As part of a study of corporate employees, the director of human resources for PNC Inc. wants to compare the distance traveled to work by employees at its office in downtown Cincinnati with the distance for those in downtown Pittsburgh. A sample of 35 Cincinnati employees showed they travel a mean of 370 miles per month. A sample of 40 Pittsburgh employees showed they travel a mean of 380 miles per month. The population standard deviations for the Cincinnati and Pittsburgh employees are 30 and 26 miles, respectively. At the 0.05 significance level, is there a difference in the mean number of miles traveled per month between Cincinnati and Pittsburgh employees?

Comparing the p value with the significance level [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and a wouldn't be a significant difference in the average of the two groups

Step-by-step explanation:

1) Data given and notation

[tex]\bar X_{cin}=370[/tex] represent the mean for the sample of Cincinnati

[tex]\bar X_{pit}=380[/tex] represent the mean for the sample Pittsburgh

[tex]\sigma_{cin}=30[/tex] represent the population standard deviation for the sample Cincinnati

[tex]\sigma_{pit}=26[/tex] represent the population standard deviation for the sample Pittsburgh

[tex]n_{cin}=35[/tex] sample size for the group Cincinnati

[tex]n_{pit}=40[/tex] sample size for the group Pittsburgh

z would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

Null hypothesis:[tex]\mu_{cin}=\mu_{pit}[/tex]

Alternative hypothesis:[tex]\mu_{cin} \neq \mu_{pit}[/tex]

We have the population standard deviation's, so for this case is better apply a z test to compare means, and the statistic is given by:

[tex]z=\frac{\bar X_{cin}-\bar X_{pit}}{\sqrt{\frac{\sigma^2_{cin}}{n_{cin}}+\frac{\sigma^2_{pit}}{n_{pit}}}}[/tex] (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

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

[tex]z=\frac{370-380}{\sqrt{\frac{30^2}{35}+\frac{26^2}{40}}}}=-1.532[/tex]

4) Statistical decision

Since is a bilateral test the p value would be:

[tex]p_v =2*P(Z<-1.532)=0.126[/tex]

Comparing the p value with the significance level [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and a wouldn't be a significant difference in the average of the two groups