Question

8. An experiment was conducted to investigate possible differences among four hardness testing machines. Five different specimens were used and a reading was obtained from each machine on each specimen. The resulting data was analyzed as a randomized block with the following partially completed ANOVA table: F Mean Square Source Degrees of Freedom Machines 3 Specimens Error 12 Total 119 Sum of Squares 32.18 9.16 39.12 80.46 With respect to the hardness testing machines, what conclusion can be drawn at the 5% significance level from this randomized block analysis? (Please complete the parts of the ANOVA table that are necessary to answer this question.)

Answer 1

completed

ANOVA TABLE

SOURCE	DEGREE OF FREEDOM	SUM OF SQUARES	MEAN SQUARES	F
MACHINES	3	32.18	10.726	3.29
SPECIMENS	4	9.16	2.29	0.702
ERROR	12	39.12	3.26
TOTAL	19	80.46

CONCLUSION- By F-dist table at 5% level of significance. F(3,12)=3.29     ,     F(4,12)=3.26.

Hence F tabulated is less than F calculated. With respect to the hardness testing machines we conclude that null hypothesis that all the machines resulting same , maybe retained at 5% level of significance. All the 4 machines are homogeneous. There is no significant difference between machines. supporting results and formulas are given in attached pictures.
ANOVA TABLE FROM R-BD. Source of variation degrees of freedom Sum of Squares Mean Sumos Squaress Variance Ratio z = 57 Treatments t-1 fx = 1 Blocks r-1 (n-1) Error (t-1)(x-1) (t-1) (6-1) Total rt-1 Solution : + Null hypothesis it Hot : E = Tz = Cz= Ty Sifuge Noen Hob : bi = b 2 = bz = by = 65 le; the treatments as well as blocks are homogeneneous. in respect to this question all 4 machines results in a same realing as well as specimens. 4 Alternative hypothesis : Hit: At least two machines give different reading. Hib: At least two specimens give different reading. I Now we complete the parts of the ANOVA table by using formulas given in above table. o completed ANOVA TABLE Degrees of sum of mean freedom Squares 32.18 33.18 = 10.726 Ft = 10-726 9.16 9.16 = 2.291 = 3.29 2:29 4:12 - 3.26 Total 3.26 80.46 source Squares Machines Specimens Error MJ No 39.12 10.702

conclusion :- By F. out table F3, 12 (0.05) - 3.49 2 F4,12 (0.05) = 3.26 & dola since under Hot Ft F (3, 12) and under Hob For F (4,12).2) (1 F = 3.29 < 3.49 = F3,1210.05). fo = 0.702 < 3:26 = fu, iz 10.05) we see that Ft is not significant as well as fois also not significant at 5% level of significance. Hence with respect to the hardness testing machines, we can draw the following conclusions. * £ (variance ratio for machines) is accepted at S1 level of significance. * Mot may be retained at 5 % level of significance treatments are homogeneous. Readings given by a all the 4 machines are homogeneous. machines function accurately same with each other. * treatment means does not differ significantly. above conclusion is toue in the case of fb (variance Ratio specimens) also Blocks are also homogeneous: 8.se Saruna No3 CEIP . 8