I am learning about the student t-test.
I am struggling, however, to be given a reasonable explanation why the standard deviation of the standard normal distribution curve is 1.
It says "The Standard Normal Variable is denoted Z and has mean 0 and S.D 1..."
"... this is written as Z ~ N(0,1^2)
Can someone please explain? Very much appreciate it.
It's hard to be sure what your question means. Suppose $X$ is normally distributed and has expected value $\mu$ and standard deviation $\sigma$. Let $Z = \dfrac {X-\mu} \sigma,$ so that $X = \mu + \sigma Z.$ Then $Z$ has expected value $0$ and standard deviation $1$ and is normally distributed. The one member of this family of distributions that has expected value $0$ and standard deviation $1$ is the one we call the "standard" one.
The distribution of $X$ is $$ \frac 1 {\sqrt{2\pi}} e^{-(1/2)\Big((x-\mu)/\sigma\Big)^2} \, \frac{dx} \sigma. $$ If you let $z= \dfrac{x-\mu}\sigma$, then we have $dz=\dfrac{dx}\sigma$ and the measure becomes $$ \frac 1 {\sqrt{2\pi}} e^{-(1/2) z^2} \, dz. $$ The expected value of that is $$ \int_{-\infty}^\infty z \left( \frac 1{\sqrt{2\pi}} e^{-(1/2)z^2} \, dz \right) $$ and this is clearly $0$ because an odd function is integrated over an interval symmetric about $0$. The variance takes some work. Let's call it $\tau^2:$ $$ \tau^2 = \operatorname{var}(Z) = \int_{-\infty}^\infty z^2 \left( \frac 1{\sqrt{2\pi}} e^{-(1/2)z^2} \, dz \right). $$ We have $$ \operatorname{E}(X) = \operatorname{E}(\mu+\sigma Z) = \mu + \sigma\operatorname{E}(Z) = \mu+\sigma\cdot0 = \mu, $$ and $$ \operatorname{var}(X) = \operatorname{var}(\mu+\sigma Z) = \sigma^2 \operatorname{var}(Z) = \sigma^2\tau^2. $$ If your question is how did we conclude that $\tau^2=1$, say so and maybe I'll post some more on that.
The "standard normal distribution" has mean 0 and standard deviation 1 because that is in fact the definition of the "standard normal distribution". To understand this, what I think you need to understand is what a "normal distribution" in general is.
A "normal distribution" is a distribution of "relative frequency" $\Phi(x)$ with respect to a variable $x$ that is in a very specific shape, specifically \begin{align*} \Phi(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}, \end{align*} where $\mu$ and $\sigma>0$ are parameters for the mean and standard deviation, respectively.
As you've learned in statistics, the normal distribution has a "bell curve shape" (and you'll find $\Phi(x)$ has exactly the shape you've seen).
And like I said, the only thing special about the "standard normal distribution" is that the parameters are $\mu=0$ and $\sigma=1$, hence having the equation $\Phi_s(z)=1/\sqrt{2\pi} \cdot e^{-z^2/2}$. I think the real question is "why is the standard normal distribution so useful?" It's useful because it gives an eloquent "$z$-score" measurement that captures how much "standard devations" a given point is from the mean.
Let's say the distribution $\Psi(x)$ of height (in feet) of all people in a classroom was the following normal distribution \begin{align*} \Psi(x)=\frac{1}{\sqrt{2\pi}(.5)}e^{-\frac{(x-5.5)^2}{2(.5)^2}}, \end{align*} in other words $\mu=5.5$ and $\sigma=.5$. A height like $x=6$ ft may seem like quite a deviation from the mean, but how do we quantify how much of a deviation is it really? This is where the "standard normal distribution" comes in. Any point on a distribution with given mean and standard deviation parameters can be "normalized" into a $z$-score using the equation \begin{align*} z=\frac{x-\mu}{\sigma}, \end{align*} which quantifies what the point is on the standard normal distribution $\Phi_s(z)$ (since the z-score by the defined equation above has $\mu_z=0$ and $\sigma_z=1$). And again, by design, the z-score tells us how many "standard deviations away" something is. So in our example, $6$ ft would be a height that has a $z$-score of $z=1$, meaning it is a single standard deviation from the mean.
Hope that helps. And here's the Wikipedia page for more information (the introduction has an intuitive explanation worth the read). And in case you're curious as to how the $t$-distribution is analogous to the normal distribution, let's just say it converges to the normal distribution as an "additional parameter" (the degrees of freedom) gets very large. But I won't get ahead of myself, and wait for you to ask that if you're curious about that.
