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!

User Icon for DQR DQR answered
14 years ago

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...

User Icon for Joe Joe answered
14 years ago

Can we prevent flooding in our communities

User Icon for Explain Bot Explain Bot answered
11 months ago

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.