November 3

Student’s t-Distribution

A continuous, symmetric distribution used for inference about a population mean when the population standard deviation is unknown and sample size is small. It has heavier tails than the normal distribution, which accounts for extra uncertainty from estimating variance.

Notation: X ~ t(ν) or t_ν
Parameter: ν (degrees of freedom). Commonly ν = n − 1 for a sample of size n.

Key Formulas

Probability Density Function (PDF)

For ν > 0 and x ∈ ℝ:

\[f(x) = \frac{\Gamma\!\left(\tfrac{\nu+1}{2}\right)}{\sqrt{\nu\pi}\,\Gamma\!\left(\tfrac{\nu}{2}\right)} \left(1 + \frac{x^2}{\nu}\right)^{-\frac{\nu+1}{2}}\]

Relationship to Normal and Chi-square

If Z ~ N(0,1) and U ~ χ²(ν) are independent, then

\[T = \frac{Z}{\sqrt{U/\nu}} \sim t(\nu).\]

This is the origin of the t-statistic used in practice.

Mean and Variance

Mean: 0 for ν > 1 (undefined when ν ≤ 1).
Variance: ν / (ν − 2) for ν > 2 (infinite when 1 < ν ≤ 2; undefined when ν ≤ 1).
As ν → ∞, t(ν) → N(0,1).

Use in Inference

Commonly used for confidence intervals and hypothesis tests about a population mean μ when σ is unknown.

t-statistic for a sample (X̄ sample mean, s sample standard deviation, n sample size):

\[t = \frac{\overline{X} - \mu}{s / \sqrt{n}} \sim t(n-1)\]

Critical values and p-values are obtained from the t-distribution with ν = n − 1 degrees of freedom.

Examples

1) (Construct a 95% CI for a mean — show work)
Given: n = 10, x̄ = 5.0, s = 2.0, confidence level = 95%.
Find: 95% CI for μ.

Solution:

Degrees of freedom: ν = n − 1 = 9.
Critical value: t_{0.975,9} ≈ 2.262.
Standard error: SE = s / √n = 2 / √10 ≈ 0.6325.
Margin of error: ME = t* × SE ≈ 2.262 × 0.6325 ≈ 1.431.
95% CI: (x̄ − ME, x̄ + ME) = (5.0 − 1.431, 5.0 + 1.431) ≈ (3.569, 6.431).

Answer: 95% CI ≈ (3.569, 6.431).

2) (Two-sided hypothesis test — show test statistic, p-value, decision)
Problem: Test H0: μ = 4.0 versus Ha: μ ≠ 4.0 using the same sample (n = 10, x̄ = 5.0, s = 2.0). Use α = 0.05.

Solution:

Test statistic: t = (x̄ − μ0) / (s / √n) = (5.0 − 4.0) / (2 / √10) ≈ 1.000 / 0.6325 ≈ 1.581.
Degrees of freedom: ν = 9.
Two-sided p-value: p ≈ 2·P(T_9 ≥ 1.581) ≈ 0.14–0.15 (approx. 0.148).
Decision at α = 0.05: p > 0.05 ⇒ fail to reject H0.

Conclusion: Insufficient evidence to conclude μ ≠ 4.0 (at the 5% level).