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Essentials of Mathematical Probability and Statistics

John Travis

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*}
A random sample of \(n\) measurements was selected from a population with standard deviation \(\sigma = 10.2\) and unknown mean \(\mu\text{.}\) Calculate a \(90\) % confidence interval for \(\mu\) for each of the following situations:
(a) \(n = 55, \ \overline{x} = 83.3\)
\(\leq \mu \leq\)
(b) \(n = 70, \ \overline{x} = 83.3\)
\(\leq \mu \leq\)
(c) \(n = 85, \ \overline{x} = 83.3\)
\(\leq \mu \leq\)
(d) In general, we can say that for the same confidence level, increasing the sample size the margin of error (width) of the confidence interval. (Enter: ''DECREASES'', ''DOES NOT CHANGE'' or ''INCREASES'', without the quotes.)
Answer 1.
\(81.0375\)
Answer 2.
\(85.5625\)
Answer 3.
\(81.2945\)
Answer 4.
\(85.3055\)
Answer 5.
\(81.4801\)
Answer 6.
\(85.1199\)
Answer 7.
\(\text{DECREASES}\)
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.
Use the given data to find the 95% confidence interval estimate of the population mean \(\mu\text{.}\) Assume that the population has a normal distribution.
IQ scores of professional athletes:
Sample size \(n = 25\)
Mean \(\overline{x} = 103\)
Standard deviation \(s = 12\)
\(\lt \mu \lt \)
Answer 1.
\(98.04664\)
Answer 2.
\(107.95336\)
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 \(\sigma\text{.}\))
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.
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