Do I have to standardize my data to calculate variance? I have 2 groups/samples. 
Correct me if I'm wrong, but before doing an independent-group t-test we have to verify the homogeneity of variance with Hartley's F-max test.
When doing this test, we have to divide the variance of one group by that of the other. When I did this, I got a value of 31, which is very big compared with my critical value of 2.26. To calculate the variance, I did :
Variance $= ∑(x_i-\text{mean})^2 / (N-1)$
Was I supposed to standardize all of my $x$ values to calculate the variance?
If so, do I also use the standardized values to calculate the means, median, spread (so everything)?
 A: Some preliminaries: 


*

*It is generally not recommended to use the two-stage statistical testing procedure (test for violations of assumptions; if those were non-significant, run parametric test).  For more on that see: Is normality testing 'essentially useless'?, & A principled method for choosing t test or non-parametric, e.g. Wilcoxon in small samples.  

*Hartley's $F$-max test is an $F$-test.  It is not a $t$-test.  To conduct Hartley's $F$-max test, you estimate your variances and compute the quotient of the largest group variance divided by the smallest group variance.  This ratio of variances is an $F$-statistic; you compare your realized value to the $F$ distribution with the degrees of freedom associated with the numerator and denominator variances (presumably $n_j-1$).  

*Hartley's $F$-max test is not recommended.  It is better to use Levene's test / Brown-Forsythe.  For more on that, see my answer here: Why Levene test of equality of variances rather than F ratio?

The answer to your specific question: 
Standardizing your variables causes the means and variances to become exactly equal.  This is not something to do before you test if the variances are equal.  This wouldn't even work to simulate the properties of the null hypothesis: you are guaranteed to always have $0$ type I errors.  
