The distribution of t according to the Central Limit Theorem

Last updated on 2024-03-12 | Edit this page

Estimated time: 7 minutes

Overview

Questions

  • What is the central limit theorem?
  • What does it predict for the distribution of \(t\)?

Objectives

  • Introduce the Central limit theorem.
  • Explain that it gives a theoretical distribution for \(t\).

Calculate a p-value using the central limit theorem

Let me sum up what we know now:

  • We have a sample of mice weights and a weight \(\mu_0\) that we compare these weights to. We want to know whether the average weight differs from \(\mu_0\).
  • We calculated \(t\) for this specific sample. \(t\) is a scaled difference between sample mean and \(\mu_0\). In case the sample mean is the same as \(\mu_0\), \(t\) should be close to zero.
  • Moreover, the central limit theorem tells us that, if high fat diet mice in general don’t differ from \(\mu_0\) in their weight, we can expect \(t\) to come from a standard normal distribution.

This means, that if we sample again and again (i.e. the experiment where 20 mice were fed with special diet and weighed, and a \(t\) was calculated), we should measure a different \(t\) each time, and plotting the a histogram or density of \(t\)s will give a Gaussian bell shape.
This is enough knowledge to calculate a p-value. You can do this :)

The distribution of t

We start with the mouse weights and the resulting \(t\) from the last section.

  1. What is the probability of observing a \(t\) with value at least as high as that seen here in a standard normal distribution (normal distribution with mean 0 and standard deviation of 1)?

  2. Remember that in this scenario the question is not whether the weight is “higher than”, but “different from” \(\mu_0\). How do you have to adapt the above calculation to make this a two-sided test?

R 

weights <- c(31.41, 28.29, 22.82, 26.07, 31.97, 22.60, 31.47, 29.18, 22.98, 23.26, 23.48, 20.88, 28.44, 30.34, 23.14, 22.80, 24.47, 39.73, 25.71, 22.74)
mu0 <- 23.89338
tstat <- (mean(weights) - mu0) *sqrt(length(weights)) / sd(weights) 
pnorm(tstat, mean=0, sd=1,lower.tail=FALSE)

OUTPUT 

[1] 0.005122719
  1. Just multiply by 2: that gives the probability to get something larger than the observed t, or smaller than -t. Multiplying works, because the normal distribution is symmetric.

R 

pnorm(tstat, mean=0, sd=1,lower.tail=FALSE)*2

OUTPUT 

[1] 0.01024544

Congrats…

…the worst part is over! If you could follow until here, you understood the essence of the t-test. Everything that follows is just practical details.

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