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Confidence Intervals for Means · Tutorial 582 of 996

Building an Interval from Summary Statistics

Use summary statistics to build and report a one-sample t confidence interval for a population mean.

Intermediate 10 min read

What You'll Learn

  • Identify the summary statistics needed for a one-sample t interval
  • Find the degrees of freedom and t critical value for a stated confidence level
  • Calculate the standard error and margin of error from summary data
  • Construct the lower and upper confidence limits with consistent rounding
  • Recognize the assumptions and common calculation errors involved

From Summary Statistics to an Interval

A sample mean gives a point estimate of a population mean, but it does not show how much uncertainty comes from sampling. A confidence interval adds a margin of error around that estimate. When the population standard deviation is unknown and the sample standard deviation is used instead, the standard interval for a mean uses the t distribution.

In many reports, the individual observations are not provided. Instead, the report gives a sample mean, a sample standard deviation, a sample size, and a desired confidence level. These summaries are enough to calculate the one-sample t interval. The calculation is compact, but each input has a particular role: the sample mean centers the interval, the standard deviation and sample size determine its standard error, and the confidence level and degrees of freedom determine the t critical value.

Definition: Given a sample mean \(\bar{x}\), sample standard deviation \(s\), sample size \(n\), and confidence level \(C\), the one-sample t confidence interval for the population mean \(\mu\) is $$ \bar{x}\ \pm\ t^*_{n-1}\frac{s}{\sqrt{n}}, $$ where \(t^*_{n-1}\) is the positive t critical value with \(n-1\) degrees of freedom that leaves probability \((1-C)/2\) in each tail. The quantity \(s/\sqrt{n}\) is the standard error \(SE\), and \(t^*_{n-1}SE\) is the margin of error.

Inputs, Assumptions, and the Calculation

The interval estimates the population mean \(\mu\) from a quantitative sample. Its usual model assumes that the observations are independent and come from a population with a common mean and variance. For an exact t procedure, the observations are normally distributed. In practice, a t interval is also often reasonable for a sufficiently large sample when the data do not have severe skewness or extreme outliers. The sample should be obtained in a way that supports inference to the population of interest; a calculation cannot fix a biased sample.

The summary statistics do not reveal every feature of the data. In particular, \(\bar{x}\), \(s\), and \(n\) alone cannot show whether there are extreme outliers or serious departures from the model. Those conditions need to be considered using information about how the data were collected and, when available, the observations themselves. Here, the examples explicitly state that the hypothetical sampling and data shape are adequate for using the t interval.

The calculation follows a fixed sequence. First, calculate \(df=n-1\). Then use the confidence level to determine the two-sided t critical value for those degrees of freedom. Next, calculate \(SE=s/\sqrt{n}\), multiply by \(t^*\) to obtain the margin of error, and subtract and add that margin to \(\bar{x}\). A t table or statistical software can supply \(t^*\); the confidence level alone is not enough, because the critical value also depends on the degrees of freedom.

1
Find the degrees of freedom.
For a one-sample t interval, use \(df=n-1\).
2
Find the critical value.
For a two-sided confidence level \(C\), use the t value with probability \((1-C)/2\) above it.
3
Calculate standard error and margin of error.
Compute \(SE=s/\sqrt{n}\), then \(ME=t^*SE\).
4
Form the interval.
Report \((\bar{x}-ME,\ \bar{x}+ME)\), with units and sensible rounding.

Why the Formula Uses a t Critical Value

The reason for the t critical value can be seen by starting with the standardized sample mean. If \(X_1,\ldots,X_n\) are independent observations from a normal population with mean \(\mu\) and standard deviation \(\sigma\), then standardizing \(\bar{x}\) using the unknown \(\sigma\) gives a standard normal quantity. In practice, \(\sigma\) is usually unknown, so it is replaced by \(s\). This additional estimation of spread leads to a t statistic with \(n-1\) degrees of freedom:

$$ T=\frac{\bar{X}-\mu}{S/\sqrt{n}}\sim t_{n-1}. $$

Choose \(t^*\) so that the central probability under this t distribution is \(C\). Equivalently, the probability of being between \(-t^*\) and \(t^*\) is \(C\). Rearranging this event gives the interval:

$$ \begin{aligned} P\left(-t^*\leq \frac{\bar{X}-\mu}{S/\sqrt{n}}\leq t^*\right)&=C,\\ P\left(\bar{X}-t^*\frac{S}{\sqrt{n}}\leq \mu\leq \bar{X}+t^*\frac{S}{\sqrt{n}}\right)&=C. \end{aligned} $$

The rearrangement multiplies by \(S/\sqrt{n}\), which is positive, and then isolates \(\mu\). In repeated samples under the model, the method produces intervals that contain the fixed population mean a proportion \(C\) of the time. It does not mean that after one interval has been calculated, there is probability \(C\) that this fixed \(\mu\) lies in that particular interval.

Worked Example: A 95% Interval for a Mean Completion Time

In a hypothetical study, a random sample of 16 independent trail users has a mean completion time of 52.4 minutes and a sample standard deviation of 8.1 minutes. Assume the population distribution is sufficiently close to normal, with no severe outliers. Construct a 95% t confidence interval for the population mean completion time.

The sample size is \(n=16\), so the degrees of freedom are \(df=16-1=15\). For a two-sided 95% interval, 5% lies outside the central area, or 2.5% in each tail. The t critical value is approximately \(t^*_{15}=2.131\).

The standard error is

$$ SE=\frac{s}{\sqrt{n}}=\frac{8.1}{\sqrt{16}}=\frac{8.1}{4}=2.025\text{ minutes}. $$

The margin of error is

$$ ME=t^*SE=(2.131)(2.025)=4.315275\text{ minutes}\approx4.315\text{ minutes}. $$

Therefore, the interval is

$$ 52.4\pm4.315275=(48.084725,\ 56.715275)\text{ minutes}, $$

or approximately (48.08, 56.72) minutes. The rounding is applied after calculating with the displayed critical value and standard error. The interval estimates the population mean completion time, not the completion time of every individual trail user.

Worked Examples with Different Inputs

Worked Example: A 90% Interval from Reported Summaries

Suppose a hypothetical report gives a sample mean assay concentration of 74.2 milligrams per liter, a sample standard deviation of 12.5 milligrams per liter, and a sample size of 25. Assume the observations are independent, the sample is appropriately collected, and the data shape is reasonable for a t interval. Construct a 90% confidence interval for the population mean concentration.

First, \(df=25-1=24\). A 90% two-sided interval has 5% in each tail. The critical value is approximately \(t^*_{24}=1.711\). The standard error is

$$ SE=\frac{12.5}{\sqrt{25}}=\frac{12.5}{5}=2.5\text{ milligrams per liter}. $$

Thus,

$$ ME=(1.711)(2.5)=4.2775\text{ milligrams per liter}. $$

The interval is

$$ 74.2\pm4.2775=(69.9225,\ 78.4775)\text{ milligrams per liter}. $$

Rounded to two decimal places, the 90% interval is (69.92, 78.48) milligrams per liter. Notice that this interval uses the 90% critical value, not the 95% value; a different confidence level changes the critical value and therefore the margin of error.

Worked Example: A 99% Interval with a Small Sample

Consider a hypothetical sample of 10 independent measurements of a chemical response time. The sample mean is 18.6 hours and the sample standard deviation is 4.2 hours. Suppose the underlying measurement distribution is reasonably normal and has no severe outliers. Construct a 99% t confidence interval for the population mean response time.

There are \(df=10-1=9\) degrees of freedom. For a 99% two-sided interval, each tail has probability 0.005. The critical value is approximately \(t^*_{9}=3.250\). The standard error is

$$ SE=\frac{4.2}{\sqrt{10}}\approx1.32816\text{ hours}. $$

Using the unrounded standard error in the multiplication,

$$ ME=(3.250)\left(\frac{4.2}{\sqrt{10}}\right)\approx4.31651\text{ hours}. $$

The interval is

$$ 18.6\pm4.31651=(14.28349,\ 22.91651)\text{ hours}. $$

Rounded to two decimal places, the interval is (14.28, 22.92) hours. The critical value is relatively large because the requested confidence level is high and the sample has only 9 degrees of freedom. Both features affect the width; the higher confidence level requires more coverage, while the small sample leaves more uncertainty about the population spread.

Reading and Checking a Constructed Interval

A useful check is to verify the interval’s center and half-width. The midpoint of the lower and upper bounds should equal \(\bar{x}\), apart from rounding, and the distance from \(\bar{x}\) to either endpoint should equal \(ME\). If the interval is symmetric around the sample mean but its endpoints do not agree with the reported margin of error, recheck the arithmetic or rounding.

The units of the standard error, margin of error, and interval endpoints are the same as the units of the original measurements. The sample size and degrees of freedom are counts and have no measurement units. Include the measurement units when reporting the interval so the estimated quantity is clear.

QuantityHow it is obtainedRole in the interval
CenterSample mean \(\bar{x}\)Point estimate of the population mean
Standard error\(s/\sqrt{n}\)Estimated sampling variability of the sample mean
Critical valueFrom confidence level and \(df=n-1\)Sets the number of standard errors on each side
Margin of error\(t^*SE\)Distance from the center to either endpoint

Common Mistakes

  • Using \(s\) instead of the standard error. The interval is not \(\bar{x}\pm t^*s\). The standard deviation describes variation among observations; the standard error \(s/\sqrt{n}\) describes estimated variation in the sample mean.
  • Using the wrong degrees of freedom. For this one-sample t interval, \(df=n-1\), not \(n\). The degrees of freedom are needed to choose the correct critical value.
  • Using the wrong tail probability. For a two-sided confidence level \(C\), each tail has probability \((1-C)/2\). A 95% interval therefore uses 0.025 in each tail.
  • Changing the method without a reason. When the population standard deviation is unknown, the one-sample t interval uses \(s\) and a t critical value. Do not substitute a normal critical value merely because it is more familiar.
  • Rounding intermediate calculations too early. Keep calculator precision through the standard error and margin of error, then round the endpoints for reporting. If rounded intermediate values are displayed, the stated arithmetic should still match those values.
  • Treating the confidence level as a probability about this fixed interval. The confidence level describes the long-run success rate of the method under its assumptions, not the probability that a fixed population mean changes or falls inside a particular computed interval.
Key takeaway: From \(\bar{x}\), \(s\), \(n\), and a confidence level, calculate \(df=n-1\), obtain the appropriate t critical value, compute \(SE=s/\sqrt{n}\), and form \(\bar{x}\pm t^*SE\). The numerical result is meaningful only when the sampling and distribution assumptions are reasonable.

Check Your Understanding

Try these questions before moving on.

  1. A hypothetical sample has \(\bar{x}=31.6\) seconds, \(s=5.4\) seconds, and \(n=36\). What are the degrees of freedom and standard error?
  2. For a 95% two-sided t interval with \(n=12\), what degrees of freedom are used, and what probability belongs in each tail?
  3. Why is the standard error, rather than the sample standard deviation itself, multiplied by the t critical value?
  4. A student uses \(df=n\) to select a critical value for a one-sample t interval. Identify and correct the mistake.
  5. What does a 90% confidence level describe about the interval-building method over repeated samples?