t-Distribution

The t-distribution is a bell-shaped curve, centered at 0, with heavier tails than the standard normal curve. You use it instead of z when the population standard deviation σ is unknown and you estimate it with the sample standard deviation s. Its shape depends on the degrees of freedom, df = n − 1.

Before this lesson: Sampling Distribution of a Sample Mean, Standard Normal Distribution and Confidence Intervals

Why we need t instead of z

For a sample mean, the z-score uses σ/√n. But in real studies, σ is almost never known. So you replace σ with s, the sample standard deviation, and get the standard error s/√n.

That swap adds extra uncertainty, because s changes from sample to sample. With a small sample, s can be far from σ. The resulting statistic is no longer exactly normal. It follows a t-distribution, which is more spread out to account for the guesswork in s.

The degrees of freedom tell you which t-curve to use. For one sample, df = n − 1. As df grows, s becomes a better estimate of σ, and the t-curve gets closer to the standard normal curve.

t-distribution formula

t = \dfrac{\bar{x} - \mu}{s / \sqrt{n}} \qquad df = n - 1

• Like z, the t statistic counts how many standard errors x̄ is from μ.
• The t-curve is symmetric about 0 and bell-shaped.
• It has heavier tails than z, so t critical values are larger than z critical values.
• As df increases, t critical values get closer to z: at df = ∞, t* = z*.

If your df is not in a printed table, use the next smaller df. That gives a slightly larger t*, which is the safe choice.

Calculator: on a TI-84, invT(left area, df) gives a critical value; for 95%, use invT(0.975, df). tcdf(lower, upper, df) gives an area. In Desmos, use tdist(df).inversecdf(0.975).

t-distribution examples

Example 1: Find df and t*

A random sample of 16 students is used to build a 95% confidence interval for a mean. Find the degrees of freedom and the critical value t*.

df = 16 - 1 = 15

1 - 0.95 = 0.05

\dfrac{0.05}{2} = 0.025

t^* = \text{invT}(0.975, 15) \approx 2.131

Answer: df = 15 and t* ≈ 2.131. Compare that with z* = 1.96: the t value is larger.

Example 2: Compute a t statistic

A random sample of 25 batteries has mean life x̄ = 52.4 hours and standard deviation s = 6 hours. The maker claims μ = 50 hours. Find the t statistic and its degrees of freedom.

\sqrt{25} = 5

\text{SE} = \dfrac{6}{5} = 1.2

t = \dfrac{52.4 - 50}{1.2} = \dfrac{2.4}{1.2}

= 2.0

df = 25 - 1 = 24

Answer: t = 2.0 with df = 24. The sample mean is 2 standard errors above the claimed mean.

Example 3: How much wider is t?

For a sample of n = 10, compare the 95% critical values t* and z*. How much wider is a t-interval than a z-interval with the same standard error?

df = 10 - 1 = 9

t^* = \text{invT}(0.975, 9) \approx 2.262

\dfrac{2.262}{1.96} \approx 1.154

Answer: t* ≈ 2.262, so the t-interval is about 15% wider. With only 10 values, s is a shaky estimate of σ, and the wider interval makes up for it.

Example 4: Find a tail area

Use Example 2. If the true mean is 50 hours, what is the probability of a t statistic of 2.0 or more?

P(t \ge 2.0) = \text{tcdf}(2.0, 10^{99}, 24)

\approx 0.0285

Answer: about 0.0285. The standard normal curve gives about 0.0228 for z ≥ 2, so the t-curve puts a little more area in the tail.

Example 5: Checking a wait-time claim (word problem)

A coffee shop says customers wait 4 minutes on average. A random sample of 12 customers has x̄ = 4.9 minutes and s = 1.5 minutes. Find the t statistic, and the probability of a t this large or larger if the claim is true.

\sqrt{12} \approx 3.464

\text{SE} = \dfrac{1.5}{3.464} \approx 0.4330

t = \dfrac{4.9 - 4}{0.4330} = \dfrac{0.9}{0.4330}

\approx 2.079

df = 12 - 1 = 11

P(t \ge 2.079) \approx 0.031

Answer: t ≈ 2.079 with df = 11, and the probability is about 0.031. A sample mean this high would be fairly unusual if the true mean were 4 minutes. Hypothesis tests later in the course make this reasoning formal.

Common mistakes

Related methods

t vs. z for proportions

The t-distribution is used for means. For proportions, the standard error comes from p̂ alone, so intervals and tests for proportions still use z (see confidence interval for a proportion).

Margin of error with t

A t-interval has margin of error t^* \cdot s/\sqrt{n}. It works like the margin of error with z, but t* depends on the sample size through df. The next lesson builds the full confidence interval for a mean.

Practice problems

1. A sample has n = 40. Find the degrees of freedom.
2. Find t* for a 90% confidence interval with df = 12.
3. Find t* for a 99% confidence interval with df = 20.
4. A sample of 16 has x̄ = 18 and s = 4. Find the t statistic for μ = 20.
5. Which is larger: t* for 95% with df = 5, or with df = 30?
6. What value does t* for 95% confidence approach as df grows very large?
7. A sample of 36 has s = 9. Find the standard error of x̄.
8. A sample of 25 has x̄ = 103 and s = 10. Find the t statistic for μ = 100, and its df.
9. Use Problem 8. Find P(t \ge 1.5).
10. Why does the t-distribution have heavier tails than the standard normal distribution?

df = 40 - 1 = 39

1 - 0.90 = 0.10

\dfrac{0.10}{2} = 0.05

t^* = \text{invT}(0.95, 12) \approx 1.782

1 - 0.99 = 0.01

\dfrac{0.01}{2} = 0.005

t^* = \text{invT}(0.995, 20) \approx 2.845

\text{SE} = \dfrac{4}{\sqrt{16}} = \dfrac{4}{4}

= 1

t = \dfrac{18 - 20}{1} = \dfrac{-2}{1}

= -2

2.571 \gt 2.042

\text{SE} = \dfrac{9}{\sqrt{36}} = \dfrac{9}{6}

= 1.5

\text{SE} = \dfrac{10}{\sqrt{25}} = \dfrac{10}{5}

= 2

t = \dfrac{103 - 100}{2} = \dfrac{3}{2}

= 1.5

df = 25 - 1 = 24

P(t \ge 1.5) = \text{tcdf}(1.5, 10^{99}, 24) \approx 0.0733

t-distribution FAQ

When do you use a t-distribution instead of z?

Use t for inference about a mean when the population standard deviation σ is unknown and you use the sample standard deviation s instead. In practice, that is almost always.

What are degrees of freedom in a t-distribution?

Degrees of freedom set which t-curve to use. For a one-sample t procedure, df = n − 1. Smaller df means heavier tails and larger critical values.

How do you find a t critical value?

Use a t-table with the right df and confidence level, or invT(left area, df) on a TI-84. For 95% confidence, the left area is 0.975.

Is the t-distribution the same as the normal distribution?

No. Both are symmetric and bell-shaped, but the t-distribution has heavier tails. As df increases, it gets closer and closer to the standard normal distribution.

Summary

• Use t when σ is unknown and the standard error is s/√n.
• The t statistic is t = (\bar{x} - \mu)/(s/\sqrt{n}), with df = n − 1.
• t-curves are symmetric with heavier tails than z, so t* is larger than z*.
• As df grows, t* approaches z*.
• Find t* with a t-table or invT(left area, df).

Previous lesson: Margin of Error

Next lesson: Confidence Interval for a Mean

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