An experiment was conducted to investigate the effect of extrusion pressure (P)

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: x,-(P-200) 10. x,-(T-400) 25. The n=10 data points are listed in the table 2 2 5.2 0.3 0.1 1.2 1.1 2.2 2 6.2 2 2 (a) Find the least-squares regression equation of the form11 A A1 1 + x2 + (you may use a calculator or any software that can perform operations with matrices) (b) Does the model contribute information for the prediction of? Test using -0.05 (c) Suppose that we are interested in an interval estimate for the strength of plastic for the following combination of pressure and temperature: x/--1 and x2-1. Would a confidence interval or a prediction interval be more appropriate in this situation? Calculate this interval using a-0.05.

Explanation / Answer

we shall do this using the open source statistical package R , the complete R snippet is as follows

y<- c(5.2,5,.3,-.1,-1.2,-1.1,2.2,2,6.2,6.1)
x1<- c(-2,-2,-1,-1,0,0,1,1,2,2)
x2 <- c(2,2,-1,-1,-2,-2,-1,-1,2,2)

## create a dataframe
data.df <- data.frame(y,x1,x2)

## fir the model
fit <- lm(y~., data=data.df)

## summarise the model
summary(fit)

newdat<- data.frame(x1=-1,x2=1)

## predict

predict(fit,newdata=newdat,interval = "prediction")

The results are

summary(fit)

Call:
lm(formula = y ~ ., data = data.df)

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

as the r2 value is 0.9691 , this means that the model is able to explain about 96.91% variation in the data

Hence the regression fit the data well

The regression equation is

the regression equation is

y = 2.46 + 0.40*x1 +1.614*x2

The prediction interval is given as

> predict(fit,newdata=newdat,interval = "prediction")
fit lwr upr
1 3.664286 2.145253 5.183319

we know that the confidence interval of the prediction presents a range for the mean rather than the distribution of individual data points.

A prediction interval is a range that is likely to contain the response value of a single new observation given specified settings of the predictors in your model.

Hence we shalluse a prediction interval in this case

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