# 26.4 - Student's t Distribution

26.4 - Student's t Distribution

We have just one more topic to tackle in this lesson, namely, Student's t distribution. Let's just jump right in and define it!

Definition. If $$Z\sim N(0,1)$$ and $$U\sim \chi^2(r)$$ are independent, then the random variable:

$$T=\dfrac{Z}{\sqrt{U/r}}$$

follows a $$t$$-distribution with $$r$$ degrees of freedom. We write $$T\sim t(r)$$. The p.d.f. of T is:

$$f(t)=\dfrac{\Gamma((r+1)/2)}{\sqrt{\pi r} \Gamma(r/2)} \cdot \dfrac{1}{(1+t^2/r)^{(r+1)/2}}$$

for $$-\infty<t<\infty$$.

By the way, the $$t$$ distribution was first discovered by a man named W.S. Gosset. He discovered the distribution when working for an Irish brewery. Because he published under the pseudonym Student, the $$t$$ distribution is often called Student's $$t$$ distribution.

History aside, the above definition is probably not particularly enlightening. Let's try to get a feel for the $$t$$ distribution by way of simulation. Let's randomly generate 1000 standard normal values ($$Z$$) and 1000 chi-square(3) values ($$U$$). Then, the above definition tells us that, if we take those randomly generated values, calculate:

$$T=\dfrac{Z}{\sqrt{U/3}}$$

and create a histogram of the 1000 resulting $$T$$ values, we should get a histogram that looks like a $$t$$ distribution with 3 degrees of freedom. Well, here's a subset of the resulting values from one such simulation: Note, for example, in the first row:

$$T(3)=\dfrac{-2.60481}{\sqrt{10.2497/3}}=-1.4092$$

Here's what the resulting histogram of the 1000 randomly generated $$T(3)$$ values looks like, with a standard $$N(0,1)$$ curve superimposed:

Hmmm. The $$t$$-distribution seems to be quite similar to the standard normal distribution. Using the formula given above for the p.d.f. of $$T$$, we can plot the density curve of various $$t$$ random variables, say when $$r=1, r=4$$, and $$r=7$$, to see that that is indeed the case:

In fact, it looks as if, as the degrees of freedom $$r$$ increases, the $$t$$ density curve gets closer and closer to the standard normal curve. Let's summarize what we've learned in our little investigation about the characteristics of the t distribution:

1. The support appears to be $$-\infty<t<\infty$$. (It is!)
2. The probability distribution appears to be symmetric about $$t=0$$. (It is!)
3. The probability distribution appears to be bell-shaped. (It is!)
4. The density curve looks like a standard normal curve, but the tails of the $$t$$-distribution are "heavier" than the tails of the normal distribution. That is, we are more likely to get extreme $$t$$-values than extreme $$z$$-values.
5. As the degrees of freedom $$r$$ increases, the $$t$$-distribution appears to approach the standard normal $$z$$-distribution. (It does!)

As you'll soon see, we'll need to look up $$t$$-values, as well as probabilities concerning $$T$$ random variables, quite often in Stat 415. Therefore, we better make sure we know how to read a $$t$$ table.

## The $$t$$ Table

If you take a look at Table VI in the back of your textbook, you'll find what looks like a typical $$t$$ table. Here's what the top of Table VI looks like (well, minus the shading that I've added): The $$t$$-table is similar to the chi-square table in that the inside of the $$t$$-table (shaded in purple) contains the $$t$$-values for various cumulative probabilities (shaded in red), such as 0.60, 0.75, 0.90, 0.95, 0.975, 0.99, and 0.995, and for various $$t$$ distributions with $$r$$ degrees of freedom (shaded in blue). The row shaded in green indicates the upper $$\alpha$$ probability that corresponds to the $$1-\alpha$$ cumulative probability. For example, if you're interested in either a cumulative probability of 0.60, or an upper probability of 0.40, you'll want to look for the $$t$$-value in the first column.

Let's use the $$t$$-table to read a few probabilities and $$t$$-values off of the table:

Let's take a look at a few more examples.

## Example 26-6

Let $$T$$ follow a $$t$$-distribution with $$r=8$$ df. What is the probability that the absolute value of $$T$$ is less than 2.306?

#### Solution

The probability calculation is quite similar to a calculation we'd have to make for a normal random variable. First, rewriting the probability in terms of $$T$$ instead of the absolute value of $$T$$, we get:

$$P(|T|<2.306)=P(-2.306<T<2.306)$$

Then, we have to rewrite the probability in terms of cumulative probabilities that we can actually find, that is:

$$P(|T|<2.306)=P(T<2.306)-P(T<-2.306)$$

Pictorially, the probability we are looking for looks something like this:

But the $$t$$-table doesn't contain negative $$t$$-values, so we'll have to take advantage of the symmetry of the $$T$$ distribution. That is:

$$P(|T|<2.306)=P(T<2.306)-P(T>2.306)$$

Can you find the necessary $$t$$-values on the $$t$$-table?  The $$t$$-table tells us that $$P(T<2.306)=0.975$$ and $$P(T>2.306)=0.025$$. Therefore:

$$P(|T|>2.306)=0.975-0.025=0.95$$

What is $$t_{0.05}(8)$$?

#### Solution

The value $$t_{0.05}(8)$$ is the value $$t_{0.05}$$ such that the probability that a $$T$$ random variable with 8 degrees of freedom is greater than the value $$t_{0.05}$$ is 0.05. That is:

Can you find the value $$t_{0.05}$$ on the $$t$$-table?  We have determined that the probability that a $$T$$ random variable with 8 degrees of freedom is greater than the value 1.860 is 0.05.

## Why will we encounter a $$T$$ random variable?

Given a random sample $$X_1, X_2, \ldots, X_n$$ from a normal distribution, we know that:

$$Z=\dfrac{\bar{X}-\mu}{\sigma/\sqrt{n}}\sim N(0,1)$$

Earlier in this lesson, we learned that:

$$U=\dfrac{(n-1)S^2}{\sigma^2}$$

follows a chi-square distribution with $$n-1$$ degrees of freedom. We also learned that $$Z$$ and $$U$$ are independent. Therefore, using the definition of a $$T$$ random variable, we get:

It is the resulting quantity, that is:

$$T=\dfrac{\bar{X}-\mu}{s/\sqrt{n}}$$

that will help us, in Stat 415, to use a mean from a random sample, that is $$\bar{X}$$, to learn, with confidence, something about the population mean $$\mu$$.

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