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.

Question

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.

PsyDAG · Answer

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.