The mean income per person in the United States is 40500 and

The mean income per person in the United States is $40,500, and the distibution of incomes follows a normal distribution. A random sample of 18 residents of Wilmington, Delaware, had a mean of $51,500 with a standard deviation of $10,200. At the .05 level of siginificance,is that enough evidence to conclude that residents of Wilmington,Delware, have more income than the national average?

(a) State the decision rule for .05 significance level.(Round your answer to 3 decimal places.)

Solution

Set Up Hypothesis
Null, Delware, have less or equal national average H0: U=40500
Alternate,Delware, have more income than the national average H1: U>40500
Test Statistic
Population Mean(U)=40500
Sample X(Mean)=51500
Standard Deviation(S.D)=10200
Number (n)=18
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =51500-40500/(10200/Sqrt(17))
to =4.575
| to | =4.575
Critical Value
The Value of |t | with n-1 = 17 d.f is 1.74
We got |to| =4.575 & | t | =1.74
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Right Tail - Ha : ( P > 4.5754 ) = 0.00013
Hence Value of P0.05 > 0.00013,Here we Reject Ho