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In order to predict with 90% confidence the expected value of y for a given value of x in a simple linear regression problem, a random sample of 10 observations is taken. Which of the following t-table values listed below would be used?

a. 2.228
b. 2.306
c. 1.860
d. 1.812

Answer :

cchilabert

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!

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