Question

Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 30 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 10 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: SST = 10,800; SSTR = 4560.
Required:
a. Set up the ANOVA table for this problem.
b. Use α= .05 to test for any significant difference in the means for the three assembly methods.

Answer 1

Answer:

H0 is rejected.There difference in the means for the three assembly methods.

Step-by-step explanation:

Calculating gives the following results.

ANOVA Table

Source        DF                  Sum                Mean            F Statistic      

                                       of Square         Square

Treatments       2               4560               2280             9.865

(between groups)

 

Error            27                  6240                231.111                                  

Total         29                   10,800  

Error SS= SST -SSTR = -10,800-4560=6240    

d.f for SSTr = r-1= 3-1

D.f for SST= n-r= 30-3= 27

Total D.f= d.f for SSTr+D.f for SST= 27+29

MSTR= SSTR/d.f= 4560/2= 2280

MSError= Error SS/ d.f= 6240/27=231.111

F test= MSTR/ MsError= 2280/231.11= 9.865

The P- value < 0.01 which  is less than 0.05 therefore H0 is rejected.

There is sufficient evidence to reject H0.