The goal is to check if a population’s mean is as claimed given that we have no knowledge on the population variance.

Suppose the sample size is nn. The main test statistic is

t=xˉ−μ0s/nt=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}

H0H_0 H1H_1 pp-value Devision Rule, Reject H0H_0 if …
μ=μ0\mu=\mu_0 or μ≤μ0\mu\le\mu_0 μ>μ0\mu\gt\mu_0 P(T>t)\mathbb{P}(T\gt t) t>tn−1,αt\gt t_{n-1,\alpha}
μ=μ0\mu=\mu_0 or μ≥μ0\mu\ge\mu_0 μ<μ0\mu\lt\mu_0 P(T<t)\mathbb{P}(T\lt t) t<−tn−1,αt\lt -t_{n-1,\alpha}
μ=μ0\mu=\mu_0 μ≠μ0\mu\ne \mu_0 P(∣T∣>∣t∣)\mathbb{P}(\vert T\vert \gt \vert t \vert) ∣t∣>tn−1,α/2\vert t\vert > t_{n-1,\alpha/2}

Here, tdf,αt_{df, \alpha} can be retrieved from tt-table. Notation dfdf means degree of freedom, which is usually n−1n-1.

tt-distribution

If ZZ is a standardized normal r.v. Z∼N(0,1)Z\sim \mathcal{N}(0,1), and r.v. XX has a χ2\chi^2 distribution with vv degrees of freedom, i.e., X∼χv2X\sim \chi_v^2, independent of ZZ, then

ZX/v∼tv\frac{Z}{\sqrt{X/v}}\sim t_v

i.e., follows a tt-distribution with vv degrees of freedom.

Confidence Interval

In testing H0:μ=μ0,H1:μ≠μ0H_0:\mu=\mu_0, H_1:\mu\ne\mu_0, the (1−α)(1-\alpha) CI for μ\mu is

[xˉ−tn−1,α/2sn,xˉ+tn−1,α/2sn]\Big[ \bar{x}-t_{n-1,\alpha/2}\frac{s}{\sqrt{n}}, \bar{x}+t_{n-1,\alpha/2}\frac{s}{\sqrt{n}} \Big]

  • ME=tn−1,α/2snME=t_{n-1,\alpha/2}\frac{s}{\sqrt{n}}
  • θ=μ,θ^=xˉ\theta=\mu, \hat\theta=\bar{x}