
Statistics help!
Question:
10 Honda Civic owners and 12 Toyota Corolla owners were randomly selected, and asked about their cars gas mileage. Yielded the following:
Honda  28.5, 31.0, 29.6, 26.8, 36.2, 32.4, 30.8, 27.4, 26.6, 29.7.
Toyota  31.2, 24.8, 28.5, 27.1, 34.9, 26.1, 25.7, 28.8, 29.6, 19.8, 35.0, 26.6.
Assuming that the standard deviations of mph are equal.
A) Find a 99% CI for the difference of their means of mpg.
B) Test the claim that Honda's have better mph with level 0.05.
I'm not just looking for answers however solutions (step by step) to the problems in which I can apply and use for future similar problems. The textbook I am using doesn't quite clearly explain how to set this up and solve.
Thanks in advance! I really appreciate the assistance!

Calculate the Means.
Calculate the Variances.
Let's see if you can get that far. In the cae of statistics, the stepbystep solution IS the answer. Learning the process is the plan.

I was hoping for a more detailed response so I can study the relationship of the data to the problem per an upcoming exam (per urgent homework help).
However,
XBAR1 (Honda)  28.5, 31.0, 29.6, 26.8, 36.2, 32.4, 30.8, 27.4, 26.6, 29.7 = 299
299/n
n=10
29.9
Deviation for each number and square of each deviation:
28.529.9 = 1.4 & 1.96
31.029.9 = 1.0 & 1.21
29.629.9 = .3 & .09
26.829.9 = 3.1 & 9.61
36.229.9 = 6.3 & 39.69
32.429.9 = 2.5 & 6.25
30.829.9 = .9 & .81
27.429.9 = 2.5 & 6.25
26.629.9 = 3.3 & 10.89
29.729.9 = .2 & .04
Sum of all squares of deviations = 67.16
67.16 / n  1
67.16 / 9
7.46222
XBAR2 (Toyota)  31.2, 24.8, 28.5, 27.1, 34.9, 26.1, 25.7, 28.8, 29.6, 19.8, 35.0, 26.6 = 338.1
338.1/n
n=12
28.175
...

I'll be out the door in about 30 minutes. =/