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Section 10.5 Interval Estimates - Confidence Interval for μ

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 μ.

Definition 10.5.1 Confidence Interval for One Mean

Given a sample mean ¯x, a two-sided confidence interval for the mean with confidence level 1α is an interval

¯xE1<μ<¯x+E2

such that

P(¯xE1<μ<¯x+E2)=1α.

Generally, the interval is symmetrical of the form ¯x±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

P(¯xE<μ<¯x+E)=1α.

is equivalent to

P(Eσ/n<¯xμσ/n<Eσ/n)=1α

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

P(Eσ/n<Z<Eσ/n)=1α.

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

Φ(zα/2)=Φ(Eσ/n)=P(Z<Eσ/n)=1α2

and so to determine E again requires the inverse of the standard normal distribution function. Using an appropriate zα/2 (as determine in a manner described in the previous section) gives a confidence interval for the mean

¯xzα/2σn<μ<¯x+zα/2σn

with confidence level 1α and margin of error

E=zα/2σn.

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 μ is not known...indeed the goal is to approximate that mean using ¯x...but that σ is known. This is often not the case. It can however be shown that if n is larger than 30, replacing σ 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 σ.)

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.