Standard Distribution Results - Statistics Graduate Level

1. Under $H_0:\mu_X-\mu_Y=\delta_0$, which statistic has an exact standard $t$ distribution?
   Answer: $\frac{\bar X-\bar Y-\delta_0}{S_p\sqrt{1/8+4/10}}\sim t_{16}$, correctly accounting for the unequal population variances When both samples share a common variance parameter $\sigma^2$ but with different multipliers, you need to construct a pooled estimator that recovers exactly one $\sigma^2$, then build the standard error accordingly. Here, $\text{Var}(X_i)=\sigma^2$ and $\text{Var}(Y_j)=4\sigma^2$, so $\frac{Y_j-\mu_Y}{2}$ has variance $\sigma^2$. This means $\frac{1}{4}\sum_{j=1}^{10}(Y_j-\bar{Y})^2$ is a chi-squared quantity scaled by $\sigma^2$ with 9 degrees of freedom. Combined with $\sum_{i=1}^{8}(X_i-\bar{X})^2/\sigma^2 \sim \chi^2_7$, the pooled $S_p^2$ satisfies $16S_p^2/\sigma^2 \sim \chi^2_{16}$, giving 16 degrees of freedom. Now consider the standard error of $\bar{X}-\bar{Y}$: $\text{Var}(\bar{X}-\bar{Y})=\frac{\sigma^2}{8}+\frac{4\sigma^2}{10}=\sigma^2\!\left(\frac{1}{8}+\frac{4}{10}\right)$. Dividing by $S_p$ (which estimates $\sigma$) yields an exact $t_{16}$ distribution, confirming B is correct. A is wrong because it uses $\sqrt{1/8+1/10}$, which ignores the factor of 4 in $\text{Var}(Y_j)$, underestimating the true standard error. C uses the correct standard error but claims $t_{17}$ degrees of freedom — there is no justification for $n_X+n_Y-1=17$; the degrees of freedom come from the chi-squared structure of $S_p^2$, which is 16. D flips the variance multiplier, incorrectly applying 4 to the $X$ sample and 1 to the $Y$ sample, producing the wrong standard error entirely. Your study tip: whenever variances are proportional ($k\sigma^2$), rescale each sample's sum of squares by $1/k$ before pooling — this forces everything back to a common $\sigma^2$ scale and keeps your degrees of freedom and standard error both correct.
2. What is the distribution of $Q$?
   Answer: A noncentral chi-square distribution, $\chi'^2_8(2)$ Whenever you see a sum of squared standardized normal variables, ask yourself two questions: how many terms are there, and are those terms truly standard normal? Here, each term in $Q = \sum_{i=1}^8 (X_i - \mu_0)^2/\sigma^2$ has the form $\left(\frac{X_i - \mu_0}{\sigma}\right)^2$. Since $X_i \sim N(\mu, \sigma^2) = N(\mu_0 + \sigma/2, \sigma^2)$, we can write $\frac{X_i - \mu_0}{\sigma} = \frac{X_i - \mu}{\sigma} + \frac{\mu - \mu_0}{\sigma} = Z_i + \frac{1}{2}$, where $Z_i \sim N(0,1)$. So each term is a squared normal with a nonzero mean of $1/2$, not a standard normal. By definition, a sum of $k$ independent squared normals $N(\lambda_i, 1)$ follows a noncentral chi-square $\chi'^2_k(\delta)$, where $\delta = \sum \lambda_i^2$ is the noncentrality parameter. Here $k = 8$ terms and $\delta = \sum_{i=1}^8 (1/2)^2 = 8 \cdot 1/4 = 2$. This confirms the distribution is $\chi'^2_8(2)$, making D correct. A is wrong because $\chi^2_7$ has only 7 degrees of freedom — nothing here subtracts a mean estimate, which would cost one degree of freedom. B gets the degrees of freedom right (8 terms = 8 df) but incorrectly assumes centrality, ignoring that $\mu \neq \mu_0$. C compounds both errors: it uses 7 degrees of freedom and correctly identifies noncentrality, but loses a degree of freedom with no justification. Your key rule: degrees of freedom equal the number of squared terms, and noncentrality arises whenever those normals have nonzero means — compute $\delta = \sum \lambda_i^2$ from those means.
3. Which identity follows from the reciprocal property of the $F$ distribution?
   Answer: $\frac{1}{c}=q_{0.025}(11,7)$ Whenever you see a question involving quantiles of the $F$ distribution, the reciprocal property is the key tool. It states that if $X \sim F(a,b)$, then $1/X \sim F(b,a)$. This has a direct consequence for quantiles: $q_p(a,b) = \frac{1}{q_{1-p}(b,a)}$, because flipping the random variable swaps the tail probabilities. Here, $c = q_{0.975}(7,11)$ means $P(X \leq c) = 0.975$ for $X \sim F(7,11)$. Since $1/X \sim F(11,7)$, the event $X \leq c$ is equivalent to $1/X \geq 1/c$, so $P(1/X \geq 1/c) = 0.975$, which means $P(1/X \leq 1/c) = 0.025$. Therefore $1/c = q_{0.025}(11,7)$, confirming answer D. Now consider why the other choices fail. A claims $1/c = q_{0.975}(11,7)$, but that would mean $1/c$ sits at the upper 2.5% tail of $F(11,7)$, not the lower — the reciprocal operation flips which tail you're in. B claims $1/c = q_{0.025}(7,11)$, which keeps the degrees of freedom in the original order $(7,11)$ — but the reciprocal property swaps the degrees of freedom. C claims $c = q_{0.025}(11,7)$, which correctly swaps the degrees of freedom and correctly identifies the 0.025 quantile, but forgets to take the reciprocal of $c$. A clean memory aid: reciprocal → swap d.f. AND flip tail probability ($p \to 1-p$). Missing either half of that transformation leads directly to one of the wrong answers.
4. Under the null hypothesis $\mu=\mu_0$, what is the distribution of $W=\frac{12(\bar X-\mu_0)^2}{S^2}$?
   Answer: $W\sim F_{1,11}$ because it is the square of a $t_{11}$ variable Whenever you see a ratio involving a squared standardized mean and a sample variance, your first instinct should be to connect it to the relationship between the $t$, $F$, and $\chi^2$ distributions. Start by recognizing that under $H_0: \mu = \mu_0$, the statistic $T = \frac{\bar{X} - \mu_0}{S/\sqrt{12}}$ follows a $t_{11}$ distribution, since the sample mean is standardized by the estimated standard error with $n-1 = 11$ degrees of freedom. Now notice that squaring $T$ gives exactly $W$: $T^2 = \frac{(\bar{X}-\mu_0)^2}{S^2/12} = \frac{12(\bar{X}-\mu_0)^2}{S^2} = W$ A fundamental identity in distribution theory states that if $T \sim t_\nu$, then $T^2 \sim F_{1,\nu}$. Therefore $W \sim F_{1,11}$, confirming B is correct. Choice A is tempting but wrong: $\frac{12(\bar{X}-\mu_0)^2}{\sigma^2}$ would be $\chi^2_1$ if $\sigma^2$ were known, but here $S^2$ replaces $\sigma^2$, fundamentally changing the distribution. Choice C confuses the degree-of-freedom ordering — the $11$ comes from the denominator (the variance estimate), not the numerator, so the correct $F$ has $1$ numerator df and $11$ denominator df, not $F_{11,1}$. Choice D is wrong because $W$ is a squared quantity and therefore cannot be negative, ruling out the symmetric $t$ distribution entirely. Your key takeaway: memorize that $T^2 \sim F_{1,\nu}$ when $T \sim t_\nu$. This identity bridges three core distributions and appears frequently on graduate-level exams, especially in ANOVA and regression contexts.
5. Which expression is an exact equal-tail $95\%$ confidence interval for $\rho$?
   Answer: $\left[\frac{2.4}{q_{0.975}(9,15)},\frac{2.4}{q_{0.025}(9,15)}\right]$ Whenever you see a confidence interval for a ratio of variances, your anchor is the pivot built from the F-distribution. Since $(S_1^2/S_2^2)/\rho \sim F(9,15)$ (with degrees of freedom $n_1-1=9$ and $n_2-1=15$), you can write the probability statement $P\!\left(q_{0.025}(9,15) \leq \frac{S_1^2/S_2^2}{\rho} \leq q_{0.975}(9,15)\right) = 0.95.$ Inverting all parts of the inequality to isolate $\rho$ — remembering that dividing by $\rho$ and flipping inequalities gives you $\rho$ sandwiched between $\frac{S_1^2/S_2^2}{q_{0.975}(9,15)}$ and $\frac{S_1^2/S_2^2}{q_{0.025}(9,15)}$ — and plugging in $S_1^2/S_2^2 = 2.4$ yields exactly Answer A: $\left[\frac{2.4}{q_{0.975}(9,15)},\,\frac{2.4}{q_{0.025}(9,15)}\right].$ This is correct because dividing by the larger quantile gives the smaller bound, preserving proper interval orientation. B swaps the degrees of freedom to $(15, 9)$, which would apply if the ratio were inverted — a common mix-up when students confuse which sample is "1" versus "2." C multiplies rather than divides by the quantiles, which inverts the algebraic manipulation incorrectly; this would only make sense if $\rho$ appeared in the numerator of the pivot. D divides the quantiles by $2.4$ instead of dividing $2.4$ by the quantiles — a simple algebra reversal that flips numerator and denominator. The key study tip: always write the pivot explicitly, then invert the inequalities step by step. The degrees of freedom always follow the order $(n_1-1, n_2-1)$ matching the ratio $S_1^2/S_2^2$, and dividing flips the quantile positions.
6. Which statement about $R$ and $W$ is correct?
   Answer: $R\sim F_{5,12}$, and $R$ is independent of $W$ Whenever you see a ratio of chi-squared variables scaled by their degrees of freedom, think F-distribution — and whenever you see a sum of independent chi-squared variables, think about whether the ratio and sum share any information. Recall the definition: if $X \sim \chi^2_m$ and $Y \sim \chi^2_n$ are independent, then $\frac{X/m}{Y/n} \sim F_{m,n}$. Here, $R = \frac{12U}{5V} = \frac{U/5}{V/12}$, which matches exactly this form with $m=5$ and $n=12$, so $R \sim F_{5,12}$. That confirms the distribution in choice A. For independence, the key insight comes from a classical result: for independent chi-squared variables, the ratio $U/V$ (equivalently $R$) is independent of the sum $W = U + V$. This is analogous to how the sample variance and sample mean are independent in normal sampling — the "direction" and "magnitude" of the vector $(U, V)$ carry separate information. A formal proof uses the Jacobian transformation: writing $(U, V)$ in terms of $(W, R)$ shows the joint density factors into a function of $W$ alone times a function of $R$ alone, establishing independence. So A is correct. Choice B correctly identifies the distribution but wrongly claims dependence — the factorization argument shows independence holds. Choice C reverses the degrees of freedom; since $U \sim \chi^2_5$ is in the numerator scaled by 5, the first index must be 5, not 12. Choice D misidentifies the distribution entirely — a scaled ratio of chi-squared variables is not itself chi-squared. As a study tip: memorize the F-distribution definition carefully, paying attention to which degree of freedom goes in the numerator. The order $F_{m,n}$ always corresponds to $\frac{\chi^2_m/m}{\chi^2_n/n}$, and swapping them gives a different distribution.
7. For the directional alternative that the coefficient is less than its null value, what is the exact one-sided $p$-value?
   Answer: $\frac{p_F}{2}$, because the observed $t$ statistic has the alternative-consistent sign Whenever you see a question linking a two-sided $t$ test to its equivalent $F$ test, the key insight is understanding what squaring the $t$ statistic actually destroys: the sign information. That sign is precisely what determines which one-sided tail is relevant. The relationship between the two distributions is exact: if $T \sim t_\nu$, then $T^2 \sim F_{1,\nu}$. This means $P(F_{1,\nu} \ge t_{\text{obs}}^2) = P(|T| \ge |t_{\text{obs}}|)$, which is the two-sided $p$-value. So $p_F$ already equals the two-sided $p$-value. For a one-sided alternative, you simply take half of that — but only when the observed statistic falls in the direction consistent with the alternative. Here, the alternative is that the coefficient is less than its null value, and $t_{\text{obs}} = -2.4$ is indeed negative (alternative-consistent). The exact one-sided $p$-value is therefore $p_F/2$, confirming C. Choice A is wrong because $p_F$ is the two-sided $p$-value disguised as a one-tailed quantity — the $F$ test is inherently two-sided since it only uses $F \ge 0$. Choice B, $2p_F$, would actually double the two-sided $p$-value, yielding something greater than 1 in extreme cases — clearly nonsensical. Choice D, $1 - p_F/2$, corresponds to the probability in the wrong tail of $t_\nu$ (i.e., $t > 0$), which would be relevant if the alternative were that the coefficient is greater than its null value. Study tip: Always check whether the sign of $t_{\text{obs}}$ aligns with your directional alternative before halving $p_F$. If it does, one-sided $p = p_F/2$; if it doesn't, one-sided $p = 1 - p_F/2$.
8. Which statistic has the stated distribution?
   Answer: $\frac{9S_X^2}{S_Y^2}\sim F_{9,13}$ Whenever you see a question involving ratios of sample variances from normal populations, your goal is to construct an $F$-statistic by making each chi-squared component standardized — that is, divided by its true population variance. Recall the key result: if $S^2$ is the sample variance from a normal population with true variance $\sigma^2$ and sample size $n$, then $\frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1}$. An $F$-statistic is formed as a ratio of two independent chi-squared variables, each divided by their degrees of freedom: $F = \frac{\chi^2_{m}/m}{\chi^2_{k}/k} \sim F_{m,k}$ For the first sample (size 10, variance $\sigma^2$): $\frac{9S_X^2}{\sigma^2} \sim \chi^2_9$. For the second sample (size 14, variance $9\sigma^2$): $\frac{13S_Y^2}{9\sigma^2} \sim \chi^2_{13}$. Forming the ratio of these (each divided by their df) and canceling $\sigma^2$: $\frac{(9S_X^2/\sigma^2)/9}{(13S_Y^2/9\sigma^2)/13} = \frac{9S_X^2}{S_Y^2} \sim F_{9,13}$ This confirms A is correct. B flips the numerator and denominator, which would give $F_{13,9}$, not $F_{9,13}$. C uses $\frac{S_X^2}{9S_Y^2}$, which fails to properly standardize by the true variances — the factor of 9 is misplaced, so the $\sigma^2$ terms don't cancel cleanly. D uses degrees of freedom 10 and 14 instead of the correct $n-1$ values of 9 and 13 — a classic off-by-one error. Your study tip: always subtract 1 from each sample size for the degrees of freedom, and carefully track which population variance belongs to which sample before constructing the ratio.
9. What is the distribution of $T=\frac{(Z'AZ+Z'CZ)/6}{Z'BZ/4}$?
   Answer: $T\sim F_{6,4}$ because the numerator is a rank-$6$ orthogonal projection Whenever you see a ratio of quadratic forms involving a standard normal vector, your instinct should be to check whether the forms correspond to independent chi-squared random variables — that's the foundation of the F-distribution. Here, since $A$ is symmetric idempotent of rank 3, $Z'AZ \sim \chi^2_3$. Similarly, $B$ is symmetric idempotent of rank 4, so $Z'BZ \sim \chi^2_4$. Now consider $C = I_{10} - A - B$. Because $AB = 0$ and both matrices are symmetric idempotent, $C$ is also symmetric idempotent with rank $10 - 3 - 4 = 3$, so $Z'CZ \sim \chi^2_3$. Crucially, the condition $AB = 0$ implies mutual orthogonality among $A$, $B$, and $C$, which guarantees that $Z'AZ$, $Z'BZ$, and $Z'CZ$ are mutually independent (by the Craig–Cochran theorem). Therefore, $Z'AZ + Z'CZ \sim \chi^2_{3+3} = \chi^2_6$, and the ratio becomes: $T = \frac{(Z'AZ + Z'CZ)/6}{Z'BZ/4} \sim F_{6,4}$ This confirms C is correct — the numerator pools two orthogonal rank-3 projections for a total rank of 6, giving an $F_{6,4}$ distribution. A is wrong because it ignores $Z'CZ$'s contribution, treating only $A$ as generating numerator degrees of freedom. B inverts the degree-of-freedom parameters entirely — the numerator rank (6) always comes first. D is wrong because while $Z'AZ + Z'BZ + Z'CZ = Z'Z \sim \chi^2_{10}$, the ratio $T$ is not itself a chi-squared random variable. Study tip: Always verify idempotency and mutual orthogonality ($AB = 0$ for all pairs) before invoking Cochran's theorem — these two conditions together unlock independence of quadratic forms.
10. What is the general linear hypothesis test statistic, and what is its null distribution?
    Answer: $3.00$, with null distribution $F_{2,24}$ When you encounter a general linear hypothesis test, the key quantities are the number of restrictions $q$, the residual degrees of freedom from the full model $n - p$, and the two residual sums of squares. Here, $n = 30$, $p = 6$ (rank of full design matrix), and $q = 2$ restrictions. The F-statistic is constructed as: $F = \frac{(RSS_R - RSS_F)/q}{RSS_F/(n-p)} = \frac{(150 - 120)/2}{120/(30-6)} = \frac{30/2}{120/24} = \frac{15}{5} = 3.00$ Under the null hypothesis, this statistic follows an $F_{q,\, n-p} = F_{2,24}$ distribution. That confirms C is correct. Looking at why each distractor fails: A gets the F-statistic wrong — dividing $30/2 = 15$ by $120/24 = 5$ gives $3.00$, not $6.00$. Answer A likely results from forgetting to divide the numerator by $q = 2$, treating the raw difference $RSS_R - RSS_F = 30$ over $5$ instead. B correctly computes the F-statistic as $3.00$ but uses $n - p - 1 = 23$ as the denominator degrees of freedom — a confusion with the slope-only degrees of freedom adjustment sometimes seen in simple regression. The denominator df is simply $n - p$, not $n - p - 1$. D inverts the F-ratio entirely, yielding $1/3 \approx 0.333$, and also reverses the degrees of freedom order to $F_{24,2}$, which is a fundamental structural error. A reliable memory anchor: the denominator degrees of freedom equal $n - \text{rank}(X)$, and the numerator degrees of freedom equal the number of restrictions $q$. Never subtract an extra 1 unless the model has a specific intercept adjustment built in separately.

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