Question

Department of Labor reported the average hourly earnings for production workers to be $15.23 per hour in 2001. A sample of 75 production workers during 2003 showed a sample mean of $15.86 per hour. Assuming the population standard deviation is $1.50, can we conclude that an increase occurred in the mean hourly earnings since 2001? Use α = .05

Answer 1

Solution:

Given:

Department of Labor reported the average hourly earnings for production workers to be $15.23 per hour in 2001.

That is: Population mean =

Sample size = n = 75

Sample mean =

Population Standard Deviation =

Level of Significance = α = 0.05

We have to test if an increase occurred in the mean hourly earnings since 2001.

That is we have to test if

Step 1) State H0 and H1:

        Vs    

Step 2) Find test statistic:

Step 3) Find z critical value:

Since    , it is > type, so this is right tailed test.

Thus we look for Area = 1 - α = 1 - 0.05 = 0.95

and find corresponding z value.

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus Z_critical = 1.645

Step 4) Decision rule:
Reject H0, if z test statistic value > Z_critical = 1.645 , otherwise we fail to reject H0.

Since z test statistic value = 3.64 > Z_critical = 1.645 , we reject H0.

Step 5) Conclusion:

Since we have rejected null hypothesis H0, we can conclude that an increase occurred in the mean hourly earnings since 2001.