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

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

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.