A group of industrial engineers used regression to estimate the relationship of

Question

A group of industrial engineers used regression to estimate the relationship of two independent variables (xi) and (x2) to a dependent variable (y). They are considering the second-order model for (y) (a) MINITAB was used to fit the model to the data. Locate the least squares prediction equation for y on the printout shown below. (b) Is the overall model statistically useful for predicting (y)? Test using = .01. (c) Based on the MINITAB results, is there evidence to indicate that (y) is curvi linearly related to (x)? Explain. The regression equation is = 80.2 + 156 xi - s2 X1SQ + 273 X2 + 760 X1X2 47 X1SQX2 Predict Coef SE Coef Constant X1 xisa X2 X1X2 X1SQX2 80.22 156.5 -42.3 272.84 760.1 47.0 30.39 2.64 0.013 128.6 1.22 0.233 123.40.34 0.734 42.98 6.35 .000 181.8 4.18 0.000 174.5 0.27 0.790 S= 54.4116 R-Sq 98.6% R-Sq (adj ) =98.3% - Analy313 ot Variance DE MS Source Regression Residual Error 30 Total 5 6173670 1234734 417.05 0.000 2961 88819 35 6262489

Explanation / Answer

Answers in detail below:

a. The least squares prediction equation for y should be:

Y = 80.2+156*X1-42*(X1^2)+273*(X2) +760*(X1*X2) + 47X1^2*(X2)

b. Yes, it is. The p-value in ANalysis of Variance table is 0.000
Therefore, the p-value is less than .01 and hence significant

c. No. See the p-value of X1SQ ( curvilinear function of X1), it is .734. It is more than .01 and hence
p-value is more than .01 which means that X1SQ is not significant

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