hw2.lyx

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\begin_body

\begin_layout Standard
John Hancock
\end_layout

\begin_layout Standard
CEN 6405
\end_layout

\begin_layout Standard
Homework 2
\end_layout

\begin_layout Standard
May 20th, 2014
\end_layout

\begin_layout Part*
13.2
\end_layout

\begin_layout Standard
Exercise 13.2 requires us to make several calculations using the data given
 in exercise 12.11.
 The data is a list of disk I/O's per program: 
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
{23, 33, 14, 15, 42, 28, 33, 45, 23 , 34, 39, 21, 36, 23, 24, 36, 25, 9,
 11, 19, 35, 24, 31, 29, 16, 23, 34, 24, 38, 15, 13, 35 , 28}
\end_layout

\begin_layout Section*
a.
 10-percentile and 90-percentile from the sample
\end_layout

\begin_layout Standard
We learned in chapter 12 that, "The a-quantiles can be estimated by sorting
 the observations and taking the 
\begin_inset Formula $\left[\left(n-1\right)\alpha+1\right]^{th}$
\end_inset

 element in the ordered set." Therefore in order to compute the 10-percentile
 and 90 percentile from the sample, we must first sort the sample.
 Here is the sorted version of the above list of disk I/O's: {9, 11, 13,
 14, 15, 15, 16, 19, 21, 23, 23, 23, 23, 24, 24, 24, 25, 28, 28, 29, 31,
 33, 33, 34, 34, 35, 35, 36, 36, 38, 39, 42, 45}
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\begin_layout Standard
The number of elements in the list is 33, so we will use 
\begin_inset Formula $n=33$
\end_inset

 in the formula in the paragraph above.
 We are required to calculate the 10 and 90 percentile from the sample,
 so we will have two values of 
\begin_inset Formula $\alpha$
\end_inset

 for the formula: 
\begin_inset Formula $0.1$
\end_inset

 and 
\begin_inset Formula $0.9$
\end_inset

 .
 Substituting these values for 
\begin_inset Formula $n=33$
\end_inset

 and 
\begin_inset Formula $\alpha=0.1$
\end_inset

 into the formula we get: 
\begin_inset Formula \[
\left[\left(33-1\right)0.1+1\right]=\left[3.2+1\right]=\left[4.2\right]=4\]

\end_inset

 Which implies the 10-percentile from the sample is the fourth element in
 the sorted list, which is 14.
\end_layout

\begin_layout Standard
Note: our text uses the notation 
\begin_inset Formula $\left[.\right]$
\end_inset

 to denote rounding the value in the square brackets to the nearest integer
 value.
 
\end_layout

\begin_layout Standard
Similarly, substuting 
\begin_inset Formula $n=33$
\end_inset

 and 
\begin_inset Formula $\alpha=0.9$
\end_inset

 into the formula we get:
\begin_inset Formula \[
\left[\left(33-1\right)0.9+1\right]=\left[28.8+1\right]=\left[29.8\right]=30\]

\end_inset


\end_layout

\begin_layout Standard
This implies the thirtieth element of the sorted list is the 90-percentlie
 from the sample, which is 38.
\end_layout

\begin_layout Section*
b.
 Mean number of disk I/O's per Program
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\begin_layout Standard
Our text states, 
\begin_inset Quotes eld
\end_inset


\series bold
Sample mean
\series default
 is obtained by taking the sum of all observations and dividing this sum
 by the number of observations in the sample.
\begin_inset Quotes erd
\end_inset

 The sum of the number of disk I/O's in the list from exercise 12.11 is 878.
 We have 33 observations in the list so the sample mean is: 
\begin_inset Formula $\frac{878}{33}\approx26.6$
\end_inset

.
\end_layout

\begin_layout Section*
c.
 90% confidence interval for the mean
\end_layout

\begin_layout Standard
Box 13.1, item 1.
 b.
 in our text gives a formula for the 
\begin_inset Formula $100\left(1-\alpha\right)$
\end_inset

 two sided confidence interval for a sample mean.
 
\begin_inset Formula \[
\bar{x}\mp z_{1-\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\]

\end_inset


\end_layout

\begin_layout Standard
We know to use this formula because we have more than 30 elements in our
 sample.
 That is, 
\begin_inset Formula $n=33$
\end_inset

.
\end_layout

\begin_layout Standard
Section 13.2 explains that 
\begin_inset Formula $\bar{x}$
\end_inset

 is the sample mean.
 We computed the sample mean in exercise 13.2 b.
 above, so we already know 
\begin_inset Formula $\bar{x}=26.6$
\end_inset

 .
\end_layout

\begin_layout Standard
\begin_inset Formula $z_{1-\frac{\alpha}{2}}$
\end_inset

 is the 
\begin_inset Formula $1-\frac{\alpha}{2}$
\end_inset

 - quantile of a normal unit variate.
 The current excercise requires us to compute a 90% confidence interval,
 so we know 
\begin_inset Formula $100\left(1-\alpha\right)=90$
\end_inset

.
 We can solve for 
\begin_inset Formula $\alpha$
\end_inset

 to determine 
\begin_inset Formula $a=0.1$
\end_inset

.
 Hence, we know we must use the 
\begin_inset Formula $z_{1-\frac{0.1}{2}}=z_{0.95}$
\end_inset

 value from appendix A.2 in our text.
 This value is: 
\begin_inset Formula $1.645$
\end_inset

.
 
\end_layout

\begin_layout Standard
\begin_inset Formula $s$
\end_inset

 is the sample standard deviation.
 We used a spreadsheet to calculate the standard deviation of the sample.
 The result we obtained for 
\begin_inset Formula $s$
\end_inset

 is 
\begin_inset Formula $s=9.4$
\end_inset

.
\end_layout

\begin_layout Standard
Substituting all these values in the formula for the confidence interval
 above we obtain the interval:
\begin_inset Formula \[
\left(26-1.645\frac{9.4}{\sqrt{33}},26+1.645\frac{9.4}{\sqrt{33}}\right)\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
\approx\left(23.3,28.7\right)\]

\end_inset


\end_layout

\begin_layout Section*
d.
 Fraction of programs making less or equal to 25 I/O's, and the 90% confidence
 interval for the fraction
\end_layout

\begin_layout Standard

\end_layout

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