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Topics Covered

Standard Normal Area Property Population & Sample Parameter / Statistic Sampling Distribution Student's t F-Distribution Chi-square Standard Error Distribution of s² Moments & Mode Limiting Forms
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  1. 1. Standard Normal Distribution
  2. 2. Population, Sample, Parameter, Statistic
  3. 3. Chi-square (χ²) Distribution
  4. 4. Student's t-Distribution
  5. 5. F-Distribution (Snedecor's F)
  6. 6. Relationships among Sampling Distributions
  7. 7. Limiting Cases of the Sampling Distributions
  8. Key Take-aways

1. Standard Normal Distribution

DEFINITION

If \(X \sim N(\mu, \sigma^2)\), the standardized variable \(Z = \dfrac{X - \mu}{\sigma}\) follows the Standard Normal distribution \(N(0, 1)\) with PDF

\[ \phi(z) \;=\; \dfrac{1}{\sqrt{2\pi}} e^{-z^2/2}, \quad -\infty < z < \infty. \]

MGF, Mean & Variance

\[ M_Z(t) = e^{t^2/2}, \qquad E(Z) = 0, \qquad \text{Var}(Z) = 1. \]

CDF

\(\Phi(z) = P(Z \le z) = \int_{-\infty}^{z}\phi(u)\,du\). Tabulated; no closed form.

Area Property (Empirical Rule)

−3σ −2σ −1σ μ +1σ +2σ +3σ 68.27% 95.45% 99.73% Standard Normal Z ~ N(0, 1)
Fig 5.1 — Area under the standard normal curve: the empirical rule (±1σ, ±2σ, ±3σ)

Worked Examples

EXAMPLE 1

If \(X \sim N(50, 25)\), find \(P(45 \le X \le 60)\).

\(z_1 = -1,\; z_2 = 2\). \(P = \Phi(2) - \Phi(-1) = 0.9772 - 0.1587 = 0.8185\).

EXAMPLE 2

Find \(P(|Z| > 1.96)\) — relevant for two-sided 5 % significance level.
\(P(|Z| > 1.96) = 2(1 - \Phi(1.96)) = 2(1 - 0.975) = 0.05\).

2. Population, Sample, Parameter, Statistic

DEFINITIONS

Sampling Distribution

The probability distribution of a statistic, computed over all possible samples of a fixed size \(n\), is called its sampling distribution. Its standard deviation is the standard error (SE).

Examples:

Parameters and statistics in symbols

Write the population units as \(X_1, X_2, \ldots, X_N\) and the sample units as \(x_1, x_2, \ldots, x_n\).

QuantityPopulation (parameter)Sample (statistic)
Mean\(\mu = \frac{1}{N}\sum_{i=1}^{N} X_i\)\(\bar x = \frac{1}{n}\sum_{i=1}^{n} x_i\)
Variance\(\sigma^2 = \frac{1}{N}\sum_{i=1}^{N} (X_i - \mu)^2\)\(s^2 = \frac{1}{n}\sum_{i=1}^{n} (x_i - \bar x)^2\)
Standard deviation\(\sigma = \sqrt{\sigma^2}\)\(s = \sqrt{s^2}\)
Proportion\(P = X/N\)\(p = x/n\)

Here \(X\) and \(x\) in the last row count the units with the attribute. This unit also uses the sample variance with divisor \(n - 1\),

\[ S^2 = \frac{1}{n-1}\sum_{i=1}^{n} (x_i - \bar x)^2, \qquad \text{so that} \qquad n s^2 = (n-1) S^2 . \]

A statistic is random because the sample is drawn at random: a different sample gives a different value. That is why every statistic has a sampling distribution.

Standard errors

The standard error (S.E.) of a statistic is the standard deviation of its sampling distribution. For a random sample (with replacement, or from a large population):

FORMULAS \[ \text{S.E.}(\bar x) = \frac{\sigma}{\sqrt n}, \qquad \text{S.E.}(\bar x_1 - \bar x_2) = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}, \] \[ \text{S.E.}(p) = \sqrt{\frac{PQ}{n}}, \qquad \text{S.E.}(p_1 - p_2) = \sqrt{\frac{P_1Q_1}{n_1} + \frac{P_2Q_2}{n_2}}, \qquad Q = 1 - P . \]

The formulas for differences need the two samples to be independent: the variance of a difference of independent statistics is the sum of their variances.

What “exact” sampling distribution means

For large samples the central limit theorem gives an approximate normal distribution for many statistics. For a sample from a normal population some statistics have a distribution that can be written down exactly for every sample size, however small. These are the exact sampling distributions:

  1. the chi-square (\(\chi^2\)) distribution,
  2. Student's and Fisher's \(t\) distribution,
  3. Snedecor's \(F\) distribution,
  4. Fisher's \(z\) distribution (\(z = \tfrac12 \log_e F\)).

Because they are exact, they are the basis of the small sample tests. Sections 3 to 5 study the first three.

3. Chi-square (χ²) Distribution

DEFINITION

If \(Z_1, Z_2, \ldots, Z_n\) are independent \(N(0,1)\) variables, then

\[ \chi^2 \;=\; Z_1^2 + Z_2^2 + \cdots + Z_n^2 \]

follows a chi-square distribution with \(n\) degrees of freedom (df).

PDF

\[ f(\chi^2) \;=\; \dfrac{1}{2^{n/2}\,\Gamma(n/2)}\, (\chi^2)^{n/2 - 1}\, e^{-\chi^2/2}, \quad \chi^2 > 0. \]

This is a Gamma distribution \(G(n/2, 1/2)\).

Properties

Deriving the \(\chi^2\) density through the MGF

Setting. Let \(X_1, \ldots, X_n\) be independent with \(X_i \sim N(\mu_i, \sigma_i^2)\), and put \(z_i = (X_i - \mu_i)/\sigma_i\). Each \(z_i \sim N(0, 1)\), the \(z_i\) are independent, and \(\chi^2 = \sum z_i^2\).

  1. MGF of one \(z^2\). By definition, for \(t < \tfrac12\), \[ M_{z^2}(t) = E\big(e^{tz^2}\big) = \int_{-\infty}^{\infty} e^{tz^2}\,\frac{1}{\sqrt{2\pi}}e^{-z^2/2}\,dz = \frac{2}{\sqrt{2\pi}}\int_0^{\infty} e^{-\frac{(1-2t)}{2}z^2}\,dz , \] using that the integrand is even.
  2. Substitute \(y = z^2\), so \(dz = dy/(2\sqrt y)\) and \(y\) runs from 0 to \(\infty\): \[ M_{z^2}(t) = \frac{1}{\sqrt{2\pi}}\int_0^{\infty} e^{-\frac{(1-2t)}{2}y}\,y^{\frac12 - 1}\,dy = \frac{1}{\sqrt{2\pi}}\cdot\frac{\Gamma(\frac12)}{\big(\frac{1-2t}{2}\big)^{1/2}} = (1 - 2t)^{-1/2}, \] by the gamma integral \(\int_0^\infty e^{-ay}y^{k-1}dy = \Gamma(k)/a^k\) and \(\Gamma(\frac12) = \sqrt\pi\). The condition \(t < \frac12\) keeps \(a = (1-2t)/2\) positive.
  3. Multiply. The MGF of a sum of independent variables is the product of their MGFs: \[ M_{\chi^2}(t) = \prod_{i=1}^{n} (1 - 2t)^{-1/2} = (1 - 2t)^{-n/2} . \]
  4. Recognise it. \((1 - t/a)^{-\lambda}\) is the MGF of the gamma density \(\frac{a^\lambda}{\Gamma(\lambda)}e^{-ax}x^{\lambda-1}\). Here \(a = \frac12\) and \(\lambda = \frac n2\). By the uniqueness theorem of MGFs, \(\chi^2\) has the density stated above, \(\frac{(1/2)^{n/2}}{\Gamma(n/2)}e^{-\chi^2/2}(\chi^2)^{n/2-1}\). \(\blacksquare\)

Cumulants, skewness and kurtosis

The cumulant generating function is \(K(t) = \log M(t) = -\frac n2 \log(1 - 2t)\). Use the series \(-\log(1 - x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots\) with \(x = 2t\):

\[ K(t) = \frac n2\Big[2t + \frac{(2t)^2}{2} + \frac{(2t)^3}{3} + \frac{(2t)^4}{4} + \cdots\Big] = nt + n t^2 + \frac{4n}{3}t^3 + 2n t^4 + \cdots \]

The \(r\)-th cumulant \(\kappa_r\) is the coefficient of \(t^r/r!\), so multiply each coefficient by \(r!\):

\[ \kappa_1 = n, \qquad \kappa_2 = 2n, \qquad \kappa_3 = 8n, \qquad \kappa_4 = 48n . \]

Both \(\beta_1 \to 0\) and \(\beta_2 \to 3\) as \(n \to \infty\), the values for a normal curve, which agrees with the limiting result of section 7.

Characteristic function. The same gamma integral with \(it\) in place of \(t\) gives \(\phi(t) = E(e^{it\chi^2}) = (1 - 2it)^{-n/2}\). Unlike the MGF, it exists for every real \(t\).

Mode of \(\chi^2\)

Up to a constant, \(\log f(x) = \big(\frac n2 - 1\big)\log x - \frac x2\). Setting the derivative to zero,

\[ \frac{d}{dx}\log f(x) = \frac{n/2 - 1}{x} - \frac12 = 0 \quad\Longrightarrow\quad x = n - 2 . \]

The second derivative, \(-(n/2 - 1)/x^2\), is negative for \(n > 2\), so the mode is \(n - 2\) when \(n > 2\). For \(n = 1\) and \(n = 2\) the density decreases from \(x = 0\), so the mode is at 0.

Additive (reproductive) property, proved

Let \(\chi_1^2, \ldots, \chi_k^2\) be independent with \(n_1, \ldots, n_k\) degrees of freedom. The MGF of their sum is the product of their MGFs:

\[ M_{\chi_1^2 + \cdots + \chi_k^2}(t) = (1-2t)^{-n_1/2}\cdots(1-2t)^{-n_k/2} = (1 - 2t)^{-(n_1 + \cdots + n_k)/2}, \]

which is the MGF of \(\chi^2\) with \(n_1 + \cdots + n_k\) degrees of freedom. By uniqueness, \(\chi_1^2 + \cdots + \chi_k^2 \sim \chi^2_{n_1 + \cdots + n_k}\).

EXAMPLE 3 — the formulas at \(n = 10\)

For \(\chi^2_{10}\): mean \(= 10\), variance \(= 20\), \(\mu_3 = 80\), \(\mu_4 = 12 \times 10 \times 14 = 1680\), \(\beta_1 = 8/10 = 0.8\), \(\beta_2 = 3 + 12/10 = 4.2\), mode \(= 10 - 2 = 8\). Integrating the density numerically gives the same values.

Applications

  1. Test for population variance.
  2. Goodness of fit test.
  3. Test of independence in contingency tables.
  4. Inference about variance of a normal population: \((n-1)S^2/\sigma^2 = ns^2/\sigma^2 \sim \chi^2_{n-1}\).
  5. Testing the homogeneity of several independent estimates of a population variance.
  6. Testing the homogeneity of several independent estimates of a population correlation coefficient (through Fisher's \(z\)-transformation).
EXAMPLE 1

If \(\chi^2_{10}\) is observed, find the value below which 95 % of the area lies.

From tables, \(\chi^2_{0.95, 10} = 18.31\).

EXAMPLE 2

Variance of 25 sample observations from \(N(\mu, 16)\): \(S^2 = 18\). Test statistic \((n-1)S^2/\sigma^2 = 24 \cdot 18 / 16 = 27.0 \sim \chi^2_{24}\).

Three results from the \(\chi^2\) density

RESULT 1

If \(X \sim \chi^2_n\), then \(Y = X/2\) has the gamma density \(\dfrac{e^{-y}y^{n/2-1}}{\Gamma(n/2)}\), \(y \ge 0\): a gamma variate with parameter \(n/2\).

Proof. \(x = 2y\), so \(|J| = |dx/dy| = 2\). Then \(f_Y(y) = f_X(2y)\cdot 2 = \dfrac{e^{-y}(2y)^{n/2-1}\cdot 2}{2^{n/2}\Gamma(n/2)} = \dfrac{e^{-y}y^{n/2-1}}{\Gamma(n/2)}\), since \(2^{n/2-1}\cdot 2 = 2^{n/2}\). \(\blacksquare\)

RESULT 2

If \(X \sim \chi^2_{n_1}\) and \(Y \sim \chi^2_{n_2}\) are independent, then \(U = X/Y\) is a beta variate of the second kind with parameters \(n_1/2\) and \(n_2/2\):

\[ f(u) = \frac{1}{B(\frac{n_1}{2}, \frac{n_2}{2})}\,\frac{u^{n_1/2 - 1}}{(1 + u)^{(n_1+n_2)/2}}, \qquad 0 \le u < \infty . \]
  1. Transformation. Put \(u = x/y\), \(v = y\), so \(x = uv\), \(y = v\), and \(J = \begin{vmatrix} v & u \\ 0 & 1 \end{vmatrix} = v\).
  2. Joint density (independence lets us multiply): \[ f(u, v) = f_X(uv)\,f_Y(v)\,v = \frac{u^{n_1/2-1}\,v^{(n_1+n_2)/2 - 1}\,e^{-(1+u)v/2}}{2^{(n_1+n_2)/2}\,\Gamma(\frac{n_1}{2})\Gamma(\frac{n_2}{2})} . \]
  3. Integrate out \(v\) with the gamma integral, \(a = (1+u)/2\), \(k = (n_1+n_2)/2\): \[ f(u) = \frac{u^{n_1/2-1}}{2^{(n_1+n_2)/2}\Gamma(\frac{n_1}{2})\Gamma(\frac{n_2}{2})}\cdot\frac{\Gamma(\frac{n_1+n_2}{2})\,2^{(n_1+n_2)/2}}{(1+u)^{(n_1+n_2)/2}} , \] and \(\Gamma(\frac{n_1}{2})\Gamma(\frac{n_2}{2})/\Gamma(\frac{n_1+n_2}{2}) = B(\frac{n_1}{2}, \frac{n_2}{2})\) gives the result. \(\blacksquare\)
RESULT 3

With \(X\) and \(Y\) as in Result 2, \(U = X/(X+Y)\) is a beta variate of the first kind with parameters \(n_1/2\) and \(n_2/2\):

\[ f(u) = \frac{1}{B(\frac{n_1}{2}, \frac{n_2}{2})}\,u^{n_1/2-1}(1-u)^{n_2/2-1}, \qquad 0 \le u \le 1 . \]
  1. Transformation. \(u = x/(x+y)\), \(v = x + y\), so \(x = uv\), \(y = (1-u)v\), and \(J = \begin{vmatrix} v & u \\ -v & 1-u \end{vmatrix} = v(1-u) + uv = v\).
  2. Joint density. The exponents add: \(e^{-uv/2}e^{-(1-u)v/2} = e^{-v/2}\), so \[ f(u, v) = \frac{u^{n_1/2-1}(1-u)^{n_2/2-1}\,v^{(n_1+n_2)/2-1}e^{-v/2}}{2^{(n_1+n_2)/2}\Gamma(\frac{n_1}{2})\Gamma(\frac{n_2}{2})} . \]
  3. Integrate out \(v\): \(\int_0^\infty v^{(n_1+n_2)/2-1}e^{-v/2}dv = \Gamma(\frac{n_1+n_2}{2})\,2^{(n_1+n_2)/2}\), which leaves the stated density.
  4. Range. \(x \ge 0\) and \(y \ge 0\) give \(uv \ge 0\) and \((1-u)v \ge 0\) with \(v > 0\), so \(0 \le u \le 1\). \(\blacksquare\)

Distribution of the sample variance

THEOREM

If \(x_1, \ldots, x_n\) is a random sample from \(N(\mu, \sigma^2)\) and \(s^2 = \frac1n\sum(x_i - \bar x)^2\), then

\[ \frac{ns^2}{\sigma^2} = \frac{\sum (x_i - \bar x)^2}{\sigma^2} \sim \chi^2_{n-1} . \]
  1. Split the sum of squares. Add and subtract \(\bar x\): \[ \sum (x_i - \mu)^2 = \sum (x_i - \bar x)^2 + n(\bar x - \mu)^2 + 2(\bar x - \mu)\sum (x_i - \bar x) , \] and the last sum is 0, because the deviations from the mean add to zero.
  2. Divide by \(\sigma^2\) and name the three pieces: \[ \underbrace{\sum \Big(\frac{x_i - \mu}{\sigma}\Big)^2}_{V} = \underbrace{\frac{ns^2}{\sigma^2}}_{W} + \underbrace{\Big(\frac{\bar x - \mu}{\sigma/\sqrt n}\Big)^2}_{Z^2} . \]
  3. Identify \(V\) and \(Z^2\). Each \((x_i - \mu)/\sigma \sim N(0, 1)\), independently, so \(V \sim \chi^2_n\). Since \(\bar x \sim N(\mu, \sigma^2/n)\), the ratio \((\bar x - \mu)/(\sigma/\sqrt n) \sim N(0, 1)\), so \(Z^2 \sim \chi^2_1\).
  4. Use independence. For a normal sample, \(\bar x\) and \(s^2\) are independent (proved by the Helmert transformation in Distribution Theory, Unit 3), so \(W\) and \(Z^2\) are independent and \(M_V(t) = M_W(t)\,M_{Z^2}(t)\).
  5. Solve for \(M_W\): \[ M_W(t) = \frac{(1 - 2t)^{-n/2}}{(1 - 2t)^{-1/2}} = (1 - 2t)^{-(n-1)/2}, \] the MGF of \(\chi^2_{n-1}\). By uniqueness, \(W = ns^2/\sigma^2 \sim \chi^2_{n-1}\). One degree of freedom is lost because the deviations \(x_i - \bar x\) satisfy one linear restriction. \(\blacksquare\)

Density of \(s^2\). Put \(y = ns^2/\sigma^2\), so \(|dy/ds^2| = n/\sigma^2\), and use the \(\chi^2_{n-1}\) density of \(y\):

\[ f(s^2) = \frac{\big(\frac{n}{2\sigma^2}\big)^{(n-1)/2}}{\Gamma\big(\frac{n-1}{2}\big)}\,e^{-ns^2/(2\sigma^2)}\,(s^2)^{(n-3)/2}, \qquad 0 \le s^2 < \infty . \]

Mean and variance of \(s^2\). From \(E(W) = n - 1\) and \(\text{Var}(W) = 2(n-1)\), with \(W = ns^2/\sigma^2\):

\[ E(s^2) = \frac{n-1}{n}\sigma^2 = \Big(1 - \frac1n\Big)\sigma^2, \qquad \text{Var}(s^2) = \frac{2(n-1)}{n^2}\sigma^4 = \frac{2\sigma^4}{n}\Big(1 - \frac1n\Big) . \]

So \(s^2\) slightly underestimates \(\sigma^2\) (which is why \(S^2\), with divisor \(n - 1\), is used for estimation), and for large \(n\), \(E(s^2) \approx \sigma^2\) and \(\text{Var}(s^2) \approx 2\sigma^4/n\).

EXAMPLE 4 — samples of 10 from \(N(\mu, 4)\)

\(E(s^2) = \frac{9}{10}\times 4 = 3.6\) and \(\text{Var}(s^2) = \frac{2 \times 9}{100}\times 16 = 2.88\). A simulation of 200,000 samples of 10 from \(N(0, 4)\) gives a mean of 3.598 and a variance of 2.865 for \(s^2\), in agreement.

4. Student's t-Distribution

DEFINITION

If \(Z \sim N(0,1)\) and \(\chi^2 \sim \chi^2_n\) are independent, then

\[ t \;=\; \dfrac{Z}{\sqrt{\chi^2 / n}} \;\sim\; t_n \]

follows the Student's t-distribution with \(n\) degrees of freedom.

PDF

\[ f(t) \;=\; \dfrac{1}{\sqrt{n}\, B(1/2, n/2)}\,\left(1 + \dfrac{t^2}{n}\right)^{-(n+1)/2}, \quad -\infty < t < \infty. \]

Properties

Student's \(t\) and Fisher's \(t\)

Student's \(t\) is a particular case of Fisher's. Take \(\xi = \dfrac{\bar x - \mu}{\sigma/\sqrt n} \sim N(0,1)\) and \(\chi^2 = \dfrac{ns^2}{\sigma^2} \sim \chi^2_{n-1}\), which are independent. Then

\[ \frac{\xi}{\sqrt{\chi^2/(n-1)}} = \frac{(\bar x - \mu)\sqrt n/\sigma}{\sqrt{ns^2/(\sigma^2(n-1))}} = \frac{\bar x - \mu}{\sqrt{ns^2/(n(n-1))}} = \frac{\bar x - \mu}{S/\sqrt n}, \]

using \(ns^2 = (n-1)S^2\). The unknown \(\sigma\) cancels, which is what makes \(t\) usable in practice. So Student's \(t\) has Fisher's \(t\) distribution with \(n - 1\) degrees of freedom.

Deriving the \(t\) density

  1. Start from the independent densities \(f(\xi) = \frac{1}{\sqrt{2\pi}}e^{-\xi^2/2}\) and \(f(u) = \frac{1}{2^{n/2}\Gamma(n/2)}e^{-u/2}u^{n/2-1}\), where \(u = \chi^2\).
  2. Transformation. \(t = \xi/\sqrt{u/n}\) and \(u = u\), so \(\xi = t\sqrt{u/n}\) and \(J = \begin{vmatrix} \sqrt{u/n} & \frac{t}{2\sqrt{nu}} \\ 0 & 1 \end{vmatrix} = \sqrt{u/n}\).
  3. Joint density. \(f(t, u) = f\big(t\sqrt{u/n}\big)\,f(u)\,\sqrt{u/n}\); collecting the powers of \(u\) and the exponentials, \[ f(t, u) = \frac{1}{\sqrt{2\pi n}\;2^{n/2}\,\Gamma(\frac n2)}\;u^{\frac{n+1}{2}-1}\,e^{-\frac u2\left(1 + \frac{t^2}{n}\right)} . \]
  4. Integrate out \(u\) with the gamma integral, \(a = \frac12(1 + t^2/n)\), \(k = \frac{n+1}{2}\): \[ f(t) = \frac{\Gamma(\frac{n+1}{2})\,2^{(n+1)/2}}{\sqrt{2\pi n}\;2^{n/2}\,\Gamma(\frac n2)}\Big(1 + \frac{t^2}{n}\Big)^{-\frac{n+1}{2}} = \frac{\Gamma(\frac{n+1}{2})}{\sqrt{n\pi}\;\Gamma(\frac n2)}\Big(1 + \frac{t^2}{n}\Big)^{-\frac{n+1}{2}} . \]
  5. Write it with the beta function. \(B(\frac12, \frac n2) = \Gamma(\frac12)\Gamma(\frac n2)/\Gamma(\frac{n+1}{2})\) and \(\Gamma(\frac12) = \sqrt\pi\), so the constant is \(1/(\sqrt n\,B(\frac12, \frac n2))\), as stated above. \(\blacksquare\)

Another route. \(t^2/n = \xi^2/\chi^2\) is a ratio of independent \(\chi^2_1\) and \(\chi^2_n\) variates, so by Result 2 of section 3 it is a beta variate of the second kind, \(\beta_2(\frac12, \frac n2)\). Changing variable from \(t^2/n\) to \(t\) needs care: each value of \(t^2\) comes from two values, \(t\) and \(-t\), and the density is shared equally between them. That halving cancels the factor 2 in \(|d(t^2/n)/dt| = 2|t|/n\) and gives the same density.

Moments of \(t\)

No MGF. Only the moments of order less than \(n\) exist, and \(E(e^{st})\) is infinite for every \(s \ne 0\), because \(e^{st}\) grows faster than any power of \(t\) while the density falls only like \(|t|^{-(n+1)}\). So the \(t\) distribution has no moment generating function.

EXAMPLE 3 — the formulas at \(n = 10\)

\(\mu_2 = 10/8 = 1.25\), \(\mu_4 = 300/(8 \times 6) = 6.25\), \(\beta_2 = 3 \times 8/6 = 4\). Integrating \(t^2 f(t)\) and \(t^4 f(t)\) numerically gives 1.25 and 6.25.

Applications

  1. Inference about mean when SD is unknown (single mean, paired samples, two means with equal variances).
  2. Confidence intervals: \(\bar x \pm t_{\alpha/2, n-1}\, S/\sqrt n\).
  3. Tests of significance of a regression coefficient.
  4. Tests of significance of an observed sample correlation coefficient and of a partial correlation coefficient. (A multiple correlation coefficient is tested with \(F\), section 5.)
EXAMPLE 1

Sample of 16 from \(N(\mu, \sigma^2)\): \(\bar x = 50,\; s = 8\). Test \(H_0: \mu = 45\) at 5 %.

\(t = (50 - 45)/(8/\sqrt{16}) = 5/2 = 2.5\). df = 15. \(t_{0.025, 15} = 2.13\). Since \(2.5 > 2.13\), reject \(H_0\).

EXAMPLE 2

For df = 20, find probability that \(|t| > 2.086\). From tables, this is 0.05.

5. F-Distribution (Snedecor's F)

DEFINITION

If \(\chi^2_{n_1}\) and \(\chi^2_{n_2}\) are independent, then

\[ F \;=\; \dfrac{\chi^2_{n_1}/n_1}{\chi^2_{n_2}/n_2} \;\sim\; F_{n_1, n_2}. \]

PDF

\[ f(F) \;=\; \dfrac{(n_1/n_2)^{n_1/2}}{B(n_1/2,\, n_2/2)}\cdot \dfrac{F^{n_1/2 - 1}}{(1 + n_1 F/n_2)^{(n_1+n_2)/2}}, \quad F > 0. \]

Properties

Deriving the \(F\) density

  1. Start from independent \(X \sim \chi^2_{n_1}\) and \(Y \sim \chi^2_{n_2}\), and \(F = \dfrac{X/n_1}{Y/n_2}\).
  2. Transformation. \(F\) and \(u = y\), so \(x = \frac{n_1}{n_2}Fu\), \(y = u\), and \(J = \begin{vmatrix} \frac{n_1}{n_2}u & \frac{n_1}{n_2}F \\ 0 & 1 \end{vmatrix} = \frac{n_1}{n_2}u\).
  3. Joint density. \(f(F, u) = f_X\big(\frac{n_1}{n_2}Fu\big)\,f_Y(u)\,\frac{n_1}{n_2}u\), which collects to \[ f(F, u) = \frac{(n_1/n_2)^{n_1/2}\,F^{n_1/2-1}}{2^{(n_1+n_2)/2}\,\Gamma(\frac{n_1}{2})\Gamma(\frac{n_2}{2})}\;u^{\frac{n_1+n_2}{2}-1}\,e^{-\frac u2\left(1 + \frac{n_1}{n_2}F\right)} . \]
  4. Integrate out \(u\) with the gamma integral, \(a = \frac12(1 + \frac{n_1}{n_2}F)\), \(k = \frac{n_1+n_2}{2}\). The powers of 2 cancel and the gammas form \(B(\frac{n_1}{2}, \frac{n_2}{2})\), leaving the density stated above. \(\blacksquare\)

Moments of \(F\)

Substitute \(y = \frac{n_1}{n_2}F\) in \(\mu'_r = \int_0^\infty F^r f(F)\,dF\). The integral becomes \(\big(\frac{n_2}{n_1}\big)^r \int_0^\infty \frac{y^{n_1/2 + r - 1}}{(1+y)^{(n_1+n_2)/2}}\,dy\big/B(\frac{n_1}{2}, \frac{n_2}{2})\), a beta integral of the second kind:

\[ \mu'_r = \Big(\frac{n_2}{n_1}\Big)^r \frac{B(\frac{n_1}{2} + r, \frac{n_2}{2} - r)}{B(\frac{n_1}{2}, \frac{n_2}{2})} = \Big(\frac{n_2}{n_1}\Big)^r \frac{\Gamma(\frac{n_1}{2} + r)\,\Gamma(\frac{n_2}{2} - r)}{\Gamma(\frac{n_1}{2})\,\Gamma(\frac{n_2}{2})}, \qquad n_2 > 2r . \]

Mode of \(F\)

Up to a constant, \(\log f = \big(\frac{n_1}{2} - 1\big)\log F - \frac{n_1+n_2}{2}\log\big(1 + \frac{n_1}{n_2}F\big)\). Setting the derivative to zero and multiplying through by \(2F(1 + n_1F/n_2)\):

\[ (n_1 - 2)\Big(1 + \frac{n_1}{n_2}F\Big) = (n_1 + n_2)\frac{n_1}{n_2}F \quad\Longrightarrow\quad F = \frac{n_2(n_1 - 2)}{n_1(n_2 + 2)}, \qquad n_1 > 2 . \]

Both factors \(\frac{n_1-2}{n_1}\) and \(\frac{n_2}{n_2+2}\) are below 1, so the mode is always less than 1, while the mean \(\frac{n_2}{n_2-2}\) is above 1: the distribution is positively skewed.

Reciprocal property, proved

If \(F = \dfrac{X/n_1}{Y/n_2} \sim F_{n_1, n_2}\), then \(\dfrac1F = \dfrac{Y/n_2}{X/n_1}\) is again a ratio of independent \(\chi^2\) variates, each divided by its degrees of freedom, now with \(Y\) on top. So \(1/F \sim F_{n_2, n_1}\). This is why tables give only upper percentage points: the lower point is \(F_{1-\alpha}(n_1, n_2) = 1/F_{\alpha}(n_2, n_1)\).

EXAMPLE 3 — the formulas at \((n_1, n_2) = (5, 10)\)

Mean \(= 10/8 = 1.25\); \(\mu'_2 = \dfrac{100 \times 7}{5 \times 8 \times 6} = \dfrac{35}{12} = 2.917\); variance \(= \dfrac{2 \times 100 \times 13}{5 \times 64 \times 6} = \dfrac{65}{48} = 1.354\); mode \(= \dfrac{10 \times 3}{5 \times 12} = 0.5\). Numerical integration of the density gives the same values.

Applications

  1. Test of equality of two population variances.
  2. One-way and two-way ANOVA.
  3. Test of overall significance in regression.
  4. Testing the linearity of regression, and the significance of an observed multiple correlation coefficient or correlation ratio.
EXAMPLE 1

For two samples with \(s_1^2 = 16\) (\(n_1 = 11\)) and \(s_2^2 = 9\) (\(n_2 = 13\)): \(F = 16/9 = 1.78\) on df (10, 12). Critical value at 5 % is \(F_{0.05} = 2.75\). Since \(1.78 < 2.75\), variances are not significantly different.

EXAMPLE 2

For df (5, 10), \(F_{0.05} = 3.33\). The probability that \(F > 3.33\) is 0.05.

χ² (n df) 0 5 10 15 20 n = 2 (mode 0) n = 4 (mode 2) n = 8 (mode 6) t (n df) and N(0, 1) -4 -2 0 2 4 N(0, 1) t, n = 2 t, n = 10 F (n₁, n₂ df) 0 1 2 3 4 F(5, 10), mode 0.5 F(10, 30), mode 0.75
Fig 5.2 — The three exact sampling distributions, drawn from their densities. \(\chi^2\) is right-skewed with mode \(n - 2\) and moves right as \(n\) grows; \(t\) is symmetric with heavier tails than \(N(0, 1)\) and approaches it as \(n\) grows; \(F\) is right-skewed with its mode below 1.

6. Relationships among Sampling Distributions

Proof that \(t^2 \sim F_{1, n}\)

Write \(t = \xi/\sqrt{\chi^2/n}\). Squaring,

\[ t^2 = \frac{\xi^2/1}{\chi^2/n} . \]

The numerator \(\xi^2\) is the square of a standard normal variate, so \(\xi^2 \sim \chi^2_1\); it is independent of \(\chi^2 \sim \chi^2_n\). So \(t^2\) is a ratio of independent \(\chi^2\) variates, each divided by its degrees of freedom: \(t^2 \sim F_{1, n}\). In tables, \(t_{\alpha/2}(n)^2 = F_{\alpha}(1, n)\); for example \(2.228^2 = 4.96 = F_{0.05}(1, 10)\).

7. Limiting Cases of the Sampling Distributions

STUDENT'S t → NORMAL

As the degrees of freedom \(n \to \infty\), Student's \(t\) tends to the standard normal:

\[ t_n \;=\; \dfrac{Z}{\sqrt{\chi^2_n / n}} \;\xrightarrow{d}\; N(0, 1). \]

Reason: \(\chi^2_n / n\) is the mean of \(n\) i.i.d. terms \(Z_i^2\) (each with mean 1), so by the law of large numbers \(\chi^2_n / n \to 1\); the denominator \(\to 1\) and hence \(t_n \to Z\). The heavy tails of \(t\) thin out to match the normal (in practice \(t_n \approx Z\) for \(n \ge 30\)).

CHI-SQUARE → NORMAL

As \(n \to \infty\),

\[ \dfrac{\chi^2_n - n}{\sqrt{2n}} \;\xrightarrow{d}\; N(0, 1), \qquad\text{equivalently}\qquad \chi^2_n \approx N(n,\, 2n). \]

Reason: \(\chi^2_n = \sum_{i=1}^{n} Z_i^2\) is a sum of \(n\) i.i.d. variables, each with mean 1 and variance 2, so the Central Limit Theorem applies. (Fisher's sharper approximation: \(\sqrt{2\chi^2_n} \approx N(\sqrt{2n - 1},\, 1)\).)

F → CHI-SQUARE

For fixed numerator degrees of freedom \(n_1\), as the denominator df \(n_2 \to \infty\),

\[ n_1\, F_{n_1, n_2} \;\xrightarrow{d}\; \chi^2_{n_1}. \]

Reason: \(F_{n_1, n_2} = \dfrac{\chi^2_{n_1}/n_1}{\chi^2_{n_2}/n_2}\); as \(n_2 \to \infty\), \(\chi^2_{n_2}/n_2 \to 1\), so \(F_{n_1, n_2} \to \chi^2_{n_1}/n_1\). If in addition \(n_1 \to \infty\), then \(F_{n_1, n_2} \to 1\).

The limits proved from the generating function and the densities

\(\chi^2 \to\) normal, through the CGF. Let \(z = (\chi^2 - n)/\sqrt{2n}\). Then \(M_z(t) = e^{-tn/\sqrt{2n}}M_{\chi^2}(t/\sqrt{2n}) = e^{-t\sqrt{n/2}}\big(1 - t\sqrt{2/n}\big)^{-n/2}\), and

\[ K_z(t) = -t\sqrt{\tfrac n2} + \frac n2\Big[t\sqrt{\tfrac 2n} + \frac{t^2}{2}\cdot\frac 2n + \frac{t^3}{3}\Big(\frac 2n\Big)^{3/2} + \cdots\Big] = \frac{t^2}{2} + \frac{t^3}{3}\sqrt{\frac 2n} + \cdots \]

The first two terms cancel, and every term after \(t^2/2\) has a power of \(\sqrt n\) in its denominator, so \(K_z(t) \to t^2/2\), the CGF of \(N(0, 1)\). By the continuity theorem for MGFs, \(z\) tends in distribution to \(N(0, 1)\).

LEMMA

For fixed \(K\), \(\dfrac{\Gamma(n + K)}{\Gamma(n)\,n^K} \to 1\) as \(n \to \infty\); that is, \(\Gamma(n + K)/\Gamma(n)\) grows like \(n^K\).

Proof. Stirling's formula, \(\Gamma(m) \approx \sqrt{2\pi}\,m^{m - 1/2}e^{-m}\), gives

\[ \frac{\Gamma(n+K)}{\Gamma(n)\,n^K} \approx \frac{(n+K)^{n+K-\frac12}e^{-K}}{n^{n-\frac12}\,n^K} = \Big(1 + \frac Kn\Big)^{n}\Big(1 + \frac Kn\Big)^{K - \frac12}e^{-K} \to e^K\cdot 1\cdot e^{-K} = 1, \]

using \((1 + K/n)^n \to e^K\) and \((1 + K/n)^{c} \to 1\) for a fixed exponent \(c\). (With an exponent that grows with \(n\) the second limit fails: \((1 + 1/n)^n \to e\), not 1.) \(\blacksquare\)

\(t \to\) normal, through the density. In \(f(t) = \dfrac{\Gamma(\frac{n+1}{2})}{\sqrt{n\pi}\,\Gamma(\frac n2)}\big(1 + \frac{t^2}{n}\big)^{-n/2}\big(1 + \frac{t^2}{n}\big)^{-1/2}\):

So \(f(t) \to \frac{1}{\sqrt{2\pi}}e^{-t^2/2}\), the standard normal density.

\(n_1F \to \chi^2_{n_1}\), through the density. In the \(F\) density, by the lemma with \(n_2/2\) and \(K = n_1/2\), \(\Gamma(\frac{n_1+n_2}{2})/\Gamma(\frac{n_2}{2}) \approx (\frac{n_2}{2})^{n_1/2}\), so \((n_1/n_2)^{n_1/2}(n_2/2)^{n_1/2} = (n_1/2)^{n_1/2}\). Also \(\big(1 + \frac{n_1F}{n_2}\big)^{-n_2/2} \to e^{-n_1F/2}\) and \(\big(1 + \frac{n_1F}{n_2}\big)^{-n_1/2} \to 1\). Hence

\[ f(F) \to \frac{(n_1/2)^{n_1/2}}{\Gamma(n_1/2)}F^{n_1/2-1}e^{-n_1F/2} . \]

Now put \(x = n_1F\), with \(|dF/dx| = 1/n_1\): the density of \(x\) is \(\frac{(1/2)^{n_1/2}}{\Gamma(n_1/2)}e^{-x/2}x^{n_1/2-1}\), the \(\chi^2_{n_1}\) density.

EXAMPLE — watching the limits happen
Quantity\(n = 10\)\(n = 100\)\(n = 10\,000\)Limit
\(t\) density at \(t = 1\)0.23040.24080.2420\(\phi(1) = 0.2420\)
\(P\big(\frac{\chi^2_n - n}{\sqrt{2n}} \le 1\big)\)0.84750.84210.8414\(\Phi(1) = 0.8413\)

Key Take-aways