Mathwords logoMathwords

t-Interval — Definition, Formula & Examples

A t-interval is a confidence interval for a population mean μ\mu. You use it when you know the sample mean and sample standard deviation but not the population standard deviation σ\sigma.

For a simple random sample of size nn with mean xˉ\bar{x} and standard deviation ss, a level-CC one-sample t-interval is xˉ±t∗sn\bar{x} \pm t^* \frac{s}{\sqrt{n}}. Here t∗t^* is the critical value of the t-distribution with n−1n-1 degrees of freedom that leaves a central area of CC. The procedure requires three conditions: the data come from a random sample, n≤10%n \le 10\% of the population when sampling without replacement, and the population is approximately normal or n≥30n \ge 30 (for small samples, the data show no strong skew or outliers).

Key Formula

xˉ±t∗ sn,df=n−1\bar{x} \pm t^*\,\frac{s}{\sqrt{n}}, \qquad df = n - 1
Where:
  • xˉ\bar{x} = sample mean, the point estimate of μ\mu
  • t∗t^* = t critical value for the chosen confidence level and n−1n-1 degrees of freedom
  • ss = sample standard deviation
  • nn = sample size
  • sn\frac{s}{\sqrt{n}} = standard error of the sample mean

Worked Example

Problem: A random sample of 16 students at a large school reports nightly study time. The sample mean is 50 minutes and the sample standard deviation is 8 minutes. A dotplot shows no strong skew or outliers. Construct a 95% t-interval for the true mean study time μ\mu.
Step 1: Check the conditions. The sample is random, and 16 students is far less than 10% of a large school. The sample is small, but the data show no strong skew or outliers. Find the degrees of freedom.
df=n−1=16−1=15df = n - 1 = 16 - 1 = 15
Step 2: Find the critical value from a t-table or calculator for 95% confidence with 15 degrees of freedom. Then compute the standard error.
t∗=2.131,SE=sn=816=84=2t^* = 2.131, \qquad SE = \frac{s}{\sqrt{n}} = \frac{8}{\sqrt{16}} = \frac{8}{4} = 2
Step 3: Multiply to get the margin of error, then add and subtract it from the sample mean.
ME=2.131×2=4.262,50±4.262  ⇒  (45.738, 54.262)ME = 2.131 \times 2 = 4.262, \qquad 50 \pm 4.262 \;\Rightarrow\; (45.738,\ 54.262)
Answer: The 95% t-interval is about (45.7,54.3)(45.7, 54.3) minutes. You are 95% confident that this interval captures the true mean nightly study time of all students at the school.

Why It Matters

In real studies you almost never know σ\sigma, so the t-interval, not the z-interval, is the standard tool for estimating a mean. On the AP Statistics exam, free-response questions on means expect you to name this procedure, check its conditions, and interpret the result in context.

Common Mistakes

Mistake: Using z∗=1.96z^* = 1.96 instead of t∗t^*, or looking up t∗t^* with df=ndf = n.
Correction: When you use ss in place of σ\sigma, the standardized sample mean follows a t-distribution with n−1n-1 degrees of freedom. For small samples, t∗t^* is noticeably larger than z∗z^*. Here 2.131>1.962.131 > 1.96, which gives a wider interval that accounts for the extra uncertainty in ss.

Related Terms