Question

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). 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 eight 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.The value of the test statistic isA. z=-0.33B. t35=-0.33C. z=-2.00D. t35=-2.00

Answer 1

Answer:

$z=\frac{8-12}{\frac{12}{\sqrt{36}}}=-2$    

C. z=-2.00

Step-by-step explanation:

Data given and notation  

$\bar X=8$ represent the sample mean

$\sigma=12$ represent the population standard deviation

$n=36$ sample size  

$\mu_o =12$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the ture mean is equal to 12 or no, the system of hypothesis would be:  

Null hypothesis:$\mu =12$  

Alternative hypothesis:$\mu \neq 12$  

If we analyze the size for the sample is > 30 and know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$z=\frac{8-12}{\frac{12}{\sqrt{36}}}=-2$    

C. z=-2.00