Interval Estimates - Confidence Interval for $\mu$

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

$\mu\text{.}$

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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.

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Once again, utilize the Central Limit Theorem. Notice that the symmetrical confidence interval

$\begin{equation*} P(\overline{x} - E \lt \mu \lt \overline{x} + E) = 1-\alpha. \end{equation*}$

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is equivalent to

$\begin{equation*} P \left ( \frac{-E}{\sigma / \sqrt{n}} \lt \frac{\overline{x} - \mu}{\sigma / \sqrt{n}} \lt \frac{E}{\sigma / \sqrt{n}} \right ) = 1 - \alpha \end{equation*}$

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in which the middle term can be approximated using a standard normal variable and therefore this statement is approximately

$\begin{equation*} P \left ( \frac{-E}{\sigma / \sqrt{n}} \lt Z \lt \frac{E}{\sigma / \sqrt{n}} \right ) = 1 - \alpha. \end{equation*}$

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Using the symmetry of the standard normal distribution about Z=0 gives

$\begin{equation*} \Phi (z_{\alpha/2} ) = \Phi \left ( \frac{E}{\sigma / \sqrt{n}} \right ) = P \left ( Z \lt \frac{E}{\sigma / \sqrt{n}} \right ) = 1 - \frac{\alpha}{2} \end{equation*}$

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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

$\begin{equation*} \overline{x} - z_{\alpha / 2} \frac{\sigma}{\sqrt{n}} \lt \mu \lt \overline{x} + z_{\alpha / 2} \frac{\sigma}{\sqrt{n}} \end{equation*}$

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with confidence level

$1-\alpha$ and margin of error

$\begin{equation*} E = z_{\alpha /2} \frac{\sigma}{\sqrt{n}}. \end{equation*}$

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Checkpoint 10.5.2. WebWork - Confidence Interval for the Mean.

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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.

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

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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.

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Theorem 10.5.4. 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

$\begin{equation*} n \gt \left ( z_{\alpha /2} \frac{\sigma}{E} \right )^2 \end{equation*}$

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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.

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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

$\sigma\text{.}$ )

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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.

Section 10.5 Interval Estimates - Confidence Interval for μ\mu