To test our hypothesis: Họ : Bị = Ba against H1 : B1 + Ba We first need to find the least squares estimates of the parameters. We will write our model as: y - X3 + e, e~ N(0, o².1) Then 3 = (X'X)-'(X'y) We also know that, MSE B~ N(3,0°(X'X)-'), Note that the degrees of freedom are n-5 because there are 5 parameters (%, 31,... , B5) that are estimated. .. B - Ba- N(31 – Ba, (s" + s* – 2s" J²) where s is the ()-ta element of (X'Xy!. and they are independent. ..T = n tn-5 under Ho VMSEVS" + 3 – 2s3 Using the above test statistic, we reject Ho at level of significance a if observed |t| > taçn-5

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kindly explain the terms and formulas used in this picture!

(c)
To test our hypothesis:
Họ : B1 = B3 against H1 : B1 + 33
We first need to find the least squares estimates of the parameters. We will write our model
as:
y = XB+ €, €n N(0,o².I)
Then
3 = (X'X)-'(X'y)
We also know that,
MSE
B~ N(8,02(X'X)-'),
~ Xn-5
Note that the degrees of freedom are n-5 because there are 5 parameters (90, B1,..., B5) that
are estimated.
. B1 - B - N(3, – B3, (s" + s* - 2s³ )o²) where s" is the (i)-th element of
(XX)!.
and they are independent.
..T =
tn-5 under Ho
VMSEVS"
25
Using the above test statistic, we reject Ho at level of significance a if
observed |t| > ta;n-5
Transcribed Image Text:(c) To test our hypothesis: Họ : B1 = B3 against H1 : B1 + 33 We first need to find the least squares estimates of the parameters. We will write our model as: y = XB+ €, €n N(0,o².I) Then 3 = (X'X)-'(X'y) We also know that, MSE B~ N(8,02(X'X)-'), ~ Xn-5 Note that the degrees of freedom are n-5 because there are 5 parameters (90, B1,..., B5) that are estimated. . B1 - B - N(3, – B3, (s" + s* - 2s³ )o²) where s" is the (i)-th element of (XX)!. and they are independent. ..T = tn-5 under Ho VMSEVS" 25 Using the above test statistic, we reject Ho at level of significance a if observed |t| > ta;n-5
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