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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*}
Checkpoint 10.5.2. WebWork - Confidence Interval for the Mean.

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. If not, then use replace the z-score with a corresponding value from the t-distribution.

Checkpoint 10.5.3. WebWork - Confidence Interval with t-scores.

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.

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 10.5.5. 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.