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:

ROW | Z | CHISQ (3) | T(3) |
---|---|---|---|

1 | -2.60481 | 10.2497 | -1.4092 |

2 | 2.92321 | 1.6517 | 3.9396 |

3 | -0.48633 | 0.1757 | -2.0099 |

4 | -0.48212 | 3.8283 | -0.4268 |

5 | -0.04150 | 0.2422 | -0.1461 |

6 | -0.84225 | -0.0903 | -4.8544 |

7 | -0.31205 | 1.6326 | -0.4230 |

8 | 1.33068 | 5.2224 | 1.0086 |

9 | -0.64104 | 0.9401 | -1.1451 |

10 | -0.05110 | 2.2632 | -0.0588 |

11 | 1.61601 | 4.6566 | 1.2971 |

12 | 0.81522 | 2.1738 | 0.9577 |

13 | 0.38501 | 1.8404 | 0.4916 |

14 | -1.63426 | 1.1265 | -2.6669 |

...and so on... | |||

994 | -0.18942 | 3.5202 | -0.1749 |

995 | 0.43078 | 3.3585 | 0.4071 |

996 | -0.14068 | 0.6236 | -0.3085 |

997 | -1.76357 | 2.6188 | -1.8876 |

998 | -1.02310 | 3.2470 | -0.9843 |

999 | -0.93777 | 1.4991 | -1.3266 |

1000 | -0.37665 | 2.1231 | -0.4477 |

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**:

- The support appears to be \(-\infty<t<\infty\). (It is!)
- The probability distribution appears to be symmetric about \(t=0\). (It is!)
- The probability distribution appears to be bell-shaped. (It is!)
- 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.
- 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 Section

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):

\[P(T \leq t)=\int_{-\infty}^{t} \frac{\Gamma[(r+1) / 2]}{\sqrt{\pi r} \Gamma(r / 2)\left(1+w^{2} / r\right)^{(r+1) / 2}} d w\]

\[P(T \leq-t)=1-P(T \leq t)\]

P(T≤ t) | |||||||

0.60 | 0.75 | 0.90 | 0.95 | 0.975 | 0.99 | 0.995 | |

r | ^{t}_{0.40}(r) | ^{t}_{0.25}(r) | ^{t}_{0.10}(r) | ^{t}_{0.05}(r) | ^{t}_{0.025}(r) | ^{t}_{0.01}(r) | ^{t}_{0.005}(r) |

1 | 0.325 | 1.000 | 3.078 | 6.314 | 12.706 | 31.821 | 63.657 |

2 | 0.289 | 0.816 | 1.886 | 2.920 | 4.303 | 6.965 | 9.925 |

3 | 0.277 | 0.765 | 1.638 | 2.353 | 3.182 | 4.541 | 5.841 |

4 | 0.271 | 0.741 | 1.533 | 2.132 | 2.776 | 3.747 | 4.604 |

5 | 0.267 | 0.727 | 1.476 | 2.015 | 2.571 | 3.365 | 4.032 |

6 | 0.265 | 0.718 | 1.440 | 1.943 | 2.447 | 3.143 | 3.707 |

7 | 0.263 | 0.711 | 1.415 | 1.895 | 2.365 | 2.998 | 3.499 |

8 | 0.262 | 0.706 | 1.397 | 1.860 | 2.306 | 2.896 | 3.355 |

9 | 0.261 | 0.703 | 1.383 | 1.833 | 2.262 | 2.821 | 3.250 |

10 | 0.260 | 0.700 | 1.372 | 1.812 | 2.228 | 2.764 | 3.169 |

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 Section

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?

P(T≤ t) | |||||||

0.60 | 0.75 | 0.90 | 0.95 | 0.975 | 0.99 | 0.995 | |

r | ^{t}_{0.40}(r) | ^{t}_{0.25}(r) | ^{t}_{0.10}(r) | ^{t}_{0.05}(r) | ^{t}_{0.025}(r) | ^{t}_{0.01}(r) | ^{t}_{0.005}(r) |

1 | 0.325 | 1.000 | 3.078 | 6.314 | 12.706 | 31.821 | 63.657 |

2 | 0.289 | 0.816 | 1.886 | 2.920 | 4.303 | 6.965 | 9.925 |

3 | 0.277 | 0.765 | 1.638 | 2.353 | 3.182 | 4.541 | 5.841 |

4 | 0.271 | 0.741 | 1.533 | 2.132 | 2.776 | 3.747 | 4.604 |

5 | 0.267 | 0.727 | 1.476 | 2.015 | 2.571 | 3.365 | 4.032 |

6 | 0.265 | 0.718 | 1.440 | 1.943 | 2.447 | 3.143 | 3.707 |

7 | 0.263 | 0.711 | 1.415 | 1.895 | 2.365 | 2.998 | 3.499 |

8 | 0.262 | 0.706 | 1.397 | 1.860 | 2.306 | 2.896 | 3.355 |

9 | 0.261 | 0.703 | 1.383 | 1.833 | 2.262 | 2.821 | 3.250 |

10 | 0.260 | 0.700 | 1.372 | 1.812 | 2.228 | 2.764 | 3.169 |

P(T≤ t) | |||||||

0.60 | 0.75 | 0.90 | 0.95 | 0.975 | 0.99 | 0.995 | |

r | ^{t}_{0.40}(r) | ^{t}_{0.25}(r) | ^{t}_{0.10}(r) | ^{t}_{0.05}(r) | ^{t}_{0.025}(r) | ^{t}_{0.01}(r) | ^{t}_{0.005}(r) |

1 | 0.325 | 1.000 | 3.078 | 6.314 | 12.706 | 31.821 | 63.657 |

2 | 0.289 | 0.816 | 1.886 | 2.920 | 4.303 | 6.965 | 9.925 |

3 | 0.277 | 0.765 | 1.638 | 2.353 | 3.182 | 4.541 | 5.841 |

4 | 0.271 | 0.741 | 1.533 | 2.132 | 2.776 | 3.747 | 4.604 |

5 | 0.267 | 0.727 | 1.476 | 2.015 | 2.571 | 3.365 | 4.032 |

6 | 0.265 | 0.718 | 1.440 | 1.943 | 2.447 | 3.143 | 3.707 |

7 | 0.263 | 0.711 | 1.415 | 1.895 | 2.365 | 2.998 | 3.499 |

8 | 0.262 | 0.706 | 1.397 | 1.860 | 2.306 | 2.896 | 3.355 |

9 | 0.261 | 0.703 | 1.383 | 1.833 | 2.262 | 2.821 | 3.250 |

10 | 0.260 | 0.700 | 1.372 | 1.812 | 2.228 | 2.764 | 3.169 |

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?

P(T≤ t) | |||||||

0.60 | 0.75 | 0.90 | 0.95 | 0.975 | 0.99 | 0.995 | |

r | ^{t}_{0.40}(r) | ^{t}_{0.25}(r) | ^{t}_{0.10}(r) | ^{t}_{0.05}(r) | ^{t}_{0.025}(r) | ^{t}_{0.01}(r) | ^{t}_{0.005}(r) |

1 | 0.325 | 1.000 | 3.078 | 6.314 | 12.706 | 31.821 | 63.657 |

2 | 0.289 | 0.816 | 1.886 | 2.920 | 4.303 | 6.965 | 9.925 |

3 | 0.277 | 0.765 | 1.638 | 2.353 | 3.182 | 4.541 | 5.841 |

4 | 0.271 | 0.741 | 1.533 | 2.132 | 2.776 | 3.747 | 4.604 |

5 | 0.267 | 0.727 | 1.476 | 2.015 | 2.571 | 3.365 | 4.032 |

6 | 0.265 | 0.718 | 1.440 | 1.943 | 2.447 | 3.143 | 3.707 |

7 | 0.263 | 0.711 | 1.415 | 1.895 | 2.365 | 2.998 | 3.499 |

8 | 0.262 | 0.706 | 1.397 | 1.860 | 2.306 | 2.896 | 3.355 |

9 | 0.261 | 0.703 | 1.383 | 1.833 | 2.262 | 2.821 | 3.250 |

10 | 0.260 | 0.700 | 1.372 | 1.812 | 2.228 | 2.764 | 3.169 |

P(T≤ t) | |||||||

0.60 | 0.75 | 0.90 | 0.95 | 0.975 | 0.99 | 0.995 | |

r | ^{t}_{0.40}(r) | ^{t}_{0.25}(r) | ^{t}_{0.10}(r) | ^{t}_{0.05}(r) | ^{t}_{0.025}(r) | ^{t}_{0.01}(r) | ^{t}_{0.005}(r) |

1 | 0.325 | 1.000 | 3.078 | 6.314 | 12.706 | 31.821 | 63.657 |

2 | 0.289 | 0.816 | 1.886 | 2.920 | 4.303 | 6.965 | 9.925 |

3 | 0.277 | 0.765 | 1.638 | 2.353 | 3.182 | 4.541 | 5.841 |

4 | 0.271 | 0.741 | 1.533 | 2.132 | 2.776 | 3.747 | 4.604 |

5 | 0.267 | 0.727 | 1.476 | 2.015 | 2.571 | 3.365 | 4.032 |

6 | 0.265 | 0.718 | 1.440 | 1.943 | 2.447 | 3.143 | 3.707 |

7 | 0.263 | 0.711 | 1.415 | 1.895 | 2.365 | 2.998 | 3.499 |

8 | 0.262 | 0.706 | 1.397 | 1.860 | 2.306 | 2.896 | 3.355 |

9 | 0.261 | 0.703 | 1.383 | 1.833 | 2.262 | 2.821 | 3.250 |

10 | 0.260 | 0.700 | 1.372 | 1.812 | 2.228 | 2.764 | 3.169 |

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? Section

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 \(\mu\).