Please show work! Thank you! As part of a study of corporate employees, the dire
ID: 2931568 • Letter: P
Question
Please show work! Thank you!
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 55 Cincinnati employees showed they travel a mean of 380 miles per month. A sample of 50 Pittsburgh employees showed they travel a mean of 410 miles per month. The population standard deviation for the Cincinnati and Pittsburgh employees are 32 and 28 miles, respectively. At the .04 significance level, is there a difference in the mean number of miles traveled per month between Cincinnati and Pittsburgh employees?
a. State the appropriate null and alternative hypotheses.
b. Calculate the test statistic.
c. Determine the critical value assuming an a = 0.02.
d. Should you reject the null hypothesis?—Provide a basis for your decision. Is there a difference in the mean number of miles traveled per month between Cincinnati and Pittsburgh employees?
Explanation / Answer
Given that,
mean(x)=380
standard deviation , 1 =32
number(n1)=55
y(mean)=410
standard deviation, 2 =28
number(n2)=50
null, Ho: u1 = u2
alternate, H1: 1 != u2
level of significance, = 0.04
from standard normal table, two tailed z /2 =2.054
since our test is two-tailed
reject Ho, if zo < -2.054 OR if zo > 2.054
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=380-410/sqrt((1024/55)+(784/50))
zo =-5.12
| zo | =5.12
critical value
the value of |z | at los 0.04% is 2.054
we got |zo | =5.123 & | z | =2.054
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -5.12 ) = 0
hence value of p0.04 > 0,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: 1 != u2
b.
test statistic: -5.12
critical value: -2.054 , 2.054
decision: reject Ho
p-value: 0
c.
Given that,
mean(x)=380
standard deviation , 1 =32
number(n1)=55
y(mean)=410
standard deviation, 2 =28
number(n2)=50
null, Ho: u1 = u2
alternate, H1: 1 != u2
level of significance, = 0.02
from standard normal table, two tailed z /2 =2.326
since our test is two-tailed
reject Ho, if zo < -2.326 OR if zo > 2.326
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=380-410/sqrt((1024/55)+(784/50))
zo =-5.12
| zo | =5.12
critical value
the value of |z | at los 0.02% is 2.326
we got |zo | =5.123 & | z | =2.326
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -5.12 ) = 0
hence value of p0.02 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 != u2
test statistic: -5.12
critical value: -2.326 , 2.326
decision: reject Ho
p-value: 0
Difference in the mean number of miles traveled per month between Cincinnati and Pittsburgh employees
we have enough evidence support the claim