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Chapter11_exercises.lyx
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Chapter11_exercises.lyx
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#LyX 2.3 created this file. For more info see http://www.lyx.org/
\lyxformat 544
\begin_document
\begin_header
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\use_default_options true
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theorems-ams
eqs-within-sections
figs-within-sections
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\end_header
\begin_body
\begin_layout Part*
11.
\end_layout
\begin_layout Chapter*
11.
Least squares error
\end_layout
\begin_layout Exercise*
1.
Prove that
\begin_inset Formula $\sum_{i=1}^{n}(c_{1}x_{i}+c_{2})^{2}=c_{1}^{2}\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}+n(c_{1}\bar{x}+c_{2})^{2}.$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\sum_{i=1}^{n}(c_{1}x_{i}+c_{2})^{2}=\sum_{i=1}^{n}(c_{1}x_{i}-c_{1}\bar{x}+c_{1}\bar{x}+c_{2})^{2}=c_{1}\sum_{i=1}^{n}(x_{i}-\bar{x}_{n})^{2}+2\sum_{i=1}^{n}(c_{1}x_{i}-c_{1}\bar{x})(c_{1}\bar{x}+c_{2})+(c_{1}\bar{x}+c_{2})^{2}$
\end_inset
\end_layout
\begin_layout Standard
We consider
\begin_inset Formula $2\sum_{i=1}^{n}(c_{1}x_{i}-c_{1}\bar{x})(c_{1}\bar{x}+c_{2})=2(c_{1}\bar{x}+c_{2})(c_{1}\sum_{i=1}^{n}x_{i}-c_{1}\sum_{i=1}^{n}\bar{x})=2(c_{1}\bar{x}+c_{2})0=0$
\end_inset
.
\end_layout
\begin_layout Standard
So concluded.
\end_layout
\begin_layout Exercise*
3.
Show that the least-squares line
\begin_inset Formula $y=\hat{\beta}_{0}+\hat{\beta}_{1}x\text{ }(1)$
\end_inset
passes
\end_layout
\begin_deeper
\begin_layout Standard
through the point (
\begin_inset Formula $\bar{x},\bar{y})$
\end_inset
.
From Eq.
(11.1.1)
\begin_inset Formula $\hat{\beta}_{0}=\bar{y}-\hat{\beta}_{1}\bar{x}$
\end_inset
, replace
\begin_inset Formula $(\bar{x},\bar{y})$
\end_inset
to Eq.
(1),
\begin_inset Formula $\bar{y}=\bar{y}-\hat{\beta}_{1}\bar{x}+\hat{\beta}_{1}\bar{x}$
\end_inset
, It's clear that's true.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
8.
Prove error can be 0, so it will fit all
\begin_inset Formula $k+1$
\end_inset
points.
\end_layout
\begin_layout Exercise*
Maybe :
\begin_inset Formula $y=\sum_{i=1}^{k}\beta_{i}x^{k}$
\end_inset
will have
\begin_inset Formula $k-1$
\end_inset
extreme value, so it will have ability pass through
\begin_inset Formula $k+1$
\end_inset
point: parabol having
\begin_inset Formula $1$
\end_inset
extreme value, can fit perfectly to
\begin_inset Formula $3$
\end_inset
points.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Chapter*
11.2 Regression
\end_layout
\begin_layout Exercise*
1.
Show that the M.L.E.
of
\begin_inset Formula $\sigma^{2}$
\end_inset
is given by Eq.
(11.2.3)
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Chapter*
11.3 Statistical Inference In Simple Linear Regression
\end_layout
\begin_layout Exercise*
1.
Problem of simple linear regression,
\begin_inset Formula $10$
\end_inset
pairs of observed values of
\begin_inset Formula $x_{i}$
\end_inset
and
\begin_inset Formula $y_{i}$
\end_inset
given Table 11.9 are obtained.
The the following hypotheses at the level of significance
\begin_inset Formula $0.05$
\end_inset
:
\end_layout
\begin_layout Exercise*
\begin_inset Formula
\[
H_{0}:\beta_{0}=0.7
\]
\end_inset
\end_layout
\begin_layout Exercise*
\begin_inset Formula
\[
H_{1}:\beta_{0}\neq0.7
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Standard
From
\begin_inset Formula $(11.3.18)$
\end_inset
, this hypotheses are same Eq.
\begin_inset Formula $(11.3.13)$
\end_inset
if we make the substitutions
\begin_inset Formula $c_{0}=1,c_{1}=0,c_{*}=\beta_{0}^{*}$
\end_inset
.
And we obtrain the following random variable,
\begin_inset Formula $U_{0}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
U_{0}=\frac{\hat{\beta}_{0}-\beta_{0}^{*}}{\sigma^{'}[\frac{1}{n}+\frac{\bar{x}^{2}}{s_{x}^{2}}]^{1/2}}\text{ (11.3.19)}
\]
\end_inset
\end_layout
\begin_layout Standard
which has
\begin_inset Formula $t$
\end_inset
distribution with
\begin_inset Formula $n-2$
\end_inset
degrees of freedom if
\begin_inset Formula $H_{0}$
\end_inset
is true.
\end_layout
\begin_layout Standard
\begin_inset Formula $\hat{\beta}_{0}=\bar{Y}-\bar{x}\hat{\beta}_{1}=0.4352$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\hat{\beta}_{1}=\frac{\sum_{i=1}^{n}(x_{i}-\bar{x})Y_{i}}{s_{x}^{2}}=0.14721$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $s_{x}=(\sum_{i=1}^{n}(x_{i}-\bar{x})^{2})^{1/2}=0.916$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\sigma^{'}=(\frac{S^{2}}{n-2})^{1/2}=4.02$
\end_inset
from (11.3.12)
\end_layout
\begin_layout Standard
\begin_inset Formula $\beta_{0}^{*}=0.7$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $S^{2}=\sum_{i=1}^{n}(Y_{i}-\hat{\beta}_{0}-\hat{\beta}_{1}x_{i})^{2}=129.26$
\end_inset
\end_layout
\begin_layout Standard
So
\begin_inset Formula $U_{0}=-6.695$
\end_inset
(coded)
\end_layout
\begin_layout Standard
With
\begin_inset Formula $\alpha=0.05$
\end_inset
, we have
\begin_inset Formula $T_{7}^{-1}(1-\alpha/2)=T_{7}^{-1}(0.975)=2.365$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Formula $|U_{0}|=6.695<2.365$
\end_inset
, so rejected
\end_layout
\begin_layout Exercise*
2.
Test regression line passes through origin of xyplane .
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Standard
Making use of Eq.
(11.3.20), this test is equivalent :
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
H_{0}:\beta_{0}=0
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
H_{1}:\beta_{0}\neq0
\]
\end_inset
\end_layout
\begin_layout Standard
Similar to Exercise 1, with replacing
\begin_inset Formula $\beta_{0}^{*}=0$
\end_inset
\end_layout
\begin_layout Exercise*
3.
Slope
\begin_inset Formula $=1$
\end_inset
, that is related to Eq.
(11.3.21)
\end_layout
\begin_layout Exercise*
\begin_inset Formula
\[
H_{0}:\beta_{1}=\beta_{1}^{*}
\]
\end_inset
\end_layout
\begin_layout Exercise*
\begin_inset Formula
\[
H_{1}:\beta_{1}=\beta_{1}^{*}\text{ (11.3.21)}
\]
\end_inset
\end_layout
\begin_layout Exercise*
So we have the statistic
\begin_inset Formula $U_{1}=s_{x}\frac{\hat{\beta}_{1}-\beta_{1}^{*}}{\sigma^{'}}$
\end_inset
have
\begin_inset Formula $t$
\end_inset
distribution with
\begin_inset Formula $n-2$
\end_inset
degrees of freedom if
\begin_inset Formula $H_{0}$
\end_inset
is true.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
4.
Testing the hypothesis that the regression line is horizontal, It means
that
\end_layout
\begin_layout Exercise*
\begin_inset Formula
\[
H_{0}:\beta_{1}=0
\]
\end_inset
\end_layout
\begin_layout Exercise*
\begin_inset Formula
\[
H_{1}:\beta_{1}\neq0
\]
\end_inset
\end_layout
\begin_layout Exercise*
We again use statistic from (11.3.22)
\begin_inset Formula $U_{1}=s_{x}\frac{\hat{\beta_{1}}-\beta_{1}^{*}}{\sigma^{'}}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
5.
Test the following hypotheses at the level of significance 0.1:
\end_layout
\begin_layout Exercise*
\begin_inset Formula
\[
H_{0}:\beta_{1}=5\beta_{0}
\]
\end_inset
\end_layout
\begin_layout Exercise*
\begin_inset Formula
\[
H_{1}:\beta_{1}\neq5\beta_{0}
\]
\end_inset
\end_layout
\begin_layout Standard
It is equivalent to
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
H_{0}:5\beta_{0}-\beta_{1}=0
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
H_{1}:5\beta_{0}-\beta_{1}\neq0
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Standard
This test of hypotheses is about a linear combination of
\begin_inset Formula $\beta_{0}$
\end_inset
and
\begin_inset Formula $\beta_{1}$
\end_inset
\end_layout
\begin_layout Standard
For this type, we apply Theorem 11.3.3
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
U_{01}=[\frac{c_{0}^{2}}{n}+\frac{(c_{0}\bar{x}-c_{1})^{2}}{s_{x}^{2}}]^{-1/2}(\frac{c_{0}\hat{\beta}_{0}+c_{1}-c_{*}}{\sigma^{'}})\text{ (11.3.14)}
\]
\end_inset
\end_layout
\begin_layout Standard
has
\begin_inset Formula $T_{n-2}^{-1}$
\end_inset
is the quantile function of
\begin_inset Formula $t$
\end_inset
distribution with
\begin_inset Formula $n-2$
\end_inset
degrees of freedom.
\end_layout
\begin_layout Standard
\begin_inset Formula $c_{0}=5$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $c_{1}=1$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $c_{*}=0$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $U_{01}=0.664$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Formula $T_{7}^{-1}(0.95)=1.895$
\end_inset
.
So
\begin_inset Formula $|U_{01}|<T_{7}^{-1}(1-\alpha_{0}/2)$
\end_inset
.
So don't reject test
\end_layout
\begin_layout Exercise*
6.
Similar to
\begin_inset Formula $5$
\end_inset
, with hypothesis:
\end_layout
\begin_layout Exercise*
\begin_inset Formula
\[
H_{0}:\beta_{1}+\beta_{0}=1
\]
\end_inset
\end_layout
\begin_layout Exercise*
\begin_inset Formula
\[
H_{1}:\beta_{1}+\beta_{0}\neq1
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
7.
\begin_inset Formula $D=\hat{\beta}_{0}+\hat{\beta}_{1}\bar{x}$
\end_inset
.
\begin_inset Formula $Cov(D,\hat{\beta}_{1})=E[D\hat{\beta}_{1}]-E[D]E[\hat{\beta}_{1}]$
\end_inset
\begin_inset Formula $=E[\hat{\beta}_{0}\hat{\beta}_{1}+\hat{\beta}_{1}^{2}\bar{x}]-E[\hat{\beta}_{0}+\hat{\beta}_{1}\bar{X}]E[\hat{\beta}_{1}]$
\end_inset
\end_layout
\begin_layout Exercise*
\begin_inset Formula $=E[\hat{\beta}_{0}\hat{\beta}_{1}]+\bar{X}E(\hat{\beta_{1}^{2}})-\beta_{0}\beta_{1}-\beta_{1}^{2}\bar{x}$
\end_inset
\end_layout
\begin_layout Exercise*
\begin_inset Formula $=Cov(\hat{\beta}_{0},\hat{\beta}_{1})+E[\hat{\beta}_{1}]E[\hat{\beta}_{0}]+\bar{x}E(\hat{\beta}_{1}^{2})-\beta_{0}\beta_{1}-\beta_{1}^{2}\bar{x}$
\end_inset
\end_layout
\begin_layout Exercise*
\begin_inset Formula $=\frac{-\bar{x}\sigma^{2}}{s_{x}^{2}}+\beta_{0}\beta_{1}+\bar{x}\frac{\sigma^{2}}{s_{x}^{2}}-\beta_{0}\beta_{1}=0$
\end_inset
\end_layout
\begin_layout Exercise*
(Applied
\begin_inset Formula $Cov(\hat{\beta_{1}},\hat{\beta}_{0})$
\end_inset
from Theorem 11.2.2)
\end_layout
\begin_layout Exercise*
Now, we want to prove
\begin_inset Formula $D$
\end_inset
and
\begin_inset Formula $\hat{\beta}_{1}$
\end_inset
are independent .
First, from Theorem 5.6.7 and
\begin_inset Formula $\hat{\beta}_{i}$
\end_inset
has normal distribution so
\begin_inset Formula $D$
\end_inset
have normal distribution.
\end_layout
\begin_layout Exercise*
suppose
\begin_inset Formula $\rho$
\end_inset
is corelation of
\begin_inset Formula $D$
\end_inset
and
\begin_inset Formula $\hat{\beta}_{1}$
\end_inset
.
So
\begin_inset Formula $\rho=0$
\end_inset
.
\end_layout
\begin_layout Exercise*
From Theorem
\begin_inset Formula $5.10.1$
\end_inset
, we can easily prove exists
\begin_inset Formula $Z_{1},Z_{2}$
\end_inset
such th at
\begin_inset Formula $X_{1}=D$
\end_inset
,
\begin_inset Formula $X_{2}=\hat{\beta}_{1}$
\end_inset
have parameters
\begin_inset Formula $\mu_{1},\mu_{2},\sigma_{1},\sigma_{2},\rho=0$
\end_inset
.
So joint distribution of
\begin_inset Formula $D,\hat{\beta}_{1}$
\end_inset
\begin_inset Formula $f(D,\hat{\beta}_{1})$
\end_inset
have form of bivariate normal distribution, Moreover covariance
\begin_inset Formula $\rho=0$
\end_inset
, So these 2 variables are independent.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
8.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Standard
b.
Because
\begin_inset Formula $D$
\end_inset
is normal distribution so, from Corrolary 8.2.1,
\begin_inset Formula $\frac{Q^{2}}{\sigma^{2}}$
\end_inset
has chisquare distribution with 2 degree of freedom.
\end_layout
\begin_layout Exercise*
9.
From eq.
(11.3.32)
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
10,11, 12.
From Theorem 11.3.5
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
13, 14.
From Theorem 11.3.6
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
16.
Let
\begin_inset Formula $U$
\end_inset
be as degined by Eq.
(11.3.32) and let
\begin_inset Formula $\gamma$
\end_inset
be fixed positive constant.
For all observed values
\begin_inset Formula $(x_{i},y_{i})$
\end_inset
for
\begin_inset Formula $i=1,...,n$
\end_inset
the set points
\begin_inset Formula $(\beta_{0}^{*},\beta_{1}^{*})$
\end_inset
such that
\begin_inset Formula $U^{2}<\gamma$
\end_inset
is the interior of an ellipse in the
\begin_inset Formula $\beta_{0}^{*}\beta_{1}^{*}-$
\end_inset
plane.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $U^{2}<\gamma\rightarrow Q^{2}<a\rightarrow n(\hat{\beta}_{0}-\beta_{0}^{*})^{2}+(\sum_{i=1}^{n}x_{i}^{2})(\hat{\beta}_{1}-\beta_{1}^{*})^{2}+2n\bar{x}(\hat{\beta}_{0}-\beta_{o}^{*})(\hat{\beta}_{1}-\beta_{1}^{*})$
\end_inset
\end_layout
\begin_layout Standard
We can see final equation has form of hyperbola equation.
So not ellipse ?????
\end_layout
\begin_layout Exercise*
17.
From data of table 11.9, we replace value for each term in
\begin_inset Formula $U^{2}=\frac{\frac{1}{2}Q^{2}}{\sigma^{'^{2}}}$
\end_inset
.
and from confidence coefficient 0.95, we have