# Why convert to a T distribution rather than use the standard normal distribution?

Let $Y_1, Y_2, . . . , Y_n$ denote independent, normally distributed random variables such that $Y_j$ has mean $\mu$ and variance $\sigma_2,\:j = 1, 2, \ldots, n$. Let $\theta = 3\mu − 1$ and consider constructing a confidence interval for $\theta$.

1. $\overline{Y}$ is normally distributed with mean $\mu$ and variance $\sigma^2/n$. It follows that $3\overline{Y} − 1$ is normally distributed with mean $\theta$ and variance $9\sigma^2/n$. Hence, consider

\begin{align} T &=\frac{3\overline Y − 1 − \theta}{3}\!\cdot\! \frac{S}{\sqrt n}\\ &=\frac{3\overline Y − 1 − \theta}{3\sigma /\sqrt{n}}\!\cdot\!\frac{1}{S/\sigma} \end{align}

Note that the numerator of this expression has a standard normal distribution and the denominator is of the form of the square root of a $\chi^2$ random variable divided by its degrees-of-freedom. Hence, $T$ has a $t$–distribution with $n − 1$ degrees of freedom.

So my question is, why set it up this way? In my calculations I get up to $\quad\dfrac{3\overline Y − 1 − \theta}{3\sigma /\sqrt{n}}$.

Which, as stated in the answer key, has a standard normal distribution. Is there any good reason to continue for a $t$ distribution? And why divide by the quotient of the sample and population standard deviation? Does that statistic hold some significance I'm unaware of?

• Welcome to Mathematics! Please use Mathjax. See what I did on your first paragraph for an example, but please read the link. – suomynonA Mar 11 '17 at 3:30
• $\sigma$ is an estimate of the standard deviation of the underlying $Y_1,\cdots Y_n$ and as an estimate has error. The t-distribution accounts for this error, and returns a fatter tailed distribution then the normal. As the quantity of data increases, the estimation error falls, and the t-distribution looks increasing like a normal distribution. – Doug M Mar 11 '17 at 4:00

Student's $t$ distribution, with $n-1$ degrees of freedom, is the distribution of $$\frac{\bar{Y}-\mu}{\hat{\sigma}/\sqrt{N}},$$ where $\hat{\sigma}$ is the sample standard deviation of $Y_1, Y_2, \dots, Y_n$, given by $$\hat{\sigma}^2 = \frac{1}{n-1}\sum_{i=1}^n(Y_i-\mu)^2.$$
This is used when the variance of your normally i.i.d. random variables $Y_i$ is unknown. In your case, knowing the variance of $Y_i$, you are right in only using the normal distribution.