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The \(t\) distribution

Peter Ralph

28 September 2021 – Advanced Biological Statistics

Stochastic minute: the \(t\) distribution

The \(t\) statistic

The \(t\) statistic computed from a collection of \(n\) numbers is the sample mean divided by the estimated standard error of the mean, which is the sample SD divided by \(\sqrt{n}\).

If \(x_1, \ldots, x_n\) are numbers, then \[\begin{aligned} \text{(sample mean)} \qquad \bar x &= \frac{1}{n}\sum_{i=1}^n x_i \\ \text{(sample SD)} \qquad s &= \sqrt{\frac{1}{n-1}\sum_{i=1}^n (x_i - \bar x)^2} \end{aligned}\] so \[\begin{equation} t(x) = \frac{\bar x}{s / \sqrt{n}} . \end{equation}\]

Consistency check

n <- 20
x <- rnorm(n)
c(t.test(x)$statistic, 
  mean(x) / (sd(x) / sqrt(n)))
##        t          
## 1.318919 1.318919

The \(t\) approximation

Fact: If \(X_1, \ldots, X_n\) are independent random samples from a distribution with mean \(\mu\), then \[\begin{equation} t(X - \mu) = \frac{\bar x - \mu}{s/\sqrt{n}} \approx \StudentsT(n-2) , \end{equation}\] as long as \(n\) is not too small and the distribution isn’t too wierd.

A demonstration

Let’s check this, by doing:

find the sample \(t\) score of 100 random draws from some distribution

lots of times, and looking at the distribution of those \(t\) scores.

Claim: no matter\({}^*\) the distribution we sample from, the sampling distribution of the \(t\) statistics should look close to the \(t\) distribution.

One sample

n <- 20
x <- 2 * runif(n) - 1
hist(x, xlab='value', col=grey(0.5),
     main=sprintf("t=%f", mean(x)*sqrt(n)/sd(x)))
abline(v=0, lwd=2, lty=3)
abline(v=mean(x), col='red', lwd=2)

plot of chunk r t_one_smaple

More samples

plot of chunk r t_more_samples

Distribution of 1,000 sample \(t\) scores

xm <- replicate(1000, {
            x <- 2 * runif(n) - 1;
            mean(x) * sqrt(n) / sd(x) })
xh <- hist(xm, breaks=40, main=sprintf('t of %d samples', n), col='red')

plot of chunk r t_sampling_dist

Distribution of 1,000 sample \(t\) scores

plot(xh, main=sprintf('t of %d samples', n), col='red')
xx <- xh$breaks
polygon(c(xx[-1] - diff(xx)/2, xx[1]),
        c(length(xm)* diff(pt(xx, df=(n-1))), 0),
        col=adjustcolor("blue", 0.4))

plot of chunk r t_smpling_dist2

Exercise:

Do this again (use my code) except using

x <- rexp(n) - 1

instead of 2 * runif(n) - 1.

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