The One-Sample t Test Statistic

The t Statistic Puts the Sample Mean on a Standardized Scale

In “The Logic of a Significance Test for a Mean,” you saw that a test asks how surprising a sample result would be if the null hypothesis were true. The one-sample t test statistic helps answer that question by measuring how far the observed sample mean \(\bar{x}\) is from the null value \(\mu_0\), relative to the estimated standard error of the sample mean.

The difference \(\bar{x}-\mu_0\) is measured in the original units of the quantitative variable. The estimated standard error \(s/\sqrt{n}\) is measured in those same units. Dividing one by the other expresses the difference as a number of estimated standard errors. This standardized distance is the t statistic.

Formula: For a one-sample t test of \(H_0:\mu=\mu_0\), the observed test statistic is

t=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}.

Here, \(\bar{x}\) is the sample mean, \(\mu_0\) is the value specified by the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size. The denominator \(s/\sqrt{n}\) is the estimated standard error of the sample mean, as introduced in “Computing the Standard Error of a Sample Mean.”

The statistic is calculated from sample data, so it describes the observed sample mean’s position relative to the null value. When reasoning under \(H_0\), we treat \(\mu_0\) as the center of the sampling distribution of the sample mean. The t statistic then expresses the observed distance from that center in estimated standard-error units.

Reading the Sign and Size of t

The numerator determines the sign of \(t\). If \(\bar{x}>\mu_0\), the numerator and statistic are positive: the observed sample mean is above the null value. If \(\bar{x}<\mu_0\), the statistic is negative: the observed sample mean is below the null value. If \(\bar{x}=\mu_0\), then \(t=0\).

The absolute value \(|t|\) describes the distance from \(\mu_0\) in estimated standard errors, without regard to direction. For instance, \(t=2\) means that \(\bar{x}\) is two estimated standard errors above \(\mu_0\); \(t=-2\) means it is two estimated standard errors below \(\mu_0\). The two statistics have the same distance from the null value but point in opposite directions.

Interpretation: The observed t statistic tells how many estimated standard errors the sample mean is above or below the null value. Its sign gives the direction, and its absolute value gives the size of the standardized distance.

A t statistic has no units: the units in \(\bar{x}-\mu_0\) cancel with the units in \(s/\sqrt{n}\). Do not interpret \(t=2\) as a difference of two units in the original measurement. It is a difference of two estimated standard errors.

The statistic is not itself a probability and does not, by itself, tell us whether to reject \(H_0\). To assess how unusual the result would be under the null model, a test uses the appropriate t distribution and the alternative hypothesis to find a p-value. The t distribution depends on degrees of freedom, a topic developed in the next tutorial. The alternative determines which direction or directions count as evidence; it does not change the formula for the observed t statistic.

Calculate the Statistic Carefully

A reliable calculation has three parts: find the difference \(\bar{x}-\mu_0\), calculate the estimated standard error \(s/\sqrt{n}\), and divide the difference by that standard error. Keep parentheses around the numerator and denominator when entering the calculation. This helps prevent a common order-of-operations error.

Find the signed difference.
Subtract the null value from the sample mean: \(\bar{x}-\mu_0\). Do not reverse the subtraction.

Calculate the estimated standard error.
Divide the sample standard deviation by the square root of the sample size: \(s/\sqrt{n}\).

Standardize the difference.
Divide the signed difference by the estimated standard error. The result is the observed t statistic.

Interpret the result.
State the direction and number of estimated standard errors between \(\bar{x}\) and \(\mu_0\), in context.

Worked Examples

Worked Example: A Sample Mean Above a Target

A fictional greenhouse randomly selects 36 seedling trays from a production run of 900 trays. The mean time to sprout is 42.6 hours, with a sample standard deviation of 6 hours. The greenhouse is checking a null value of 40 hours. Calculate and interpret the t statistic.

State. Let \(\mu\) be the true mean time to sprout, in hours, for seedlings from this production run. The null hypothesis is \(H_0:\mu=40\) hours. The sample has \(\bar{x}=42.6\) hours, \(s=6\) hours, and \(n=36\).

Plan and check conditions. The trays were randomly selected, supporting the Random condition. Because the sample was taken without replacement, check the 10% condition: \(0.10(900)=90\), and \(36\leq90\), so the sample is no more than 10% of the run and independence is reasonable. Since \(n=36\geq30\), the large-sample route supports the Normal/Large Sample condition for inference about the mean.

Do. First find the difference and estimated standard error:

\bar{x}-\mu_0=42.6-40=2.6\text{ hours}, \qquad \frac{s}{\sqrt{n}}=\frac{6}{\sqrt{36}}=\frac{6}{6}=1\text{ hour}.

Now divide the difference by the estimated standard error:

t=\frac{42.6-40}{6/\sqrt{36}} =\frac{2.6}{1} =2.60.

Conclude. The observed sample mean sprouting time is 2.60 estimated standard errors above the null value of 40 hours. This describes the sample’s standardized distance from the null value. It does not, by itself, establish whether there is convincing evidence that the population mean is greater than 40 hours; that decision requires the p-value and a stated significance level.

Worked Example: A Sample Mean Below a Claimed Value

A fictional bike-share program randomly selects 16 bicycles and measures the pressure in their tires. The sample mean is 31.5 pounds per square inch (psi), and the sample standard deviation is 5.6 psi. The program checks the null value of 34 psi. A quality report indicates that tire pressures are approximately Normally distributed, and a plot of the sample shows no pronounced outliers. Find and interpret the t statistic.

State. Let \(\mu\) be the true mean tire pressure, in psi, for bicycles in this program. The null hypothesis is \(H_0:\mu=34\) psi.

Plan and check conditions. The 16 bicycles were randomly selected, supporting the Random condition. There are 500 bicycles in the program, and \(16\leq0.10(500)=50\), so the 10% condition supports independence when sampling without replacement. The sample size is \(16<30\), so the large-sample route is not met. For this small sample, the report of an approximately Normal population and the plot with no pronounced outliers support the Normal/Large Sample condition.

Do. The difference is \(31.5-34=-2.5\) psi. The estimated standard error is

\frac{s}{\sqrt{n}}=\frac{5.6}{\sqrt{16}}=\frac{5.6}{4}=1.4\text{ psi}.

Therefore,

t=\frac{31.5-34}{5.6/\sqrt{16}} =\frac{-2.5}{1.4} \approx-1.79.

Conclude. The sample mean tire pressure is about 1.79 estimated standard errors below 34 psi. The negative sign indicates that the sample mean is below the null value; it does not mean that the pressure itself is negative.

Worked Example: The Same Difference With Different Sample Sizes

Consider two fictional audits of a production process that claims a mean package mass of 100 grams. In Audit A, a random sample of 9 packages has a mean mass of 106 grams and a standard deviation of 9 grams. In Audit B, a separate random sample of 36 packages has the same sample mean and standard deviation. Compare the t statistics.

State. For each audit, let \(\mu\) be the true mean package mass, in grams, for packages from the process. The null hypothesis is \(H_0:\mu=100\) grams.

Plan and check conditions. Both samples were selected randomly, supporting the Random condition. Audit A sampled 9 packages from a run of 300, and \(9\leq0.10(300)=30\); Audit B sampled 36 from a run of 1,200, and \(36\leq0.10(1200)=120\). Thus, the 10% condition supports independence in both audits. For Audit A, \(n=9<30\), so we also need information about shape; the process is described as approximately Normal, with no pronounced outliers in the sample. For Audit B, \(n=36\geq30\), so the large-sample route supports the Normal/Large Sample condition.

Do. Both audits have the same difference from the null value: \(106-100=6\) grams. For Audit A,

\frac{s}{\sqrt{n}}=\frac{9}{\sqrt{9}}=3\text{ grams}, \qquad t=\frac{106-100}{9/\sqrt{9}}=\frac{6}{3}=2.00.

For Audit B,

\frac{s}{\sqrt{n}}=\frac{9}{\sqrt{36}}=1.5\text{ grams}, \qquad t=\frac{106-100}{9/\sqrt{36}}=\frac{6}{1.5}=4.00.

Conclude. In Audit A, the sample mean is 2 estimated standard errors above 100 grams; in Audit B, it is 4 estimated standard errors above 100 grams. The raw difference is the same, but the larger sample has a smaller estimated standard error, so its standardized difference is larger. These statistics alone are not test decisions; each test’s p-value also depends on its t distribution and degrees of freedom.

Common Mistakes and AP Exam Tips

When showing the t statistic, write the formula, substitute the values, and explain the result in standard-error units. A numerical answer without that interpretation may leave unclear whether you understand what the statistic measures.

Reversing the subtraction. The numerator is \(\bar{x}-\mu_0\), not \(\mu_0-\bar{x}\). Reversing it changes the sign and gives the wrong direction.
Dividing by \(s\) instead of \(s/\sqrt{n}\). The denominator is the estimated standard error of the sample mean. The sample standard deviation \(s\) describes variability among individual observations, not the estimated variability of sample means.
Interpreting t in the original units. A statistic of \(t=2.60\) does not mean the sample mean is 2.60 hours above the null value. In the first example, the difference is 2.6 hours, while the t statistic is 2.60 estimated standard errors.
Dropping the sign. Reporting only \(|t|\) loses whether the sample mean is above or below the null value. Retain the sign when reporting the observed statistic.
Treating t as a p-value or decision. A t statistic is a standardized distance, not a probability. To determine whether the result is surprising under \(H_0\), use the appropriate t distribution, degrees of freedom, and alternative hypothesis to find a p-value.
Changing the statistic to match the alternative. The same sample values give the same signed t statistic whether the alternative is lower-sided, upper-sided, or two-sided. The alternative affects which outcomes count when finding the p-value, not the calculation of t.

A full-credit interpretation states the direction and standardized distance in context. For example: “The sample mean tire pressure is about 1.79 estimated standard errors below the null value of 34 psi.” If asked whether this result provides convincing evidence for an alternative, the t statistic alone is not enough; complete the test using the appropriate p-value and significance level.

Key takeaway: The one-sample t statistic is \((\bar{x}-\mu_0)/(s/\sqrt{n})\). Its sign shows whether the sample mean is above or below the null value, and its absolute value gives the distance in estimated standard errors. A t statistic is not itself a probability or a test decision.

Check Your Understanding

For each question, focus on the formula and what the signed statistic communicates.

A sample has \(\bar{x}=27\), \(\mu_0=30\), \(s=6\), and \(n=9\). Calculate \(t\) and interpret its sign and size in standard-error units.
What does \(t=0\) tell you about the sample mean and the null value?
Why is the denominator of the one-sample t statistic \(s/\sqrt{n}\), rather than \(s\)?
Two samples have the same difference \(\bar{x}-\mu_0\), but one has a smaller estimated standard error. Which sample has the larger absolute t statistic? Explain.
Why can you not decide to reject \(H_0\) from the t statistic alone?