Sunday

March 1, 2015

March 1, 2015

Posted by **tracy12** on Sunday, June 23, 2013 at 8:41pm.

x 7 10 19 9 20 22 6 13 18 15

y 29 27 28 33 24 15 25 23 21 18

(a) Find the equation of the line of best fit.

y hat = -0.56 + 1.78 x

(b) Find se2.

36.00

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**Kuai**, Sunday, June 23, 2013 at 9:58pmY hat = -.57x + 32.20

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**tracy12**, Sunday, June 23, 2013 at 10:17pmSo what did I do wrong on the part b? That one is wrong also.

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**Kuai**, Sunday, June 23, 2013 at 10:20pmWhat Do you mean se2

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**Kuai**, Sunday, June 23, 2013 at 10:30pmWhat Do you mean by this se2

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**tracy12**, Sunday, June 23, 2013 at 10:49pmI am sorry that is Se squared. That is the way it is written on the test.

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**MathMate**, Monday, June 24, 2013 at 5:54amSE² is given by:

[ Σ(y-yBar)^2 - [Σ(x-xBar)(y-yBar)]²/Σ(x-xBar)² ] /(n-2)

I get:

Σ(y-yBar)²=258.1

Σ(x-xBar)(y-yBar) = -168.7

Σ(x-xBar)²=296.9

n=10

to get

se²=20.28

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**tracy12**, Monday, June 24, 2013 at 9:45amI have a question where did you get 258.1. I must of had my numbers wrong, because I had those other two.

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**MathMate**, Monday, June 24, 2013 at 10:08amI got it from summing (y-yBar)^2, and I had yBar=24.3.

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**tracy12**, Monday, June 24, 2013 at 12:20pmOk, thank you, I had the wrong numbers for the yBar

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**MathMate**, Monday, June 24, 2013 at 1:05pmYou're welcome!

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