Answer:
a) Level of significance α=0.05
Two-tailed test, with null and alternative hypothesis:
b) Student's t distribution. We assume equal variances for both populations, independent sampled values and populations normally distributed.
Test statistic t=-2.4
c) P-value = 0.018
d) Rejection of the null hypothesis.
The data is statistically significant.
e) There is evidence to conclude there is significant difference in average off-schedule times between the bus lines. The difference we see in the samples seems not due to pure chance.
Step-by-step explanation:
This is a hypothesis test for the difference between populations means.
The claim is that there is a significant difference in average off-schedule times for this bus lines.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample 1 (bus line A), of size n1=51 has a mean of 53 and a standard deviation of 17.
The sample 2 (bus line B), of size n2=60 has a mean of 60 and a standard deviation of 13.
The difference between sample means is Md=-7.
The estimated standard error of the difference between means is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this test are:
This test is a two-tailed test, with 109 degrees of freedom and t=-2.4, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.018) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that there is a significant difference in average off-schedule times for this bus lines.