For this problem the rst column is the plastic hardness (Y ) and the second colu

Question

For this problem the rst column is the plastic hardness (Y ) and the second column is the elapse time (X). if you could please use SAS and show the code. Use the SAS programming language to find questions (e), (f), and (g).

199.0 16.0
205.0 16.0
196.0 16.0
200.0 16.0
218.0 24.0
220.0 24.0
215.0 24.0
223.0 24.0
237.0 32.0
234.0 32.0
235.0 32.0
230.0 32.0
250.0 40.0
248.0 40.0
253.0 40.0
246.0 40.0

(e) Construct the condence interval for the mean plastic hardness when the elapse time is 20.

(f) Construct the prediction interval for the plastic hardness when a particular prod-uct has the elapse time is 20. Is this interval the same as the previous? Explain.

(g) What percent of the variation in hardness is explained by elapsed time?

(h) Using the information you’ve gathered about these data so far, comment on whether you think this will be a good predictive model for hardness.

Explanation / Answer

we cannot provide solutions using commercial softwares such as sas. However , we shall provide the solution using the open source alternative R .

The R snippet is as follows

# read the data into R dataframe
data.df<- read.csv("C:\Users\586645\Downloads\Chegg\plastic.csv",header=TRUE)
str(data.df)

# perform anova analysis
fit<- lm(Y~ X,data=data.df)

#summarise the results
summary(fit)

# confidence interval for elapse time = 20
new <- data.frame(X=20)   

predict(fit,new,interval="confidence")

# prediction interval

predict(fit,new,interval="predict")

The results are

> summary(fit)

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

Residuals:
Min 1Q Median 3Q Max
-5.1500 -2.2188 0.1625 2.6875 5.5750

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 168.60000 2.65702 63.45 < 2e-16 ***
X 2.03438 0.09039 22.51 2.16e-12 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 3.234 on 14 degrees of freedom
Multiple R-squared: 0.9731,   Adjusted R-squared: 0.9712 , # as the r2 value is 0.9731 , so model is able to explain 97.31% variation of the data
F-statistic: 506.5 on 1 and 14 DF, p-value: 2.159e-12 , as the p value is less than 0.05 , hence we can conclude that the model is signifcant

> predict(fit,new,interval="confidence")
fit lwr upr
1 209.2875 206.961 211.614
> predict(fit,new,interval="predict")
fit lwr upr
1 209.2875 201.9714 216.6036

as the p value is signifcant and the r2 value is high , hence we can say that the model is a good predictive model for hardness

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