Question

Problem 4 An experiment was conducted to investigate the effect of extrusion pressure (P) and temperature at extrusion (T) on the strength y of a new type of plastic. Two plastic specimens were prepared for each of five combinations of pressure and temperature. The specimens were then tested in a random order and the breaking strength for each specimen was recorded. The independent variables were coded (transformed) as follows to simplify the calculations: x) = (P-200)/10, X2 = (T-400)/25. The n=10 data points are listed in the table: NN-LOOSEN NNSNN SANN (a) Find the least-squares regression equation of the form y = Bo + Bxy + Box: + € (you may use a calculator or any software that can perform operations with matrices). (b) Does the model contribute information for the prediction of y? Test using -0.10. (c) Suppose that we are interested in an interval estimate for the strength of plastic for the following combination of pressure and temperature: x17 and x, 1. Would a confidence interval or a prediction interval be more appropriate in this situation? Calculate this interval using a 0.01

Answer 1

> model<-1m(y-x1+x2, data=data) > summary(model) Call: Im(formula - y - x1 + x2, data = data) Residuals: Min 10 Median -0.5357 -0.3893 -0.2221 30 0.2814 Max 0.9443 Coefficients : Estimate std. Error t value Pr (>t) (Intercept) 2.4600 0 .1866 13.186 3. 37e-06 www 0.4100 0.13193.108 0.0171 * 1.6143 0.1115 14.479 1.79e-06 *** X1 X2 signif. codes: ****' 0.001 ***' 0.01 *' 0.05 '. ' 0.1''1 Residual standard error: 0.5899 on 7 degrees of freedom Multiple R-squared: 0.9691, Adjusted R-squared: 0.9602 F-statistic: 109.7 on 2 and 7 DF, p-value: 5. 205-06

$Equation\rightarrow y=2.46+0.41x_1+1.6143x_2$

The adjusted $R^2$ of the model is 96% which is a very good prediction estimate and the pvalue is less than 0.1.

Hence, yes the model contributes for the prediction of y.

c)The prediction interval should be used in this case because we are predicting a result. Confidence interval can be used for a sampled point estimate.

for x1=1,x2=1->y=4.4843

I got the answer for the below from the R console

> predict(model, newdata = newdata,level=0.99, interval="prediction") fit lwr upr 1 4.484286 2.23622 6.732352