Or maybe we could stop being such douches and actually give the guy/girl some help!
I don't pretend to be the best at this and i'm sure there are more qualified people on here but considering I have my Stats 2 Second year undegrad exam on the 29th I thought id generate a few thought processes for both our sakes.. that may or may not be right... depending on how well im actually going to do in MY exam!
Your initial assumption would probably be that the data are normally distributed and the variances are equal (Homogeneity of variances) i.e the unknown parameters are variances. That opens up a whole load of potential tests. Probably a t test would be best where H1 and Halternative are equal means and means not equal respectively. So in other words if the means arent equal you know one of the products performs better than the other - However to determine which this was would require a post-hoc test of some kind.
The first thing to do is calculate the means for each sample. Take for example Product A: n=6, Sum of times=99.06, Hence Xbar(mean of sample)=Sum/n => 16.51. You then want to find the expected deviation about the mean. To do this you take each value for x from Xbar and square it and sum these values i.e. the first is (16.61-16.51)^2 = .01 second is (15.38-16.51)^2 = 1.2769 and so on and you add these all up and keep this number handy .Do exactly the same for Product B. You now have two means and two expected deviations[i.e. variances], one for A, and one for B.
To calculate the unbiased pooled estimate of the variances you need to sum the two variances and divide by (n1 + n2 - 2). This is because of the unbiased aspect as each variance has denominator (ni-1). So what you have so far as an equation is [Var(x1)+Var(x2)]/[n1+n2-2] Which gives you your pooled unbiased estimate of the variances. Otherwise nown as s^2 or sigma squared.
These are all preliminary steps required to calculate a figure which, when tested against certain critical values found in the t tables, will test the hypothesis to discern whether your Products have the same effect.
The t value itself will now be calculated. Note the ( - 0) in the equation that follows. This is the part that asserts the Null Hypothesis that the means are equal as it provides out t value GIVEN THAT the means are equal i.e. same effect per product [think]. Ok NOW we'll calculate our test statistic! This is the formula t =
[(X1bar - X2bar) - 0
]/SQRT
{(S^2)*[(1/n1)+(1/n2)]
} .. Ok i hope that works for you - the colour is just (hopefully) to make it easier to read!
So once you have the value t all you need to do is find a t table, available online or in books or handouts etc and remembering that your degrees of freedom are n1+n2-2.. so here (6+5-2) = 9 You decide on a suitable significance level to test at, say 5%, you find the value of 2.26. You then compare your value of t with that of the critical value of 2.26 and if it is larger you
reject the null hypothesis and accept the alternative and conclude that there is EVIDENCE to suggest one product does better than the other. Most likely the one with the lower mean but as i said a post-hoc test would be needed to more confidently confirm this.
To construct a 95% confidence level just google it.. its fairly straightforward compared to the rest of the calculations and i dont have the provisions to show the theory behind it here.. or maybe i do there seems to be a picture symbol but hey this is only my .. what 3rd..4th post?
But anyway thats about all I can be bothered with now. To be honest I can pretty much say with about 99% confidence (no coincidental puns intended) that this is right now that I did a little research aswel. If anyone spots any mistakes please feel free to correct me and please accept my apologies
Good luck with the rest I need food and sleep!
Russ0707