Asked by Moe

5 Part Question:
6 timed runs were made along route A with an average of 95 minutes and a standard deviation of 10.5 minutes. Route B had 8 test runs with an average of 93 minutes and a standard deviation of 5.2 minutes. Determine if there sufficient evidence that route B is more consistent with 95% confidence.

Part 1: Which test is used?
F, Student-t, Chi-square, Normal
(I'm pretty sure this is a "t" test)
Part 2: What is the critical test value?
4.88, 3.58, 3.97, 4.08
(This is where I need the help,I'm not
getting anywhere close to these)
Part 3: What is the calculated value?
4.88, 3.58, 3.97, 4.08
(same situation as above)
Part 4: Is this a one-tail test or two?
(I'm fairly sure this is a two tail
because the question is looking for
any change at all...hi or low)
Part 5: Do we reject the null hypothesis
Fail to reject
Reject (route B is more consistent)

Thanks for any help you can give!
Answered by DQR
Actually, it's not a t-test. The question doesn't ask whether the route is shorter or longer, but whether the route is more consistent - and that means whether its less variable. To find THAT out, you need to ask whether the standard deviation for Route B is significantly smaller than the standard deviation for Route A.

You can do that with an F test on the two variances (i.e. the squares of the standard deviations), by dividing the Route A variance by the Route B variance (note that this is a ONE-tailed test, because you're only considering the possibility that B is more consistent than A, not whether there's any difference at all). The test statistic (i.e. the calculated value) is therefore (10.5)^2 / (5.2)^2 = 4.08. Aha! It looks as though we're on the right track, because that's Part 3 answer (d).

But we've skipped Part 2: what's the critical test value? We need a set of F tables to answer that one, namely the 95% cutoff value for an F statistic on (N1-1) and (N2-2) degrees of freedom. N1 = 6, and N2 = 8. But I think perhaps its time you took over now...
Answered by Joe
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