Question

Problem 1: (7 pointsl A study was performed on wear of a bearing Y and its relationship to XI - oil viscosity and X2 load. The following data were obtained. Use the Minitab output on the next page to answer questions (a) to (e) XI 1.6 230 15.5 816 172 22.0 1058 293 91 113 125 43.0 1201 33.0 1357 40.0 1115 polnsl Find the regression equation that links the bearing Y to the oil viscosity and the load. ) P polatsd Is the regression model in part (a) valid? (Test the joint effect of XI and X2 to Y). Use a 5% significance level. b) pintl Which independent variable is linearly related to Y c) I pointsl How is the fit of the model? d) e)I pintl Estimate the bearing if xi, X2 are 25 and 900 respectively.

Answer 1

a) The regression equation estimated using MINITAB is given below
Y = 384 - 3.64 X1 - 0.112 X2

b) We will set the null hypothesis that

$H_{0}:$ Regression Model is not Valid.

$H_{1}:$ Regression Model is Valid.

F test is used to check the Validity of the model at 0.05 significance level. The results of ANOVA Table Obtained from MINITAB is given below.

Source	DF	SS	MS	F	P.Value
Regression	2	29787	14894	97.59	0.002
Residual Error	3	458	153
Total	5	30245

Since calculated P.Value = 0.002 which is less then 0.05 hence we will reject our null hypothesis and concludes that Regression Model is Valid.

c) We will set the null hypothesis that

$H_{0}:\beta _{i}=0$; i = 1.2 or in other word regression coefficients ($\beta _{1}$ and $\beta _{2}$) are not linearly related.

$H_{0}:\beta _{i}\neq0$; i = 1.2 or in other word regression coefficients ($\beta _{1}$ and $\beta _{2}$) are linearly related.

Under the null hypothesis the test in this case is t.test at 0.05 significance level. The results of t.test Obtained from MINITAB is given below.

Predictor	Coef	SE Coef	T	P.Value
Constant	383.8	36.22	10.6	0.002
X1	-3.6381	0.5665	-6.42	0.008
X2	-0.11168	0.04338	-2.57	0.082

Since calculated P.Value = 0.008 for X1 which is less then 0.05 hence we will reject our null hypothesis and concludes X1 is linearly related to Y. Also calculated P.Value = 0.082 for X2 which is greater then 0.05 hence we will accept our null hypothesis and concludes X2 is not linearly related to Y.

d) How a model fits the data is checked with R². R² is also called as the coefficient of determination. It varies from 0 to 1. 0 means model is very weak and 1 model is best fit. The R²​​​​​​​ obtained using MINITAB for the above model is R² = 98.5% which means the data is best fit.

e) The regression equation is
Y = 384 - 3.64 X1 - 0.112 X2

for X1 = 25, X2=900 we have

Y = 384 - 3.64 * 25 - 0.112 * 900

Y = 192.2