Confidence Intervals for One Mean: Unknown Population Variance

When constructing a confidence interval for the population mean $\mu$, the formula $\bar{x} \pm z^* \cdot \dfrac{\sigma}{\sqrt{n}}$ requires knowing the population standard deviation $\sigma$. In practice $\sigma$ is almost never known, so we replace it with the sample standard deviation $s$.

This substitution introduces extra uncertainty, so we cannot use the normal critical value $z^*$. Instead, we use a critical value from the Student's t-distribution with $n-1$ degrees of freedom:

$$\bar{x} \pm t^*_{n-1} \cdot \dfrac{s}{\sqrt{n}}$$

Here:

• $\bar{x}$ is the sample mean,
• $s$ is the sample standard deviation,
• $n$ is the sample size,
• $t^*_{n-1}$ is the critical value from the t-distribution with $\text{df} = n-1$ degrees of freedom.

The quantity $\dfrac{s}{\sqrt{n}}$ is the estimated standard error, and $t^* \cdot \dfrac{s}{\sqrt{n}}$ is the margin of error.

For example, suppose a random sample of $n = 16$ measurements has $\bar{x} = 10$ and $s = 4$. For a 95% confidence interval, $\text{df} = 15$ and $t^*_{15} = 2.131$. The margin of error is

$$\text{ME} = 2.131 \cdot \dfrac{4}{\sqrt{16}} = 2.131 \cdot 1 = 2.131,$$

so the confidence interval is $(10 - 2.131,\ 10 + 2.131) = (7.869,\ 12.131)$.