Three different methods for assembling a product were proposed by an industrial

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 = 10,830; SSTR = 4,570.

Set up the ANOVA table for this problem (to 2 decimals, if necessary).

Use  = .05 to test for any significant difference in the means for the three assembly methods.

Calculate the value of the test statistic (to 2 decimals).


The p-value is

What is your conclusion?

Source of Variation Sum of Squares Degrees of Freedom Mean Square F Treatments ? ? ? ? Error ? ? ? Total ? ?

Explanation / Answer

Ans:

a)

Explanation for table:

n=3*10=30

df for numerator=k-1=3-1=2

df for denominator=n-k=30-3=27

Total df=n-1=30-1=29 or 2+27=29

SST (total sum of squares) = SSE(error sum of squares) +SSTR (sum of squares for treatments)

SST=10830

SSTR=4570

SSE=10400-4570=6260

MSR=4570/2=2285

MSE=6260/27=231.85

Test statistic:

F=2285/231.85=9.86

p-value=FDIST(9.86,2,27)=0.00061

b)p-value approach:

As,p-value<0.05,We reject null hypothesis.

critical value approach:

F-critical==FINV(0.05,2,27)=3.35

As,F=9.86>3.35,We reject null hypothesis.

There is sufficient evidence that there is significant difference in means for three assembly methods.

df SS MS F p-value Regression 2 4570 2285 9.86 0.00061 Error 27 6260 231.85 Total 29 10830

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.