What kind of test should I use for two different objects? Say I'm trying to compare the speeds of 2 different cars in 5 different climates and roads.
I've got the mean of each one and the standard deviation. Assuming all required assumptions are made, what test should I use to conclude if there's either enough or not enough evidence that 1 car is faster than the other?
 A: The best thing to do is to take differences between the cars within each set of conditions, and then test whether the population difference differs from 0*, giving you a paired test. 
*(A possible alternative would be to compute ratios rather than differences and test whether the population-ratios differ significantly from 1. This is equivalent to considering the log-ratio -- i.e. the difference in logs of speeds -- and testing whether that is different from zero.)
The usual contenders would be a paired t-test and a Wilcoxon signed-rank test.
That is, you don't just calculate the mean and standard deviation within cars, that throws away a lot of the information.
e.g. if you had data like this:
         Table of speeds
Condition   Veh1    Veh2      
    A       86.4    96.8      
    B       31.6    27.5       
    C       92.0    73.7       
    D       76.8    55.0       
    E       56.8    56.6       

You'd first compute the differences::
         Table of speeds
Condition   Veh1    Veh2      V1-V2
    A       86.4    96.8      -10.4
    B       31.6    27.5        4.1
    C       92.0    73.7       18.3
    D       76.8    55.0       21.8
    E       56.8    56.6        0.2

and then test whether the differences in speed themselves differ from 0. This removes the variation in the values due to the conditions, just leaving the differences due to the vehicles (and random noise).
For example, a paired t-test would look at the mean and standard deviation of the differences:
          V1-V2
          -10.4
            4.1
           18.3
           21.8
            0.2

   mean     6.80
  std.dev  13.27

Then for the hypothesis $H_0: \mu_1 - \ mu_2 = 0$ against the alternative $H_1: \mu_1 - \ mu_2 \neq 0$, the test statistic is
$$ t = \frac{\overline{d} - 0}{s_d/\sqrt{n}} $$
where $n$ is the number of pairs (5 in this case).
So for our data, $t = 6.80/(13.27/\sqrt{5}) = 1.146$ (I kept the s.d. to more figures than shown here). This would be compared with the two-tailed critical value for 4 degrees of freedom at the desired significance level (2.78 for a 5% test), or equivalently, the p-value could be computed (0.316) and compared with the significance level.
In the first case the null would be rejected when the test statistic is at least as extreme as the critical value, while in the second case the null would be rejected if the p-value was no bigger than the significance level. (In our example, we fail to reject the null; we find no difference that couldn't be accounted for by random variation.)
Also see here
