The t-Distribution¶

What do we do if we want to use the statistical techniques that we've learned to this point to model data, but our samples just aren't large enough? We use a new distribution (called the $t$ -distribution) that's specifically designed to account for the extra variability inherent in smaller samples.

This is essentially section 7.1 of our textbook.

An example problem¶

Suppose we have data on the level of mercury in dolphin muscle arising from 19 dolphins. We'd like a confidence interval for the average amount of mercury in dolphins. Unfortunately, we can't measure mercury in dolphin muscle without killing the dolphin. While it might be important for researchers to assess the threat of mercury in the ocean, they do not want to go kill more dolphins to get that data. What can they conclude from this data set, even though it's a bit too small to use a normal distribution?

Some basic comments¶

The normal distribution, as awesome as it is, requires that we work with large sample sizes - at least 30 and more is better.
The $t$ -distribution is similar but better suited to small sample sizes.
Just as with the normal distribution, there's not just one $t$ -distribution but, rather, a family of distributions.
Like all continuous distributions, we compute probabilities with the $t$ -distribution by computing the area under a curve. We do so using either a computer or a table.

Some more detailed comments¶

The mean of the $t$ -distribution is zero and its variance is related to the degrees of freedom $\nu$ by $\sigma^2 = \frac{\nu}{\nu-2}.$

Unlike the normal distribution, there's no easy way to translate from a $t$ -distribution with one standard deviation to another standard one. As a result, it's less common to use tables and more common to use software than it is with the normal.

Given a particular number of degrees of freedom, however, there is a standard way to derive a $t$ -score that's analogous to the $z$ -score for the normal distribution. This $t$ -score (which is also called a test statistic) is a crucial thing that you need to know when using tables for the $t$ -distribution.

Finding a confidence interval¶

Finding a confidence interval using a $t$ -distribution is a lot like finding one using the normal. It'll have the form $[\overline{x}-ME, \overline{x}+ME],$ where the margin of error $ME$ is $ME = t^* \frac{\sigma}{\sqrt{n}}.$ Note that the familiar $z^*$ multiplier has been replaced with a $t^*$ -multiplier; it plays the exact same role but it comes from the $t$ -distribution, rather than the normal distribution.

More details on the dolphin example¶

Here's the actual data recording the mercury content in dolphin muscle of 19 Risso’s dolphins from the Taiji area in Japan:

d = [
    2.57,4.43,2.09,7.68,4.77,2.12,5.13,5.71,5.33,3.31,
    7.49,4.91,2.58,1.08,6.60,3.91,3.97,6.18,5.90
]

Let's try to use this to compute a 95% confidence interval for the average mercury content in Risso's dolphins.

Computations with Python¶

Here's how we would use Python to compute the mean and standard deviation for out data:

import numpy as np
m = np.mean(d)
s = np.std(d, ddof=1)
[m,s]

[4.513684210526317, 1.8806388114630852]

The crazy looking ddof parameter forces np.std to compute the sample standard deviation, rather than the population standard deviation - i.e. it uses an $n-1$ in the denominator, rather than an $n$ .

Alternatively, you can get the mean and standard deviation with our basic calculator for numeric data

The $t^*$ -multiplier¶

Next, the multiplier $t^*$ can be computed using t.ppf from the scipy.stats module:

from scipy.stats import t
tt = t.ppf(0.975,df=18)
tt

2.10092204024096

Note that the $t^*>2$ and $2$ , of course, would be the multiplier for the normal distribution. This makes some sense because the $t$ -distribution is more spread out than the normal.

Alternatively, you can try our new t-distribution calculator.

A hypothesis test¶

We can also use the $t$ -distribution to do hypothesis tests with small sample size. When running a hypothesis test for the mean of numerical data, we again have a hypothesis test that looks like so:

\begin{aligned} (1) & H_{0} : μ & = μ_{0} \\ (2) & H_{A} : μ & \neq μ_{0} (or > or <) . \end{aligned}

$\begin{align} H_0 : \mu&=\mu_0 \\ H_A : \mu&\neq \mu_0 \: (\text{or} > \text{or} <). \end{align}$

As before, $\mu$ denotes the actual population mean and $\mu_0$ is an assumed, specific value. The question is, whether recently collected data supports the alternative hypothesis (two-sided here, though it could be one-sided).

The test-statistic¶

As before, we compute something that looks like a $Z$ -score, though more generally, it's typically called a test statistic. Assuming the data has sample size $n$ , mean $\bar{x}$ , and standard deviation $s$ , then the test-statistic is $\frac{\bar{x}-\mu_0}{s/\sqrt{n}}.$ We then evaluate the test statistic using the appropriate $t$ -distribution to determine whether to reject the null hypothesis or not.