Question

Question 3. - Please do c, d and e

An engineer suspects that the surface finish of a metal part is influenced by the feed rate and the depth of cut of a particular manufacturing machine. An experiment is conducted and the data can be found in SurfaceFinish.jmp on eCampus.

a. State the two-way ANOVA model corresponding to this data. Be sure to define each term in your model and list any assumptions that are made.

b. Perform the test to determine if the model from part a) is significant.

c. Prepare a profile plot for the cell mean surface finish scores. Does it appear that any interaction effects are present? Explain.

d. Test whether or not interaction effects are present.

e. Given that low surface finish scores are desirable, which combination(s) of feed rate and depth of cut do you recommend?

Depth of Cut (in) = 0.15 0.15 0.15 0.18 0.18 0.18 0.2 0.2 0.2 0.25 0.25 0.25 0.15 0.15 0.15 0.18 0.18 0.18 0.2 0.2 0.2 0.25 0.25 0.25 0.15 0.15 0.15 0.18 0.18 0.18 0.2 0.2 0.2 0.25 0.25 0.25

Feed Rate (in/min) = 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3

Surface Finsih Score = 74 64 60 79 68 73 82 88 92 99 104 96 92 86 88 98 104 88 99 108 95 104 110 99 99 98 102 104 99 95 108 110 99 114 107 111

Depth of Cut (in)	Feed Rate (in/min)	Surface Finish Score
0.15	0.2	74
0.15	0.2	64
0.15	0.2	60
0.18	0.2	79
0.18	0.2	68
0.18	0.2	73
0.2	0.2	82
0.2	0.2	88
0.2	0.2	92
0.25	0.2	99
0.25	0.2	104
0.25	0.2	96
0.15	0.25	92
0.15	0.25	86
0.15	0.25	88
0.18	0.25	98
0.18	0.25	104
0.18	0.25	88
0.2	0.25	99
0.2	0.25	108
0.2	0.25	95
0.25	0.25	104
0.25	0.25	110
0.25	0.25	99
0.15	0.3	99
0.15	0.3	98
0.15	0.3	102
0.18	0.3	104
0.18	0.3	99
0.18	0.3	95
0.2	0.3	108
0.2	0.3	110
0.2	0.3	99
0.25	0.3	114
0.25	0.3	107
0.25	0.3	111

Answer 1

Randomized Block design: Two-way ANOVA model DepthA 0.15 0.18 73.33333333 96.66666667 99.33333333 0.2 87.33333333 100.6666667 105.6666667 of cuts, 0.25 99.66666667 104.3333333 110.6666667 88.66666667 99.66666667 Mean values of the surface finishes for each of the four different depth [66 88.66666667 99.66666667] 73.33333333 96.66666667 99.33333333 Data:- ntd i- l 87.33333333 100.6666667 105.6666667 L99.66666667 104.3333333 110.6666667 Number of levels of factor A r-rows (Data)-4 i0,1..r-1 Number of levels of factor B c:-cols (Data)-3

Number of values (replicates) for each depth of cut X :Data i,j Total Variation in the Randomized Block Design c-1 r-1 Σ(x,-GM)-=1947.5556 尸0.0 ss-Σ GM- =94.333 Among-Group Variation in the Randomized Block Design Γ 81.5831 197.583| 103.833 SSA := r. Σ(X-i-GM〉-1053.5 0.2 81.583 .25 97.583 .3 103.833 Plot of the mean surface finishes of the three feed rates;

.3 103.833 81.5 0.2 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29 0.3 Among-Block Variation in the Randomized Block Design 「 84.7781 | 89.7781 Σ (Xi,j) T 97.889 i 104.889 -' c Random Variation in the Randomized Block Design SSE := SST _ S SA-S SBL 185.685

Degrees of Freedom. The Mean Squares in the Randomized Block Design SSE 0.9475 SSBL SSA 236.1235 MSE-3 35A -52 6.75 MSBL MSA- J SE dfBL Randomized Block FSTAT Statistic MSA = 17.021 Fstat MSHE FSTATStatistic for Block Effects .MSBL statMSE

Analysis-of-Variance Table for the Randomized Block Design Sum of Source Degrees of freedomm Mean Squares (Variance) squares SSA-1053.5MSA-526.75 SSBL=708.37 MSBL=236.123 MSA 17.021 MSE Among gropus(A) Among Blocks(BL) Erron dfA-2 stat- MSBL 7.63 stat MSE dfse-6 SSE-185.685 MSE-30.948 Total dfr-11 SST 1947.556 Testing for factor and interaction effects: Null hypothesis Alternative hypothesis: Hi Not all are equal

MSA MSE F test for differences among the restaurants: Fstat 17.0207 Fais the upper tail critical value from an F distribution with dfA-2 and dfsE-6 degrees of freedonm Level of significance test: α=0.05 Upper-tail critical value of F distribution: Fa qF ((1-a),dfA,dfsE) 5.1433 P-value: Reject the null hypothesis at α= 0.05, level of significance. Because Fstat-17.0207 > ,-5.143 and P<α. You conclude that there is evidence of a difference in the mean values among the different surface finishes. F test for a difference among the depth of cuts MSBL = 7.6298 Fstat:- α : 0.05 MSE

Fais the upper tail critical value from an F distribution with r-1 and rc(n'-1) degrees of freedom Upper-tail critical value of F distribution: P,-qF(1-a),dfBL , df E-47571 Reject the null hypothesis at a 0.05, level of significance. Because Fstat 7.6298 > F ,-4.757 and P〈α. You conclude that there is evidence of a difference among the depth of cuts and that the blocking has been advantageous in reducing the random error. Multiple comparisons: Turkey-Kramer procedure Objective: to determine which groups are significantly different from the others. Compute the absolute mean difference: SにData 5)

Critical range for the Turkey Kramer.procedure 4.34 Upper-tail critical value from a Studentized range distribution having c: 3 degrees of freedom in the numerator and (r-1). (c-1)-6 degrees of freedom in the denominator. lable E. TABLE E.7 Critical Values of the Studentized Range, O Upper 5% Points( -0.05) Numerator, df Denominator, df 10 12 18.00 27.00 32.80 37.10 40.40 43.10 45.40 47.40 49.10 50.60 52.00 6.09 8.30 9.80 10.90 11.70 12.40 13.00 13.50 14.00 14.40 14.70 4.50 5.91 6.82 7.50 8.04 8.48 8.85 9.18 9.46 9.72 9.95 3.93 5.04 5.76 6.29 6.71 7.05 7.35 7.60 7.83 8.03 8.21 3.64 4.60 5.22 5.67 6.03 6.33 6.58 6.80 6.99 7.17 7.32 4 3.46 4.34 4.90 5.31 5.63 5.89 6.12 6.32 6.49 6.65 6.79

MSE -12.072 There is a significant difference between the surface finishes except B and C