Question

An aircraft firm is considering three different alloys for use in the wing construction of a new airplane. Each alloy can be produced in four different thicknesses (1 = thinnest, 4 = thickest). Two test samples are constructed for each combination of alloy type and thickness, then each of the 24 test samples is subjected to a laboratory device the severely flexes it until failure occurs. For each test sample the number of flexes before failure is recorded, with the results shown in the Aircraft worksheet in the ANOVA HW data workbook on Moodle.

a) Use JMP to fit a two-way ANOVA to the data. Using alpha = 0.05 draw conclusions for the ANOVA. Make sure you state your conclusion in the context of the problem.

b) Does there appear to be a need to include the alloy type / thickness interaction term?

c) Use JMP to fit a two-way ANOVA with interaction to the data. Using alpha= 0.05 draw conclusions for the ANOVA. Make sure you state your conclusion in the context of the problem.

Alloy    Thickness    Flexes

Alloy A Thickness 1    804

Alloy A Thickness 1    816

Alloy A    Thickness 2 819

Alloy A    Thickness 2 813

Alloy A    Thickness 3    820

Alloy A    Thickness 3    821

Alloy A Thickness 4 806

Alloy A Thickness 4    805

Alloy B    Thickness 1 836

Alloy B Thickness 1 828

Alloy B    Thickness 2    844

Alloy B    Thickness 2    836

Alloy B Thickness 3    814

Alloy B Thickness 3    811

Alloy B Thickness 4 811

Alloy B    Thickness 4 806

Alloy C    Thickness 1    804

Alloy C    Thickness 1 808

Alloy C Thickness 2 807

Alloy C Thickness 2    819

Alloy C    Thickness 3 819

Alloy C    Thickness 3    829

Alloy C Thickness 4    827

Alloy C    Thickness 4 835

Answer 1

Factor B, alloy type,j 1 to 3 Means for levels of factor A 804 816 836 828 804 808 816.00 The grand mean is the mean of all the data values, or 804 816+836+828++ 827 +835 24 818.25 flexes 819 813 807 819 823.00 836 Factor A, thickness i-1 to 4 820 821 814 811 819 829 819.00 There are 4 levels of factor A and 3 levels of factor B, leading to 4 x3-12 combinations, or cells. Within each cell, there are r- 2 observations, or replications. For example, the cell for (i- 2,j -3) contains the following observations: X2807 flexes and X232 819 flexes. 806 805 811 806 827 835 815.00 813.00 823.25 818.50 Means for levels of factor B The mean for thej- 2 level of factor B is 836+ 828 844+ ..+811+806 8 231 - 823.25 flexes The mean for the i 4 level of factor A is 806805 811 806+ 827 +835 815.00 flexes

Factor A Sum of Squares, SSA There are r = 2 replications within each cell and b = 3 levels for factor B. SSA is based on differences between the grand mean x-818.25) and the respective means for the a-4 levels of factor A oC (816.00 818.25)2(823.00 818.25)2 (819.00 818.25)2 (815.00 818.25)2 - 2(3) 6(38.75) and SA= 232.50

Factor B Sum of Squares, SSB There are r = 2 replications within each cell and a = 4 levels for factor A. SSB is based on differences between the grand mean-818.25) and the respective means for the b - 3 levels of factor B. -2(4)[(8 13.00-81 8.292 + (823.25-81 8.292 + (818.50-818.25%] 8(52.625) and SSB 421.00

Error Sum of Squares, SSE In this calculation, each observation is compared to the mean of its own cell. For example, the mean of the (i = 2, j = 3) cell is x23.-(8074 819)/2, or 813.00. (836 -832.00)2 +(816 - 810.002+(828 832.0(808 806.00)-2 (844-840.00) +(813 - 816.00)2(836 - 840.002(819 - 813.00)-2 (814 - 812.50)2 +1821-820.50)2「[+(811-8 12.50)2] ' L+1829-824.00)2 (811 - 808.50)2 (805-80.50+(806 808.50)(835 831.00)2 (804 810.00)2 (819 816.00)2 (820 820.50)2 (806 - 805.50)2 (804 - 806.00)2 (807-8 13.00)21 (819 824.00)2 (827-831.00)2 and SSE 334.00

Total Sum of Squares, SST This calculation compares each observation with the grand mean ( with the differences squared and summed: 818.25), (836 -818.25)2 +18 16-81 8.25 )2] . T+1828-81 8.25)2] . T+1808-81 8.25% (844 818.25)2 +1813-81 8.25 )2 ' |+1836-8 18.25)2| ' |+1819-81 8292 (814 -818.25)2 +1821-818.292「[+(811-81 8.292] . T+829-81 8.25% (81 1-81 8.25 +1805-81 8.25 )2] . T+1806-8 18.25)2] " L+1835-818292 (804 818.25)2 (819 818.25)2 (820-8 18.25% (806 818.25)21 (804-81 8.25% (807-8 18.25% (819-818.25% (827 -818.25)21 and SST 3142.50

Interaction Sum of Squares, SSAB Having calculated the other sum of squares terms, we can obtain SSAB by sub- tracting the other terms from SST, or SR-S142.3- 32.50-41.0- 14.00 and SMA8-385.0 3142.50-232.50 -421.00 334.00 andSSAB 2155.00

The Mean Square Terms As part C of Table 12.9 shows, each sum of squares term is divided by the num ber of degrees of freedom with which it is associated. There are a 4 levels for factor A, b-3 levels for factor B, and r-2 replications per cell, and the mean square terms are as follows: Factor A: SSA232.50 MSA1 41 and MSA 77.5 Factor B: SSB 421.00 b-1 3- 1 and | MSB= 210.50 = Interaction, AB: SSAB 2155.00 MSAB- and MSAB 359.17 (a 1(b- (4- 1)(3 - 1) Error, E: 334.00 and | MSE= 27.83 ab(r - 1 4(3) (2-1)

here are: three sts of null and alternative hypoiheseo be evaluated. In each case, the calculated F is compared to the critical lisited in thedistribution table The denominator of the F-ratio for each test is MSE, which has ab(r-1), or 4(3)(2-1)-12 degrees of freedom.

The Test Statistics, the Critical Values, and the Decisions 1. Testing for main effects, factor A The calculated Fis MSA 77.50_ MSE 27.83 The critical F for the 0.05 level is or F(0.05, 3, 12) 3.49 Because the calculated value of F (2.78) does not exceed the critical value (3.49), Ho: all a0 cannot be rejected. Our conclusion is that none of the thicknesses has any effect on the number of flexes a test unit will withstand before it fails.

The Test Statistics, the Critical Values, and the Decisions 2. Testing for main effects, factor B The calculated Fis MSB 210.50 MSE 27.83 7.56 The critical F for the 0.05 level is FI0.05,(b 1), ab(r 1)] SII F(0.05, 2, 12) 3.89 In this case, the calculated F (7.56) is greater than the critical value (3.89), and H,: all B, 0 is rejected. At least one of the alloys has an effect on the number of flexes a test unit will withstand before it fails. The calculated F also exceeds the critical values for both the 0.025 level (5.10) and 0.01 level (6.93), so the null hypothesis would be rejected at these levels as well. From the F table listings, the most accu rate statement we can make about the p-value for this test is that the p-value is less than 0.01 10"

The Test Statistics, the Critical Values, and the Decisions 3. Testing for interaction effects between levels of factors A and B The calculated Fis MSAB 359.17 27.8312.91 The critical F for the 0.05 level is FI0.05, (a 1)(b 1), ab(r 1)] F(0.05, 6, 12) 3.00 In the test for interaction effects, the calculated F (12.91) exceeds the critical value (3.00) and Ho: all (aB0 is rejected. The factors are not operating inde- pendently, and there is some relationship between the levels of thickness (factor A) and alloy type (factor B) in determining how many flexes a test unit will with stand before failure. The calculated F also exceeds the critical values for both the 0.025 level (3.73) and the 0.01 level (4.82), so the null hypothesis would be re- jected at these levels as well. As with the test for the main effects of factor B, the most accurate statement we can make based on our tables is that the p-value for this test is less than 0.01

Alloy 1 Alloy 2 Alloy 3 Data Analysis Thickness1 804836804nalysis Tools 816828808 OK Anova: Single Factor Anova: Two-Factor With Replication nova:Two-Factor Without Replication Cancel Thickness 2 819844807Coreianon Thickness 3| 820 814819 Fourier Analysis Thickness 480681827 Help 813836 819nothing F-Test Two-Sample for Va 821811829 805806835

preparing data for ,,statistica" 1D Alloy Thickness Flexes 804 816 2 819 2 813 3 820 3 821 4 806 4805 1836 828 2 844 2 836 3 814 3 811 4 811 4 806 1804 1808 2 807 2 819 3 819 3 829 4827 4835 Alloy 1 Alloy 2 Alloy 3 804 836804 816 828808 Thickness 2 819844807 813836819 Thickness 3 820814819 821 811829 Thickness 480681827 805 806835 Thickness1 2 2

Thickness; LS Means Current effect: F(3, 12)-2,7844, p ,08641 Effective hypothesis decomposition Vertical bars denote 0,95 confidence intervals 830 828 826 824 822 O 820 L 818 816 814 812 810 808 4 Thickness