Effect of Sample Size and Confidence Level on the Interval

What Determines an Interval’s Width?

In Margin of Error for a Difference in Proportions, the margin of error was described as the distance from the center of a two-proportion confidence interval to either endpoint. The interval’s total width is twice that distance. For the same sample data, the point estimate \(\hat{p}_1-\hat{p}_2\) stays fixed when the confidence level changes, but the margin of error—and therefore the width—changes.

Two features directly affect the margin of error: the critical value \(z^*\) for the chosen confidence level and the estimated standard error, which depends on both sample proportions and both sample sizes. The formula from the earlier tutorial makes these effects visible:

Formula: For a two-proportion confidence interval, the margin of error and total width are:

ME=z^*\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} +\frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \qquad \text{Width}=2ME

The interval is \((\hat{p}_1-\hat{p}_2)\mathbin{\pm}ME\). Its center is the point estimate; its width is the distance between its endpoints.

A higher confidence level uses a larger \(z^*\). That increases the margin of error and makes the interval wider. This is the cost of seeking a higher level of confidence: the interval must cover a broader range of plausible values.

A larger sample size generally makes the standard error smaller, so the interval becomes narrower, provided the sample proportions and confidence level are held fixed. If both sample sizes are multiplied by the same factor \(k\), while the proportions stay fixed, each variance term is divided by \(k\). The standard error—and thus the margin of error and width—is divided by \(\sqrt{k}\). For example, quadrupling both sample sizes cuts the width in half, not to one-quarter of its original value.

Key relationship: When both sample sizes are multiplied by \(k\), with the sample proportions and confidence level held fixed, the standard error, margin of error, and width are each multiplied by \(1/\sqrt{k}\). When the confidence level changes with the data held fixed, the center stays the same while the width changes in proportion to \(z^*\).

These are comparisons that hold some features constant to isolate the effect being studied. In real, separate samples, the observed proportions may also change when sample sizes change, so the center and standard error may both differ. A larger sample does not guarantee that every pair of actual intervals will be narrower than every interval from a smaller sample; the general comparison is clearest when the observed proportions and confidence level are held fixed.

Comparing Confidence Levels

The confidence level determines the critical value used in the interval. For the common two-sided confidence levels below, a higher confidence level has a larger critical value:

Confidence level	Approximate \(z^*\)
90%	1.645
95%	1.960
99%	2.576

With the same two samples, each confidence interval has the same center, \(\hat{p}_1-\hat{p}_2\). The 99% interval is wider than the 95% interval, which is wider than the 90% interval. A higher confidence level does not mean that the sample difference is larger, nor does it change the sample data. It changes how far the interval extends on either side of that same estimate.

As in Interpreting a Confidence Interval for \(p_1-p_2\), interpret the endpoints as plausible values for the population difference in the stated group order. A wider interval gives a broader range of plausible differences; it does not necessarily indicate a larger difference between the populations.

Worked Examples

Worked Example: Changing the Confidence Level

Setting: Imagine independent random samples of growers in two regions. The outcome is whether a grower used a particular irrigation-monitoring tool last season. In Region 1, 84 of 140 growers used it; in Region 2, 55 of 125 did. Compare the 90%, 95%, and 99% confidence intervals for Region 1 minus Region 2.

State: Let \(p_1\) and \(p_2\) be the true proportions of growers who used the tool in Regions 1 and 2. The parameter is \(p_1-p_2\).

Plan: The two samples are described as independent random samples, and growers belong to separate groups. Assume each region has at least 10 times its sample size in growers, so the 10% condition holds: \(140\leq0.10(1400)\) and \(125\leq0.10(1250)\). For Large Counts, Region 1 has 84 successes and \(140-84=56\) failures; Region 2 has 55 successes and \(125-55=70\) failures. All four counts are at least 10, so the Large Counts condition holds.

Do: The sample proportions are \(\hat{p}_1=84/140=0.60\) and \(\hat{p}_2=55/125=0.44\). The point estimate is \(0.60-0.44=0.16\). The estimated standard error, which is unchanged across these confidence levels, is:

\begin{aligned} SE_{\hat{p}_1-\hat{p}_2} &=\sqrt{\frac{0.60(0.40)}{140}+\frac{0.44(0.56)}{125}}\\ &=\sqrt{0.001714286+0.001971200}\\ &=\sqrt{0.003685486}\approx0.060708 \end{aligned}

For 90% confidence, \(z^*\approx1.644854\); for 95%, \(z^*\approx1.959964\); and for 99%, \(z^*\approx2.575829\). Multiplying each critical value by the same standard error gives the following margins and widths:

\begin{aligned} ME_{90}&=1.644854(0.060708)\approx0.099856, &\quad \text{width}_{90}&=2(0.099856)\approx0.1997\\ ME_{95}&=1.959964(0.060708)\approx0.118986, &\quad \text{width}_{95}&=2(0.118986)\approx0.2380\\ ME_{99}&=2.575829(0.060708)\approx0.156374, &\quad \text{width}_{99}&=2(0.156374)\approx0.3127 \end{aligned}

The corresponding intervals, rounded to four decimal places, are \(0.16\mathbin{\pm}0.099856=(0.0601,\ 0.2599)\) for 90%, \(0.16\mathbin{\pm}0.118986=(0.0410,\ 0.2790)\) for 95%, and \(0.16\mathbin{\pm}0.156374=(0.0036,\ 0.3164)\) for 99%.

Conclude: All three intervals have the same center, 0.16, because the sample data are unchanged. The 99% interval is widest, and the 90% interval is narrowest. Here, the higher confidence level increases the margin of error, producing a broader range of plausible values for the difference in regional population proportions.

Worked Example: Quadrupling Both Sample Sizes

Setting: Use the grower comparison above as a controlled illustration of sample size. Suppose the sample sizes are quadrupled while both observed proportions remain 0.60 and 0.44: Region 1 has 336 successes out of 560 growers, and Region 2 has 220 successes out of 500. Compare the 95% interval width with the original 95% interval.

State: The parameter remains \(p_1-p_2\), the difference between the true proportions of growers who used the tool in the two regions.

Plan: Treat these as independent random samples. Assume each target region has at least 10 times the larger sample size, so the 10% condition holds for both samples. For Large Counts, Region 1 has 336 successes and \(560-336=224\) failures; Region 2 has 220 successes and \(500-220=280\) failures. All four counts are at least 10.

Do: The sample proportions remain \(336/560=0.60\) and \(220/500=0.44\), so the point estimate is still 0.16. The 95% standard error is:

\begin{aligned} SE_{\hat{p}_1-\hat{p}_2} &=\sqrt{\frac{0.60(0.40)}{560}+\frac{0.44(0.56)}{500}}\\ &=\sqrt{0.000428571+0.000492800}\\ &=\sqrt{0.000921371}\approx0.030354 \end{aligned}

Using \(z^*\approx1.959964\), the margin and width are:

ME=1.959964(0.030354)\approx0.059493 \qquad \text{Width}=2(0.059493)\approx0.1190

The interval is \(0.16\mathbin{\pm}0.059493=(0.1005,\ 0.2195)\), rounded to four decimal places. The original 95% interval had width about 0.2380. The new width, about 0.1190, is half as large, as expected when both sample sizes are quadrupled: \(1/\sqrt{4}=1/2\).

Conclude: With both sample proportions and the confidence level held fixed, quadrupling both sample sizes halves the interval’s width. The center does not change in this controlled comparison because the sample proportions remain unchanged.

Worked Example: Increasing Only One Sample Size

Setting: Imagine independent random samples comparing whether clients in two fictional exercise programs meet a weekly activity goal. Initially, 70 of 100 clients in Program 1 and 50 of 100 clients in Program 2 meet the goal. For a controlled comparison, consider increasing only Program 1’s sample to 280 of 400 while keeping Program 2’s sample at 50 of 100. Compare the 95% widths.

State: Let \(p_1\) and \(p_2\) be the true proportions of clients meeting the goal in Programs 1 and 2. The parameter is \(p_1-p_2\).

Plan: In both comparisons, assume independent random samples from the two programs, with separate clients in each group. Assume each program’s target population has at least 10 times the sample size used, satisfying the 10% condition. For the original samples, the success and failure counts are 70 and 30 in Program 1, and 50 and 50 in Program 2. With the larger Program 1 sample, its counts are 280 successes and \(400-280=120\) failures; Program 2 still has 50 successes and 50 failures. Every count is at least 10, so Large Counts holds for both comparisons.

Do: The proportions are held at \(\hat{p}_1=0.70\) and \(\hat{p}_2=0.50\), making the point estimate 0.20 in each comparison. With 100 clients in each group, the standard error is:

\begin{aligned} SE_{\text{original}} &=\sqrt{\frac{0.70(0.30)}{100}+\frac{0.50(0.50)}{100}}\\ &=\sqrt{0.0021+0.0025} =\sqrt{0.0046}\approx0.067823 \end{aligned}

After increasing only Program 1’s sample size to 400, the standard error is:

\begin{aligned} SE_{\text{larger group 1}} &=\sqrt{\frac{0.70(0.30)}{400}+\frac{0.50(0.50)}{100}}\\ &=\sqrt{0.000525+0.0025} =\sqrt{0.003025}=0.055 \end{aligned}

At 95% confidence, \(z^*\approx1.959964\). The original margin is \(1.959964(0.067823)\approx0.132931\), giving width \(2(0.132931)\approx0.2659\). The new margin is \(1.959964(0.055)\approx0.107798\), giving width \(2(0.107798)\approx0.2156\).

Conclude: Increasing one sample size narrows the interval, but it does not halve the width. The uncertainty contribution from Program 2 remains unchanged, so the combined standard error does not shrink as much as it would if both sample sizes were increased. With the proportions held fixed, both intervals have the same center, 0.20.

Common Mistakes and AP Exam Tips

Saying higher confidence makes an interval narrower: A higher confidence level uses a larger \(z^*\), so it increases the margin of error and produces a wider interval.
Confusing width with margin of error: The margin is the distance from the center to one endpoint. The total width is twice the margin. State which one you are comparing.
Claiming the point estimate changes when only confidence level changes: For the same data, the center \(\hat{p}_1-\hat{p}_2\) stays fixed. The endpoints move outward or inward as the margin changes.
Claiming that quadrupling the sample sizes quarters the width: With proportions fixed, the standard error and width are divided by \(\sqrt{4}=2\), not by 4.
Ignoring one group’s uncertainty: Both group-specific terms remain in the standard-error formula. Increasing one sample size does not remove the other group’s contribution.
Making an unconditional claim about actual samples: The simple width comparisons hold the sample proportions fixed. In real samples, the proportions may change too, affecting both the center and standard error.

AP Exam Tip: Identify what is held constant before comparing intervals. For a confidence-level comparison using the same data, keep the center and standard error fixed and change \(z^*\). For a sample-size comparison, explain whether the sample proportions are assumed fixed, then show how the sample-size terms change the standard error and width.

Key Takeaway

A two-proportion interval gets wider when the confidence level rises because the critical value increases. With the sample proportions and confidence level held fixed, larger sample sizes reduce the standard error and narrow the interval. Increasing both sample sizes by a factor of \(k\) reduces the width by a factor of \(\sqrt{k}\); increasing only one sample size leaves the other group’s uncertainty contribution in place.

Key takeaway: The interval’s width is \(2z^*\sqrt{\hat{p}_1(1-\hat{p}_1)/n_1+\hat{p}_2(1-\hat{p}_2)/n_2}\). A larger \(z^*\) widens the interval; larger sample sizes generally narrow it when the proportions are held fixed.

Check Your Understanding

Use the relationships between critical value, sample size, margin of error, and width to answer these questions.

For the same two samples, what happens to the interval’s center and width when the confidence level increases from 90% to 99%?
If both sample sizes are multiplied by 9 while the sample proportions and confidence level stay fixed, by what factor does the interval width change?
Why does increasing only one group’s sample size not eliminate all uncertainty in a two-proportion interval?
An interval has a margin of error of 0.07. What is its total width?
When comparing intervals from different actual samples, why might the interval from the larger sample not always be narrower?