Lesson 73 – Learning from “Student”

On the fifteenth day of July 2018, Jenny and Joe discussed the confidence interval for the mean of the population.

The $100(1-\alpha)$ % confidence interval for the true mean $\mu$ is $[\bar{x} - Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}} \le \mu \le \bar{x} + Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}]$ .

$\bar{x}$ is the mean of a random sample of size n. The assumption is that the sample is drawn from a population with a true mean $\mu$ and true standard deviation $\sigma$ .

The end-points $[\bar{x} - Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}, \bar{x} + Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}]$ are called the lower and the upper confidence limits.

While developing the confidence interval of the mean water quality at the beach with a sample of 48, Jenny pointed out that the value for the true standard deviation, $\sigma$ is not known.

Joe collected a much larger sample (all data points available for the Rockaway beach) and computed an estimate for the true standard deviation. Based on the principle of consistency, he suggested that this estimate is close to the truth, so can be used as $\sigma$ .

Jenny rightly pointed out that not always we will have such a large sample. Most often, the data is limited.

Joe’s suggestion was to use sample standard deviation, s, i.e., the estimated standard deviation from the limited sample in place of $\sigma$ . However, Jenny was concerned that this will introduce more error into the estimation of the intervals.

March 1908

This was a concern for W. S. Gosset (aka “Student”) in 1908. He points out that one way of dealing with this difficulty is to run the experiment many times, i.e., collect a large enough sample so that the standard deviation can be computed once for all and used for subsequent similar experiments.

He further points out that there are numerous experiments that cannot easily be repeated often. In the situation where a large sample cannot be obtained, one has to rely on the results from the small sample.

The standard deviation of the population ( $\sigma$ ) is not known a priori.

The confidence interval Joe and Jenny derived is based on the conjecture that the distribution of the sample mean ( $\bar{x}$ ) tends to a normal distribution. $\bar{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ , and if the sample size is large enough, it would be reasonable to substitute sample standard deviation s in place of the population standard deviation $\sigma$ . But it is not clear how “large” the sample size should be.

The sample standard deviation can be computed as $s=\sqrt{\frac{1}{n-1}{\displaystyle \sum_{i=1}^{n} (x_{i}-\bar{x})^{2}}}$ . While s is a perfectly good estimate of $\sigma$ , it is not equal to $\sigma$ .

“Student” pointed out that when we substitute s for $\sigma$ , we cannot just assume that $\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}$ will tend to a normal distribution just like $Z = \frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1)$ .

He derived the frequency distribution of $\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}$ in his landmark paper “The Probable Error of a Mean.”

This distribution came to be known as the Student’s t-distribution.

In his paper, he did not use the notation t though. He referred to it as quantity z, obtained by dividing the distance between the mean of a sample and the mean of the population by the standard deviation of the sample.

He derived the distribution of the sample variance and the sample standard deviation, showed that there is no dependence between s and $\bar{x}$ and used this property to derive the joint distribution of $\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}$ .

Today, we will go through the process of deriving the probability distributions for the sample variance $s^{2}$ , sample standard deviation s and the quantity t.

$t = \frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}$

It is important that we know these distributions. They will be a recurring phenomenon from now on and we will be using them in many applications.

I am presenting these derivations using standard techniques. There may be simpler ways to derive them, but I found this step by step thought and derivation process enriching.

During this phase, I will refer back to “Student’s” paper several times. I will also use the explanations given by R.A Fisher in his papers on “Student.”

You can follow along these steps using a pen and paper, or you can just focus on the thought process and skip the derivations, either way, it is fun to learn from “Student.” Trust me.

$t = \frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}$

To know the distribution of t, we should know the distributions of $\bar{x}$ and s.

We already know that $\bar{x}$ tends to a normal distribution; $\bar{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ , but what is the limiting distribution of s, i.e., what is the probability distribution function f(s) of the sample standard deviation?

To know this, we should first know the limiting distribution of the sample variance $s^{2}$ , from which we can derive the distribution of s.

What is the frequency distribution of the sample variance?

Expressing variance as the sum of squares of normal random variables.

Let’s take the equation of the sample variance $s^{2}$ and see if there is a pattern in it.

$s^{2} = \frac{1}{n-1} \sum(x_{i}-\bar{x})^{2}$

Move the $n-1$ over to the left-hand side and do some algebra.

$(n-1)s^{2} = \sum(x_{i}-\bar{x})^{2}$

$(n-1)s^{2} = \sum(x_{i} - \mu -\bar{x} + \mu)^{2}$

$(n-1)s^{2} = \sum((x_{i} - \mu) -(\bar{x} - \mu))^{2}$

$(n-1)s^{2} = \sum[(x_{i} - \mu)^{2} + (\bar{x} - \mu)^{2} -2(x_{i} - \mu)(\bar{x} - \mu)]$

$(n-1)s^{2} = \sum(x_{i} - \mu)^{2} + \sum (\bar{x} - \mu)^{2} -2(\bar{x} - \mu)\sum(x_{i} - \mu)$

$(n-1)s^{2} = \sum(x_{i} - \mu)^{2} + n (\bar{x} - \mu)^{2} -2(\bar{x} - \mu)(\sum x_{i} - \sum \mu)$

$(n-1)s^{2} = \sum(x_{i} - \mu)^{2} + n (\bar{x} - \mu)^{2} -2(\bar{x} - \mu)(n\bar{x} - n \mu)$

$(n-1)s^{2} = \sum(x_{i} - \mu)^{2} + n (\bar{x} - \mu)^{2} -2n(\bar{x} - \mu)(\bar{x} - \mu)$

$(n-1)s^{2} = \sum(x_{i} - \mu)^{2} + n (\bar{x} - \mu)^{2} -2n(\bar{x} - \mu)^{2}$

$(n-1)s^{2} = \sum(x_{i} - \mu)^{2} - n (\bar{x} - \mu)^{2}$

Let’s divide both sides of the equation by $\sigma^{2}$ .

$\frac{(n-1)s^{2}}{\sigma^{2}} = \frac{1}{\sigma^{2}}(\sum(x_{i} - \mu)^{2} - n (\bar{x} - \mu)^{2})$

$\frac{(n-1)s^{2}}{\sigma^{2}} = \sum(\frac{x_{i} - \mu}{\sigma})^{2} - \frac{n}{\sigma^{2}} (\bar{x} - \mu)^{2}$

$\frac{(n-1)s^{2}}{\sigma^{2}} = \sum(\frac{x_{i} - \mu}{\sigma})^{2} - (\frac{\bar{x} - \mu}{\sigma/\sqrt{n}})^{2}$

The right-hand side now looks like the sum of squared standard normal distributions.

$\frac{(n-1)s^{2}}{\sigma^{2}} = Z_{1}^{2} + Z_{2}^{2} + Z_{3}^{2} + ... + Z_{n}^{2} - Z^{2}$

Sum of squares of (n – 1) standard normal random variables.

Does that ring a bell? Sum of squares is the language of the Chi-square distribution. We learned this in lesson 53.

If there are n standard normal random variables, $Z_{1}, Z_{2}, ..., Z_{n}$ , their sum of squares is a Chi-square distribution with n degrees of freedom. Its probability density function is $f(\chi)=\frac{\frac{1}{2}*(\frac{1}{2} \chi)^{\frac{n}{2}-1}*e^{-\frac{1}{2}*\chi}}{(\frac{n}{2}-1)!}$ for $\chi > 0$ and 0 otherwise.

Sine we have $\frac{(n-1)s^{2}}{\sigma^{2}} = Z_{1}^{2} + Z_{2}^{2} + Z_{3}^{2} + ... + Z_{n}^{2} - Z^{2}$

$\frac{(n-1)s^{2}}{\sigma^{2}}$ follows a Chi-square distribution with (n-1) degrees of freedom.

$\frac{(n-1)s^{2}}{\sigma^{2}} \sim \chi^{2}_{n-1}$ with a probability distribution function

$f(\frac{(n-1)s^{2}}{\sigma^{2}}) = \frac{\frac{1}{2}*(\frac{1}{2} \chi)^{\frac{n-1}{2}-1}*e^{-\frac{1}{2}*\chi}}{(\frac{n-1}{2}-1)!}$

⇔

The frequency distribution of the sample variance.

It turns out that, with some modification, this equation is the frequency distribution $f(s^{2})$ of the sample variance.

$f(\frac{(n-1)s^{2}}{\sigma^{2}}) = \frac{\frac{1}{2}*(\frac{1}{2} \chi)^{\frac{n-1}{2}-1}*e^{-\frac{1}{2}*\chi}}{(\frac{n-1}{2}-1)!}$

$f(\frac{(n-1)s^{2}}{\sigma^{2}}) = \frac{\frac{1}{2}(\frac{(n-1)s^{2}}{2\sigma^{2}})^{\frac{n-3}{2}}}{(\frac{n-3}{2})!} e^{-\frac{1}{2}*\frac{(n-1)s^{2}}{\sigma^{2}}}$

$f(\frac{(n-1)s^{2}}{\sigma^{2}}) = \frac{1}{2}\frac{1}{(\frac{n-3}{2})!}(\frac{n-1}{\sigma^{2}})^{\frac{n-1}{2}}\frac{1}{\frac{n-1}{\sigma^{2}}}\frac{1}{2^{\frac{n-3}{2}}}(s^{2})^{\frac{n-3}{2}}e^{-\frac{1}{2}*\frac{(n-1)s^{2}}{\sigma^{2}}}$

$f(\frac{(n-1)s^{2}}{\sigma^{2}}) = \frac{1}{\frac{n-1}{\sigma^{2}}}\frac{1}{2}\frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n-1}{2}}\frac{1}{2^{\frac{n-3}{2}}}\frac{s^{n-3}}{\sigma^{n-1}}e^{-\frac{1}{2}*\frac{(n-1)s^{2}}{\sigma^{2}}}$

$f(\frac{(n-1)s^{2}}{\sigma^{2}}) = \frac{1}{\frac{n-1}{\sigma^{2}}}[\frac{1}{2}\frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n-1}{2}}\frac{1}{2^{\frac{n-3}{2}}}\frac{s^{n-3}}{\sigma^{n-1}}e^{-\frac{1}{2}*\frac{(n-1)s^{2}}{\sigma^{2}}}]$

The above equation can be viewed as $f(aX) = \frac{1}{a}f(X)$ where $X = s^{2}$ and $a = \frac{n-1}{\sigma^{2}}$ .

These few steps will clarify this further.

Let $Y = aX$

$P(Y \le y) = P(aX \le y)$

$P(Y \le y) = P(X \le \frac{1}{a}y)$

$F_{Y}(y) = F_{X}(\frac{1}{a}y)$

$\frac{d}{dy}(F(y)) = \frac{d}{dy}F_{X}(\frac{1}{a}y)$

Applying the fundamental theorem of calculus and chain rule together, we get,

$f(y) = \frac{1}{a}f(\frac{1}{a}y)$

$f(aX) = \frac{1}{a}f(X)$

Hence, the frequency distribution of $s^{2}$ is

$f(s^{2}) = \frac{1}{2}\frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n-1}{2}}\frac{1}{2^{\frac{n-3}{2}}}\frac{s^{n-3}}{\sigma^{n-1}}e^{-\frac{1}{2}*\frac{(n-1)s^{2}}{\sigma^{2}}}$

From $f(s^{2})$ , we can derive the probability distribution of s, i.e., $f(s)$ .

WHAT IS THE FREQUENCY DISTRIBUTION OF THE SAMPLE Standard Deviation?

Here, I refer you to this elegant approach by “Student.”

“The distribution of s may be found from this ( $s^{2}$ ), since the frequency of s is equal to that of $s^{2}$ and all that we must do is to compress the base line suitably.”

Let $f_{1} = f(s^{2})$ be the distribution of $s^{2}$ .

Let $f_{2} = f(s)$ be the distribution of s.

Since the frequency of $s^{2}$ is equal to that of s, we can assume,

$f_{1}ds^{2} = f_{2}ds$

In other words, the probability of finding $s^{2}$ in an infinitesimal interval $ds^{2}$ is equal to the probability of finding s in an infinitesimal interval $ds$ .

We can reduce this using substitution as

$2sf_{1}ds = f_{2}ds$

$f_{2} = 2sf_{1}$

The frequency distribution of s is equal to 2s multiplied by the frequency distribution of $s^{2}$ .

Hence, the frequency distribution of s is

$f(s) = 2s\frac{1}{2}\frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n-1}{2}}\frac{1}{2^{\frac{n-3}{2}}}\frac{s^{n-3}}{\sigma^{n-1}}e^{-\frac{1}{2}*\frac{(n-1)s^{2}}{\sigma^{2}}}$

$f(s) = \frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n-1}{2}}\frac{1}{2^{\frac{n-3}{2}}}\frac{s^{n-2}}{\sigma^{n-1}}e^{-\frac{1}{2}*\frac{(n-1)s^{2}}{\sigma^{2}}}$

WHAT IS THE FREQUENCY DISTRIBUTION OF t?

We are now ready to derive the frequency distribution of t.

Some of the following explanation can be found in R. A. Fisher’s 1939 paper. I broke it down step by step for our classroom.

$t = \frac{\bar{x}-\mu}{s/\sqrt{n}}$

$\bar{x} - \mu = \frac{st}{\sqrt{n}}$

For a given value of s, $d\bar{x} = \frac{s}{\sqrt{n}}dt$

The frequency distribution of $\bar{x}$ is

$\frac{\sqrt{n}}{\sqrt{2\pi}}\frac{1}{\sigma}e^{-\frac{n}{2\sigma^{2}}(\bar{x}-\mu)^{2}}$

Remember, $\bar{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ .

Substituting $\bar{x} - \mu = \frac{st}{\sqrt{n}}$

The distribution becomes

$\frac{\sqrt{n}}{\sqrt{2\pi}}\frac{1}{\sigma}e^{-\frac{n}{2\sigma^{2}}(st/\sqrt{n})^{2}}$

$\frac{\sqrt{n}}{\sqrt{2\pi}}\frac{1}{\sigma}e^{-\frac{t^{2}}{2\sigma^{2}}s^{2}}$

For a given value of s, the probability that $\bar{x}$ will be in $d\bar{x}$ is

$df = \frac{\sqrt{n}}{\sqrt{2\pi}}\frac{1}{\sigma}e^{-\frac{t^{2}}{2\sigma^{2}}s^{2}}d\bar{x}$

Substituting $d\bar{x} = \frac{s}{\sqrt{n}}dt$ , we can get

$df = \frac{\sqrt{n}}{\sqrt{2\pi}}\frac{1}{\sigma}e^{-\frac{t^{2}}{2\sigma^{2}}s^{2}} \frac{s}{\sqrt{n}}dt$

$df = \frac{1}{\sqrt{2\pi}}\frac{1}{\sigma}e^{-\frac{t^{2}}{2\sigma^{2}}s^{2}} s dt$

Fisher points out that for all values of s, we can substitute the frequency distribution of s and integrate it over the interval 0 and $\infty$ . This can be done because $\bar{x}$ and s are independent.

So the joint distribution becomes

$df = \frac{1}{\sqrt{2\pi}}\frac{1}{\sigma} s dt e^{-\frac{t^{2}}{2\sigma^{2}}s^{2}} \int_{0}^{\infty}\frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n-1}{2}}\frac{1}{2^{\frac{n-3}{2}}}\frac{s^{n-2}}{\sigma^{n-1}}e^{-\frac{1}{2}*\frac{(n-1)s^{2}}{\sigma^{2}}}ds$

In the next few steps, I will rearrange some terms to convert the integral into a recognizable form.

$df = \frac{1}{\sqrt{2\pi}}\frac{1}{\sigma} dt \int_{0}^{\infty}\frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n-1}{2}}\frac{1}{2^{\frac{n-3}{2}}}s \frac{s^{n-2}}{\sigma^{n-1}}e^{-\frac{t^{2}}{2\sigma^{2}}s^{2}}e^{-\frac{1}{2}*\frac{(n-1)s^{2}}{\sigma^{2}}}ds$

$df = \frac{1}{\sqrt{2\pi}}\frac{1}{\sigma} \frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n-1}{2}}\frac{1}{2^{\frac{n-3}{2}}} \frac{1}{\sigma^{n-1}}dt \int_{0}^{\infty}s^{n-1} e^{-\frac{t^{2}}{2\sigma^{2}}s^{2}}e^{-\frac{1}{2}*\frac{(n-1)s^{2}}{\sigma^{2}}}ds$

$df = \frac{1}{\sqrt{2\pi}}\frac{1}{\sigma^{n}} \frac{1}{(\frac{n-3}{2})!}\frac{(n-1)^{\frac{n}{2}}}{(n-1)^{1/2}}\frac{1}{2^{\frac{n-3}{2}}} dt \int_{0}^{\infty}s^{n-1} e^{-\frac{t^{2}}{2\sigma^{2}}s^{2}}e^{-\frac{1}{2}*\frac{(n-1)s^{2}}{\sigma^{2}}}ds$

$df = \frac{1}{\sqrt{2\pi(n-1)}}\frac{1}{\sigma^{n}} \frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n}{2}}\frac{1}{2^{\frac{n-3}{2}}} dt \int_{0}^{\infty}s^{n-1} e^{-\frac{t^{2}}{2\sigma^{2}}s^{2}}e^{-\frac{1}{2}*\frac{(n-1)s^{2}}{\sigma^{2}}}ds$

$df = \frac{1}{\sqrt{\pi(n-1)}}\frac{1}{\sigma^{n}} \frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n}{2}}\frac{1}{2^{\frac{n-2}{2}}} dt \int_{0}^{\infty}s^{n-1} e^{-\frac{t^{2}}{2\sigma^{2}}s^{2}}e^{-\frac{1}{2}*\frac{(n-1)s^{2}}{\sigma^{2}}}ds$

$df = \frac{1}{\sqrt{\pi(n-1)}}\frac{1}{\sigma^{n}} \frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n}{2}}\frac{2}{2^{\frac{n}{2}}} dt \int_{0}^{\infty}s^{n-1} e^{-\frac{n -1 + t^{2}}{2\sigma^{2}}s^{2}}ds$

Hang in there. We will need some more concentration!

Let $k = \frac{n - 1 + t^{2}}{2\sigma^{2}}$

The equation becomes

$df = \frac{1}{\sqrt{\pi(n-1)}}\frac{1}{\sigma^{n}} \frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n}{2}}\frac{2}{2^{\frac{n}{2}}} dt \int_{0}^{\infty}s^{n-1} e^{-ks^{2}}ds$

Some more substitutions.

Let $p = ks^{2}$

Then $dp = 2ksds$

The equation can be reduced as

$df = \frac{1}{\sqrt{\pi(n-1)}}\frac{1}{\sigma^{n}} \frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n}{2}}\frac{2}{2^{\frac{n}{2}}} dt \int_{0}^{\infty}s^{n-1} e^{-p}\frac{1}{2ks}dp$

$df = \frac{1}{\sqrt{\pi(n-1)}}\frac{1}{\sigma^{n}} \frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n}{2}}\frac{2}{2^{\frac{n}{2}}} dt \int_{0}^{\infty}\frac{1}{2k}s^{n-2} e^{-p}dp$

Since $s = \frac{p^{1/2}}{k^{1/2}}$

The equation becomes,

$df = \frac{1}{\sqrt{\pi(n-1)}}\frac{1}{\sigma^{n}} \frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n}{2}}\frac{2}{2^{\frac{n}{2}}} dt \frac{1}{2k^{n/2}}\int_{0}^{\infty}p^{\frac{n-2}{2}} e^{-p}dp$

Replacing $k = \frac{n - 1 + t^{2}}{2\sigma^{2}}$

The equation becomes

$df = \frac{1}{\sqrt{\pi(n-1)}}\frac{1}{\sigma^{n}} \frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n}{2}}\frac{2}{2^{\frac{n}{2}}} dt \frac{1}{2(\frac{n - 1 + t^{2}}{2\sigma^{2}})^{n/2}}\int_{0}^{\infty}p^{\frac{n-2}{2}} e^{-p}dp$

Some more reduction …

$df = \frac{1}{\sqrt{\pi(n-1)}}\frac{1}{\sigma^{n}} \frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n}{2}}\frac{2}{2^{\frac{n}{2}}} dt \frac{(2\sigma^{2})^{n/2}}{2(n - 1 + t^{2})^{n/2}}\int_{0}^{\infty}p^{\frac{n-2}{2}} e^{-p}dp$

$df = \frac{1}{\sqrt{\pi(n-1)}}\frac{1}{\sigma^{n}} \frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n}{2}}\frac{2}{2^{\frac{n}{2}}} dt \frac{(2^{n/2}\sigma^{n})}{2(n - 1 + t^{2})^{n/2}}\int_{0}^{\infty}p^{\frac{n-2}{2}} e^{-p}dp$

$df = \frac{1}{\sqrt{\pi(n-1)}} \frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n}{2}} dt \frac{1}{(n - 1 + t^{2})^{n/2}}\int_{0}^{\infty}p^{\frac{n-2}{2}} e^{-p}dp$

$df = \frac{1}{\sqrt{\pi(n-1)}} \frac{1}{(\frac{n-3}{2})!} dt \frac{1}{(\frac{n-1+t^{2}}{n-1})^{\frac{n}{2}}}\int_{0}^{\infty}p^{\frac{n-2}{2}} e^{-p}dp$

$df = \frac{1}{\sqrt{\pi(n-1)}} \frac{1}{(\frac{n-3}{2})!} dt \frac{1}{(1+\frac{t^{2}}{n-1})^{\frac{n}{2}}} \int_{0}^{\infty}p^{\frac{n-2}{2}} e^{-p}dp$

The integral you see on the right is a Gamma function that is equal to $(\frac{n-2}{2})!$ for positive integers.

There we go, the t-distribution emerges

$df = \frac{(\frac{n-2}{2})!}{\sqrt{\pi(n-1)}} \frac{1}{(\frac{n-3}{2})!} \frac{1}{(1+\frac{t^{2}}{n-1})^{\frac{n}{2}}} dt$

The probability distribution of t is

$f(t) = \frac{(\frac{n-2}{2})!}{\sqrt{\pi(n-1)}} \frac{1}{(\frac{n-3}{2})!} \frac{1}{(1+\frac{t^{2}}{n-1})^{\frac{n}{2}}}$

It is defined as t-distribution with (n-1) degrees of freedom. As you can see, the function only contains n as a parameter.

The probability of t within any limits is fully known if we know n, the sample size of the experiment.

“Student” also derived the moments of this new distribution as

$E[T] = 0$

$V[T] = \frac{n-1}{n-3}$

The function is symmetric and resembles the standard normal distribution Z.

The t-distribution has heavier tails, i.e. it has more probability in the tails than the normal distribution. As the sample size increases (i.e., as $n \to \infty$ ) the t-distribution approaches Z.

You can look at the equation and check out how it converges to Z in the limit when $n \to \infty$ .

These points are illustrated in this animation.

I am showing the standard Z in red color. The grey color functions are the t-distributions for different values of n. For small sample sizes, the t-distribution has fatter tails — as the sample size increases, there is little difference between T and Z.

😫😫😫

I am pretty sure you do not have the energy to go any further for this week. I don’t too.

So here is where we stand.

$f(s^{2}) = \frac{1}{2}\frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n-1}{2}}\frac{1}{2^{\frac{n-3}{2}}}\frac{s^{n-3}}{\sigma^{n-1}}e^{-\frac{1}{2}*\frac{(n-1)s^{2}}{\sigma^{2}}}$

$f(s) = \frac{1}{(\frac{n-3}{2})!}(n-1)^{\frac{n-1}{2}}\frac{1}{2^{\frac{n-3}{2}}}\frac{s^{n-2}}{\sigma^{n-1}}e^{-\frac{1}{2}*\frac{(n-1)s^{2}}{\sigma^{2}}}$

and finally

The probability distribution of t is

$f(t) = \frac{(\frac{n-2}{2})!}{\sqrt{\pi(n-1)}} \frac{1}{(\frac{n-3}{2})!} \frac{1}{(1+\frac{t^{2}}{n-1})^{\frac{n}{2}}}$

See you next week.

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March 1908

What is the frequency distribution of the sample variance?

WHAT IS THE FREQUENCY DISTRIBUTION OF THE SAMPLE Standard Deviation?

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