The accompanying data resulted from a 23 experiment with three replications per combination of treatments designed to study the effects of concentration of detergent (A), concentration of sodiumcarbonate (B), and concentration of sodium carboxymethyl cellulose (C) on the cleaning ability of a solution in washing tests (a larger number indicates better cleaning ability than a smallernumber).Factor LevelsA12ACY AC12121211B с Condition(1)ababсacbcabc=112211111212222 2(a) After obtaining cell totals Xijk. compute estimates of ₁, YAC11, and yAC 21. (Round your answers to two decimal places.)B₁ =unitsunitsunits(b) Use the cell totals along with Yates's method to compute the effect contrasts and sums of squares. (Round your answers to two decimal places.)(contrast)224TreatmentConditionObservations106, 93, 116199, 200, 214197, 202, 185329, 331, 307149, 169, 136243, 247, 220255, 234, 252383, 360, 364(1) = X111.a=X211.b = x121.ab=X221.C = X112ac X212.EffectContrastSS=Ibc=X122.abc = X222.Construct an ANOVA table. (Round your answers to two decimal places.) Construct an ANOVA table. (Round your answers to two decimal places.)Source dfABсABACBCABCErrorTotalSSTest all appropriate hypotheses using a = 0.05. Find the critical value. (Round your answer to two decimal places.)F12Which main effects are significant at level 0.05? (Select all that apply.)A (detergent)OB (sodium carbonate)C (sodium carboxymethyl cellulose)There are no significant factors at level 0.05.Which interactions are significant at level 0.05? (Select all that apply.)AB (detergent and sodium carbonate)AC (detergent and sodium carboxymethyl cellulose)BC (sodium carbonate and sodium carboxymethyl cellulose)ABC (detergent, sodium carbonate, and sodium carboxymethyl cellulose)There are no significant interactions at level 0.05.ConditionMSd(c) Suppose a low water temperature has been used to obtain the data. The entire experiment is then repeated with a higher water temperature to obtain the following data. Use Yates's algorithmon the entire set of 48 observations to obtain the sums of squares and ANOVA table, and then test appropriate hypotheses at level 0.05. (Round your answers to two decimal places.)Observations144, 154, 1582 Rounding Numbers Calculator

Factors Level ABCConditionObservations111(1)106, 93, 116211a199, 200, 214121b197, 202, 185221ab329, 331, 307112c149, 169, 136212ac243, 247, 220122bc255, 234, 252222abc383, 360, 364a) Factors Level ABCConditionObservationsCell total(Xijk)111(1)106, 93, 116315211a199, 200, 214613121b197, 202, 185584221ab329, 331, 307967112c149, 169, 136454212ac243, 247, 220710122bc255, 234, 252741222abc383, 360, 3641107b)Use the Yate's method to assign sign for all the condition treatments: ConditionObservationsCell total(Xijk)ABCACBCABABC-1106, 93, 116315---+++-a199, 200, 214613+---+-+b197, 202, 185584-+-+--+ab329, 331, 307967++---+-c149, 169, 136454--+--++ac243, 247, 220710+-++---bc255, 234, 252741-++-+--abc383, 360, 3641107+++++++Determine the contrast and sum of squares for condition treatment using the below formula: ConditionCell total(Xijk)ABCACBCABABC-1315---+++-a613+---+-+b584-+-+--+ab967++---+-c454--+--++ac710+-++---bc741-++-+--abc1107+++++++Constrast13031307533-596119525Sum of squares707427117711837.04145.0417155.04171584.37526.04167b)R-code:1# Create a data frame with the provided data 2data <- data.frame( 3 A = c(1, 2, 1, 2, 1, 2, 1, 2), 4 B = c(1, 1, 2, 2, 1, 1, 2, 2), 5 C = c(1, 1, 1, 1, 2, 2, 2, 2), 6 Condition = c("(1)", "a", "b", "ab", "c", "ac", "bc", "abc"), 7 Observations = c(106, 199, 197, 329, 149, 243, 255, 383, 8 93, 200, 202, 331, 169, 247, 234, 360, 9 116, 214, 185, 307, 136, 220, 252, 364) 10) 11 12# Perform ANOVA for the factorial experiment 13model <- aov(Observations ~ A * B * C, data = data) 14summary(model)Output:1 Df Sum Sq Mean Sq F value Pr(>F) 2A 1 70742 70742 461.738 3.16e-13 *** 3B 1 71177 71177 464.577 3.01e-13 *** 4C 1 11837 11837 77.261 1.60e-07 *** 5A:B 1 1584 1584 10.341 0.0054 ** 6A:C 1 145 145 0.947 0.3450 7B:C 1 155 155 1.012 0.3294 8A:B:C 1 26 26 0.170 0.6856 9Residuals 16 2451 153 10--- 11Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Anova table: Source dfSSMSfA17074270742461.74B17117771177464.58C1118371183777.26A:B11584158410.34A:C11451450.95B:C11551551.01A:B:C126260.17Error162451153Total23158117Determine the critical value using the excel formula, "=F.INV.RT(0.05,1,16)":Fα,v1,v2=4.493998Fα,v1,v2=4.50The critical value at 5% significance level is 4.50.Main Effects:Any main effects having F-statistics greater than 4.50 would be considered significant effect.All the main effects Factors A, B, and C are significant at a 0.05 level because their F-values are greater than 4.50.Interaction Effects:Any interaction effects having F-statistics greater than 4.50 would be considered significant effect.Among all the interaction effects, only A:B interaction has F-value(10.34) greater than 4.50 whereas all other interactions has F-value less than 4.50. Therefore A:B interaction is significant at a 0.05 level while the other interaction effects (A:C, B:C, A:B:C) are not significant.a) b) Use the Yate's method to assign sign for all the condition treatments: ConditionObservationsCell total(Xijk)ABCACBCABABC-1106, 93, 116315---+++-a199, 200, 214613+---+-+b197, 202, 185584-+-+--+ab329, 331, 307967++---+-c149, 169, 136454--+--++ac243, 247, 220710+-++---bc255, 234, 252741-++-+--abc383, 360, 3641107+++++++Determine the contrast and sum of squares for condition treatment using the formula: ConditionCell total(Xijk)ABCACBCABABC-1315---+++-a613+---+-+b584-+-+--+ab967++---+-c454--+--++ac710+-++---bc741-++-+--abc1107+++++++Constrast13031307533-596119525Sum of squares707427117711837.04145.0417155.04171584.37526.04167R-code:1# Create a data frame with the provided data 2data <- data.frame( 3 A = c(1, 2, 1, 2, 1, 2, 1, 2), 4 B = c(1, 1, 2, 2, 1, 1, 2, 2), 5 C = c(1, 1, 1, 1, 2, 2, 2, 2), 6 Condition = c("(1)", "a", "b", "ab", "c", "ac", "bc", "abc"), 7 Observations = c(106, 199, 197, 329, 149, 243, 255, 383, 8 93, 200, 202, 331, 169, 247, 234, 360, 9 116, 214, 185, 307, 136, 220, 252, 364) 10) 11 12# Perform ANOVA for the factorial experiment 13model <- aov(Observations ~ A * B * C, data = data) 14summary(model)Output:1 Df Sum Sq Mean Sq F value Pr(>F) 2A 1 70742 70742 461.738 3.16e-13 *** 3B 1 71177 71177 464.577 3.01e-13 *** 4C 1 11837 11837 77.261 1.60e-07 *** 5A:B 1 1584 1584 10.341 0.0054 ** 6A:C 1 145 145 0.947 0.3450 7B:C 1 155 155 1.012 0.3294 8A:B:C 1 26 26 0.170 0.6856 9Residuals 16 2451 153 10--- 11Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Anova table: Source dfSSMSfA17074270742461.74B17117771177464.58C1118371183777.26A:B11584158410.34A:C11451450.95B:C11551551.01A:B:C126260.17Error162451153Total23158117b) Fα,v1,v2=4.50The critical value at 5% significance level is 4.50.Main Effects:Any main effects having F-statistics greater than 4.50 would be considered significant effect.All the main effects Factors A, B, and C are significant at a 0.05 level because their F-values are greater than 4.50.Interaction Effects:Any interaction effects having F-statistics greater than 4.50 would be considered significant effect.Among all the interaction effects, only A:B interaction has F-value(10.34) greater than 4.50 whereas all other interactions has F-value less than 4.50. Therefore A:B interaction is significant at a 0.05 level while the other interaction effects (A:C, B:C, A:B:C) are not significant.