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9.3. Precision of a Confidence Interval

In the previous section, we established that confidence intervals convey the degree of uncertainty through their width. A narrow interval suggests high precision, while a wide interval reflects greater uncertainty. This naturally raises two key questions:

  • What factors determine the width of a confidence interval?

  • What can we do to make an interval more precise (narrower)?

We will explore the answers to these questions in this lesson.

Road Map 🧭

  • Identify the three building blocks of a margin of error and understand how each influences the precision of a confidence interval.

  • Understand why the sample size is the only component of an ME that researchers can actively control.

  • For a desired level of precision, compute the mininum required sample size, and incorporate this into experiment planning.

9.3.1. What Makes a Confidence Interval Precise?

The width of a confidence interval is equal to 2ME

Fig. 9.5 The width of a confidence interval is equal to 2ME

The width of a confidence interval is entirely determined by its margin of error (ME) as shown in Fig. 9.5. Specifically,

\[\text{width of a CI} = 2ME = 2z_{\alpha/2}\frac{\sigma}{\sqrt{n}}.\]

From its mathematical definition, we can identify three elements that determine the size of an ME: \(\alpha\) (and thus \(C\)), \(\sigma\), and \(n\). Let us examine how each affects the width of a confidence interval.

Component of ME

Effect on precision

Confidence level \(C\)

Higher \(C \to\) the central region of area \(C\) expands \(\to\) \(z_{\alpha/2}\) pushed to the right \(\to\) increased ME & lower precision

pdf of standard normal with the z-critical value marked

Fig. 9.6 Standard normal PDF

Population standard deviation \(\sigma\)

Higher \(\sigma \to\) increaseed ME & lower precision

Sample size \(n\)

Higher \(n \to\) decreased ME & higher precision

Controlling the Precision

Of the three components of a margin of error, researchers can practically control only the sample size, \(n\). The population standard deviation \(\sigma\) is a fixed property of the population, and the confidence level \(C\) must be chosen to reflect the desired rigor of the experiment; it is not meaningful to lower confidence merely to obtain a narrower interval.

This leads to an important practical question: “How do we control the sample size to meet a target precision?”

9.3.2. Sample Size Planning for a Target Precision

Maximum Width

Suppose we want to construct a confidence interval whose width is at most \(w_{max}\). That is, we want

\[\text{width} = 2z_{\alpha/2}\frac{\sigma}{\sqrt{n}} \leq w_{max}.\]

Rearranging to isolate \(n\),

\[\frac{2z_{\alpha/2}\sigma}{w_{max}} \leq \sqrt{n}.\]

Square both sides and bring \(n\) to the left hand side:

\[n \geq \left(\frac{2 z_{\alpha/2} \sigma}{w_{max}}\right)^2.\]

Since \(n\) must be a whole number, we take the the smallest integer on or above the lower bound. That is, the smallest required sample size \(n^*\) for a target width \(w_{max}\) is computed as:

\[n^* = \Bigg\lceil \left(\frac{2 z_{\alpha/2} \sigma}{w_{max}}\right)^2 \Bigg\rceil.\]

Maximum Margin of Error

The desired level of precision can also be given as a maximum margin of error, \(\text{ME}_{max}\). In this case, we take \(n^*\) as:

\[n^* = \Bigg\lceil\left(\frac{z_{\alpha/2} \sigma}{\text{ME}_{max}}\right)^2 \Bigg\rceil\]

through a similar flow of logic. You are encouraged to confirm this result as a quick exercise.

Example 💡: Weights of American Adult Males, Continued

We continue with the study of weights of American male adults from Section 9.2. Suppose the researchers want to compute a 99% confidence interval for the mean weight with a margin of error of at most 1. What is the smallest sample size they need?

The building blocks

  • The maximum ME: \(\text{ME}_{max} = 1\).

  • The population standard deviation: \(\sigma=49.26\) lbs

  • Confidence level: \(99\% \, (C=0.99, \alpha=0.01)\)

  • The z-critical value: \(z_{\alpha/2} = z_{0.005} = 2.5758\)

The goal

We would like to compute the smallest integer \(n\) satisfying

\[ME = z_{\alpha/2}\frac{\sigma}{\sqrt{n}} \leq \text{ME}_{max}\]

Rearranging and isolating \(n\),

\[n \geq \left(\frac{z_{\alpha/2}\sigma}{ME_{max}}\right)^2 = \left(\frac{(2.5758)(49.26)}{1}\right)^2 = 16099.53\]

The smallest sample size needed for the desired level of precison is \(n^* = \lceil 16099.53\rceil =16,100\).

9.3.3. Bringing It All Together

Key Takeaways 📝

  • The margin of error quantifies the precision of a confidence interval. The three factors of margin of error that control precision are: confidence level \(C\), population variance \(\sigma^2\) (or population sd \(\sigma\)), and sample size \(n\).

  • The smallest required sample size for a given maximum ME can be calculated using: \(n^* = \Big\lceil \left(\frac{z_{\alpha/2}\,\sigma}{\text{ME}_{max}}\right)^2\Big\rceil\).

Exercises

  1. Sample Size Planning: A battery manufacturer wants to estimate the mean battery voltage with a margin of error of \(0.05 V\) at 99% confidence. Past data suggest \(\sigma = 0.21 V\). Calculate the required sample size.

  2. Cost-Effectiveness vs Precision: Why is it generally not cost-effective to seek extremely narrow confidence intervals (e.g., reducing margin of error below 1% of the estimate)? Provide both statistical and practical justifications for your answer.