The value of A such that P(-A < T < A) = .9 for a T random var with 16 degrees of freedom
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The value of A such that P(-A < T < A) = .9 for a T random var with 16 degrees of freedom

[From: ] [author: ] [Date: 12-06-16] [Hit: ]
..Actually, you want to find (1-0.9) / 2 = 0.05 in each tail.......
What is the value of A such that P(-A < T < A) = .90 for a T random variable with 16 degrees of freedom? Answer to three decimal places.

I know I'm supposed to use a T-Table for this, and what I assume is that I divide .9 by 2 to get .45 for a single tail, but then I'm totally lost as to how to reference this in the T-table...

-
fmchris -

Actually, you want to find (1-0.9) / 2 = 0.05 in each tail.

Look up in the T-table (16 d.f.) to find an area equal to 0.05 in each tail. Depending upon the table you use, the value for T should be either +/- 1.746

Answer: A = 1.746

Hope that makes sense!
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