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Quiz 3_solution

# Quiz 3_solution - .12(a(4 pts Compute the SSE(Pure Error...

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Quiz 3 Name: Observations of the yield of a chemical reaction taken at various temperatures were recorded as follows: x ( o C) y (%) Sample variance 150 77.4 76.7 78.2 0.5633 200 84.1 84.5 83.7 0.1600 250 88.9 89.2 89.7 0.1633 300 94.8 94.7 95.9 0.4433 Estimate the linear model, ε β + + = x Y 1 0 , via Minitab and obtain the following output: The regression equation is y = 60.3 + 0.117 x Predictor Coef SE Coef T P Constant 60.2633 0.7444 80.96 0.000 x 0.116533 0.003211 36.29 0.000 S = 0.621772 R-Sq = 99.2% R-Sq(adj) = 99.2% Analysis of Variance Source DF SS MS F P Regression 1 509.25 509.25 1317.25 0.000 Residual Error 10 3.87 0.39 Lack of Fit 2 1.21 0.605 1.82 0.224 Pure Error 8 2.66 0.3325
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Unformatted text preview: Total 11 513.12 (a) (4 pts) Compute the SSE (Pure Error) and MSE (Pure Error) , and fill in the blanks. Hint: âˆ§ = = 2 2 ) ( Ïƒ s PureError MSE is the best estimate of the variance of Y at a given x. (0.5633+0.1600+0.1633+0.4433)/4 = 0.3325 (b) (3 pts) Is the linear relationship between the yield ( y ) and the temperature ( x ) significant at a 0.05 significance level? Why? Yes, because the P_value is very low. 0.000<0.05. (c) (3 pts) Is there any quadratic or higher-order relationship between y and x at a 0.05 significance level? Why? No, because the P_value is 0.224>0.05....
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