Raising the bar

Chapter 8. Estimation: One or Two Populations

Unit 8.3.1. Confidence interval for the population mean. Case of a known variance

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

Step 1. Select the significance level $latex \alpha$.

Step 2. Using Table 1 in the Appendix to NCT (or Excel and Table 14.1 in this book) find a number $latex z_{\alpha/2} > 0$ such that

$latex P(-z_{\alpha/2} < z < z_{\alpha/2}) = 1 - \alpha. \qquad (\textrm{Eq. 8.1})$

Definition 8.4 (Critical value). $latex z_{\alpha/2}$ is called a critical value of the $latex z$ statistic corresponding to $latex \alpha$ (for a two-tail confidence interval). $latex \alpha/2$ is the probability of the left and right tails and the critical value could be equivalently defined by $latex \alpha/2 = P(z < -z_{\alpha/2}) = P(z > z_{\alpha/2})$.

Step 3. Let us standardize the sample mean

$latex z = \frac{\bar X - \mu_{\bar X}}{\sigma_{\bar X}} = \frac{\bar X - \mu_X}{\sigma_X/\sqrt n}. \qquad (\textrm{Eq. 8.2})$

The second equality here uses Eq. (5.13). We want to prove that this is really a standard normal.

Theorem 8.2. A sum of independent normal variables is normal.

Exercise 8.4. A linear transformation of a normal variable is normal.

Proof. Suppose $latex X$ is normal, $latex X = \sigma z + \mu$ where $latex z$ is standard normal, and $latex Y$ is its linear transformation, $latex Y = aX + b$. Then $latex Y = a(\sigma z + \mu) + b = a \sigma z + (a\mu + b)$ is normal as a linear transformation of a standard normal. $latex \blacksquare$

From Theorem 8.2 it follows that $latex S_n = X_1 + \ldots + X_n$ is normal. The sample mean is a linear transformation of $latex S_n$, so it is normal too. The variable in Eq. (8.2) is a linear transformation of $latex \bar X$ so it's normal. And we know that, being a standardized variable, it has mean zero and variance 1. Thus, Eq. (8.2) is standard normal.

Now we plug Eq. (8.2) in Eq. (8.1) and use equivalent transformations to get

$latex 1 - \alpha = P\left( -z_{\alpha/2} < \frac{\bar X - \mu_X}{\sigma_X/\sqrt n} < z_{\alpha/2} \right)$

$latex = P\left( -z_{\alpha/2} \frac{\sigma_X}{\sqrt n} < \bar X - \mu_X < z_{\alpha/2} \frac{\sigma_X}{\sqrt n} \right)$

$latex = P\left( \bar X - z_{\alpha/2} \frac{\sigma_X}{\sqrt n} < \mu_X < \bar X + z_{\alpha/2} \frac{\sigma_X}{\sqrt n} \right).$

Definition 8.5 (Margin of error). The quantity $latex ME = z_{\alpha/2} \sigma_X / \sqrt n$ is called a margin of error.

We have obtained a confidence interval for the population mean

$latex 1 - \alpha = P(\bar X - ME < \mu_X < \bar X + ME). \qquad (\textrm{Eq. 8.3})$

Eq. (8.3) can be written in a different way:

$latex 1 - \alpha = P(\mu_X - ME < \bar X < \mu_X + ME). \qquad (\textrm{Eq. 8.4})$

These two forms reflect different points of view:

1. In Eq. (8.3) the population mean is unknown and estimated from the sample data.
2. In Eq. (8.4) the population mean is known and the purpose is to obtain ex-ante bounds for the sample mean.

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