Statistics Graduate Level Quiz: Standard Distribution Results

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Standard Distribution ResultsQuestion 1 of 10

Let $X_1,\ldots,X_{12}$ be independent observations from $N(\mu,\sigma^2)$ , and let $S^2$ be the usual unbiased sample variance.

Under the null hypothesis $\mu=\mu_0$ , what is the distribution of $W=\frac{12(\bar X-\mu_0)^2}{S^2}$ ?

W\sim\chi^2_1

because the numerator is a squared standard normal variable

W\sim F_{1,11}

because it is the square of a

t_{11}

variable

W\sim F_{11,1}

because the variance estimate contributes

11

numerator degrees of freedom

W\sim t_{11}

because the unknown variance is replaced by the sample variance

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Standard Distribution Results

Practice Standard Distribution Results in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Standard Distribution Results, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Let $X_1,\ldots,X_{12}$ be independent observations from $N(\mu,\sigma^2)$ , and let $S^2$ be the usual unbiased sample variance.

Under the null hypothesis $\mu=\mu_0$ , what is the distribution of $W=\frac{12(\bar X-\mu_0)^2}{S^2}$ ?

$W\sim\chi^2_1$ because the numerator is a squared standard normal variable
$W\sim F_{1,11}$ because it is the square of a $t_{11}$ variable (correct answer)
$W\sim F_{11,1}$ because the variance estimate contributes $11$ numerator degrees of freedom
$W\sim t_{11}$ because the unknown variance is replaced by the sample variance

Explanation: 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

\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.

Question 2

Two independent normal samples satisfy $X_i\sim N(\mu_X,\sigma^2)$ for $i=1,\ldots,8$ and $Y_j\sim N(\mu_Y,4\sigma^2)$ for $j=1,\ldots,10$ . Define $S_p^2=\frac{\sum_{i=1}^8(X_i-\bar X)^2+\frac14\sum_{j=1}^{10}(Y_j-\bar Y)^2}{16}$ .

Under $H_0:\mu_X-\mu_Y=\delta_0$ , which statistic has an exact standard $t$ distribution?

$\frac{\bar X-\bar Y-\delta_0}{S_p\sqrt{1/8+1/10}}\sim t_{16}$ , treating both population variances as equal
$\frac{\bar X-\bar Y-\delta_0}{S_p\sqrt{1/8+4/10}}\sim t_{16}$ , correctly accounting for the unequal population variances (correct answer)
$\frac{\bar X-\bar Y-\delta_0}{S_p\sqrt{1/8+4/10}}\sim t_{17}$ , using total sample size minus one as degrees of freedom
$\frac{\bar X-\bar Y-\delta_0}{S_p\sqrt{4/8+1/10}}\sim t_{16}$ , assigning the variance multiplier of $4$ to the wrong sample

Explanation: 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.

Question 3

In a normal linear model with $30$ observations, the full design matrix has rank $6$ . A null hypothesis imposes two linearly independent restrictions on the regression coefficients. The residual sums of squares from the full and restricted models are $120$ and $150$ , respectively.

What is the general linear hypothesis test statistic, and what is its null distribution?

$6.00$ , with null distribution $F_{2,24}$
$3.00$ , with null distribution $F_{2,23}$
$3.00$ , with null distribution $F_{2,24}$ (correct answer)
$0.333$ , with null distribution $F_{24,2}$

Explanation: 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

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.

Question 4

Suppose $X_1,\ldots,X_8$ are independent with distribution $N(\mu,\sigma^2)$ . A researcher evaluates $Q=\sum_{i=1}^8 (X_i-\mu_0)^2/\sigma^2$ , where the true mean satisfies $\mu=\mu_0+\sigma/2$ .

What is the distribution of $Q$ ?

A central chi-square distribution, $\chi^2_7$
A central chi-square distribution, $\chi^2_8$
A noncentral chi-square distribution, $\chi'^2_7(2)$
A noncentral chi-square distribution, $\chi'^2_8(2)$ (correct answer)

Explanation: 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.

Question 5

Let $U\sim\chi^2_5$ and $V\sim\chi^2_{12}$ be independent. Define $W=U+V$ and $R=\frac{12U}{5V}$ .

Which statement about $R$ and $W$ is correct?

$R\sim F_{5,12}$ , and $R$ is independent of $W$ (correct answer)
$R\sim F_{5,12}$ , but $R$ is not independent of $W$
$R\sim F_{12,5}$ , and $R$ is independent of $W$
$R\sim\chi^2_5$ , and $R$ is independent of $W$

Explanation: 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.

Question 6

Independent normal samples of sizes $10$ and $16$ yield a sample-variance ratio $S_1^2/S_2^2=2.4$ . Let $\rho=\sigma_1^2/\sigma_2^2$ , and let $q_p(a,b)$ denote the lower-tail probability- $p$ quantile of $F_{a,b}$ .

Which expression is an exact equal-tail $95\%$ confidence interval for $\rho$ ?

$\left[\frac{2.4}{q_{0.975}(9,15)},\frac{2.4}{q_{0.025}(9,15)}\right]$ (correct answer)
$\left[\frac{2.4}{q_{0.975}(15,9)},\frac{2.4}{q_{0.025}(15,9)}\right]$
$\left[2.4q_{0.025}(9,15),2.4q_{0.975}(9,15)\right]$
$\left[\frac{q_{0.025}(9,15)}{2.4},\frac{q_{0.975}(9,15)}{2.4}\right]$

Explanation: 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.

Question 7

Let $Z\sim N_{10}(0,I_{10})$ . Matrices $A$ and $B$ are symmetric idempotent matrices of ranks $3$ and $4$ , respectively, and satisfy $AB=0$ . Define $C=I_{10}-A-B$ .

What is the distribution of $T=\frac{(Z'AZ+Z'CZ)/6}{Z'BZ/4}$ ?

$T\sim F_{3,4}$ because only the quadratic form involving $A$ contributes numerator degrees of freedom
$T\sim F_{4,6}$ because the denominator rank becomes the first degrees-of-freedom parameter
$T\sim F_{6,4}$ because the numerator is a rank- $6$ orthogonal projection (correct answer)
$T\sim\chi^2_{10}$ because the three quadratic forms sum to $Z'Z$

Explanation: 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.

Question 8

A regression coefficient is tested using a statistic having distribution $t_\nu$ under the null hypothesis. The observed statistic is $t_{\mathrm{obs}}=-2.4$ . The equivalent one-restriction general linear hypothesis statistic is $F_{\mathrm{obs}}=(-2.4)^2=5.76$ . Let $p_F=P(F_{1,\nu}\ge 5.76)$ .

For the directional alternative that the coefficient is less than its null value, what is the exact one-sided $p$ -value?

$p_F$ , because the $F$ statistic already represents the relevant one-sided tail
$2p_F$ , because squaring the $t$ statistic removes its observed sign
$\frac{p_F}{2}$ , because the observed $t$ statistic has the alternative-consistent sign (correct answer)
$1-\frac{p_F}{2}$ , because the lower tail of $t_\nu$ corresponds to the upper tail of $F_{1,\nu}$

Explanation: 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

Question 9

For an $F$ distribution, let $q_p(a,b)$ denote its lower-tail probability- $p$ quantile with numerator and denominator degrees of freedom $a$ and $b$ . Suppose $c=q_{0.975}(7,11)$ .

Which identity follows from the reciprocal property of the $F$ distribution?

$\frac{1}{c}=q_{0.975}(11,7)$
$\frac{1}{c}=q_{0.025}(7,11)$
$c=q_{0.025}(11,7)$
$\frac{1}{c}=q_{0.025}(11,7)$ (correct answer)

Explanation: 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.

Question 10

Two independent samples are drawn from normal populations. The first has size $10$ and variance $\sigma^2$ ; the second has size $14$ and variance $9\sigma^2$ . Let $S_X^2$ and $S_Y^2$ denote the usual unbiased sample variances.

Which statistic has the stated distribution?

$\frac{9S_X^2}{S_Y^2}\sim F_{9,13}$ (correct answer)
$\frac{S_Y^2}{9S_X^2}\sim F_{9,13}$
$\frac{S_X^2}{9S_Y^2}\sim F_{9,13}$
$\frac{9S_X^2}{S_Y^2}\sim F_{10,14}$

Explanation: 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.