Tutorials › AP Statistics › How Sample Size Changes Interval Width

Home  /  Tutorials  /  One-sample t confidence intervals

One-sample t confidence intervals · Tutorial 632 of 1000

How Sample Size Changes Interval Width

Learn how sample size affects both parts of a one-sample t interval’s margin of error, and compare the resulting interval widths with careful attention to conditions.

Intermediate 9 min read

What You'll Learn

  • Explain how sample size changes the estimated standard error \(s/\sqrt{n}\)
  • Describe why \(t^*\) also decreases as sample size and degrees of freedom increase
  • Compare one-sample t interval widths while holding confidence level and sample standard deviation fixed
  • Calculate how a change in sample size affects margin of error and interval width
  • Check the conditions for a one-sample t interval, including random sampling and the 10% condition
  • Distinguish a predictable design comparison from the variable results of actual samples

Why Sample Size Changes Interval Width

In “How Confidence Level Changes Interval Width,” we held the sample fixed and saw that a higher confidence level increases \(t^*\), making the interval wider. Now we will hold the confidence level fixed and examine a different factor: sample size.

A one-sample t interval depends on both the estimated standard error, \(s/\sqrt{n}\), and the critical value, \(t^*\). When \(n\) increases, the estimated standard error generally decreases if the sample standard deviation \(s\) is similar. Also, the degrees of freedom increase because \(df=n-1\), and the \(t^*\) for a fixed confidence level becomes a little smaller. Both changes make the interval narrower.

Key idea: At a fixed confidence level and with comparable sample standard deviations, a larger sample size narrows a one-sample t interval through two related changes: \(s/\sqrt{n}\) gets smaller, and \(t^*\) decreases as the degrees of freedom increase.

Recall from “The Form of a One-Sample t Interval” that the interval is centered at \(\bar{x}\), with a margin of error on either side. Its full width is twice that margin of error:

$$ \text{interval width}=2t^*\left(\frac{s}{\sqrt{n}}\right) $$

This formula shows why sample size matters twice. First, \(n\) is in the standard error’s denominator, under a square root. Second, \(n\) determines \(df=n-1\), which affects \(t^*\). The standard error usually accounts for most of the change in width; the \(t^*\) change adds a smaller narrowing effect, especially when comparing moderate or large samples.

For example, if the sample size is multiplied by four while \(s\) stays the same, \(s/\sqrt{n}\) is cut in half—not to one quarter. The critical value also falls somewhat at a fixed confidence level. So the resulting width is a little less than half the original width, assuming the same confidence level and comparable \(s\).

Important qualification: This comparison isolates the effect of \(n\) by keeping the confidence level and \(s\) fixed or similar. In actual samples, \(s\) can change, and a larger sample can have a different \(\bar{x}\). A larger sample does not guarantee that every realized interval will be narrower than every interval from a smaller sample.

How to Compare Widths

Use the same confidence level in both intervals so that the only planned changes are the sample size and its associated degrees of freedom. Find \(df=n-1\) and the correct \(t^*\) for each sample size. Then calculate each standard error and multiply by \(2t^*\) to get the full width. The margin of error, \(t^*(s/\sqrt{n})\), is half the width.

Do not assume that \(t^*\) stays exactly constant as \(n\) changes. It does move closer to the corresponding Normal critical value as the degrees of freedom increase, but it does not disappear from the calculation. For a fixed confidence level, the fall in \(t^*\) is usually smaller than the fall in standard error.

The width comparison does not replace checking whether a t interval is appropriate. As in “Building a One-Sample t Interval by Hand,” check that the data come from a random sample or that the sampling process is otherwise justified, that observations are independent, and that the data support a t procedure. For sampling without replacement, check the 10% condition. Random assignment alone supports conclusions about cause and effect; it does not, by itself, justify generalizing to a population.

Worked Examples: Comparing Sample Sizes

Worked Example: A Larger Sample and a Narrower Interval

For a fictional study of completion times for a type of equipment inspection, compare 95% t interval widths for samples of \(n=9\) and \(n=25\). For a clean comparison, suppose each sample has a standard deviation of 12 minutes. The population of inspections is Normal, and both samples are random samples from a population of 300 inspections. What are the widths, and how do they differ?

State. The parameter is the true mean inspection time, \(\mu\), in minutes. We will compare the widths of two 95% one-sample t intervals, using the same \(s=12\) minutes for both.

Plan. Both samples are random. The largest sample meets the 10% condition because \(25<0.10(300)=30\); the smaller sample does as well. Thus, independence is reasonable for sampling without replacement. The population is stated to be Normal, which supports using a t procedure for either sample size. Random sampling supports inference to the population of inspections.

Do. For \(n=9\), \(df=9-1=8\), and the 95% critical value is \(t^*\approx2.306\). Its estimated standard error and width are:

$$ \frac{s}{\sqrt{n}}=\frac{12}{\sqrt{9}}=4\text{ minutes}, \qquad W=2(2.306)(4)=18.448\text{ minutes}. $$

For \(n=25\), \(df=25-1=24\), and the 95% critical value is \(t^*\approx2.064\). The corresponding standard error and width are:

$$ \frac{s}{\sqrt{n}}=\frac{12}{\sqrt{25}}=2.4\text{ minutes}, \qquad W=2(2.064)(2.4)=9.9072\text{ minutes}. $$

Rounded to three decimal places, the widths are 18.448 and 9.907 minutes. The larger-sample interval is \(18.448-9.907=8.541\) minutes narrower, or about \(8.541/18.448=0.463\), a 46.3% reduction in width.

Conclude. With the confidence level and sample standard deviation held fixed, the interval for \(n=25\) is much narrower than the interval for \(n=9\). Its standard error falls from 4 to 2.4 minutes, and its \(t^*\) also falls from 2.306 to 2.064. The comparison shows the width effect; it does not predict where either interval is centered, because each sample’s \(\bar{x}\) may differ.

Worked Example: Quadrupling the Sample Size

A fictional random sample of 16 community garden plots has an average soil-moisture reading. For planning a future survey, compare it with a sample of 64 plots. Suppose the sample standard deviation is 10 percentage points in each comparison, and use 95% confidence. The plots are sampled from a population of 800; the sample data are reasonably symmetric with no outliers. How does the width change?

State. The target is the true mean soil-moisture reading, \(\mu\), in percentage points. We compare 95% interval widths for \(n=16\) and \(n=64\), assuming \(s=10\) percentage points for both.

Plan. The samples are random, and the larger sample meets the 10% condition because \(64<0.10(800)=80\); the smaller one also meets it. This supports independence for sampling without replacement. The sample plots are reasonably symmetric with no outliers, supporting a t procedure for \(n=16\); the larger sample also supports using a t interval. Random sampling supports generalizing to the garden plots in the stated population.

Do. For \(n=16\), \(df=15\), \(t^*\approx2.131\), and the standard error is \(10/\sqrt{16}=2.5\) percentage points. Therefore:

$$ W_{16}=2(2.131)(2.5)=10.655\text{ percentage points}. $$

For \(n=64\), \(df=63\), \(t^*\approx1.998\), and the standard error is \(10/\sqrt{64}=1.25\) percentage points. Therefore:

$$ W_{64}=2(1.998)(1.25)=4.995\text{ percentage points}. $$

The sample size is multiplied by four, so the standard error is halved, from 2.5 to 1.25. The \(t^*\) also falls, from 2.131 to 1.998. The width ratio is \(4.995/10.655\approx0.469\), so the larger-sample interval is about 46.9% as wide, a reduction of about 53.1%.

Conclude. Under these comparison assumptions, increasing the sample size from 16 to 64 reduces the interval width from about 10.655 to 4.995 percentage points. The width falls by slightly more than half because the standard error is halved and \(t^*\) decreases as well.

Worked Example: Comparing a Fourfold Increase at a Larger Starting Size

A fictional random sample of 25 rechargeable batteries is used to estimate the population mean run time. A future comparison considers a sample of 100 batteries. Assume \(s=8\) hours for both samples, use 95% confidence, and suppose the batteries come from a Normal population of 2,500. Compare the interval widths.

State. The parameter is the true mean battery run time, \(\mu\), in hours. We compare the widths for sample sizes 25 and 100 while holding the confidence level and sample standard deviation fixed.

Plan. Both samples are random samples. The larger sample meets the 10% condition because \(100<0.10(2500)=250\), and the smaller sample does as well. The Normal population supports the t procedure for both sample sizes. Random sampling supports generalizing to the population of batteries described.

Do. For \(n=25\), \(df=24\), \(t^*\approx2.064\), and the standard error is \(8/\sqrt{25}=1.6\) hours. Thus:

$$ W_{25}=2(2.064)(1.6)=6.6048\text{ hours}. $$

For \(n=100\), \(df=99\), \(t^*\approx1.984\), and the standard error is \(8/\sqrt{100}=0.8\) hours. Thus:

$$ W_{100}=2(1.984)(0.8)=3.1744\text{ hours}. $$

Rounded to three decimal places, the widths are 6.605 and 3.174 hours. Their ratio is \(3.1744/6.6048\approx0.481\), so the second interval is about 48.1% as wide as the first.

Conclude. Increasing \(n\) from 25 to 100 halves the standard error, and the slightly smaller \(t^*\) narrows the interval further. With the assumed standard deviation and confidence level, the 100-battery interval is about 3.174 hours wide, compared with about 6.605 hours for the 25-battery interval.

What the Comparison Does—and Does Not—Tell You

These examples isolate the effect of sample size by setting \(s\) equal across sample sizes. In real data, the sample standard deviation may rise or fall. If the larger sample happens to have a much larger \(s\), its interval might not be narrower, even though the standard error includes a larger \(n\). Likewise, sample means from different samples need not match, so their intervals may have different centers.

A larger sample size also does not fix a biased sampling method. For example, collecting many responses only from volunteers may still fail to represent the population of interest. The random-sampling or other sampling-justification requirement matters regardless of \(n\). Nor does a large sample automatically make extreme outliers harmless for a t interval.

Key takeaway: At a fixed confidence level and with the same or comparable \(s\), increasing \(n\) narrows a one-sample t interval because \(s/\sqrt{n}\) decreases and \(t^*\) decreases as \(df=n-1\) increases. In actual samples, check the conditions and remember that \(s\) and \(\bar{x}\) can also change.

Common Mistakes and AP Exam Tips

  • Saying that quadrupling \(n\) cuts the standard error to one quarter. Because \(n\) is under a square root, quadrupling \(n\) halves \(s/\sqrt{n}\), if \(s\) stays the same.
  • Holding \(t^*\) fixed without explanation. At a fixed confidence level, \(t^*\) depends on \(df=n-1\). It gets smaller as \(n\) increases, although often less dramatically than the standard error does.
  • Claiming every larger-sample interval must be narrower. That claim ignores possible changes in \(s\). A full-credit comparison states what is held fixed or similar and explains that actual samples may differ.
  • Checking only the sample size. A larger \(n\) does not remove the need to check random sampling or justify the sampling process, independence (including the 10% condition when appropriate), and whether the data support a t procedure.
  • Treating random assignment as a substitute for random sampling. Random assignment supports causal conclusions in an experiment. By itself, it does not justify generalizing results to a population.
  • Confusing width with margin of error. The margin of error is the distance from \(\bar{x}\) to one endpoint; the full width is twice that amount. Include the factor of 2 when calculating width.

For full-credit AP communication, name the quantities that change and connect them to the context. For example: “With 95% confidence and the sample standard deviation held at 10 percentage points, increasing the sample size from 16 to 64 reduces the estimated standard error from 2.5 to 1.25 percentage points. The \(t^*\) also decreases from 2.131 to 1.998, so the interval width decreases from about 10.655 to 4.995 percentage points.”

Check Your Understanding

Use the width formula and the sample-size relationship to answer each question.

  1. At a fixed confidence level and with \(s\) held constant, what happens to the estimated standard error when \(n\) is multiplied by four? Explain using the square root in its formula.
  2. Why does \(t^*\) change when the sample size changes, even when the confidence level stays fixed?
  3. A 95% t interval uses \(n=25\) and \(s=6\) units. Its \(t^*\) is 2.064. Calculate the estimated standard error and the full interval width.
  4. A student says, “The larger sample must give a narrower interval because \(n\) is larger.” What important qualification should be added to this statement?
  5. For a one-sample t interval based on a random sample without replacement, name the condition that relates \(n\) to the population size and explain what it supports.

Previous

Next

Back to Tutorials