<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\begin{document}
\parindent=0pt

QUESTION

Two firms $X$ and $Y$ manufacture batteries with a
nominal life of 1000 hrs.  It is claimed that those manufactures
by $Y$ have a greater average life.  8 batteries are chosen at
random from each manufacturer and their life-times (in hours above
1000) are:

$\begin{array} {ccrrrrrrrr} X &amp; {} &amp; 34 &amp; -9 &amp; -20 &amp; 11 &amp; -2 &amp; 15
&amp; 1 &amp; 2\\ Y &amp; {} &amp; 40 &amp; -3 &amp; 20 &amp; -2 &amp; 16 &amp; 18 &amp; 10 &amp; 13
\end{array}$

Assuming that the battery life-times are normally distributed show
that it is reasonable to assume that they come from populations
with the same variance.

Test the claim of greater average life of $Y$ batteries (a) using
an appropriate t-test, (b) using ANOVA.


ANSWER

\begin{tabular}{ccccccccc}
 X&amp;34&amp;-9&amp;-20&amp;11&amp;-2&amp;15&amp;1&amp;2\\
 Y&amp;40&amp;-3&amp;20&amp;-2&amp;16&amp;18&amp;10&amp;13
 \end{tabular}

 $H_0:\sigma_1^2=\sigma_2^2\ \ H_1:\sigma_1^2 \neq \sigma_2^2\ \
 \alpha =5\%$\\
 Test 6, comparison of two variances. Assume normal distribution
 $z=\frac{s_1^2}{s_2^2}\sim F_{n_1-1,n_2-1}$

 $\begin{array}{l}
 \overline{x}=4\\
 s_1=16.3183, \ \ n_1=8\\
 \overline{y}=14\\
 s_2=13.5962, \ \ n_2=8\\
 z=\frac{(16.3183)^2}{(13.5962)^2}=1.44\\
 \textrm{not significant}
 \end{array}
 \ \ \
 \begin{array}{c}
 \epsfig{file=641-5-5.eps, width=70mm}
 \end{array}$

 Hence accept variances equal.



  \begin{description}

   \item[(a)]
    $H_0: \mu_1=\mu_2\ \ H_1&lt;\mu_2$\\
    Test 4, comparison of two means, variances unknown but
    equal.\\
    $z=\frac{\overline{x}_1-\overline{x}_2}{\sqrt{\{s^2(\frac{1}{n_1}+\frac{1}{n_2})\}}}
    \sim t_{n_1+n_3}\\
    s^2=\frac{7\times 16.3183^2+7 \times 13.5962^2}{14}\\
    s=15.0190$

    $\begin{array}{l}
    z=\frac{4-14}{15.019\sqrt{\frac{1}{8}+\frac{1}{8}}}\\
    =1.33\\
    \textrm{not significant}\\
    \textrm{Hence accept }H_0.
    \end{array}
    \ \ \
    \begin{array}{c}
    \epsfig{file=641-5-6.eps, width=60mm}
    \end{array}$


   \item[(b)]
    $T_1=32\ \ T_2=112\ \ T=144\ \ C=\frac{(144)^2}{16}=1296\\
    \sum x^2\ \ TSS=4854-1296=3558\ \
    BSS= \frac{(32)^2}{8}+\frac{(112)^2}{8}-1296=400\ \
    WSS=3558-400=3158$. (Check that this is the numerator of
    $s^2=\sum (x-\overline{x})^2+\sum (y-\overline{y})^2$\\
    \hspace*{30mm} Anova table

    \begin{tabular}{cccc}
    source&amp;df&amp;ss&amp;ms\\
    \hline
    between groups&amp;1&amp;400&amp;400\\
    within groups&amp;14&amp;3158&amp;225.57=$\sigma^2$\\
    total&amp;15*3558
    \end{tabular}

    $H_0:\mu_1=\mu_2\ \ H_1 \neq \mu_2\ \ \alpha=10\%$ will test
    $H_0:\mu_1=\mu_2$ against $ H_1 &lt; \mu_2$ at $\alpha=5\%\\
    F_{1,14}=\frac{400}{225,57}=1.77=(1.33)^2$ which is not
    significant.

  \end{description}

\end{document}
</pre></body></html>