How Confidence Level Affects Interval Width

Confidence Level Changes the Width

A one-proportion confidence interval uses sample data to estimate a population proportion. As in Margin of Error and Its Meaning, the interval is centered at the sample proportion \(\hat{p}\), and its margin of error is the critical value multiplied by the estimated standard error. Choosing a different confidence level changes the critical value, and therefore changes the interval’s width.

For the same sample, \(\hat{p}\), \(n\), and the estimated standard error stay fixed. A higher confidence level uses a larger critical value \(z^*\), producing a larger margin of error and a wider interval. The tradeoff is that the method is designed to capture the true population proportion more often over repeated samples, but each interval gives a less precise range of plausible values.

Key relationship: For one sample, changing the confidence level changes \(z^*\), not the sample proportion or estimated standard error. A higher confidence level means a larger \(z^*\), a larger margin of error, and a wider interval.

\text{Interval width} =2ME =2z^*SE_{\hat{p}} =2z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

The factor of 2 appears because the margin extends the same distance below and above \(\hat{p}\). If the sample and estimated standard error are fixed, the width is directly proportional to \(z^*\). For common central confidence levels, the critical values are about 1.645 for 90%, 1.960 for 95%, and 2.576 for 99%, as covered in Finding Critical Values z* for Common Confidence Levels.

Confidence level and precision are related but are not the same thing. A 99% interval has a higher long-run capture rate than a 90% interval when the method’s conditions hold, but it is wider and therefore less precise in the sense that it gives a broader range of plausible values. A wider interval is not evidence that the sample proportion changed; it reflects the chosen confidence level.

Compare Three Intervals from the Same Sample

Consider a hypothetical random sample of 400 households from a town of 18,000 households. In the sample, 248 households say they compost food scraps. Let \(p\) be the true proportion of households in the town that compost food scraps. The sample proportion is:

\hat{p}=\frac{x}{n} =\frac{248}{400} =0.62

Before constructing the intervals, check the conditions for a one-proportion \(z\)-interval. The sample is stated to be random. Because it is drawn without replacement from a finite population, check the 10% condition: \(0.10(18{,}000)=1{,}800\), and \(400\leq1{,}800\). The observed success and failure counts are 248 and \(400-248=152\), both at least 10, so the Large Counts condition is met. These checks support using the interval method.

The estimated standard error is the same for all three intervals because the sample proportion and sample size do not change:

SE_{\hat{p}} =\sqrt{\frac{(0.62)(1-0.62)}{400}} =\sqrt{\frac{0.2356}{400}} =\sqrt{0.000589} \approx0.0242693

Now use the appropriate critical value for each confidence level. The following values are rounded; the endpoints and widths are calculated using the unrounded standard error and margins of error.

Confidence level	\(z^*\)	Margin of error	Interval	Width
90%	1.645	0.03992	(0.5801, 0.6599)	0.07985
95%	1.960	0.04757	(0.5724, 0.6676)	0.09514
99%	2.576	0.06252	(0.5575, 0.6825)	0.12504

The 90% interval is the narrowest, and the 99% interval is the widest. All three are centered at \(0.62\). As the confidence level rises, each endpoint moves farther from \(0.62\): the lower endpoint decreases and the upper endpoint increases. Thus the higher-confidence interval includes a broader range of plausible values for the true proportion.

Worked Example: Constructing and Comparing 90%, 95%, and 99% Intervals

Use the random sample of 400 households, with 248 reporting that they compost food scraps, to construct and compare one-proportion \(z\)-intervals at the 90%, 95%, and 99% confidence levels.

State.
Let \(p\) be the true proportion of households in the town that compost food scraps. The sample has \(x=248\) successes out of \(n=400\), so \(\hat{p}=0.62\).

Plan and check conditions.
The sample is random. The town has 18,000 households, and \(400\leq0.10(18{,}000)=1{,}800\), so the 10% condition is met. There are 248 successes and 152 failures, both at least 10, so the Large Counts condition is met. Use the one-proportion \(z\)-interval formula at each requested confidence level.

Do.
The estimated standard error is \(SE_{\hat{p}}=\sqrt{(0.62)(0.38)/400}\approx0.0242693\). For 90%, \(ME=1.645(0.0242693)\approx0.03992\), giving \(0.62\pm0.03992=(0.58008,0.65992)\), with width \(2(0.0399230)\approx0.07985\). For 95%, \(ME=1.960(0.0242693)\approx0.04757\), giving \(0.62\pm0.04757=(0.57243,0.66757)\), with width \(2(0.0475678)\approx0.09514\). For 99%, \(ME=2.576(0.0242693)\approx0.06252\), giving \(0.62\pm0.06252=(0.55748,0.68252)\), with width \(2(0.0625177)\approx0.12504\).

Conclude.
We are 90%, 95%, or 99% confident, respectively, that the true proportion of households in the town that compost food scraps is within the corresponding interval. The higher-confidence interval is wider. The confidence level describes the long-run capture rate of the method, not the probability that a particular interval contains the fixed population proportion.

The calculations show the tradeoff with the data held constant. Moving from 90% to 99% confidence increases the width from about 0.07985 to 0.12504. The 99% interval is more than 0.04 proportion units wider, even though both intervals use exactly the same sample.

How Much Wider Is a Higher-Confidence Interval?

For intervals based on the same sample, the estimated standard error is a common factor in each width. Consequently, the ratio of two widths is the ratio of their critical values. This makes it possible to compare the relative increase in width without recalculating the sample’s standard error.

\frac{\text{width at confidence level }C_2} {\text{width at confidence level }C_1} = \frac{z^*_{C_2}}{z^*_{C_1}}

For the household sample, the 95% interval is about \(1.960/1.645\approx1.1915\) times as wide as the 90% interval. That is an increase of about 19.15%. The 99% interval is about \(2.576/1.645\approx1.5660\) times as wide as the 90% interval, an increase of about 56.60%. These percentages compare widths, not endpoints or confidence levels.

The ratio calculation works because the center and standard error are fixed. It would not be appropriate to use this shortcut to compare intervals from different samples if their sample proportions or sample sizes differ: those differences can change the estimated standard error as well as the critical value.

Worked Example: Comparing Widths by Ratios

Using the intervals for the household sample, compare the width of the 95% interval with the 90% interval, and the width of the 99% interval with the 90% interval. Explain what each comparison means.

The sample proportion and sample size are the same for every interval, so the estimated standard error is unchanged. The ratio of widths therefore equals the ratio of critical values. For 95% compared with 90%:

\frac{1.960}{1.645} \approx1.1915

The 95% interval is about 1.1915 times as wide as the 90% interval. As a percentage increase, \((1.1915-1)(100\%)\approx19.15\%\). This agrees with the widths: \(0.09514/0.07985\approx1.1915\), with slight differences possible from rounding.

For 99% compared with 90%:

\frac{2.576}{1.645} \approx1.5660

The 99% interval is about 1.5660 times as wide, or about \((1.5660-1)(100\%)=56.60\%\) wider than the 90% interval. In context, choosing 99% instead of 90% gives a method with a higher long-run capture rate but a substantially broader range of plausible values for the proportion of town households that compost food scraps.

Choosing a Confidence Level Is a Tradeoff

A confidence level is selected before looking at the interval’s endpoints. The level describes the long-run proportion of intervals that would capture the true population proportion if the same method were used repeatedly under its conditions. As explained in Interpreting a 95% Confidence Level Correctly, it does not mean there is a 95% probability that the true proportion lies in the particular interval calculated from one sample.

For a fixed sample, choosing 99% rather than 90% makes the method less likely to miss the true proportion over repeated use, but the interval is wider. Choosing 90% gives a narrower, more precise interval, but the method captures the true proportion less often in repeated sampling. Neither level is automatically best for every purpose. The appropriate choice depends on how much long-run confidence is wanted and how useful a narrower range would be.

Worked Example: Choosing a Level for a Desired Width

A town analyst wants an interval for the proportion of households that compost food scraps. Using the sample of 400 households above, suppose the analyst wants an interval with a width no greater than 0.10 proportion units. Which of the 90%, 95%, and 99% intervals meets that width goal?

The original sample is random, the 10% condition is met because \(400\leq1{,}800\), and the observed success and failure counts are 248 and 152. The conditions remain the same because the data do not change. Compare the calculated widths with 0.10:

The 90% interval has width about 0.07985, which is less than 0.10.
The 95% interval has width about 0.09514, which is less than 0.10.
The 99% interval has width about 0.12504, which is greater than 0.10.

Thus the 90% and 95% intervals meet the stated width goal, while the 99% interval does not. If the analyst also wants the highest confidence level that meets the goal, the 95% interval is the choice among these three. This decision makes the precision-confidence tradeoff explicit; it does not change the observed sample proportion or remove other possible sources of error in the data collection.

Common Mistakes and AP Exam Tips

Thinking higher confidence makes the interval narrower. The critical value increases as the confidence level increases, so the margin of error and width increase.
Changing the center when only the confidence level changes. All intervals from the same sample are centered at the same \(\hat{p}\). A new confidence level changes the endpoints, not the sample estimate.
Changing the standard error without changing the sample. For these one-proportion intervals, the estimated standard error depends on \(\hat{p}\) and \(n\). If those are fixed, it stays fixed across confidence levels.
Calling a wider interval more precise. A wider interval gives a broader range of plausible values and is less precise in that sense, even though its method has a higher long-run capture rate.
Interpreting confidence as a probability for one interval. The population proportion is fixed. A 99% confidence level describes the long-run performance of the method, not a 99% chance that this particular interval contains \(p\).
Confusing percent change in width with percentage-point change in an endpoint. A width ratio compares the whole interval widths. It does not say that each endpoint moves by that percentage.

AP Exam Tip: When comparing confidence levels for the same sample, state that \(\hat{p}\) and \(SE_{\hat{p}}\) remain fixed, while \(z^*\), the margin of error, and the width increase with confidence level. Support the comparison with calculated widths or their ratio, then describe the long-run capture-rate and precision tradeoff without assigning a probability to the fixed parameter.

Key Takeaway

For the same sample, a higher confidence level produces a wider one-proportion \(z\)-interval because it uses a larger critical value. The width is \(2z^*SE_{\hat{p}}\), so the estimate and estimated standard error stay the same while the margin of error grows. Higher confidence means a higher long-run capture rate for the method, but a less precise, broader interval.

Key takeaway: For a fixed sample, compare confidence intervals by comparing their critical values: larger \(z^*\) gives larger margins of error and wider intervals. Explain the tradeoff as higher long-run confidence versus lower precision.

Check Your Understanding

Use the relationship between confidence level, critical value, margin of error, and width to answer each question.

For one sample, what stays the same when you change from a 90% to a 99% one-proportion confidence interval, and what changes?
Which interval is wider for the same sample, a 95% interval or a 99% interval? Explain using the critical values.
A 90% interval from a sample is \((0.31,0.43)\). What is its width, and what is its center?
For the same sample, a 95% interval is about 1.19 times as wide as a 90% interval. What does that ratio mean in context, and what does it not mean?
Does a 99% confidence level mean there is a 99% probability that the particular interval from one sample contains the true population proportion? Explain the correct long-run interpretation.