Raising the bar

Chapter 8. Estimation: One or Two Populations

Unit 8.3.2. Confidence interval for the population mean. Case of an unknown variance

Assumptions. A sample of size $latex n$ is drawn from a normal population. $latex \sigma_X$ is unknown and estimated by the sample standard deviation, $latex \mu_X$ is unknown and the objective is to obtain a confidence interval for $latex \mu_X$.

We apply a general statistical idea: when some parameter is unknown, replace it by its estimator. Because of Eq. (7.1) this idea leads to

$latex t = \frac{\bar X - \mu_X}{s_X/\sqrt n}. \qquad (\textrm{Eq. 8.5})$

This takes us to a new class of distributions.

Definition 8.6 ($latex t$-distribution). Let $latex z_0, z_1, \ldots, z_n$ be independent standard normal. Then the random variable

$latex t_n = \frac{z_0}{\sqrt{\sum_{i=1}^n z_i^2 / n}}$

is called a $latex t$-distribution with $latex n$ degrees of freedom. Equivalently, it can be defined as $latex t_n = z_0/\sqrt{\chi_n^2/n}$ where $latex z_0$ is standard normal, $latex \chi_n^2$ is chi-square with $latex n$ degrees of freedom and they are independent.

Property 1. Like $latex z$, $latex t_n$ is symmetric and therefore $latex Et_n = 0$.

Proof. The general definition of a symmetric (around 0) distribution is that $latex X$ and $latex -X$ have the same distribution. Since $latex z_0$ and $latex -z_0$ have the same distribution, the variables $latex t_n = z_0/\sqrt{\chi_n^2/n}$ and $latex -t_n = -z_0/\sqrt{\chi_n^2/n}$ have the same distribution. $latex \blacksquare$

Property 2. The tails of $latex t_n$ are fatter than those of $latex z$. Both have densities that are bell-shaped, see in Figure 8.2 densities of $latex t$-distribution with 3 degrees of freedom (continuous line) and of standard normal (dashed line). If the density of $latex t_n$ is higher than that of $latex z$ for large arguments, then for small arguments (around the origin) the opposite should be true because the total density is 1.

Figure 8.2. Densities of $latex t$-distribution and of standard normal.

Property 3. As $latex n \to \infty$, $latex t_n$ converges to $latex z$. This means that for values $latex n > 60$ that are not in Table 8 in the Appendix of NCT one can use values for $latex z$ from Table 1. Table 14.8 can be used for all $latex n$ because it is not limited by the book volume.

Theorem 8.3. Under independent sampling from a normal population the variables $latex \bar X$ and $latex s_X^2$ are independent.

Exercise 8.5. Prove that Eq. (8.5) is distributed as $latex t_{n-1}$.

Proof. We have proved in the previous unit (see Confidence interval for the population mean: known variance) that $latex z_0 = \frac{\bar X - \mu_X}{\sigma_X/\sqrt n}$ is standard normal. Using Theorem 7.1 we get

$latex \frac{\bar X - \mu_X}{s_X/\sqrt n} = \frac{\bar X - \mu_X}{\sigma_X/\sqrt n} \cdot \frac{1}{\sqrt{s_X^2/\sigma_X^2}} = \frac{z_0}{\sqrt{\chi_{n-1}^2/(n-1)}}.$

By Theorem 8.3 and definition of the $latex t$-distribution this is really $latex t_{n-1}$. $latex \blacksquare$

Exercise 8.6. (1) Derive the confidence interval for $latex \mu_X$ when $latex \sigma_X$ is unknown.

(2) What happens to the critical value $latex t_{n-1, \alpha/2}$ and to the confidence interval when (i) $latex n$ increases, (ii) $latex \alpha$ decreases, (iii) $latex s_X$ increases?

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