A city council is deciding whether or not to spend additiona

A city council is deciding whether or not to spend additional money to reduce the amount of traffic. The council decides that it will increase the transportation budget if the amount of waiting time for drivers exceeds 18 minutes. A sample of 29 main roads results in a mean waiting time of 21.07 minutes with a standard deviation of 6.34 minutes. Null hypothesis and alternative hypothesis of H₀: 18; H_A: > 18.

A)Compute the value of the appropriate test statistic.

B)Calculate the critical value at a 5% level of significance.

Solution

Set Up Hypothesis
Null, H0: U<=18
Alternate, H1: U>18
Test Statistic
Population Mean(U)=18
Sample X(Mean)=21.07
Standard Deviation(S.D)=6.34
Number (n)=29
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =21.07-18/(6.34/Sqrt(28))
to =2.608
| to | =2.608
Critical Value
The Value of |t | with n-1 = 28 d.f is 1.701
We got |to| =2.608 & | t | =1.701
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Right Tail - Ha : ( P > 2.6076 ) = 0.00723
Hence Value of P0.05 > 0.00723,Here we Reject Ho