Module M04 Quiz: Inference — t-Tests & F-Tests
16 questions · Graduate (ECMT5001) · MCQ, calculation, short answer
How to use
Attempt each question before clicking Show Answer. Show full working for hypothesis tests — state the test statistic, critical value, and conclusion.
Question 1
Lesson L02 · t-Statistic
The t-statistic \(t = (\hat{\beta}_j - \beta_{j0})/\text{se}(\hat{\beta}_j)\) follows a \(t_{n-k-1}\) distribution exactly when:
Type: Multiple Choice
- A) The sample size is large (\(n > 30\))
- B) The CLM assumptions hold: MLR.1–MLR.5 plus MLR.6 (normality of \(u\))
- C) The regressors are normally distributed
- D) Heteroskedasticity is absent
Show Answer
Answer: B
Under the Classical Linear Model (CLM) assumptions MLR.1–MLR.6, \(t = (\hat{\beta}_j - \beta_{j0})/\text{se}(\hat{\beta}_j) \sim t_{n-k-1}\) exactly in finite samples. MLR.6 (normality of \(u\)) is the critical additional assumption — it ensures \(\hat{\beta}_j\) is normally distributed conditional on \(X\), making the standardised ratio a \(t\) distribution. Without MLR.6, the t-statistic is asymptotically \(N(0,1)\) by the CLT (used for large samples).
Question 2
Lesson L02 · t-Test Calculation
[Calculation] \(\hat{\beta}_1 = 2.4\), \(\text{se}(\hat{\beta}_1) = 0.8\), \(n = 32\), \(k = 2\) regressors (plus intercept). Test \(H_0: \beta_1 = 0\) vs \(H_1: \beta_1 \neq 0\) at \(\alpha = 0.05\). Critical value \(t_{0.025, 29} = 2.045\).
Type: Calculation
Show Answer
Since \(|t| = 3.0 > t_{0.025, 29} = 2.045\), reject \(H_0\) at the 5% level. The coefficient is statistically significant.
Degrees of freedom: \(df = n - k - 1 = 32 - 2 - 1 = 29\).
Question 3
Lesson L02 · One-Sided Test
[Calculation] \(H_0: \beta_1 \leq 0\) vs \(H_1: \beta_1 > 0\). Observed \(t = 1.85\) with \(df = n - k - 1 = 38\). At \(\alpha = 0.05\), the one-tailed critical value \(t_{0.05, 38} = 1.686\). Reject?
Type: Calculation
Show Answer
Since \(t = 1.85 > t_{0.05, 38} = 1.686\), reject \(H_0\) at the 5% level (one-tailed).
Note: If we had used a two-tailed test, the critical value would be \(t_{0.025, 38} \approx 2.024\), and we would fail to reject. This illustrates the power advantage of a one-tailed test when the direction of the effect is known a priori. However: the one-tailed test is only valid if the directional hypothesis was specified before seeing the data.
Question 4
Lesson L03 · Confidence Interval
[Calculation] \(\hat{\beta}_1 = 3.2\), \(\text{se}(\hat{\beta}_1) = 0.9\), \(n = 60\), \(k = 3\) regressors. Construct a 95% CI for \(\beta_1\). Use \(t_{0.025, 56} \approx 2.003\).
Type: Calculation
Show Answer
Since the CI does not include zero, we reject \(H_0: \beta_1 = 0\) at the 5% level — consistent with a t-test. The interval can also be used to test any specific null value: e.g. \(H_0: \beta_1 = 5\) is rejected since 5 lies outside \([1.40, 5.00]\) — just barely.
Question 5
Lesson L02 · p-Value
A coefficient has \(t = 2.15\) with 45 degrees of freedom. For a two-tailed test, \(p < 0.05\)?
Type: Multiple Choice
- A) No — \(p > 0.05\) because \(t = 2.15 < t_{0.025,45} = 2.014\)... wait, which is correct?
- B) Yes — \(t_{0.025, 45} \approx 2.014 < 2.15\), so \(p < 0.05\)
- C) Cannot determine without knowing \(\sigma^2\)
- D) No — we need \(n\) not just df
Show Answer
Answer: B
\(t_{0.025, 45} \approx 2.014\) (from t-tables). Since \(|t| = 2.15 > 2.014\), the two-tailed p-value \(< 0.05\). More precisely, \(p = 2 \times P(T_{45} > 2.15) \approx 2 \times 0.018 = 0.036\). We reject \(H_0: \beta_j = 0\) at the 5% level.
Question 6
Lesson L05 · Economic Significance
A regression of firm profits on advertising spending gives: \(\hat{\beta}_\text{advertising} = 0.0001\) with \(p = 0.001\). Is this effect practically important?
Type: Short Answer
Show Answer
Probably not — this is a case of statistical significance without economic significance. With a very large sample (\(n\) must be large for \(p = 0.001\) on such a tiny coefficient), even trivially small effects become statistically significant.
The coefficient \(0.0001\) means $1 extra advertising spending raises profits by $0.0001 — one hundredth of a cent. That is economically negligible regardless of the p-value.
Key lesson: Always report effect sizes alongside p-values. Statistical significance tells you whether the effect is distinguishable from zero; economic significance tells you whether it matters. For policy, the magnitude and confidence interval matter far more than the p-value.
Question 7
Lesson L04 · F-Test Setup
Testing the joint hypothesis \(H_0: \beta_1 = 0 \text{ AND } \beta_2 = 0\) — how many restrictions \(q\) does this impose?
Type: Multiple Choice
- A) 1
- B) 2
- C) Depends on the sample size
- D) Depends on the number of regressors
Show Answer
Answer: B
Each "$\beta_j = $ [specific value]" is one linear restriction. Testing two coefficients simultaneously imposes \(q = 2\) restrictions. In general, \(q\) equals the number of equality constraints in \(H_0\). The F-test statistic with \(q\) and \(n-k-1\) degrees of freedom tests whether all \(q\) restrictions hold simultaneously — this cannot be replicated by running \(q\) separate t-tests (which would inflate the Type I error rate).
Question 8
Lesson L04 · F-Test Calculation
[Calculation] Restricted model (imposing \(H_0\)): \(SSR_r = 600\). Unrestricted model: \(SSR_{ur} = 480\). \(q = 2\) restrictions, \(n = 100\), \(k = 4\) regressors (unrestricted). Calculate \(F\) and compare to \(F_{2, 95, 0.05} \approx 3.09\).
Type: Calculation
Show Answer
Since \(F = 11.87 \gg F_{2, 95, 0.05} = 3.09\), reject \(H_0\) strongly. The two restricted coefficients are jointly significant.
Question 9
Lesson L04 · F vs t
For a single restriction (\(q = 1\)), \(F = t^2\). If \(t = 2.5\), what is \(F\)?
Type: Calculation
Show Answer
For a single restriction, the F-test and the two-tailed t-test are equivalent: reject \(H_0\) at 5% if \(F > F_{1, n-k-1, 0.05} \approx 3.84\) (which equals \(t_{0.025}^2 \approx 1.96^2 = 3.84\)). The F-test becomes essential (and more powerful than repeated t-tests) when testing joint hypotheses with \(q \geq 2\).
Question 10
Lesson L04 · Overall F-Test
The overall F-test tests \(H_0: \beta_1 = \beta_2 = \cdots = \beta_k = 0\) (all slopes are zero). If \(F = 8.3\) with \(p < 0.001\), the correct conclusion is:
Type: Multiple Choice
- A) At least one slope coefficient is individually significant
- B) The regressors, as a group, explain a statistically significant portion of the variation in \(y\)
- C) The model is correctly specified
- D) \(R^2\) is large
Show Answer
Answer: B
The overall F-test asks: "Is the regression as a whole useful?" — do the regressors jointly explain any variation in \(y\)? Rejecting \(H_0\) means at least one \(\beta_j \neq 0\), so the model has joint explanatory power. This does not imply any individual coefficient is significant (they could all be jointly significant due to multicollinearity), nor does it validate model specification (\(R^2\) could still be low).
Question 11
Lesson L01 · Asymptotic Normality
Why does asymptotic normality matter, and when does it apply?
Type: Short Answer
Show Answer
Why it matters: Without MLR.6 (normality of \(u\)), the t and F statistics are not exactly t- and F-distributed in finite samples. However, by the Central Limit Theorem, as \(n \to \infty\), \(\hat{\beta}_j\) is approximately normally distributed regardless of the distribution of \(u\) (under regularity conditions). This means t and F tests are asymptotically valid.
When it applies: Large samples (\(n\) large relative to \(k\)). The approximation is better when \(u\)'s distribution is not too skewed or heavy-tailed. In practice, \(n \geq 50\)–100 usually suffices for good approximations; with binary or skewed regressors, larger samples may be needed. For small \(n\) with non-normal errors, bootstrap inference is preferred.
Question 12
Lesson L05 · Interpreting Output
A regression table shows: coef = \(1.23\), se = \(0.41\), \(t = 3.00\), \(p = 0.003\), 95% CI = \([0.42, 2.04]\). Write a complete interpretation.
Type: Short Answer
Show Answer
The estimated coefficient is 1.23 [units of \(y\) per unit of \(x\)]. The standard error of 0.41 reflects sampling uncertainty. The t-statistic of 3.00 tests \(H_0: \beta = 0\); with \(p = 0.003 < 0.05\), we reject \(H_0\) — the coefficient is statistically significant at the 5% and 1% levels. The 95% confidence interval \([0.42, 2.04]\) tells us: if we repeated the sampling procedure many times, 95% of constructed intervals would contain the true \(\beta\). Since zero is outside this interval, significance at 5% is confirmed. The effect size (1.23) should be assessed for economic significance relative to the scale of \(y\) and \(x\).
Question 13
Lesson L02 · Linear Combination
You want to test \(H_0: \beta_1 = \beta_2\) (two coefficients are equal). How do you construct a t-test?
Type: Short Answer
Show Answer
Reparametrise: define \(\theta = \beta_1 - \beta_2\) and test \(H_0: \theta = 0\).
Method 1: Create a new variable \(\tilde{x} = x_1 - x_2\) and include \(x_2\) and \(\tilde{x}\) in the regression. The coefficient on \(\tilde{x}\) is \(\theta = \beta_1 - \beta_2\), and its standard t-test applies.
Method 2 (direct): \(t = (\hat{\beta}_1 - \hat{\beta}_2)/\text{se}(\hat{\beta}_1 - \hat{\beta}_2)\) where \(\text{Var}(\hat{\beta}_1 - \hat{\beta}_2) = \text{Var}(\hat{\beta}_1) + \text{Var}(\hat{\beta}_2) - 2\text{Cov}(\hat{\beta}_1, \hat{\beta}_2)\). The covariance term is available from the variance-covariance matrix of \(\hat{\beta}\), reported by most software. Never test \(H_0: \beta_1 = \beta_2\) by comparing their individual CIs — this ignores the covariance.
Question 14
Lesson L04 · ECMT5001 style
[Calculation] A regression of house price (AUD) on size (sqm), bedrooms, and distance from CBD gives \(R^2 = 0.72\), \(n = 150\). The restricted model (dropping bedrooms and distance, keeping only size and intercept) gives \(SSR_r = 2{,}400\). The unrestricted \(SSR_{ur} = 1{,}800\). Test \(H_0: \beta_\text{bedrooms} = \beta_\text{distance} = 0\) at the 5% level. \(F_{2, 146, 0.05} \approx 3.06\).
Type: Calculation
Show Answer
\(q = 2\) restrictions (two slopes set to zero). Unrestricted model has \(k = 3\) regressors (size, bedrooms, distance) \(\Rightarrow df_{ur} = n - k - 1 = 150 - 3 - 1 = 146\).
Since \(F = 24.3 \gg F_{2, 146, 0.05} = 3.06\), reject \(H_0\) strongly. Bedrooms and distance from CBD are jointly highly significant — excluding them would be a major specification error.
Question 15
Lesson L01 · Normality Assumption
MLR.6 says \(u \mid \mathbf{x} \sim N(0, \sigma^2)\). What does this assumption add over MLR.1–MLR.5, and when is it strictly needed?
Type: Multiple Choice
- A) It adds nothing — MLR.1–MLR.5 are sufficient for all econometric inference
- B) It ensures exact \(t_{n-k-1}\) and \(F_{q,n-k-1}\) distributions in finite samples; without it, these hold only asymptotically
- C) It is needed only for OLS to be unbiased
- D) It ensures that OLS is consistent
Show Answer
Answer: B
MLR.1–MLR.5 give BLUE (Gauss-Markov) and imply asymptotic normality of \(\hat{\beta}\) (so t and F tests are approximately valid in large samples). MLR.6 additionally ensures \(\hat{\beta}_j \mid X \sim N(\beta_j, \sigma^2(X'X)^{-1}_{jj})\) exactly in finite samples, making t and F statistics exactly distributed — critical for small samples. If errors are non-normal (e.g. skewed, heavy-tailed), rely on large-sample approximations or the bootstrap.
Question 16
Lesson L03 · CI Interpretation
"I am 95% confident that \(\beta_1\) lies in \([1.2,\ 3.8]\)." What is the precise frequentist meaning of this statement?
Type: Short Answer
Show Answer
Precise meaning: If we were to draw many random samples from the population and compute a 95% CI from each sample, 95% of those intervals would contain the true parameter \(\beta_1\).
Common misconception: "\(\beta_1\) is in \([1.2, 3.8]\) with probability 95%." This is wrong in the frequentist framework — \(\beta_1\) is a fixed (unknown) constant, not a random variable. A given realised interval either contains \(\beta_1\) or it doesn't; there is no probability about it. The 95% refers to the procedure (the long-run coverage rate), not the specific interval.
The Bayesian credible interval does have the direct probability interpretation, but requires a prior distribution on \(\beta_1\).