Two-way ANOVA In-class WorkTwo-way ANOVA In-class Work1. The yield of a chemical process is being studied. The two most important variables are thought to be the pressure (in pascals) and the temperature (in degrees Fahrenheit). Three levels of each factor are selected and an experiment with two replicates is performed. The yield data follow:Two-way Analysis of VarianceAnalysis of Variance for Yield Source DF SS MS F PTemperature 2 0.3011 0.1506 8.47 0.009Pressure 2 0.7678 0.3839 21.59 0.000Interaction 4 0.0689 0.0172 0.97 0.470Error 9 0.1600 0.0178Total 17 1.2978 Temperature Mean 150 90.417 160 90.250 170 90.567 Pressure Mean 200 90.367 215 90.683 230 90.183Temperature Pressure Mean 150 200 90.30150 215 90.65150 230 90.30160 200 90.20160 215 90.55160 230 90.00170 200 90.60170 215 90.85170 230 90.25PressureTemperature 200 215 230150°90.4 90.7 90.290.2 90.6 90.4160°90.1 90.5 89.990.3 90.6 90.1170°90.5 90.8 90.490.7 90.9 90.1150 160 170 23021520090.890.690.490.290.0PressureTempMeanInteraction Plot - Data Means for Chem(a) n12 = 2 x312 = 90.7 13x90.30 1x 90.367 I= 3 J=3 SSAB = 0.0689 MSB = 0.3839(b) Draw the interaction plots. (c) Conduct the test for interaction.- State the hypotheses in terms of the problem. H0: There is no interaction effect on the mean yield of the chemical processHa: There is an interaction effect on the mean yield of the chemical process- What is the value of the test statistic and p-value? F=.97, p-value = .47- What is the distribution of the test statistic when HO is true? F(4,9)- State your decision. Fail to Reject H0 since p-value > any reasonable significance level. - State your conclusion in terms of the problem. There is insufficient evidence to suggest that an interaction between temperature and pressure effects the mean yield of the chemical process.(d) Conduct the test for the Temperature main effect.- Why is it appropriate to carry out a test for the Temperature main effect?Since we failed to reject H0 in the interaction test, then it is appropriate to look for main effects.- State the hypotheses in terms of the problem. H0: Temperature does not effect the mean yield of the chemical processHa: Temperature does effect the mean yield of the chemical process- What is the value of the test statistic and p-value? F=8.47, p-value = 0.009- What is the distribution of the test statistic when HO is true? F(2,9)- State your decision. Reject H0 since p-value < any reasonable significance level- State your conclusion in terms of the problem. There is sufficient evidence to suggest that temperature effects the mean yield of the chemical process.- What is an estimate of the main effect for Temperature going from 150° to 170°? Interpret the main effect estimate in terms of the problem.15.417.90567.9013xx200 215 230 17016015090.890.690.490.290.0TempPressureMeanInteraction Plot - Data Means for ChemThus, when temperature increases from 150 degrees to 170 degrees, we estimate that the mean yield of the chemical process increases by .15 (i.e. this is our estimated effect of temperature on the mean yield of the chemical process).- What is an estimate of the main effect for Pressure going from 215 to 230 pascals? Interpret the main effect estimate in terms of the problem.5.683.90183.9023xxThus, when pressure increases from 215 pascals to 230 pascals, we estimate that the mean yield of the chemical process decreases by .5 (i.e. this is our estimated effect of pressure on the mean yield of the chemical process).2. An experiment was run to investigate how the type of glass and the type of phosphorescent coating affects the brightness of a light bulb. The response variable is the current (in microamps) to obtain a specified brightness. Phosphor TypeA B CGlass Type12782912852973042962732842882229235241259249241228225235- Calculate the cell means and marginal means. (Write the cell means in the appropriate cell of the table and the marginal means at the ends of the rows and columns.)- Make the interaction plots.Two-way Analysis of VarianceAnalysis of Variance for BrightnessSource DF SS MS F PGlass 1 11451 11451 90.01 0.000Phosphor 2 59 30 0.23 0.796Interaction 2 120 60 0.47 0.634Error 12 1527 127Total 17 13157Conduct the hypothesis test for an interaction effect.- State the hypotheses.H0: The interaction between glass type and phosphor type does not affect the mean current used by the lightbulbsPhosphor TypeA B CGlass Type1 284.7 299.0 281.72 235 249.7 229.31 2 cba300290280270260250240230phosphorglassMeanInteraction Plot - Data Means for brighta b c 21300290280270260250240230glassphosphorMeanInteraction Plot - Data Means for brightHa: The interaction between glass type and phosphor type does affect the mean current used by the lightbulbsFind the values of the test statistic and the p-value.F = .47 and p-value = .634- What is the distribution of the test statistic when H0 is true?F ~ F(2, 12)- Make a decision.Since the p-value is larger than any reasonable significance level (like .01, .02, .05, .1), we fail to reject H0.- Make a conclusion in terms of the problem.There is insufficient evidence to suggest that an interaction between glass type and phosphor affects the mean current used by the lightbulbs- Make the main effects plots.Conduct the hypothesis test for a Glass Type main effect.- Why is it appropriate to carry out a test for the Temperature main effect?Since there is an insignificant interaction (we FAILED TO REJECT the null hypothesis in the test for an interaction at p-value=0.000), then we can test for the main effect due to Glass Type.- State the hypotheses.H0: Glass Type does not affect the mean current used by the lightbulbs.Ha: Glass Type does affect the mean current used by the lightbulbs.- Find the values of the test statistic and the p-value.F = 90.01 and p-value = 0.000- What is the distribution of the test statistic when H0 is true?F ~ F(1,12)- Make a decision.Conclusively reject H0- Make a conclusion in terms of the problem.There is a ton of evidence to support the claim that Glass Type does affect the mean current used by thelightbulbsWhat is