The Form of a One-Sample t Interval

The Structure of a One-Sample t Interval

In “Finding t* Critical Values With a Table or invT,” you learned how to find $t^*$ for a chosen confidence level and degrees of freedom. Now we put that critical value into the form of a one-sample t interval for a population mean, $\mu$. The interval uses the sample mean, $\bar{x}$, as its center and extends in both directions by a margin of error.

Formula: A one-sample t interval for a population mean is $$ \bar{x}\ \pm\ t^*\left(\frac{s}{\sqrt{n}}\right). $$ The equivalent endpoint form is $$ \left(\bar{x}-t^*\frac{s}{\sqrt{n}},\ \bar{x}+t^*\frac{s}{\sqrt{n}}\right). $$

Each piece has a specific role. The sample mean $\bar{x}$ is the center of the interval and the point estimate of $\mu$. The multiplier $t^*$ is a positive critical value selected for the confidence level and degrees of freedom $df=n-1$. The quantity $s/\sqrt{n}$ is the estimated standard deviation of the sampling distribution of $\bar{x}$, called the standard error. Their product, $t^*(s/\sqrt{n})$, is the margin of error, the distance from the center to either endpoint.

Notice that the interval estimates a population mean, so its units are the same as the original measurements. The sample standard deviation $s$ and standard error have those units; $t^*$ has no units. Multiplying the standard error by $t^*$ therefore gives a margin of error in the original units.

Definition: In a one-sample t interval, $\bar{x}$ is the center, $t^*$ sets the multiplier for the selected confidence level and $df$, $s/\sqrt{n}$ is the estimated standard error, and $t^*(s/\sqrt{n})$ is the margin of error.

The formula uses $s$, not the population standard deviation $\sigma$, because $\sigma$ is unknown. As discussed in “Why We Use t Instead of z for Means,” estimating variability with $s$ is why the procedure uses a t distribution. The critical value depends on both the confidence level and $df$; it is not determined by the sample mean or the sample standard deviation.

Assembling the Interval

Before calculating endpoints, identify the parameter in context: it is the population mean $\mu$ of the quantitative variable being measured. Then identify the sample statistics $n$, $\bar{x}$, and $s$, find $df=n-1$, and use the appropriate $t^*$. Finding $t^*$ was the focus of the previous tutorial; here, it becomes the multiplier in the interval.

Identify the population mean and sample statistics.
State what $\mu$ represents, and record $n$, $\bar{x}$, and $s$ with their units.

Find $df$ and $t^*$.
For a one-sample t interval, use $df=n-1$. Use the chosen confidence level and this $df$ to obtain $t^*$.

Find the margin of error.
Use $t^*(s/\sqrt{n})$. Keep extra digits during calculations, then round the margin and endpoints consistently.

Write and interpret the endpoints.
Subtract the margin of error from $\bar{x}$ for the lower endpoint and add it for the upper endpoint. Interpret the interval for $\mu$ in context.

A one-sample t interval also requires suitable conditions. The data should come from a random sample or an appropriate randomized process; observations should be independent, with the 10% condition checked for sampling without replacement. The distribution of the measured variable should be approximately Normal, or the sample should be large enough for the procedure to be reasonably robust. With a small sample, inspect the data for strong skewness or outliers.

Worked Examples: Identifying the Pieces

Worked Example: A 95% Interval for Commuting Time

A random sample of $n=132$ commuters has a mean commute time of $\bar{x}=22.0$ minutes and a standard deviation of $s=13.0$ minutes. Construct a 95% one-sample t interval for the population mean commute time, $\mu$, for commuters in the population represented by the sample.

State. The parameter is $\mu$, the true mean commute time for that population. We want a 95% confidence interval for $\mu$.

Plan and check conditions. The problem states that the data come from a random sample. If commuters were sampled without replacement, the 10% condition is met provided the population contains at least 1,320 commuters, because $132\leq0.10(1320)$. The sample size, 132, is large; we would still look for extreme skewness or outliers that might make the t procedure unreliable. Assuming none are present, the conditions support using a one-sample t interval.

Do: identify the pieces. The sample mean is the center, $22.0$ minutes. The degrees of freedom are

$$ df=n-1=132-1=131. $$

For a 95% confidence level with 131 degrees of freedom, $t^*\approx1.9782$, as found using the method in “Finding t* Critical Values With a Table or invT.” The standard error is $s/\sqrt{n}$, so the interval’s form with the values substituted is

\bar{x}\pm t^*\left(\frac{s}{\sqrt{n}}\right) = 22.0\pm1.9782\left(\frac{13.0}{\sqrt{132}}\right)\text{ minutes}.

The standard error is approximately $1.1315$ minutes. Using the displayed $t^*$ and retaining the unrounded standard error in the multiplication gives

\text{margin of error} = 1.9782\left(\frac{13.0}{\sqrt{132}}\right) \approx2.2383\text{ minutes} \approx2.238\text{ minutes}.

Using the four-decimal margin to show the endpoints,

(22.0-2.2383,\ 22.0+2.2383) \approx(19.7617,\ 24.2383)\text{ minutes}.

Conclude. We are 95% confident that the population mean commute time is between approximately 19.7617 and 24.2383 minutes. In this interval, $22.0$ is the sample-based center, $1.9782$ is the t critical value, $13.0/\sqrt{132}$ is the estimated standard error, and about $2.2383$ minutes is the margin of error.

Worked Example: A 90% Interval for Weekly Practice Time

A random sample of 36 students at a music program reports a mean of 18.4 hours of practice per week and a standard deviation of 4.8 hours. Construct and interpret a 90% one-sample t interval for the population mean weekly practice time.

Identify the pieces. The parameter $\mu$ is the population mean weekly practice time for students represented by the sample. Here, $n=36$, $\bar{x}=18.4$ hours, and $s=4.8$ hours. The degrees of freedom are

$$ df=n-1=36-1=35. $$

For a central 90% confidence level and 35 degrees of freedom, $t^*\approx1.6896$. Thus the interval form is

\bar{x}\pm t^*\left(\frac{s}{\sqrt{n}}\right) = 18.4\pm1.6896\left(\frac{4.8}{\sqrt{36}}\right)\text{ hours}.

The standard error and margin of error are

\frac{s}{\sqrt{n}}=\frac{4.8}{6}=0.8\text{ hours}, \qquad 1.6896(0.8)=1.35168\approx1.3517\text{ hours}.

The endpoints are

(18.4-1.3517,\ 18.4+1.3517) \approx(17.0483,\ 19.7517)\text{ hours}.

Assuming the sample was selected randomly, the 10% condition is satisfied if the population contains at least 360 students. The sample of 36 is large enough for a t interval to be reasonably robust, provided there are no extreme outliers or severe skewness. We are 90% confident that the population mean weekly practice time is between approximately 17.0483 and 19.7517 hours.

Worked Example: A 95% Interval with a Smaller Sample

A random sample of 25 calibration tasks has a mean completion time of 74.2 minutes and a standard deviation of 10.0 minutes. A graph of the sample shows no strong skewness or outliers. Construct a 95% one-sample t interval for the population mean completion time.

The parameter is $\mu$, the population mean time to complete a calibration task. The sample statistics are $n=25$, $\bar{x}=74.2$ minutes, and $s=10.0$ minutes. The degrees of freedom are

$$ df=n-1=25-1=24. $$

For a 95% confidence level and 24 degrees of freedom, $t^*\approx2.0639$. The graph provides evidence that the sample is not strongly skewed and has no outliers, which is important with this smaller sample. Assuming the tasks were randomly selected and the population contains at least 250 tasks, the random and 10% conditions are also satisfied.

Substitute the sample values into the interval form:

\bar{x}\pm t^*\left(\frac{s}{\sqrt{n}}\right) = 74.2\pm2.0639\left(\frac{10.0}{\sqrt{25}}\right)\text{ minutes}.

The standard error is $10.0/5=2.0$ minutes, so the margin of error is

2.0639(2.0)=4.1278\text{ minutes}.

The endpoints are

(74.2-4.1278,\ 74.2+4.1278) = (70.0722,\ 78.3278)\text{ minutes}.

We are 95% confident that the population mean time to complete a calibration task is between approximately 70.0722 and 78.3278 minutes. The interval is centered at the sample mean, and the margin of error is the same distance, 4.1278 minutes, on either side.

Common Mistakes and AP Exam Tips

Using $\sigma$ instead of $s$. In a one-sample t interval, the population standard deviation is unknown, so the standard error uses the sample standard deviation: $s/\sqrt{n}$.
Using $n$ instead of $n-1$ for degrees of freedom. For a one-sample t interval, calculate $df=n-1$. For the sample of 132 commuters, that is 131 degrees of freedom.
Calling $t^*$ the margin of error. The margin of error is the product $t^*(s/\sqrt{n})$. The critical value alone is only the multiplier.
Adding and subtracting the standard error without multiplying by $t^*$. The interval extends by the margin of error, not by one standard error unless $t^*=1$, which is not the usual case.
Mixing up the center and the endpoints. The sample mean is the center. Subtract the margin for the lower endpoint and add it for the upper endpoint.
Dropping the units or interpreting the interval as individual values. State the interval for the population mean in context, with the measurement units. It does not describe where most individual observations fall.
Giving only a calculator result. Show the interval form, identify the values substituted for $\bar{x}$, $t^*$, $s$, and $n$, and report the margin and endpoints. Carry extra digits until the final rounding so displayed values remain consistent.

A complete AP response makes the structure visible: name the population mean, show the sample mean as the center, state the degrees of freedom and critical value, calculate the margin of error, and give the endpoints with units. Then interpret the interval in context. The conditions justify the procedure; the formula shows how the sample information and confidence level determine the interval’s width.

Key takeaway: A one-sample t interval has the form $\bar{x}\pm t^*(s/\sqrt{n})$. The sample mean is its center, $s/\sqrt{n}$ is the estimated standard error, and $t^*(s/\sqrt{n})$ is the margin of error used to find the two endpoints.

Check Your Understanding

Use the interval form to identify each component and explain what the interval estimates.

A sample has $n=40$, $\bar{x}=12.6$ liters, and $s=3.2$ liters. What parameter is being estimated, and what are the center and standard error expressions?
For a one-sample t interval with $n=40$, what are the degrees of freedom?
If $t^*=2.023$ and the standard error is 0.51 kilograms, write the margin-of-error calculation.
In the form $\bar{x}\pm t^*(s/\sqrt{n})$, which quantity determines the center, and which product determines the distance from the center to either endpoint?
Why should a conclusion describe an interval for the population mean rather than a range containing most individual observations?