Subversion Repositories programming

\documentclass[12pt,letterpaper]{report}
\parindent 0cm % don't bother with \noindent all the time!
\parskip 1ex

\usepackage{amsmath}
\usepackage{exscale}

% draws means
\newcommand{\Xbar}{\overline{X}}
\newcommand{\xbar}{\overline{x}}

\begin{document}
\sffamily % sans-serif font

\title{STA-326-02 Final}
\author{Ira W. Snyder}
\date{2006-06-06}

\maketitle

\section*{Chapter 5, Problem 108}

If the probability of a random variable is given by

\begin{equation*}
 f(x)=
\begin{cases}
 k(1-x^2) & \text{for $0<x<1$}\\
 0 & \text{elsewhere}
\end{cases}
\end{equation*}

find the value of $k$ and the probabilities that a random variable having this probability density will take on a value

\begin{enumerate}
 \item[(a)] between $0.1$ and $0.2$
 \item[(b)] greater than $0.5$.
\end{enumerate}

\subsection*{Finding $k$}

The definition of a probability density function states that

\begin{equation*}
 \int_{-\infty}^\infty f(x)\mathrm{d}x=1
\end{equation*}

For this problem, we can split this integral into 3 parts, then solve for $k$.

\begin{align}
 \int_{-\infty}^0 0\mathrm{d}x = 0\\
 \int_0^1 k(1-x^2)\mathrm{d}x = 1\\
 \int_1^\infty 0\mathrm{d}x = 0
\end{align}

We can ignore integrals 1 and 3, since they are both equal to zero. We will concentrate on integral 2.

To solve for $k$, we will first find the indefinite integral of $k(1-x^2)$.

\begin{align*}
 \begin{split}
  \int k(1-x^2) \mathrm{d}x & = k \int (1-x^2) \mathrm{d}x\\
                            & = k (x - \frac{x^3}{3})
 \end{split}
\end{align*}

Now we can easily solve for k on the interval (0,1) using the indefinite integral of integral 2, above.

\begin{align*}
 \begin{split}
  k (x - \frac{x^3}{3}) \Big|_0^1 & = k (1-\frac{1}{3}) - k (0-\frac{0}{3})\\
                                  & = \frac{2k}{3}
 \end{split}
\intertext{Which is easily solved to give}
 k = \frac{3}{2}
\end{align*}

\subsection*{Part a}

Now, to find the probability that a random variable having this probability density will take on a value between $0.1$ and $0.2$, we can integrate the original equation, using the value for $k$ that we just found.

\begin{align*}
 \begin{split}
  \int_{0.1}^{0.2} \frac{3}{2} (1-x^2) \mathrm{d}x & = \frac{3}{2} \int_{0.1}^{0.2} (1-x^2) \mathrm{d}x\\
                                                   & = \frac{3}{2} \left(x - \frac{x^3}{3}\right) \Big|_{0.1}^{0.2}\\
                                                   & = \frac{3}{2} \left(({0.2} - \frac{{0.2}^3}{3}) - ({0.1} - \frac{{0.1}^3}{3})\right)\\
                                                   & = \frac{3}{2} (0.976667)\\
                                                   & = 0.1465
 \end{split}
\end{align*}

\subsection*{Part b}

A process very similar to what we just did can be used to solve this part as well. Since we want to find the area under the curve that is greater than $0.5$, we integrate from $0.5$ to $1$.

\begin{align*}
 \begin{split}
  \int_{0.5}^1 \frac{3}{2} (1-x^2) \mathrm{d}x & = \frac{3}{2} \int_{0.5}^1 (1-x^2) \mathrm{d}x\\
                                               & = \frac{3}{2} \left((x - \frac{x^3}{3}) \Big|_{0.5}^1\right)\\
                                               & = \frac{3}{2} \left((1 - \frac{1^3}{3}) - ({0.5} - \frac{{0.5}^3}{3})\right)\\
                                               & = \frac{3}{2} \left(\frac{2}{3} - 0.45833\right)\\
                                               & = 0.3125
 \end{split}
\end{align*}

\section*{Chapter 5 Problem 116}

The probability density shown in Figure 5.6 is the log-normal distribution with $\alpha = 8.85$ and $\beta = 1.03$. Find the probability that
\begin{enumerate}
 \item[(a)] the interrequest time is more than $200$ microseconds
 \item[(b)] the interrequest time is less than $300$ microseconds
\end{enumerate}

\subsection*{Part a}

The log-normal function is given by the probability density function

\begin{equation*}
 f(x) = 
 \begin{cases}
  \frac{1}{\sqrt{2\pi}\beta} x^{-1}e^{-(\mathrm{ln}\ x - \alpha)^2 / 2\beta^2} & \text{for $x > 0$, $\beta > 0$}\\
  0                                                                           & \text{elsewhere}
 \end{cases}
\end{equation*}

The book tells us on page 167 that to find the probability that a random variable will take on a value between $a$ and $b$ for $(0 < a < b)$, we must evaluate the integral

\begin{equation*}
 \int_a^b \frac{1}{\sqrt{2\pi}\beta} x^{-1}e^{-(\mathrm{ln}\ x - \alpha)^2 / 2\beta^2} \mathrm{d}x
\end{equation*}

Which the book then simplifies to

\begin{equation*}
 F\left(\frac{\mathrm{ln}\ b - \alpha}{\beta}\right) - F\left(\frac{\mathrm{ln}\ a - \alpha}{\beta}\right)
\end{equation*}

where $F$ is the distribution function of the standard normal distribution.

Using the information above, we can figure out the correct values of $a$ and $b$, and then plug in and solve directly.

Since we want to find the probability that the interrequest time is greater than $200$ microseconds, we must solve using limits $a = 200$ and $b = \infty$. Plugging these in, we get

\begin{equation*}
 \begin{split}
  F\left(\frac{\mathrm{ln}\ \infty - 8.85}{1.03}\right) - F\left(\frac{\mathrm{ln}\ 200 - 8.85}{1.03}\right) & = F(\infty) - F(-3.45)\\
                                                                                       & = 1 - 0.0003\\
                                                                                       & = 0.9997
 \end{split}
\end{equation*}

\subsection*{Part b}

This part can be solved in a similar way to Part a above. The only difference is using different values for $a$ and $b$. Since the log-normal distribution is only valid for $x > 0$, we use $a = 0$ and $b = 300$. Using the book's equation from above, we get

\begin{equation*}
 \begin{split}
  F\left(\frac{\mathrm{ln}\ 300 - 8.85}{1.03}\right) - F\left(\frac{\mathrm{ln}\ 0 - 8.85}{1.03}\right) & = F(-3.05) - F(-\infty)\\
                                                                                  & = 0.0011 - 0.0\\
                                                                                  & = 0.0011
 \end{split}
\end{equation*}

\section*{Chapter 6 Problem 36}

If measurements of the specific gravity of a metal can be looked upon as a sample from a normal population having a standard deviation of $0.04$, what is the probability that the mean of a random sample of size $25$ will be ``off'' by at most $0.02$?

\subsection*{Solution}

From the given information in the problem, we know the following information

\begin{align*}
 \sigma & = 0.04\\
 n & = 25
\end{align*}

So now we find $z$, which is the standardized normal value. Basically, $z$ is a way to convert any normal distribution to a value that we can read off of the standard normal distribution table, which is Table 3.

\begin{equation*}
 \begin{split}
  z = \frac{\Xbar - \mu}{\sigma / \sqrt{n}} & = \frac{\pm 0.02}{0.04 / \sqrt{25}}\\
                                            & = \pm 2.5
 \end{split}
\end{equation*}

Now we need to find the area underneath the standard normal curve, on the interval $(-2.5,2.5)$. To do this, we can look up the values from Table 3 and subtract.

\begin{align*}
 F(-2.5) & = 0.0062\\
 F(2.5) & = 0.9938
\end{align*}

Now, we can subtract $F(-2.5)$ from $F(2.5)$ to get the probability that the mean of a random sample of size $25$ will be ``off'' by at most $0.02$.

\begin{equation*}
 F(2.5) - F(-2.5) = 0.9938 - 0.0062 = 0.9876
\end{equation*}

\section*{Chapter 6 Problem 42}

Explain why the following may not lead to random samples from the desired populations:

\begin{enumerate}
 \item[(a)] To determine the smoothness of shafts, a manufacturer measures the roughness of the first piece made each morning.
 \item[(b)] To determine the mix of cars, trucks, and buses in the rush hour, an engineer records the type of vehicle passing a fixed point at 1-minute intervals.
\end{enumerate}

\subsection*{Part a}

This may not lead to a random sample since since the machine making the shafts could produce a very rough first shaft, but produce much smoother shafts after it has warmed up.

\subsection*{Part b}

This may not lead to a random sample, since trucks and buses are longer than cars. This would mean that trucks and buses are more likely to get included, since they take longer to pass the sensor. This is explained in an example using logs on Page 205 in the book.

\section*{Chapter 7 Problem 80}

While performing a certain task under simulated weightlessness, the pulse rate of 32 astronaut trainees increased on the average by 26.4 beats per minute with a standard deviation of 4.28 beats per minute. What can one assert with 95\% confidence about the maximum error if $x = 26.4$ is used as a point estimate of the true average increase in the pulse rate of astronaut trainees performing the given task.

\subsection*{Solution}

We can easily solve this problem using the definition of the Maximum Error of Estimate, from Page 229 in the book

\begin{equation*}
 E = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}
\end{equation*}

We know the following information

\begin{align*}
 n & = 32\\
 \sigma & = 4.28\\
 \alpha & = 0.05
\end{align*}

The only one that is not directly given in the problem is $\alpha$, but it can easily be derived from the 95\% figure that is given. It is the area outside of the 95\% confidence range, which must be 5\%. In probability form, this is 0.05. The book provides us with estimates of $z_{\alpha/2}$ for very common confidence ranges. It tells us that for $\alpha = 0.05$, $z_{\alpha/2} = 1.96$.\\

So, using this information, we plug in and solve

\begin{equation*}
 \begin{split}
  E & = 1.96 \cdot \frac{4.28}{\sqrt{32}}\\
    & = 1.483
 \end{split}
\end{equation*}

What this information is telling us is that we can be sure that for 95\% of the astronauts trainees in the program, their heart rate will increase by $26.4 \pm 1.483$ beats per minute while they are performing the task.

\section*{Chapter 7 Problem 94}

In an air-pollution study, ozone measurements were taken in a large California city at 5:00 PM. The eight readings (in parts per million) were:

\begin{center}
 7.9\ \ 11.3\ \ 6.9\ \ 12.7\ \ 13.2\ \ 8.8\ \ 9.3\ \ 10.6
\end{center}

Assuming the population sampled is normal, construct a 95\% confidence interval for the corresponding true mean.

\subsection*{Solution}

From this information, we know, or can very easily find, four things

\begin{equation*}
 n = 8
\end{equation*}

\begin{equation*}
 \xbar = \left(\sum_{k=1}^n x_k\right) = \left(\sum_{k=1}^8 x_k\right) = \frac{80.7}{8} = 10.0875
\end{equation*}

\begin{equation*}
 \begin{split}
  s^2 & = \frac{n \cdot \sum_{i=1}^n x_i^2 - \left( \sum_{i=1}^n x_i \right)^2}{n (n-1)}\\
      & = \frac{8 \cdot \sum_{i=1}^8 x_i^2 - \left( \sum_{i=1}^8 x_i \right)^2}{8 (8-1)}\\
      & = \frac{8 \cdot \sum_{i=1}^8 x_i^2 - \left( \sum_{i=1}^8 x_i \right)^2}{8 (8-1)}\\
      & = \frac{8 (848.16) - 6496.36}{8 \cdot 7}\\
      & = 5.16
 \end{split}
\end{equation*}

Which means that $s = \sqrt{s^2} = \sqrt{5.16} = 2.2714$.

And the level of significance, $\alpha = 0.05$.

We can use $\alpha$ to look up the value $t_{\alpha/2}$ in Table 4 of the $t$ distribution. We will use $\nu = n -1$ degrees of freedom.

\begin{equation*}
 t_{\alpha/2} = t_{0.05/2} = t_{0.025} = 2.365
\end{equation*}

Now we can use the equation for small sample confidence interval for $\mu$ from Page 233 in the book. It is

\begin{equation*}
 \xbar - t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} < \mu < \xbar + t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}
\end{equation*}

Substituting in the values we found earlier, we get

\begin{equation*}
 \begin{split}
  10.0875 - 2.365 \cdot \frac{2.2714}{\sqrt{8}} <\ & \mu\ < 10.0875 + 2.365 \cdot \frac{2.2714}{\sqrt{8}}\\
  10.0875 - 1.899 <\ & \mu\ < 10.0875 + 1.899\\
  8.1885 <\ & \mu\ < 11.9865
 \end{split}
\end{equation*}


\section*{Chapter 8 Problem 18}

With reference to the example on page 250, construct a 95\% confidence interval for the true standard deviation of the breaking strength of the given kind of ribbon.

\subsection*{Example from Page 250}

The specifications for a certain kind of ribbon call for a mean breaking strength of 180 pounds. Five pieces of the ribbon (randomly selected from different rolls) have a mean breaking strength of 169.5 pounds with a standard deviation of 5.7 pounds.

\subsection*{Solution}

From the information given, we know the following three things

\begin{align*}
 \xbar & = 169.5\ \text{pounds}\\
 s & = 5.7\ \text{pounds}\\
 \alpha & = 0.05
\end{align*}

From this we can find $\chi_{\alpha/2}^2$ and $\chi_{1-\alpha/2}^2$ by using Table 5, and setting\\
$\nu = n - 1 = 4$.

\begin{align*}
 \begin{split}
  \chi_{\alpha/2}^2 & = \chi_{0.025}^2 = 11.143\\
  \chi_{1-\alpha/2}^2 & = \chi_{0.975}^2 = 0.484
 \end{split}
\end{align*}


To solve this problem, we will use the book's formula from Page 283 for the Confidence interval for $\sigma^2$. This is reproduced below

\begin{equation*}
 \frac{(n-1)s^2}{\chi_{\alpha/2}^2} < \sigma^2 < \frac{(n-1)s^2}{\chi_{1-\alpha/2}^2}
\end{equation*}

Now we can plug in and solve

\begin{equation*}
 \begin{split}
  \frac{(5-1)5.7^2}{11.143} & < \sigma^2 < \frac{(5-1)5.7^2}{0.484}\\
  11.663 & < \sigma^2 < 268.51
 \end{split}
\end{equation*}

Now we can square root the whole thing to find the confidence interval for $\sigma$ instead of $\sigma^2$.

\begin{equation*}
 3.415 < \sigma < 16.386
\end{equation*}

\section*{Chapter 8 Problem 24}

With reference to Exercise 7.69, use the 0.02 level of significance to test the assumption that the two populations have equal variances.

\subsection*{Exercise 7.69}

The follwing are the number of sales which a sample of 9 salespeople of industrial chemicals in California and a sample of 6 salespeople of industrial chemicals in Oregon made over a certain fixed period of time:

\begin{flushleft}
 \hspace{2cm} \textsl{California:} 59\ 68\ 44\ 71\ 63\ 46\ 69\ 54\ 48\\
 \hspace{2cm} \textsl{Oregon:} \hspace{6pt} 50\ 36\ 62\ 52\ 70\ 41
\end{flushleft}

\subsection*{Solution}

To solve this problem we will use the Statistic for test of equality of two variances (normal populations). This is reproduced below

\begin{equation*}
 F = \frac{S_1^2}{S_2^2}
\end{equation*}

$F$ is a random variable having the $F$ distribution with $n_1 - 1$ and $n_2 - 1$ degrees of freedom.\\

The next thing we must do is find the two sample variances. We will use the formula from Page 31, reproduced below

\begin{equation*}
 s^2 = \frac{\sum_{i=1}^n (x_i - \xbar)^2}{n-1}
\end{equation*}

Using this formula, we get the following two sample variances

\begin{align*}
 s_1^2 & = 109\\
 s_2^2 & = 160.97
\end{align*}

Now we can find $F$ using the two sample variations we just found

\begin{equation*}
 F = \frac{109}{160.97} = 0.677
\end{equation*}

Now, we are trying to test the following two hypotheses with the level of significance $\alpha = 0.02$, as given in the problem.

\begin{align*}
 H_0 \hspace{1cm} \sigma_1^2 & = \sigma_2^2\\
 H_a \hspace{1cm} \sigma_1^2 & \not= \sigma_2^2
\end{align*}

The book, on page 287, states that we can reject the null hypothesis $H_0$ if we can show that $F > F_{\alpha/2}(n_1-1, n_2-1)$. Since we found $F$ above, all we need to do now is calculate the right-hand side of the inequality. This can be looked up in Table 6(b).

\begin{equation*}
 F_{\alpha/2}(n_1-1, n_2-1) = F_{0.01}(8, 5) = 10.29
\end{equation*}

Now, we can compare and decide whether or not to reject the null hypothesis $H_0$.

\begin{equation*}
 0.677 \not> 10.29
\end{equation*}

Therefore we can not reject the null hypothesis $H_0$. This means that we can be 98\% sure that the two populations have equal variances.

\section*{Chapter 9 Problem 52}

With reference to exercise 9.51, test the null hypothesis $p = 0.20$ versus the alternative hypothesis $p < 0.20$ at the $0.05$ level of significance.

\subsection*{Exercise 9.51}

In a sample of 100 ceramic pistons made for an experimental diesel engine, 18 were cracked. Construct a 95\% confidence interval for the true proportion of cracked pistons.

\subsection*{Solution}

From the information given, we have the following information

\begin{align*}
 n & = 100\\
 X & = 18\\
 \alpha & = 0.05
\end{align*}

We also know the hypotheses that we are trying to determine

\begin{align*}
 H_0 \hspace{1cm} p & = 0.20\\
 H_a \hspace{1cm} p & < 0.20\\
\end{align*}

Now, since we have a large sample ($n = 100$) we can use the Statistic for large sample test concerning $p$ from Page 299 in the book. It is reproduced here

\begin{equation*}
 Z = \frac{X - np_0}{\sqrt{np_0(1-p_0)}}
\end{equation*}

Now we can plug in and solve. Keep in mind that $p_0$ is the value of $p$ from the null hypothesis $H_0$.

\begin{equation*}
 \begin{split}
  Z & = \frac{X - np_0}{\sqrt{np_0(1-p_0)}}\\
    & = \frac{18 - 100 \cdot 0.20}{\sqrt{100 \cdot 0.20 (1 - 0.20)}}\\
    & = -0.5
 \end{split}
\end{equation*}

The book states that we can reject the null hypothesis $H_0$ if $Z < -z_\alpha$. We will now look up $z_\alpha$ from Table 3.

\begin{equation*}
 z_\alpha = z_{0.05} = 1.645
\end{equation*}

Now, all that is left to do is to perform the comparison

\begin{equation*}
 -(-0.5) < 1.645
\end{equation*}

Since this is true, we can reject the null hypothesis $H_0$. This means that we can be 95\% sure that $p \not= 0.20$.

\section*{Chapter 9 Problem 54}

With reference to exercise 9.53, test the null hypothesis $p = 0.18$ versus the alternative hypothesis $p \not= 0.18$ at the 0.01 level of significance.

\subsection*{Exercise 9.53}

In a random sample of 160 workers exposed to a certain amount of radiation, 24 exerienced some ill effects. \ldots

\subsection*{Solution}

From the information given in the problem we have the following information

\begin{align*}
 n & = 160\\
 X & = 24\\
 \\
 H_0 \hspace{1cm} p & = 0.18\\
 H_a \hspace{1cm} p & \not= 0.18\\
 \alpha & = 0.01
\end{align*}

To complete this problem, we will use the Statistic for large sample test concerning $p$, from page 299 in the book, which is reproduced below

\begin{equation*}
 Z = \frac{X - np_0}{\sqrt{np_0(1-p_0)}}
\end{equation*}

So now we can solve for $Z$, which we will use later, in a comparison to see if we can reject the null hypothesis $H_0$

\begin{equation*}
 \begin{split}
  Z & = \frac{X - np_0}{\sqrt{np_0(1-p_0)}}\\
    & = \frac{24 - 160 \cdot 0.18}{\sqrt{160 \cdot 0.18 (1-0.18)}}\\
    & = \frac{4.92}{4.86}\\
    & = 1.0124
 \end{split}
\end{equation*}

Now the only thing remaining is to test the hypothesis itself. The book states, on Page 299, that we will reject the null hypothesis if one or both of the following conditions are true

\begin{equation*}
 \begin{split}
  Z & < -z_{\alpha/2}\\
  Z & > z_{\alpha/2}
 \end{split}
\end{equation*}

Now we will find $-z_{\alpha/2}$ and $z_{\alpha/2}$ so we can complete the comparison. We will find these by looking them up in Table 3.

\begin{equation*}
 \begin{split}
  -z_{\alpha/2} & = -2.576\\
  z_{\alpha/2} & = 2.576
 \end{split}
\end{equation*}

And now we perform the comparison

\begin{equation*}
 \begin{split}
  Z = 1.0124 & \not< -2.576 = -z_{\alpha/2}\\
  Z = 1.0124 & \not> 2.576 = z_{\alpha/2}
 \end{split}
\end{equation*}

Since at least one, and in fact both, of the comparisons are false, we can reject the null hypothesis $H_0$. This means that we can be 99\% confident that $p \not= 0.18$.

\end{document}