Question

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: x₁= (P-200)/10, x₂= (T-400)/25. The n=10 data points are listed in the table:

                y                     X₁                  X₂

              5.2                  -2                  2

                5                  -2                  2

              0.3                  -1                 -1

            -0.1                  -1                 -1

            -1.2                   0                  -2

            -1.1                   0                  -2

              2.2                  1                  -1

2                   1                  -1

6.2________2 ________2

              6.1                  2                   2

!                                                               

1. Find the least-squares regression equation of the form y = β₀ + β₁x₁ + β₂x₂ + ε (you may use a calculator or any software that can perform operations with matrices).
2. Does the model contribute information for the prediction of y? Test using α=0.10.
3. Suppose that we are interested in an interval estimate for the strength of plastic for the following combination of pressure and temperature: x₁=1 and x₂=1. Would a confidence interval or a prediction interval be more appropriate in this situation? Calculate this interval using α=0.01.

Answer 1

Given, y, x1 and x2, we input the data set into R programming software and use the summary(.), lm(.) and predict(.) functions to carry out the regression analysis. The output is given below.

Call:
lm(formula = y ~ x1 + x2)

Residuals:
Min 1Q Median 3Q Max
-0.5357 -0.3893 -0.2221 0.2814 0.9443

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.4600 0.1866 13.186 3.37e-06 ***
x1 0.4100 0.1319 3.108 0.0171 *
x2 1.6143 0.1115 14.479 1.79e-06 ***
---
Signif. codes: 0 ‘***’ 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: 5.205e-06

fit lwr upr
1 4.484286 2.23622 6.732352

(a) From the output above, we can say that the least squares regression equation is: = 2.46 + (0.41) + (1.6143) .
(b) The p-value for the test of significance of the model has come out to be 5.205e-06, which is very much less than the 0.10 significance level. Hence, we reject the null hypothesis and conclude that there is significant evidence to suggest that the model contributes information for the prediction of y.
(c) The prediction interval will be more appropriate for this situation and the 99% prediction interval is (2.236, 6.732).