Section 10.5 Interval Estimates - Confidence Interval for \(\mu\)
As with the confidence intervals above for proportions, the Central Limit Theorem also allows you to create an interval centered on a sample mean for estimating the population mean \(\mu\text{.}\)
Definition 10.5.1 Confidence Interval for One Mean
Given a sample mean \(\overline{x}\text{,}\) a two-sided confidence interval for the mean with confidence level \(1-\alpha\) is an interval
\begin{equation*}
\overline{x} - E_1 \lt \mu \lt \overline{x} + E_2
\end{equation*}
such that
\begin{equation*}
P(\overline{x} - E_1 \lt \mu \lt \overline{x} + E_2) = 1-\alpha.
\end{equation*}
Generally, the interval is symmetrical of the form \(\overline{x} \pm E\) with E again known as the margin of error. One-sided confidence intervals can be determined in the same manner as in the previous section.
Once again, utilize the Central Limit Theorem. Notice that the symmetrical confidence interval
\begin{equation*}
P(\overline{x} - E \lt \mu \lt \overline{x} + E) = 1-\alpha.
\end{equation*}
is equivalent to
\begin{equation*}
P \left ( \frac{-E}{\sigma / \sqrt{n}} \lt \frac{\overline{x} - \mu}{\sigma / \sqrt{n}} \lt \frac{E}{\sigma / \sqrt{n}} \right ) = 1 - \alpha
\end{equation*}
in which the middle term can be approximated using a standard normal variable and therefore this statement is approximately
\begin{equation*}
P \left ( \frac{-E}{\sigma / \sqrt{n}} \lt Z \lt \frac{E}{\sigma / \sqrt{n}} \right ) = 1 - \alpha.
\end{equation*}
Using the symmetry of the standard normal distribution about Z=0 gives
\begin{equation*}
\Phi (z_{\alpha/2} ) = \Phi \left ( \frac{E}{\sigma / \sqrt{n}} \right ) = P \left ( Z \lt \frac{E}{\sigma / \sqrt{n}} \right ) = 1 - \frac{\alpha}{2}
\end{equation*}
and so to determine E again requires the inverse of the standard normal distribution function. Using an appropriate \(z_{\alpha /2}\) (as determine in a manner described in the previous section) gives a confidence interval for the mean
\begin{equation*}
\overline{x} - z_{\alpha / 2} \frac{\sigma}{\sqrt{n}} \lt \mu \lt \overline{x} + z_{\alpha / 2} \frac{\sigma}{\sqrt{n}}
\end{equation*}
with confidence level \(1-\alpha\) and margin of error
\begin{equation*}
E = z_{\alpha /2} \frac{\sigma}{\sqrt{n}}.
\end{equation*}
It should be noted that the use of the Central Limit Theorem makes the use of InvNorm an approximation. It can be shown that so long as n is larger than 30 then generally this approximation is reasonable.
Additionally, this derivation assumes that \(\mu\) is not known...indeed the goal is to approximate that mean using \(\overline{x}\text{...}\)but that \(\sigma\) is known. This is often not the case. It can however be shown that if n is larger than 30, replacing \(\sigma\) with the sample standard deviation s gives an acceptable confidence interval.
Theorem 10.5.2 Sample Size needed for \(\mu\) given Margin of Error
Given confidence level \(1-\alpha\) and margin of error E, the sample size needed to determine an appropriate confidence interval satisfies
\begin{equation*}
n \gt \left ( z_{\alpha /2} \frac{\sigma}{E} \right )^2
\end{equation*}
Proof
Solve for n in the formula for E above. Notice that n must be an integer so you will need to round up. You will also need an estimate for the sample standard deviation s by using a preliminary sample.
Notice, in practice you might want to take n to be a little larger than the absolute minimum value prescribed above since you are dealing with approximations (Central Limit Theorem and the use of an estimate for s rather than the actual \(\sigma\text{.}\))
Example 10.5.3 Determining Sample Size for one Mean
Given a 95% confidence level, margin of error E=0.1, and preliminary sample with standard deviation s = 2, \(z_{\alpha / 2} = 1.96\) gives
\begin{equation*}
n \gt \left ( 1.96 \cdot \frac{2}{0.1} \right )^2 \approx 1536.64
\end{equation*}
or a sample size of at least 1537.