程序代写案例-HUDM5126

HUDM5126
Ji Wu
HW5
Oct 22nd ,2021
Question1
a)
The correlation matrix indicates that x_1 has comparatively strong positive relationsh
ip with the
brand liking (y_i).While x_2 is not highly correlated to the y_i.
b)
We can have Y= 37.650 + 4.425*x_1 + 4.375*x_2.b1 here means every one unit increase in
x_1,the predicted Y value would increase by 4.425 and vice versa.
c)
From the plot,we can conclude that it is normally distributed and no outliers observed.
d)
We can observe that only X_1X_2 skewed to the right ,Yhat, X_1,X_2 and residuals are normally
distributed.
e)
Null Hypothesis (H0): Homoscedasticity is present (the residuals are distributed with equal
variance)
Alternative Hypothesis (HA): Heteroscedasticity is present (the residuals are not distributed with
equal variance)
We have P-value 0.3599>0.01
We can’t reject H0.Thus,constant variance assumption holds here.
Question2
a)
b)
c)
d)
e)
f)
(Yh)hat=53.84715
g)
Question3
a)
SSR(X4) = 67.775,DF=1
SSR(X1|X4) = 42.275,DF=1
SSR(X2|X1, X4) = 27.857,DF=1
SSR(X3|X1, X2, X4) = 0.420,DF=1
SSE(X1, X2, X3, X4) = 98.231,DF=76
b)
Hypothesis:
Ho: β3 = 0
Ha: β3 ≠0.
We can obtain the F* value from lm(y ~ x_1+x_2+x_4+x_3).And the F*=0.3248
Also,the corresponding p-value here =0.570446
By looking up at the F-distribution table, with α = 0.01 and DF=1,DF2=76,
F =6.980578>F*
Thus we can’t reject Ho,Ho still holds here.Which means X3 is not significant if x1,x2 and x4 are
already included in the regression model.
Question4
Know that SSTO=236.5575
R2Y4=SSR(X4)/SSTO=67.775/236.5575=0.28651
R2Y1=SSR(X1)/SSTO=14.819/236.5575=0.06264
R2Y1|4=SSR(X1|X4)/SSE(X4)=42.275/168.782=0.25047
R214=SSR(X1,X4)/SSTO=(67.775+42.275)/236.5575=0.46521
R2Y2|14=SSR(X2|X1,X4)/SSE(X1,X4)=27.857/126.508=0.22020
R2Y3|124=SSR(X3|X1,X2,X4)/SSE(X1,X2,X4)=0.420/98.650=0.00426
R2=SSR/SSTO=(14.819+72.802+50.287+0.420)/236.5575=0.58475
The degree of marginal linear association between Y and X1 I think is increased when we did
the adjustment for the X4.
R Script

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