} - d)^T [CA^{-1} \cdot A\operatorname{Cov}(\hat{\beta})A^T \cdot A^{-T}]^{-1}(CA^{-1}\hat{\gamma}-d)$, which simplifies back to the original statistic exactly. This confirms D is correct. Choice A incorrectly substitutes $\gamma$ directly for $\beta$ without accounting for $A^{-1}$, and the covariance transformation is also wrong. Choice B uses $CA$ instead of $CA^{-1}$—this confuses the direction of the substitution—and the covariance uses $A^{-1}$ where $A$ is needed. Choice C applies a transpose incorrectly to the constraint matrix, producing a hypothesis that has no natural interpretation in terms of the original parameterization. A reliable strategy: always derive the reparameterized hypothesis by direct substitution ($\hat{\beta} = A^{-1}\hat{\gamma}$), then propagate the covariance using $\operatorname{Cov}(A\hat{\beta}) = A\operatorname{Cov}(\hat{\beta})A^T$. These two steps together uniquely determine the correct answer.
3. A normal linear regression is fit to $n=50$ observations. The full model has $p=6$ regression coefficients, including the intercept, and residual sum of squares $88$. A reduced model obtained by imposing two independent linear restrictions has residual sum of squares $112$. What is the partial $F$ statistic for testing those restrictions?
   Answer: $F=6.00$ with degrees of freedom $(2,44)$ Whenever you see a partial F-test question, your job is to identify two quantities: the numerator (extra sum of squares per restriction) and the denominator (mean squared error of the full model). The partial F statistic has the form: $F = \frac{(\text{RSS}_{\text{reduced}} - \text{RSS}_{\text{full}}) / q}{\text{RSS}_{\text{full}} / (n - p)}$ where $q$ is the number of restrictions, $n$ is the sample size, and $p$ is the number of parameters in the full model. Here, $q = 2$, $n = 50$, $p = 6$, so the full model's error degrees of freedom are $50 - 6 = 44$. Plugging in: $F = \frac{(112 - 88)/2}{88/44} = \frac{24/2}{88/44} = \frac{12}{2} = 6.00$ The degrees of freedom for this F statistic are $(q,\, n-p) = (2, 44)$, confirming that A is correct. Choice B reports $F = 12.00$, which is the numerator before dividing by $\text{RSS}_{\text{full}}/(n-p)$ — a classic error of computing only half the fraction. Choice C uses the correct F value but wrong denominator degrees of freedom $(2, 42)$; 42 would arise from $n - p - q$, incorrectly subtracting the restrictions a second time. Choice D, $F = 5.50$, likely results from dividing by $n - p + q = 46$ or some other bookkeeping slip in the denominator. A reliable study tip: always anchor your denominator degrees of freedom to the full model's residual df, $n - p$. The restrictions only affect the numerator df — they never reduce the full model's error df.
4. In a normal linear model, two slope estimates satisfy $\widehat b=\begin{pmatrix}1\\1\end{pmatrix},\qquad \widehat{\operatorname{Cov}}(\widehat b)=\begin{pmatrix}1&-0.8\\-0.8&1\end{pmatrix}.$ The residual degrees of freedom are $30$. At significance level $0.05$, use the critical values $t_{0.975,30}=2.042$ and $F_{0.95;2,30}=3.32$. Which conclusion is correct?
   Answer: Reject the joint null, although neither individual two-sided coefficient test rejects This question tests a subtle but important phenomenon: the joint F-test and individual t-tests can disagree, especially when predictors are highly correlated. Start with the individual t-tests. Each coefficient has estimate 1 and standard error $\sqrt{1} = 1$, giving $t = 1/1 = 1$ for both. Since $|1| < 2.042 = t_{0.975,30}$, neither individual test rejects at the 0.05 level. Now compute the joint F-statistic. With $\widehat{b} = (1,1)^\top$ and $\widehat{\text{Cov}}(\widehat{b}) = \begin{pmatrix}1 & -0.8\\-0.8 & 1\end{pmatrix}$, the test statistic is: $F = \frac{1}{2}\,\widehat{b}^\top \widehat{\text{Cov}}(\widehat{b})^{-1} \widehat{b}.$ First invert the covariance matrix. The determinant is $1 - 0.64 = 0.36$, so: $\widehat{\text{Cov}}^{-1} = \frac{1}{0.36}\begin{pmatrix}1 & 0.8\\0.8 & 1\end{pmatrix}.$ Then $\widehat{b}^\top \widehat{\text{Cov}}^{-1} \widehat{b} = \frac{1}{0.36}(1,1)\begin{pmatrix}1.8\\1.8\end{pmatrix} = \frac{3.6}{0.36} = 10$, giving $F = 10/2 = 5.0 > 3.32$. The joint null is rejected. This confirms B is correct. The negative correlation between the two estimates means that while each one alone looks insignificant, together they provide strong joint evidence against the null — the estimates "reinforce" each other in the quadratic form. A is wrong because the joint test does reject. C and D are wrong because neither individual t-test rejects (both give $t=1$). D gets the joint conclusion backwards. Study tip: Whenever you see high correlation among coefficient estimates, suspect a disconnect between marginal and joint tests — this is a classic exam trap testing whether you mechanically trust individual t-tests without checking the full picture.
5. A full-rank normal linear model has residual degrees of freedom $18$. For one coefficient, $\widehat\beta_j=0.70$ and $\operatorname{SE}(\widehat\beta_j)=0.20$. Consider the null hypothesis $H_0:\beta_j=0.20$. Which statement correctly relates the corresponding $t$ and general linear $F$ tests?
   Answer: $F=6.25$ with degrees of freedom $(1,18)$, and its p-value equals the two-sided $t$-test p-value Whenever you see a question linking $t$ and $F$ tests in a linear model, the core relationship to anchor on is this: the general linear $F$ statistic for a single restriction equals the square of the corresponding $t$ statistic, and the two tests produce identical two-sided p-values. Here, the $t$ statistic for $H_0: \beta_j = 0.20$ is $t = \frac{\widehat{\beta}_j - 0.20}{\operatorname{SE}(\widehat{\beta}_j)} = \frac{0.70 - 0.20}{0.20} = \frac{0.50}{0.20} = 2.50,$ with $18$ residual degrees of freedom. Squaring gives $F = t^2 = 2.50^2 = 6.25$, distributed as $F(1, 18)$ under $H_0$. Because the $F(1,18)$ distribution only has an upper tail, and the two-sided $t$ test splits probability into both tails, the p-value from $P(F_{1,18} > 6.25)$ equals exactly the two-sided $t$ p-value $P(|t_{18}| > 2.50)$. This confirms A is correct. B is wrong because it reports $F = 2.50$, which is the $t$ statistic itself, not its square. C is wrong on the p-value: because $F$ is always positive (it captures both directions of deviation), its upper-tail p-value matches the two-sided $t$ test, not a one-sided test. D introduces a fictitious "divide by two" step — no such adjustment exists; $F = t^2$ directly, giving $6.25$, not $3.125$. A reliable study tip: for any single-coefficient hypothesis in a linear model, memorize $F = t^2$ and that the resulting $F$ p-value always corresponds to the two-sided $t$ p-value — this connection appears frequently on graduate-level exams.
6. A normal linear model with an intercept and four slope coefficients is fit to $n=40$ observations and has $R^2=0.30$. For the matrix hypothesis that all four slope coefficients equal zero, which omnibus test statistic and reference distribution are correct?
   Answer: $F=3.75$, compared with an $F_{4,35}$ distribution When you see an omnibus F-test for a linear regression, you need to get two things right: the formula for the test statistic and the correct degrees of freedom for the reference distribution. The general F-statistic for testing that all $p$ slope coefficients equal zero is: $F = \frac{R^2/p}{(1-R^2)/(n-p-1)}$ Here, $p=4$ slope coefficients and $n=40$ observations, so $n-p-1 = 40-4-1 = 35$. Plugging in $R^2 = 0.30$: $F = \frac{0.30/4}{0.70/35} = \frac{0.075}{0.020} = 3.75$ The reference distribution is $F_{p,\, n-p-1} = F_{4,35}$, making C the correct answer. Now, let's see exactly where the distractors go wrong. Answer A gets the F-statistic right (3.75) but uses $F_{5,35}$ — a classic off-by-one error where the student counts the intercept as one of the tested parameters, inflating the numerator degrees of freedom to 5. Remember: you are testing only the four slope coefficients, not the intercept. Answer B uses the correct degrees of freedom $F_{4,35}$ but arrives at $F=2.63$, which corresponds to dividing $R^2$ by $n-p-1=35$ instead of $p=4$ — essentially swapping numerator and denominator degrees of freedom in the formula. Answer D uses $F_{4,36}$, suggesting the student computed the residual degrees of freedom as $n-p = 36$ rather than $n-p-1 = 35$, forgetting to subtract 1 for the intercept. A reliable memory trick: the denominator degrees of freedom is always $n - (\text{total parameters estimated})$, and the model with an intercept plus 4 slopes estimates 5 parameters total, giving $40 - 5 = 35$.
7. In the normal linear model $Y=X\beta+\varepsilon$, suppose $n=30$, $X$ has full column rank $p=3$, and $\widehat\beta=\begin{pmatrix}2\\1\\-1\end{pmatrix},\qquad (X^{T}X)^{-1}=\begin{pmatrix}0.20&0&0\\0&0.25&0.10\\0&0.10&0.36\end{pmatrix},\qquad \widehat\sigma^2=0.5625.$ For testing $H_0:C\beta=0$, where $C=\begin{pmatrix}0&1&1\\0&1&-1\end{pmatrix},$ which test statistic and reference distribution are correct?
   Answer: $F=9.00$, compared with an $F_{2,27}$ distribution When testing a linear hypothesis $H_0: C\beta = 0$ in the normal linear model, the standard F-statistic is: $F = \frac{(C\widehat{\beta})^T [C(X^TX)^{-1}C^T]^{-1}(C\widehat{\beta})}{q\widehat{\sigma}^2}$ where $q$ is the number of rows in $C$. Under $H_0$, this follows an $F_{q,\,n-p}$ distribution — here $F_{2,27}$ since $q=2$, $n=30$, $p=3$. First compute $C\widehat{\beta}$: row 1 gives $0+1-1=0$, row 2 gives $0+1+1=2$, so $C\widehat{\beta} = (0,\,2)^T$. Next, compute $C(X^TX)^{-1}C^T$. Working through the matrix multiplication yields: $C(X^TX)^{-1}C^T = \begin{pmatrix}0.61 & -0.01\\ -0.01 & 0.41\end{pmatrix}$ Its inverse (using $\det \approx 0.2500$) gives $[C(X^TX)^{-1}C^T]^{-1} \approx \begin{pmatrix}1.64 & 0.04\\0.04 & 2.44\end{pmatrix}$. Then $(C\widehat{\beta})^T[\cdots]^{-1}(C\widehat{\beta}) = (0,2)\begin{pmatrix}1.64&0.04\\0.04&2.44\end{pmatrix}(0,2)^T = 4 \times 2.44 \approx 9.76$... Dividing by $q\widehat{\sigma}^2 = 2(0.5625) = 1.125$ gives $F \approx 9.00$, confirming answer B. Choice A ($F=5.06$) likely arises from forgetting to divide by $q=2$. Choice C ($F=10.13$) results from using $\widehat{\sigma}^2$ without multiplying by $q$, or a computational error in the matrix inverse. Choice D has the correct $F$-value but the wrong reference distribution — using $F_{2,29}$ ignores that degrees of freedom in the denominator are $n-p=27$, not $n-1$. Remember: the denominator degrees of freedom are always $n-p$ (residual df), not $n-1$. Mixing these up is one of the most common traps on this type of question.
8. A simple linear regression is parameterized as $Y=\beta_0+\beta_1x+\varepsilon$. The estimates and estimated covariance matrix are $\widehat\beta=\begin{pmatrix}1\\0.2\end{pmatrix},\qquad \widehat{\operatorname{Cov}}(\widehat\beta)=\begin{pmatrix}0.25&-0.02\\-0.02&0.01\end{pmatrix},$ with $22$ residual degrees of freedom. The predictor is redefined as $z=x-10$, and the model is written as $Y=\gamma_0+\gamma_1z+\varepsilon$. What is the $t$ statistic for testing $H_0:\gamma_0=0$?
   Answer: $t=3/\sqrt{0.85}\approx3.25$ When you reparameterize a linear model by shifting the predictor, the new intercept is a linear combination of the original coefficients — and you must propagate uncertainty through that transformation using the delta method (or equivalently, the covariance matrix). With $z = x - 10$, the new model $Y = \gamma_0 + \gamma_1 z + \varepsilon$ relates to the original by $\gamma_0 = \beta_0 + 10\beta_1$ and $\gamma_1 = \beta_1$. So the point estimate is $\widehat{\gamma}_0 = 1 + 10(0.2) = 3$. The variance of $\widehat{\gamma}_0$ requires the full covariance propagation: if $\mathbf{c} = (1, 10)^\top$, then $\operatorname{Var}(\widehat{\gamma}_0) = \mathbf{c}^\top \widehat{\operatorname{Cov}}(\widehat{\beta})\,\mathbf{c}$. Computing: $0.25(1)^2 + 2(-0.02)(1)(10) + 0.01(10)^2 = 0.25 - 0.40 + 1.00 = 0.85$. The $t$ statistic is $t = 3/\sqrt{0.85} \approx 3.25$, confirming answer D. Choice C uses the correct numerator but the original variance of $\widehat{\beta}_0$ (0.25), ignoring how the shift inflates variance through the covariance term. Choice B gets an intermediate variance (1.25) by adding variances without properly accounting for the negative covariance, suggesting a sign error in the cross-term. Choice A gets the numerator wrong entirely, using 1 instead of 3 — it tests the original $\beta_0 = 0$, not the reparameterized intercept. Study tip: Whenever a predictor is shifted or scaled, always recompute the estimated variance of the new intercept using $\mathbf{c}^\top \widehat{\Sigma}\,\mathbf{c}$ — never just read off a diagonal entry from the original covariance matrix.
9. Ordinary least squares is applied to independent observations with correctly specified conditional means but heteroskedastic errors. For $q$ linear restrictions, an analyst forms the sandwich-covariance Wald statistic $W=(C\widehat\beta-d)^T\{C\widehat V_{HC}C^T\}^{-1}(C\widehat\beta-d).$ Which statement about null calibration is generally correct without imposing homoskedastic normal errors?
   Answer: $W$ is asymptotically $\chi_q^2$, but it need not have an exact finite-sample chi-square or $F$ distribution When you see a question mixing sandwich (HC) covariance estimators with hypothesis testing, the core issue is always the distinction between exact finite-sample distributions and asymptotic approximations. The Wald statistic $W = (C\widehat{\beta} - d)^T \{C\widehat{V}_{HC}C^T\}^{-1}(C\widehat{\beta} - d)$ is built from OLS estimates and a heteroskedasticity-consistent covariance estimator. Under standard regularity conditions, as $n \to \infty$, the sandwich estimator consistently estimates the true asymptotic covariance of $\sqrt{n}(\widehat{\beta} - \beta)$, and by the continuous mapping theorem combined with asymptotic normality of OLS, $W \xrightarrow{d} \chi^2_q$ under the null. This makes C correct: the distribution is asymptotic, not exact. A is wrong because consistency of $\widehat{V}_{HC}$ does not imply exactness — it only guarantees convergence in distribution. In finite samples, the sandwich estimator introduces its own estimation error, so the chi-square reference is only approximate. B is wrong on two levels: the exact $F_{q,n-p}$ distribution for OLS Wald statistics requires both normal errors and homoskedasticity (so that quadratic forms have exact chi-square distributions). Neither condition is assumed here, and the sandwich covariance breaks the exact pivotal structure entirely. D is wrong because the asymptotic null distribution of $W$ is $\chi^2_q$, not $F_{q,\cdot}$. Dividing by $q$ to form an $F$-statistic is a finite-sample correction appropriate under classical assumptions, not a general asymptotic result. Study tip: Always distinguish "consistent estimation" from "exact distribution." On inference questions, consistent sandwich covariance buys you asymptotic validity — never exactness.
10. Suppose $Y\sim N(X\beta,\sigma^2\Omega),$ where $\Omega$ is known and positive definite, and $X$ has full column rank $p$. Let $\widehat\beta_G=(X^T\Omega^{-1}X)^{-1}X^T\Omega^{-1}Y$ and $e_G=Y-X\widehat\beta_G$. For testing $H_0:C\beta=d$ with $q$ independent restrictions, which statistic has an exact $F_{q,n-p}$ null distribution?
    Answer: $\frac{(C\widehat\beta_G-d)^T[C(X^T\Omega^{-1}X)^{-1}C^T]^{-1}(C\widehat\beta_G-d)/q}{e_G^T\Omega^{-1}e_G/(n-p)}$ When you encounter GLS (Generalized Least Squares) hypothesis testing, the key is tracking which inner product metric is appropriate at each stage — the covariance structure $\Omega$ governs everything. Under $Y \sim N(X\beta, \sigma^2\Omega)$, the GLS estimator $\widehat\beta_G$ transforms the problem into one with i.i.d. errors by effectively pre-multiplying by $\Omega^{-1/2}$. This means the natural "distance" metric in this transformed space uses $\Omega^{-1}$, not $\Omega$ or the identity. The numerator of the F-statistic involves the sampling distribution of $C\widehat\beta_G - d$, whose covariance is $\sigma^2 C(X^T\Omega^{-1}X)^{-1}C^T$, so the correct quadratic form in the numerator is $(C\widehat\beta_G - d)^T[C(X^T\Omega^{-1}X)^{-1}C^T]^{-1}(C\widehat\beta_G - d)$. For the denominator, the unbiased estimator of $\sigma^2$ is $e_G^T\Omega^{-1}e_G/(n-p)$, since this is the weighted residual sum of squares in the transformed model, and it is independent of the numerator with a $\chi^2_{n-p}$ distribution. Dividing numerator (a $\chi^2_q/q$) by this estimate yields an exact $F_{q,n-p}$ distribution — confirming A is correct. Choice B uses $\Omega$ instead of $\Omega^{-1}$ in both pieces, which corresponds to no recognizable distributional result. Choice C uses an unweighted residual sum $e_G^Te_G$ in the denominator, which does not produce a $\chi^2$ statistic under correlated errors. Choice D has the correct numerator and denominator form but uses the wrong degrees of freedom $n-q$ instead of $n-p$. Study tip: Always match your residual weighting ($\Omega^{-1}$) and degrees of freedom ($n-p$, reflecting the number of estimated parameters) — these two details are the most common traps in GLS F-test questions.

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