Answer:
c. 1.860
Step-by-step explanation:
Hello!
For a simple linear regression the formula to make a prediction interval is:
^Y±[tex]t_{n-2;1-\alpha /2} * \sqrt{Se^2{[1+\frac{1}{n}+\frac{(x_{n+1}-X[bar])^2}{sumX^2-\frac{(sumX)^2}{n} } ]}[/tex]
The value of the t-value would be
[tex]t_{n-2;1-\alpha /2}= t_{10-2;1-0.5}= t_{8;0.95}= 1.860[/tex]
I hope it helps!
