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hw3.lyx
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#LyX 1.6.7 created this file. For more info see http://www.lyx.org/
\lyxformat 345
\begin_document
\begin_header
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\begin_body
\begin_layout Standard
John Hancock
\end_layout
\begin_layout Standard
CEN 6405
\end_layout
\begin_layout Standard
May 20th, 2014
\end_layout
\begin_layout Part*
Homework 3
\end_layout
\begin_layout Section*
14.2
\end_layout
\begin_layout Standard
In order to do exercise 14.2 it will be necessary to find a linear model
where the predictor is CPU times and the response is disk I/O's.
We can use the technique shown in example 14.1, treating the CPU times as
the
\begin_inset Formula $x$
\end_inset
values and disk I/O's as
\begin_inset Formula $y$
\end_inset
values.
The data points are the set
\begin_inset Formula \[
\left\{ \left(2,14\right),\left(5,16\right),\left(7,27\right),\left(9,42\right),\left(10,39\right),\left(13,50\right),\left(20,83\right)\right\} \]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $n=7$
\end_inset
,
\begin_inset Formula $\sum xy=3375$
\end_inset
,
\begin_inset Formula $\sum x=66$
\end_inset
,
\begin_inset Formula $\sum x^{2}=828$
\end_inset
,
\begin_inset Formula $\sum y=271$
\end_inset
,
\begin_inset Formula $\sum y^{2}=13855$
\end_inset
,
\begin_inset Formula $\bar{x}=9.43$
\end_inset
,
\begin_inset Formula $\bar{y}=38.71$
\end_inset
\end_layout
\begin_layout Standard
The formula for the slope,
\begin_inset Formula $b_{1}$
\end_inset
, of the linear model is:
\begin_inset Formula \[
b_{1}=\frac{\sum xy-n\bar{x}\bar{y}}{\sum x^{2}-n\left(\bar{x}\right)^{2}}=\frac{3375-7\left(9.43\right)\left(38.71\right)}{828-7\left(9.43\right)^{2}}=3.9986\]
\end_inset
\end_layout
\begin_layout Standard
The formula for the intercept,
\begin_inset Formula $b_{0}$
\end_inset
, of the linear model is:
\begin_inset Formula \[
b_{0}=\bar{y}-b_{1}\bar{x}=38.71-3.9986\left(9.43\right)=1.003\]
\end_inset
\end_layout
\begin_layout Standard
The linear model is therefore:
\end_layout
\begin_layout Standard
disk I/O's =
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
3.9986
\family default
\series default
\shape default
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(CPU time) + 1.003
\end_layout
\begin_layout Standard
The following table summarizes the model:
\end_layout
\begin_layout Standard
\begin_inset Tabular
<lyxtabular version="3" rows="10" columns="5">
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CPU Time
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Disk I/O's
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</cell>
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\begin_inset Text
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Estimate
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\end_inset
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\begin_inset Text
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Error
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\begin_inset Text
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Error Squared
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\begin_inset Formula $x_{i}$
\end_inset
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\begin_inset Formula $y_{i}$
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</cell>
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\begin_inset Text
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\begin_inset Formula $\hat{y_{i}}=b_{0}+b_{1}x_{i}$
\end_inset
\end_layout
\end_inset
</cell>
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\begin_inset Formula $e_{i}=y_{i}-\hat{y_{i}}$
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</cell>
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\begin_inset Formula $e_{i}^{2}$
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2
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14
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9.0002
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4.9998
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24.9980
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5
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16
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28.9932
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80.975
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2.025
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4.1006
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66
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271
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270.9286
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0.0714
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95.9927
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\end_inset
\end_layout
\begin_layout Standard
Standard deviation of errors
\begin_inset Formula $s_{e}=\sqrt{\frac{95.9927}{5}}=4.3816$
\end_inset
\end_layout
\begin_layout Subsection*
a.
Significant Parameters
\end_layout
\begin_layout Standard
In order to know which parameters are significant we must compute confidence
intervals for
\begin_inset Formula $b_{0}$
\end_inset
and
\begin_inset Formula $b_{1}$
\end_inset
and check whether or not they include 0.
\end_layout
\begin_layout Standard
First of all, we must compute the standard deviations of
\begin_inset Formula $b_{0}$
\end_inset
and
\begin_inset Formula $b_{1}$
\end_inset
, using formulas from example 14.4:
\begin_inset Formula \[
s_{b_{0}}=s_{e}\left[\frac{1}{n}+\frac{\bar{x}^{2}}{\sum x^{2}-n\bar{x}^{2}}\right]^{\frac{1}{2}}\]
\end_inset
Here,
\begin_inset Formula $n=7$
\end_inset
,
\begin_inset Formula $\bar{x}=9.43$
\end_inset
,
\begin_inset Formula $\bar{x}^{2}=88.9249$
\end_inset
,
\begin_inset Formula $\sum x^{2}=828$
\end_inset
,
\begin_inset Formula $s_{e}=4.3816$
\end_inset
so
\begin_inset Formula \[
s_{b_{0}}=4.3816\left[\frac{1}{7}+\frac{88.9249}{828-7\left(88.9249\right)}\right]^{\frac{1}{2}}=1.5880\]
\end_inset
\begin_inset Formula \[
s_{b_{1}}=\frac{s_{e}}{\left[\sum x^{2}-n\bar{x}^{2}\right]^{\frac{1}{2}}}=\frac{4.3816}{\left(828-7\left(88.9249\right)\right)^{\frac{1}{2}}}=0.3056\]
\end_inset
\end_layout
\begin_layout Standard
We have 5 degrees of freedom, and we will compute a 90% confidence interval
for our parameters, so we will use
\begin_inset Formula $t_{0.95;5}=2.015$
\end_inset
\end_layout
\begin_layout Standard
Hence the 90% confidence interval for
\begin_inset Formula $b_{0}$
\end_inset
is
\begin_inset Formula \[
b_{0}\mp t_{95;5}s_{b_{o}}=1.003\mp\left(2.015\right)1.5880=\left(-2.19682,4.20282\right)\]
\end_inset
\end_layout
\begin_layout Standard
And the 90% confidence interval for
\begin_inset Formula $b_{1}$
\end_inset
is
\end_layout
\begin_layout Standard
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\begin_inset Formula \[
b_{1}\mp t_{0.95;5}s_{b_{1}}=3.9986\mp\left(2.015\right)0.3056=\left(3.3828,4.6143\right)\]
\end_inset
\end_layout
\begin_layout Standard
The 90% confidence interval for
\begin_inset Formula $b_{0}$
\end_inset
includes 0.
We cannot reject the hypothesis that
\begin_inset Formula $b_{0}$
\end_inset
is 0 at the 0.1 significance level.
The 90% confidence interval for
\begin_inset Formula $b_{1}$
\end_inset
does not include 0.
\series bold
Therefore the parameter
\begin_inset Formula $b_{0}$
\end_inset
is not significant, and the parameter
\begin_inset Formula $b_{1}$
\end_inset
is significant.
\end_layout
\begin_layout Subsection*
b.
Percentage of variation explained by the regression
\end_layout
\begin_layout Standard
We use the calculations done in example 14.2 as a template to complete this
section of the exercise:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
SSE=\sum y^{2}-b_{0}\sum y-b_{1}\sum xy=13855-1.003\left(271\right)-3.9986\left(3375\right)=87.912\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
SST=SSY-SSO=\sum y^{2}-n\left(\bar{y}\right)^{2}=13855-7\left(38.71\right)^{2}=3365.7513\]
\end_inset
\begin_inset Formula \[
SSR=SST-SSE=3365.7513-87.912=3277.8393\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
R^{2}=\frac{SSR}{SST}=\frac{3277.8393}{3365.7513}=0.9739\]
\end_inset
Therefore our regression explains 97% of disk I/O's variation.
\end_layout
\begin_layout Subsection*
c.
Expected number of disk I/O's
\end_layout
\begin_layout Standard
Section c requires us to calculate the expected number of disk I/O's for
a program with a CPU time of 40 milliseconds.
We use the linear model to find this number:
\begin_inset Formula \[
diskI/O's=3.9986(CPUtime)+1.003\]
\end_inset
\begin_inset Formula \[
diskI/O's=3.9986(40)+1.003\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
diskI/O's=160.947\]
\end_inset
\end_layout
\begin_layout Subsection*
d.
Bounds on c.
for 10% chance of error on single program measured tomorrow
\end_layout
\begin_layout Standard
In order to do part d of this exercise, we follow the example 14.5 of our
text to compute a 90% confidence interval for a single predicted value.
\end_layout
\begin_layout Standard
First we must compute the standard deviation of a single future observation
\begin_inset Formula \[
s_{\hat{y}_{p}}=s_{e}\left[1+\frac{1}{n}+\frac{\left(x_{p}-\bar{x}\right)^{2}}{\sum x^{2}-n\bar{x}^{2}}\right]^{\frac{1}{2}}\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
=4.3816\left[1+\frac{1}{7}+\frac{\left(40-9.43\right)^{2}}{828-\left(7\right)9.43^{2}}\right]^{\frac{1}{2}}=10.4516\]
\end_inset
\end_layout
\begin_layout Standard
We will again use the
\begin_inset Formula $t$
\end_inset
value for the 0.95 quantile with 5 degrees of freedom to calculate the 90%
confidence interval for the bounds on the predicted value from section
c.
of this exercise:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
160.947\mp2.015\left(10.4516\right)=\left(139.8870,182.0070\right)\]
\end_inset
\end_layout
\begin_layout Standard
Because we computed the 90% confidence interval, we claim there is a 10%
chance that the value we observe tomorrow will be outside this interval.
\end_layout
\begin_layout Subsection*
e.
Repeat of d.
for large number of programs
\end_layout
\begin_layout Standard
We make calculations similar to those we make in section d.
of this exercise, but we use the formula for the standard deviation of
a large number of observations instead:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
s_{\hat{y}_{p}}=s_{e}\left[\frac{1}{n}+\frac{\left(x_{p}-\bar{x}\right)^{2}}{\sum x^{2}-n\bar{x}^{2}}\right]^{\frac{1}{2}}\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
=4.3816\left[\frac{1}{7}+\frac{\left(40-9.43\right)^{2}}{828-\left(7\right)9.43^{2}}\right]^{\frac{1}{2}}=9.4888\]
\end_inset
\end_layout