Statistic Problem using SAS software 4. For this problem use the “plastic hardne
ID: 3209181 • Letter: S
Question
Statistic Problem using SAS software
4. For this problem use the “plastic hardness” data which can be found in Moodle PLAS-TIC.dat . The rst column is the plastic hardness (Y ) and the second column is theelapse time (X).
(a) Plot the a scatter plot for this data. Include a regression line on the plot (i = rl).
Is the relationship approximately linear?
(b) Run the linear regression to predict hardness from time using the REG procedureto nd the estimated regression equation.
(c) Test the null hypothesis that the true slope is 0; be sure to include null, alternativehypotheses, test method, and interpret your results.
(d) Obtain a 95% condence interval for the slope and interpret your result.
(e) Construct the condence interval for the mean plastic hardness when the elapsetime 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 onwhether you think this will be a good predictive model for hardness.
Plastic.txt
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
Explanation / Answer
Sol:
data plastichardness;
infile cards;
input plastic_hardness elapsetime;
cards;
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
;
run;
PLOT Statement
Solutionb:
Proc reg data=plastichardness;
model plastic_hardness=elapsetime;
run;
output fromExcel:
regression equation is
plastichardness=168.6+2.034375elapsetime
slope=2.034375
y intercept=168.6
Solutiond:
'lower 95% slope
1.8405
upper 95% slope
Solutiong:
R2=coefficient of determination]
=0.9864
98.64%percent variation in plastichardness is explained by the model
SUMMARY OUTPUT Regression Statistics Multiple R 0.98646 R Square 0.973103 Adjusted R Square 0.971182 Standard Error 3.234027 Observations 16 ANOVA df SS MS F Significance F Regression 1 5297.513 5297.513 506.5062 2.16E-12 Residual 14 146.425 10.45893 Total 15 5443.938 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Intercept 168.6 2.657024 63.45445 1.26E-18 162.9013 174.2987 162.9013 elapsetime 2.034375 0.090394 22.50569 2.16E-12 1.8405 2.22825 1.8405Related Questions
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