Why the One-Sample t Statistic Has a Student t Distribution

Why Estimating the Standard Error Changes the Distribution

For a sample mean, knowing the population standard deviation gives a standard normal statistic. When the population standard deviation is unknown, it is replaced by the sample standard deviation. That replacement matters: the denominator is now random, so the resulting statistic does not have a standard normal distribution. Under a normal sampling model, its exact distribution is Student \(t\), with degrees of freedom determined by the sample size.

The derivation depends on two facts working together. First, the standardized sample mean is standard normal. Second, the sample variance, after scaling, has a chi-square distribution and is independent of that standardized mean. Dividing the normal variable by the square root of the independent scaled chi-square variable gives the Student \(t\) distribution.

Model and definition: Let \(X_1,\ldots,X_n\) be independent observations from \(N(\mu,\sigma^2)\), where \(\mu\) and \(\sigma^2>0\) are unknown. Define the sample mean and sample variance by

\bar X=\frac{1}{n}\sum_{i=1}^n X_i, \qquad S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar X)^2.

The one-sample statistic is \(T=(\bar X-\mu)/(S/\sqrt n)\). Under the stated independent normal model, \(T\) has a Student \(t\) distribution with \(n-1\) degrees of freedom.

The normality assumption is important for this exact result. Independence alone, or a sample that is merely large, does not make the statistic exactly Student \(t\). For some nonnormal populations, a \(t\) procedure may still be a useful approximation, especially with a sufficiently large sample, but that is a separate approximation claim and not the derivation given here.

Building the Statistic from Normal and Chi-Square Variables

Begin by standardizing every observation using the population parameters. Let \(Y_i=(X_i-\mu)/\sigma\). Under the model, the \(Y_i\) are independent standard normal variables. Their average gives the standardized sample mean:

Z=\frac{\sqrt n(\bar X-\mu)}{\sigma} =\frac{1}{\sqrt n}\sum_{i=1}^n Y_i \sim N(0,1).

The sample variance is based on deviations from \(\bar X\), rather than deviations from the fixed population mean \(\mu\). The corresponding scaled residual sum of squares is \(V=(n-1)S^2/\sigma^2\). Under normal sampling, \(V\) has a chi-square distribution with \(n-1\) degrees of freedom. Crucially, \(Z\) and \(V\) are independent.

Theorem: If \(Z\sim N(0,1)\) and \(V\sim\chi^2_\nu\) are independent, then

\frac{Z}{\sqrt{V/\nu}}\sim t_\nu.

For a normal sample of size \(n\), \(\nu=n-1\), and this ratio is exactly the one-sample statistic \(T\).

Why the Normal and Chi-Square Parts Are Independent

The independence can be seen by changing coordinates in the standardized sample. The vector \(Y=(Y_1,\ldots,Y_n)\) has a multivariate standard normal distribution. Apply an orthogonal transformation whose first row is \((1/\sqrt n,\ldots,1/\sqrt n)\). Call the transformed coordinates \(U_1,\ldots,U_n\). An orthogonal transformation preserves the standard multivariate normal distribution, so these coordinates are independent standard normal variables.

The first coordinate is the standardized sample mean, \(U_1=Z\). Orthogonal transformations preserve the sum of squares, and the other coordinates represent the part of the data left after removing the sample-mean direction. Therefore, \(\sum_{j=2}^n U_j^2\) equals the standardized residual sum of squares. It is a sum of squares of \(n-1\) independent standard normal variables, so it has a \(\chi^2_{n-1}\) distribution. It is independent of \(U_1\), because the transformed coordinates are independent.

\frac{(n-1)S^2}{\sigma^2} =\sum_{i=1}^n\left(\frac{X_i-\bar X}{\sigma}\right)^2 =\sum_{j=2}^n U_j^2 \sim\chi^2_{n-1}, \qquad Z=U_1\ \perp\ V.

Here \(\perp\) denotes independence. The omitted coordinate is not a lost degree of freedom by convention: one coordinate has been used to determine the sample mean, leaving \(n-1\) independent normal directions to contribute to the residual sum of squares.

Deriving the Student t Density

The ratio theorem can also be justified directly by finding the density of the ratio. Write \(\nu=n-1\), and let \(T=Z/\sqrt{V/\nu}\). Since \(Z\) and \(V\) are independent, their joint density is the product of their densities. For a fixed \(v>0\), the change of variable \(z=t\sqrt{v/\nu}\) has Jacobian \(\sqrt{v/\nu}\). Integrating over all possible values of \(v\) gives

\begin{aligned} f_T(t) &=\int_0^\infty \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{t^2v}{2\nu}\right) \frac{v^{\nu/2-1}e^{-v/2}}{2^{\nu/2}\Gamma(\nu/2)} \sqrt{\frac{v}{\nu}}\,dv\\ &=\frac{1}{\sqrt{2\pi\nu}\,2^{\nu/2}\Gamma(\nu/2)} \int_0^\infty v^{(\nu+1)/2-1} \exp\left[-\frac{v}{2}\left(1+\frac{t^2}{\nu}\right)\right]\,dv. \end{aligned}

Use the gamma-integral identity \(\int_0^\infty v^{a-1}e^{-bv}\,dv=\Gamma(a)/b^a\), for \(a,b>0\). Here \(a=(\nu+1)/2\) and \(b=(1+t^2/\nu)/2\). Substituting these values and simplifying yields

f_T(t)= \frac{\Gamma((\nu+1)/2)} {\sqrt{\nu\pi}\,\Gamma(\nu/2)} \left(1+\frac{t^2}{\nu}\right)^{-(\nu+1)/2}, \qquad -\infty<t<\infty.

This is the Student \(t\) density with \(\nu\) degrees of freedom. It is symmetric about zero, like the standard normal density, but has heavier tails for finite \(\nu\). As \(\nu\) increases, the \(t\) distribution approaches the standard normal distribution. The finite-sample distribution is nevertheless \(t\), not normal, when the model assumptions hold and \(\sigma\) is estimated.

Worked Examples

Worked Example: Computing the Statistic from a Small Normal Sample

A hypothetical calibration study records five independent instrument errors, in units, as \(8,10,12,14,16\). Assume these observations are from \(N(\mu,\sigma^2)\), with both parameters unknown. Find the one-sample statistic for testing the reference value \(\mu=10\), and state its exact reference distribution under that value.

Method and assumptions. Use the sample standard deviation with divisor \(n-1\). The observations are assumed independent and identically distributed normal, and the null reference value is \(\mu_0=10\). The interest rate and mortality assumption are not applicable to this statistical sampling model.

Step 1: Find the sample mean. The sum is \(8+10+12+14+16=60\), so \(\bar X=60/5=12\).

Step 2: Find the sample variance. The deviations from 12 are \(-4,-2,0,2,4\). Their squared values sum to \(16+4+0+4+16=40\). Thus \(S^2=40/(5-1)=10\), and \(S=\sqrt{10}\).

Step 3: Calculate the statistic. The estimated standard error is \(S/\sqrt n=\sqrt{10}/\sqrt5=\sqrt2\). Therefore, \(T=(12-10)/\sqrt2=2/\sqrt2=\sqrt2\), which is approximately \(1.414\) to three decimal places.

Step 4: Identify its distribution. Under the null value and the normal model, \(T\sim t_4\), since \(n-1=4\). The calculation is consistent with the derived result: the unknown \(\sigma\) is replaced by \(S\), so the statistic is compared with \(t_4\), not with a standard normal distribution.

Common mistake avoided: dividing the squared deviations by 5 would give the variance with divisor \(n\), not the sample variance used in the one-sample \(t\) statistic. The required divisor is \(n-1=4\).

Worked Example: Constructing the Ratio from Its Independent Components

Suppose a hypothetical normal sample has size \(n=6\). In the normal-coordinate representation, a particular realization has standardized mean component \(Z=1.2\) and residual sum of squares \(V=10\). Compute the corresponding one-sample statistic and state the degrees of freedom.

Method and assumptions. For independent observations from \(N(\mu,\sigma^2)\), the representation is \(T=Z/\sqrt{V/(n-1)}\), where \(Z\sim N(0,1)\), \(V\sim\chi^2_{n-1}\), and the two are independent. The given values are treated as realized components of this representation; the rate and mortality assumptions are not applicable.

Step 1: Find the degrees of freedom. Here \(\nu=n-1=6-1=5\). Thus the scaled variance component is \(V/\nu=10/5=2\).

Step 2: Form the denominator. Its value is \(\sqrt{V/\nu}=\sqrt2\), approximately \(1.414214\).

Step 3: Divide the normal component by that denominator. The statistic is \(T=1.2/\sqrt2\), approximately \(0.848528\), or \(0.849\) to three decimal places.

Step 4: State the distributional result. Before observing these particular component values, the ratio has a \(t_5\) distribution under the model. The calculation uses the scaled chi-square term \(V/5\), not \(V\) alone.

Common mistake avoided: using \(1.2/\sqrt{10}\) omits the division of \(V\) by its degrees of freedom. The Student \(t\) denominator is \(\sqrt{V/\nu}\), so the correct denominator here is \(\sqrt{10/5}\).

Worked Example: An Exact Upper-Tail Probability When n Equals Two

A hypothetical measurement process produces two independent observations from \(N(\mu,\sigma^2)\), with both parameters unknown. For the one-sample statistic \(T\), find \(P(T>1)\) under the model.

Method and assumptions. The observations are independent and normally distributed with common mean \(\mu\) and common positive variance \(\sigma^2\). Here \(n=2\), so \(T\sim t_1\). The problem concerns a probability under this distribution; interest and mortality assumptions are not applicable.

Step 1: Determine the degrees of freedom. The degrees of freedom are \(n-1=1\).

Step 2: Use the \(t_1\) density. Setting \(\nu=1\) in the Student \(t\) density gives \(f_T(t)=1/[\pi(1+t^2)]\). Its upper-tail probability at 1 is

P(T>1)=\int_1^\infty\frac{1}{\pi(1+t^2)}\,dt =\left.\frac{\arctan(t)}{\pi}\right|_1^\infty =\frac{\pi/2-\pi/4}{\pi} =\frac14.

Thus \(P(T>1)=0.2500\) to four decimal places. As a check, the \(t_1\) density is symmetric, and the area to the right of zero is \(1/2\); the area from 1 to infinity is less than that and equals \(1/4\).

Common mistake avoided: using a standard normal tail probability would give the wrong answer. The exact reference distribution here is \(t_1\), which has substantially heavier tails than the standard normal distribution.

What the Degrees of Freedom and Assumptions Tell You

The degrees of freedom are \(n-1\) because the sample mean is estimated from the same observations used to calculate the residual sum of squares. The residuals satisfy \(\sum_{i=1}^n(X_i-\bar X)=0\), so only \(n-1\) of them can vary independently. In the normal-coordinate argument, the same fact appears as one coordinate assigned to the sample mean and \(n-1\) coordinates left for the residual variation.

The exact Student \(t\) result is not merely a statement that the sample variance is random. It requires the normal-theory independence between the sample mean and sample variance. For a general population, those quantities need not be independent, and the scaled sample variance need not be chi-square. Consequently, replacing \(\sigma\) by \(S\) does not by itself prove that a statistic has a Student \(t\) distribution.

With known \(\sigma\), the standardized sample mean uses \(\sigma/\sqrt n\) and has a standard normal distribution under normal sampling.
With unknown \(\sigma\), the statistic uses \(S/\sqrt n\) and has a \(t_{n-1}\) distribution under independent normal sampling.
The \(t\) distribution is centered at zero and symmetric, but has heavier tails than the standard normal for finite degrees of freedom.
The exact result depends on both the chi-square distribution of the scaled sample variance and its independence from the standardized sample mean.

The derivation explains why the one-sample \(t\) reference distribution is appropriate when the population variance is unknown. It is aligned with the SOA VEE Mathematical Statistics syllabus because it connects normal sampling, chi-square variation, independence, and the Student \(t\) distribution in one result.

Check Your Understanding

For each problem, use the stated assumptions and distinguish the exact distributional result from a numerical approximation.

A sample of nine independent observations comes from a normal population with unknown mean and variance. In the normal-coordinate representation, \(Z=-1.4\) and \(V=8\). Compute \(T\) and state its degrees of freedom. (Answer: \(T=-1.4\), with 8 degrees of freedom.)
For an independent normal sample of size 12, identify the distribution of \((11S^2)/\sigma^2\), and state whether it is independent of \(\sqrt{12}(\bar X-\mu)/\sigma\). (Answer: \(\chi^2_{11}\), and yes, the two quantities are independent.)
A sample of four independent normal observations has \(\bar X=7\), \(S^2=3\), and the reference mean is \(\mu_0=5\). Calculate the one-sample statistic and its degrees of freedom. (Answer: \(T=2/\sqrt{3/4}=4/\sqrt3\approx2.309\), with 3 degrees of freedom.)
An analyst has \(n=5\) independent observations but only knows that their population has a finite mean and variance; normality is not assumed. Is the statistic guaranteed to have an exact \(t_4\) distribution? (Answer: No; the exact result requires the independent normal sampling model.)
For a normal sample of size two, use the \(t_1\) density to find \(P(T<-1)\). (Answer: \(0.2500\).)

What You'll Learn