When Sigma Is Not Written on the Blackboard

The textbook’s normal-confidence-interval example gives us the population standard deviation. That is convenient and, outside a classroom exercise, rather suspicious.

In actual studies, the population does not generally hand us its unknown parameter, $\sigma$, with a helpful label. We estimate it from the sample. That extra uncertainty is why the t distribution matters.

Replace unknown sigma with sample information

For a sample of $n$ observations with mean $\bar x$ and sample standard deviation $s$, the estimated standard error of the mean is

$$\widehat{\operatorname{SE}}(\bar X)=\frac{s}{\sqrt n}.$$

To test a hypothesized population mean $\mu_0$, use

$$t=\frac{\bar x-\mu_0}{s/\sqrt n}.$$

For independent observations from a normal population, this statistic follows a Student t distribution with $n-1$ degrees of freedom when the null is true. Compared with the standard normal distribution, the t distribution has heavier tails, especially when $n$ is small. As the degrees of freedom increase, it approaches the normal.

A numerical example

Five synthetic readings are 2, 4, 6, 8, and 10. Their mean is 6. Their sample standard deviation is $\sqrt{10}\approx3.1623$.

The estimated standard error is $\sqrt{10/5}=\sqrt2\approx1.4142$. Testing $H_0:\mu=4$ yields

$$t=\frac{6-4}{\sqrt2}\approx1.4142$$

with four degrees of freedom.

A two-sided 95% confidence interval is

$$\bar x\pm t_{.975,n-1}\frac{s}{\sqrt n}.$$

Using $t_{.975,4}\approx2.7764$, the interval is $6\pm3.9265$, approximately [2.07, 9.93]. It is wide because five observations cannot tell us very precisely where the population mean lies.

Why the design still matters

With small samples, the exact t reference distribution depends on the normal-population model. Outliers, extreme skew, dependence, and selection bias can make the nominal procedure inappropriate.

A larger sample can make a t procedure more robust to moderate nonnormality under appropriate conditions, but it does not turn a systematically biased sample into a representative one.

Reproduce it in R

x <- c(2, 4, 6, 8, 10)
t.test(x, mu = 4)

Read the estimate, interval, test statistic, degrees of freedom, and p-value as parts of one inferential argument. None is a replacement for the question you started with.

Check your understanding

1. Why is the t critical value at $n=5$ larger than the corresponding normal critical value?
2. How many degrees of freedom does a one-sample t test have with 20 observations?
3. Does a t test repair nonresponse bias?

Worked responses

1. The population spread must be estimated from a small sample, adding uncertainty represented by heavier tails.

2. Nineteen.

3. No. An inferential formula does not correct a defective sample-selection mechanism.

Reference: R documentation for t.test.

Why the reference distribution has heavier tails

For independent normal observations, $Z=(\bar X-\mu)/(\sigma/\sqrt n)$ is standard normal. The scaled sample variance $U=(n-1)S^2/\sigma^2$ has a chi-square distribution with $n-1$ degrees of freedom and is independent of $Z$. Their ratio $Z/\sqrt{U/(n-1)}$ is the t statistic. A small denominator can produce unusually large ratios, accounting for the heavier tails. The independence and exact distribution claims here depend on the normal-population model; they are not consequences of merely computing a sample standard deviation.

The degrees-of-freedom count also has an algebraic explanation. Once the sample mean has been estimated, the centered deviations sum to zero. Specifying $n-1$ of them determines the last. That constraint explains the variance denominator, while the normal model supplies the exact distributional theorem.

A two-sided test and its matching confidence interval invert the same inequality. Reject $H_0:\mu=\mu_0$ at level $\alpha$ when $|t|>t_{1-\alpha/2,n-1}$. Rearranging gives exactly the parameter values outside $\bar x\pm t_{1-\alpha/2,n-1}s/\sqrt n$. Thus zero inclusion and a two-sided test decision agree only when the procedures, assumptions, and confidence levels actually match.

Additional practice with solutions

1. A synthetic sample has $n=16$, mean $42$, and standard deviation $8$. Find the standard error and statistic for $H_0:\mu=38$.
2. With $t_{.975,15}\approx2.13145$, calculate a 95% interval.
3. Would this interval establish that every individual measurement lies in its range?
4. Explain why repeated readings on one instrument may not give sixteen independent observations.

Worked additional solutions

1. $SE=8/4=2$ and $t=(42-38)/2=2$, with 15 degrees of freedom.
2. $42\pm2.13145(2)$, approximately $[37.7371,46.2629]$. The value 38 lies inside, so the matching two-sided 5% test does not reject it.
3. No. The interval concerns the population mean, not individual spread or prediction.
4. Serial correlation or shared calibration error can connect readings. Their information content then differs from independent sampling; increasing the reading count alone does not remove shared error.