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!
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...
Can we prevent flooding in our communities
Part 1: The appropriate test to use in this scenario is a two-sample t-test. This is because we have two independent samples of continuous data (the run times) and we want to compare the means of these samples to determine if there is a significant difference.
Part 2: To find the critical test value, we need to consider the degrees of freedom (df) and the desired level of confidence. Since this is a two-sample t-test, the df can be calculated as follows:
df = (n1 + n2) - 2
where n1 and n2 are the sample sizes. In this case, n1 = 6 and n2 = 8. Plugging these values into the formula, we get:
df = (6 + 8) - 2 = 12
With a desired level of confidence of 95%, we divide the significance level (alpha) by 2 to account for the two tails of the distribution. The remaining probability is split equally between the two tails. Therefore, our alpha value is 1 - (0.95 / 2) = 0.975.
Now, we can find the critical test value using a t-distribution table or a statistical software. For a two-sided test with df = 12 and alpha = 0.975, the critical t-value is approximately 2.178.
Part 3: The calculated test value, also known as the t-statistic, is obtained by comparing the sample means and standard deviations. In this case, we want to compare the mean of route B (93 minutes) to the hypothesized mean of 95 minutes (route A's average). The formula for calculating the t-statistic is as follows:
t = (mean1 - mean2) / sqrt((std1^2 / n1) + (std2^2 / n2))
Plugging in the values, we get:
t = (93 - 95) / sqrt((5.2^2 / 8) + (10.5^2 / 6)) = -0.301
Part 4: Since we are interested in determining if route B is more consistent with a mean of 95 minutes, regardless of being higher or lower, this is indeed a two-tail test. We want to check for any significant difference, not just one specific direction.
Part 5: To determine if we reject or fail to reject the null hypothesis, we compare the calculated test value (t = -0.301) to the critical test value (2.178). If the calculated test value falls outside the range of the critical test value, we reject the null hypothesis.
Given that -0.301 falls within the range of -2.178 to 2.178, we fail to reject the null hypothesis. This means that there is not sufficient evidence to conclude that route B is more consistent with a 95% confidence level.