Deriving the One-Sample T Interval

From a Sampling Distribution to an Interval

A confidence interval for a population mean is not an arbitrary range centered on the sample mean. Its form comes from a probability statement about how sample means vary across repeated samples. When the population standard deviation is unknown, the sample standard deviation is used to estimate the spread of the sample mean. That substitution leads to a t statistic rather than a standard normal statistic.

The goal here is to derive the familiar one-sample t interval from the central t distribution. The key step is to begin with a probability statement that captures the middle $1-\alpha$ of that distribution, then rearrange the inequalities so that the unknown population mean is between two sample-based endpoints.

Model and Notation

Let $X_1,\ldots,X_n$ be a random sample of $n$ independent observations from a normal population with mean $\mu$ and variance $\sigma^2$. The population must be nondegenerate, meaning $\sigma>0$, and the sample size must satisfy $n\geq 2$. The population standard deviation is unknown.

The sample mean and sample standard deviation are

\bar{x}=\frac{1}{n}\sum_{i=1}^{n}x_i, \qquad s=\sqrt{\frac{\sum_{i=1}^{n}(x_i-\bar{x})^2}{n-1}}.

The denominator $n-1$ in the sample variance is also the degrees of freedom for this one-sample t procedure: $df=n-1$. The estimated standard error of the sample mean is $s/\sqrt{n}$. Since $n\geq2$ and the population is continuous and nondegenerate, the sample observations are not all identical with probability 1; thus $s>0$ with probability 1.

Model for the exact one-sample t interval: The observations are independent and identically distributed from a normal population with unknown mean $\mu$ and unknown standard deviation $\sigma>0$, and the sample size is $n\geq2$. Under this model, the central t statistic is defined and has a t distribution with $n-1$ degrees of freedom.

The Central t Statistic

The central t statistic compares the distance between the sample mean and the population mean with the estimated standard error:

T=\frac{\bar{X}-\mu}{S/\sqrt{n}}, \qquad T\sim t_{n-1}.

Here $\bar{X}$ and $S$ are random quantities across repeated samples; in a particular observed sample, they are reported as $\bar{x}$ and $s$. The statistic is called central because its distribution is centered at zero when the hypothesized value in the numerator is the true population mean.

The result $T\sim t_{n-1}$ follows from two standard facts under the stated normal-sample model. First, the standardized sample mean $Z=(\bar{X}-\mu)/(\sigma/\sqrt{n})$ has a standard normal distribution. Second, the scaled sample variance $Q=(n-1)S^2/\sigma^2$ has a chi-square distribution with $n-1$ degrees of freedom, and $Z$ and $Q$ are independent. The t distribution is defined by the ratio of an independent standard normal variable to the square root of an independent chi-square variable divided by its degrees of freedom.

T = \frac{Z}{\sqrt{Q/(n-1)}} = \frac{(\bar{X}-\mu)/(\sigma/\sqrt{n})} {\sqrt{S^2/\sigma^2}} = \frac{\bar{X}-\mu}{S/\sqrt{n}}.

In the second equality, $\sqrt{Q/(n-1)}=S/\sigma$, and $\sigma>0$. This shows why the unknown $\sigma$ disappears from the statistic, while the extra uncertainty from estimating it is reflected in the t distribution.

Deriving the Interval

For a chosen confidence level $1-\alpha$, let $t^*$ be the positive t critical value with $df=n-1$ such that the central area under the t curve is $1-\alpha$. Equivalently, each tail has area $\alpha/2$:

P(-t^*\leq T\leq t^*)=1-\alpha.

Substitute the central t statistic for $T$. The resulting event is about the random sample mean and standard deviation, along with the fixed population mean. To solve for $\mu$, multiply by the positive quantity $S/\sqrt{n}$, then rearrange both sides. The multiplier is positive with probability 1 under the model, so the directions of the inequalities do not change.

\begin{aligned} P\left(-t^*\leq\frac{\bar{X}-\mu}{S/\sqrt{n}}\leq t^*\right)&=1-\alpha,\\ P\left(-t^*\frac{S}{\sqrt{n}}\leq \bar{X}-\mu\leq t^*\frac{S}{\sqrt{n}}\right)&=1-\alpha,\\ P\left(\bar{X}-t^*\frac{S}{\sqrt{n}}\leq\mu\leq \bar{X}+t^*\frac{S}{\sqrt{n}}\right)&=1-\alpha. \end{aligned}

The last line is the confidence interval statement. Once the sample has been observed, replace $\bar{X}$ and $S$ by the observed values $\bar{x}$ and $s$. The resulting endpoints are numbers calculated from the data; the unknown mean $\mu$ is the quantity the interval is designed to cover.

Theorem: Under the independent normal-sample model with unknown $\sigma>0$ and $n\geq2$, an exact two-sided confidence interval for $\mu$ with confidence level $1-\alpha$ is $$ \bar{x}\ \pm\ t^*\frac{s}{\sqrt{n}}, $$ where $t^*$ is the positive critical value from the t distribution with $df=n-1$ satisfying $P(-t^*\leq T\leq t^*)=1-\alpha$.

Worked Applications

Each example below uses a hypothetical sample summary. The normal-population and independence assumptions are stipulated for the example; the summary statistics by themselves cannot establish those assumptions. The t critical values are given so the calculations can focus on applying the derived result.

Worked Example: A 95% Interval for a Mean Time

In a hypothetical independent random sample of $n=10$ measurements from a normal population, the sample mean is $\bar{x}=12.4$ minutes and the sample standard deviation is $s=3.0$ minutes. Assume the population is nondegenerate and its standard deviation is unknown. Construct a 95% t interval for the population mean. For $df=9$, use $t^*=2.262$.

The sample size is at least 2, and the stipulated independent normal-sample model supports the exact t procedure. The estimated standard error is

\frac{s}{\sqrt{n}}=\frac{3.0}{\sqrt{10}}=0.9487\text{ minutes}.

Substitute the sample statistics and critical value into the derived interval:

12.4\pm2.262\left(\frac{3.0}{\sqrt{10}}\right) = 12.4\pm2.146.

The endpoints are $12.4-2.146=10.254$ minutes and $12.4+2.146=14.546$ minutes. The 95% interval is approximately $(10.25,\ 14.55)$ minutes. In repeated independent samples of size 10 from this normal population, intervals constructed this way would cover the fixed population mean about 95% of the time.

Worked Example: A 90% Interval for a Mean Score

A hypothetical independent sample of $n=16$ measurements from a normal population has $\bar{x}=68.2$ points and $s=8.4$ points. Assume the population is nondegenerate and the population standard deviation is unknown. Find a 90% t interval for the population mean. For $df=15$, use $t^*=1.753$.

Here $n=16\geq2$; the stated model supplies independence and normality. First calculate the estimated standard error:

\frac{s}{\sqrt{n}}=\frac{8.4}{\sqrt{16}}=\frac{8.4}{4}=2.10\text{ points}.

The interval obtained from the central t probability statement is

68.2\pm1.753(2.10) = 68.2\pm3.6813.

Thus the lower endpoint is $68.2-3.6813=64.5187$, and the upper endpoint is $68.2+3.6813=71.8813$. The 90% interval is approximately $(64.52,\ 71.88)$ points. Its confidence level is 90%, not a probability assigned to the fixed population mean after the endpoints have been calculated.

Worked Example: A 99% Interval for a Mean Concentration

Suppose a hypothetical independent sample of $n=25$ measurements from a normal population has a sample mean of $\bar{x}=4.60$ milligrams per liter and a sample standard deviation of $s=0.75$ milligrams per liter. Assume the population is nondegenerate and its standard deviation is unknown. Construct a 99% t interval. For $df=24$, use $t^*=2.797$.

The assumptions are stipulated, and the sample size is at least 2. Calculate the estimated standard error:

\frac{s}{\sqrt{n}}=\frac{0.75}{\sqrt{25}}=\frac{0.75}{5} =0.150\text{ milligrams per liter}.

Using the 99% critical value gives

4.60\pm2.797(0.150) = 4.60\pm0.41955.

The lower endpoint is $4.60-0.41955=4.18045$, and the upper endpoint is $4.60+0.41955=5.01955$. Rounded to three decimal places, the interval is $(4.180,\ 5.020)$ milligrams per liter. The endpoints are farther from the sample mean than they would be at a lower confidence level because the selected central t area is larger.

What the Probability Statement Means

Before data are collected, $\bar{X}$ and $S$ vary from sample to sample, so the interval endpoints also vary. The derivation shows that, under the model, the probability that this random interval covers $\mu$ is $1-\alpha$. After the data are observed, the endpoints are fixed, while $\mu$ remains fixed and unknown. The confidence level describes the long-run coverage rate of the method, not the probability that the particular computed interval contains $\mu$.

The derivation also shows why the interval is centered on $\bar{x}$: the central t probability statement is symmetric around zero, and solving its two inequalities for $\mu$ places the sample mean midway between the endpoints. The distance from the center to either endpoint is determined by the t critical value and the estimated standard error.

Common Mistake: Leaving Out a Required Condition

A tempting but incomplete statement is: “If the observations come from a normal population, the one-sample t interval is valid.” This omits the requirement $n\geq2$. With $n=1$, the sample variance formula has denominator $n-1=0$; there is no defined sample standard deviation or $df=0$ t distribution for this procedure. The correction is to state the full model: independent observations from a nondegenerate normal population, unknown $\sigma$, and $n\geq2$. Under that model, the sample standard deviation is positive with probability 1 and the central t statistic is defined.

Another common error is to write “there is a 95% probability that the population mean is between these calculated endpoints.” That treats the fixed population mean as random after observing the data. Instead, say that the interval is produced by a method with 95% long-run coverage under its assumptions, and report the calculated range with its units.

Key takeaway: The one-sample t interval is obtained by inverting the central t statement. For independent observations from a nondegenerate normal population, with unknown $\sigma$ and $n\geq2$, the resulting interval is $\bar{x}\pm t^*s/\sqrt{n}$, using $df=n-1$.

Check Your Understanding

Use the derivation and assumptions to answer these questions.

In the central t statistic, what quantity replaces the unknown population standard deviation in the denominator?
Why must the sample size be at least 2 for the one-sample t interval?
For a 95% interval with $n=12$, what are the degrees of freedom, and what central probability statement defines $t^*$?
Suppose a hypothetical sample has $\bar{x}=20$, $s=4$, and $n=16$. Write the t interval expression in terms of its critical value $t^*$, without selecting that critical value.
Explain why the confidence level does not mean there is a 95% probability that the fixed population mean lies in a particular computed interval.