Homework Answers
Ans:
degree of freedom=5-2=3
critical t value=tinv(0.05,3)=3.182
y'(25)=16.5+0.162*25=20.55
95% confidence interval for average value of y
=20.55+/-3.182*3.94
=20.55+/-12.54
=(8.01,33.09)
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