A comparison is made between two bus lines to determine if arrival times of their regular buses from Denver to Durango are off schedule by the same amount of time. For 51 randomly selected runs, bus line A was observed to be off schedule by an average time of 53 minutes, with a standard deviation of 19 minutes. For 60 randomly selected runs, bus line B was observed to be off schedule an average of 62 minutes, with a standard deviation of 15 minutes. Do the data indicate a significant difference in average off schedule times? Use a 5% level of significance.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
To determine if there is a significant difference in average off schedule times between bus line A and bus line B, we can perform a hypothesis test.
The null hypothesis, denoted as H0, assumes that there is no significant difference between the two means. In this case, it means that the average off schedule times for bus line A and bus line B are the same. The alternative hypothesis, denoted as Ha, assumes that there is a significant difference between the means.
Using the information provided, let's set up the null and alternative hypotheses:
H0: μA = μB (The average off schedule times for bus line A and bus line B are the same)
Ha: μA ≠ μB (There is a significant difference in average off schedule times between bus line A and bus line B)
To perform the hypothesis test, we can use the two-sample t-test since we have two independent samples. Here are the steps to perform the test:
1. Define the significance level (α). The problem states a 5% level of significance, which means α = 0.05.
2. Calculate the test statistic. The test statistic for a two-sample t-test is given by:
t = (X1 - X2) / sqrt((s1^2 / n1) + (s2^2 / n2))
where X1 and X2 are the sample means, s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes. In this case, X1 = 53 minutes, s1 = 19 minutes, n1 = 51, X2 = 62 minutes, s2 = 15 minutes, and n2 = 60.
Plugging in these values, we can calculate the test statistic.
3. Determine the critical value(s). Since we are using a two-tailed test (Ha: μA ≠ μB), we need to find the critical values that correspond to the significance level α/2 = 0.05/2 = 0.025. For a two-sample t-test, we can use the t-distribution and find the critical values based on the degrees of freedom, which is calculated using the formula:
df = (s1^2 / n1 + s2^2 / n2)^2 / [(s1^2 / n1)^2 / (n1 - 1) + (s2^2 / n2)^2 / (n2 - 1)]
In this case, the degrees of freedom will be substituted using the sample sizes and the respective degrees of freedom for each sample.
4. Compare the test statistic with the critical value(s). If the test statistic falls within the critical region (i.e., it is greater than the positive critical value or less than the negative critical value), we reject the null hypothesis. If it falls outside the critical region (i.e., it is less than the positive critical value and greater than the negative critical value), we fail to reject the null hypothesis.
5. State the conclusion. Based on the comparison, we determine whether there is enough evidence to reject the null hypothesis and support the alternative hypothesis or fail to reject the null hypothesis.
Now, let's perform the actual calculations to evaluate the hypothesis test and draw a conclusion.