Exhibit 9-3.The Boston public school district has had difficulty maintaining on-time bus service for its students ("A Year Later,School Buses Still Late," Boston Globe,October 5,2011).Suppose the district develops a new bus schedule to help combat chronic lateness on a particularly woeful route.Historically,the bus service on the route has been,on average,12 minutes late.After the schedule adjustment,the first 36 runs were an average of 8 minutes late.As a result,the Boston public school district claimed that the schedule adjustment was an improvement-students were not as late.Assume a population standard deviation for bus arrival time of 12 minutes. Refer to Exhibit 9-3.Calculate the value of the test statistic.

Question

Quizplus · Accepted Answer

We need to conduct a one-sample t-test as we have sample mean and population standard deviation.

Given, 
Sample mean, x̄ = 8
Population standard deviation, σ = 12
Sample size, n = 36
Population mean, μ = 12

The formula for the test statistic is:

t = (x̄ - μ)/(σ/√n)

Substituting the values, we get:

t = (8 - 12)/(12/√36) = -4/2 = -2

Using a t-table with 35 degrees of freedom at α = 0.05 (assuming a two-tailed test), the critical value is ±2.03. As the calculated t-statistic (-2) falls in the rejection region (less than -2.03), we can reject the null hypothesis and conclude that the schedule adjustment has significantly improved the bus arrival time on the woeful route.