# Calculating a statistic of specific distributions and degrees of freedom

Firstly, what does it mean to calculate a statistic? Do they just want us to find the mean, variance, and standard deviation? For instance, if I'm asked to calculate the statistic of the Chi squared with three degrees of freedom or the F distribution with 1 and 2 degrees of freedom how would I go about doing that? The assumptions are that $X_i$ is independent with $N(i,i^{2})$ distributions. I'm asked to use the $X_i$ to construct the statistics.

Would i just use the formulas:

$$\sum_{n=1}^{n} (X_{1}+...+X_{n})/n = mean$$

and

$$\sum_{n=1}^{n} (X_i-mean)^{2}/n-1 = variance$$

So if you do use these formulas how can you evaluate all the $X_1$ to $X_n$ values? How is knowing just the type of distribution and its respective degrees of freedom going to help me come up with the individual random sample values? Or is my understanding completely incorrect?

• "Calculate a statistic" doesn't mean anything. Are you sure there's not more to the question you are trying to answer? – Rubarb Oct 25 '15 at 18:11
• okay i made it clearer – oxnot Oct 25 '15 at 18:37
• Again, "calculate the statistic" of the $\chi^{2}$-distribution with 3 degrees of freedom doesn't make any sense. Are you trying to use your normal random variables to find a statistic that has that $\chi^{2}$- distribution? – Rubarb Oct 25 '15 at 18:43

I'm going to assume that you have independent $X_{1}, X_{2}, \ldots, X_{n}$ with $X_{i} \sim N(i,i^{2})$ and you want to find a statistic that has, for example, a $\chi^{2}$-distribution.

A $N(0,1)$ random variable squared has a $\chi^{2}(1)$ distribution. You can show this using moment generating functions. A sum of independent $\chi^{2}$ random variables has a $\chi^{2}$ distribution with the individual degrees of freedom added up. You can show that using moment generating functions as well.

So, you could take $X_{i} \sim N(i,i^{2})$, "standardize" it to a $N(0,1)$: $$\frac{X_{i}-i}{\sqrt{i^{2}}} = \frac{X_{i}-i}{i} \sim N(0,1),$$ square it to get a $\chi^{2}$ distribution $$\left( \frac{X_{i}-i}{i} \right)^{2} \sim \chi^{2}(1),$$ and add them up (using independence) to get $$\sum_{i=1}^{n} \left( \frac{X_{i}-i}{i} \right)^{2} \sim \chi^{2}(1+1+ \cdots + 1) = \chi^{2}(n).$$

• how would you deal with the three degrees of freedom indicated here? – oxnot Oct 26 '15 at 2:44
• Yea, I ditto @oxnot's question. I'm still not quite sure how you're doing this. How would you do the second example: F distribution with 1 and 2 degrees of freedom – cambelot Oct 26 '15 at 2:47
• @Oxnot: As I said in my original 2 comments, I'm still not sure what you even want. I can tell you how to make statistics out of normal random variables that have several different $\chi^{2}$ distributions and I can go further to tell you how to make $F$ distributions (which are essentially ratios of $\chi^{2}$ random variables). However, it is not clear to me what your question is. Maybe you want Aaron Hall's answer. Do you have more context here? Thanks. – Rubarb Oct 26 '15 at 3:02
• I think I want to find a statistic that has a F distribution, like what you have shown above. – oxnot Oct 26 '15 at 13:13
• Well, if you have independent $\chi^{2}$ random variables $W_{1} \sim \chi^{2}(n_{1})$ and $W_{2} \sim \chi^{2}(n_{2})$, the random variable defined by $(W_{1}/n_{1})/(W_{2}/n_{2})$ has an $F$-distribution with $n_{1}$ and $n_{2}$ degrees of freedom. So, I suppose you could standardize some of your normals into $N(0,1)$'s, square them to get $\chi^{2}$ random variables, and make a ratio to get a random variable with an $F$ distribution. – Rubarb Oct 27 '15 at 14:58

To calculate a statistic, in your case, means to figure out the value of a test statistic, a value useful in statistical testing (though in other contexts, a statistic can also refer to the mean, standard deviation, etc. of a sample). Let's take one of the examples you mentioned: the $\chi^2$ (chi-squared) distribution. This distribution, as you may already know, is useful for determining whether differences between observed and expected values of categorical data are significant or not. The formula for the chi-squared statistic is $$\chi^2 = \sum_{i =1}^n\frac{(O_i - E_i)^2}{E_i},$$ where $n$ is the number of data points you have, $O_i$ is the $i$-th observed value, and $E_i$ is the $i$-th expected value. Now, if all you were asked for is the $\chi^2$ statistic, then you're done once you calculate this. However, the p-value that results from the $\chi^2$ statistic is much more practical, and you will usually be asked for this on a question dealing with statistical testing. And this is where degrees of freedom come in. To get the $p$-value, you can do one of two things: look it up on a table, or use a graphing calculator or a computer. You do both of these using your $\chi^2$ statistic and the number of degrees of freedom.