Problem 6 (20 Points): ical variasineer studies the viscosity of a polymer () as
ID: 3336321 • Letter: P
Question
Problem 6 (20 Points): ical variasineer studies the viscosity of a polymer () as a function of two process reacti on temperature (X) and catalyst feed rate (X) 16 experiments have been conducted, and the Residual Plots for Viscosity resulting data are summarized in the following table. Normal Probability Plo Versus Fits 2256 100 2426 2293 2330 2368 2250 12 90 Residual Fited Value Histagram Versus Oeder 94 93 12 2364 2379 2440 2364 2404 2317 309 2328 13 95 100 85 Cbsarvation Orde 12 Part (a) (5 Points): Is there sufficient evidence to reject the null bypothesis that the coefficient for xi is 0? Explain why in one or up to tao sentences Part (b) (5 Points): Is there sufficient evidence to reject the null bypothesis that the coefficient for xa is0? Explain why in one or up to two sentences Part (c) (5 Points): What is the predicted viscosity for the temperature 95 degrees and the catalyst flow rate of 10? Part (d) (5 Points): The 92.7% R-square seems lo indicate that this linear regression produced a good fit. However, in the Versus Fits plot (i.c, the Residual vs. Fitted Value plot) of the "Residual Plots for Viscosity", it seems that the variability of the residuals increases with respect to the fitted (response) value. (In other words, the large the viscosity, the larger the residuals.) A common technique to deal with this situation is to transform the original response values and study the transformed respoese as a function of the original variables. What would be a good transform to try for this problem? Explain why in one or up to two sentences. 87 12 Part of the results that Minitab has produced is as follows. You are asked to answer the following multi-part question based on these results. Model Summary S R-sq R-sq(ad) Rsqupred 1635869270% 9157% 8907% Coefficients Term Constant X1: Temperature (C) X2 Catalyst Feed Rate (lb/h)8.58 24 352 0004 Coef SE Coef T-Value P-Value 15661 616 25.43 0000 7.621 0618 12.32 0000 Regression Equation Viscosity 15661 + 7.621 X: Temperature (° + 8.58 X2:Catalyst Feed Rate ( Residual Plots for ViscosityExplanation / Answer
Solution:
(a) Let me take the level of significance = 0.05, Here we see that p-value of x1 is 0 that is less than the level of significance so we reject the null hypothesis.
Conclusion: Yes, there is sufficient evidence to reject the null hypothesis because p-value is less than the level of significance.
(b).Yes, there is sufficient evidence to reject the null hypothesis because p-value =0.004 is less than the level of significance.
(c) since Given X1=95 and X2=10
Then ,
Viscosity =1566.1+7.621*95+8.58*10 = 2375.895
Therefore,
Viscosity = 2375.895
(d) After transforming the response variable, it is often helpful to transform the predictor variable as well. Since, reciprocal transformations often work well for this purpose.
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