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      <text>41.3.1.5 The F-test continued</text>
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      <text><![CDATA[<p>We are now in a position to use the \(F\)-test to solve engineering problems. The application of the \(F\)-test will be illustrated by using the data given in a previous worked example in order to determine whether the assumption of equal variability in the samples used is realistic. This next Example was met previously. Here we test one of the underlying assumptions.</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>A manufacturer of electronic equipment has developed a circuit to feed current to a particular component in a computer display screen. While the new design is cheaper to manufacture, it can only be adopted for mass production if it passes the same average current to the component. In tests involving the two circuits, the results obtained are</p>
<table class="HELM_table">
  <thead>
    <tr><th><strong>Test Number</strong></th>
<th><strong>Circuit 1 - Current (mA)</strong></th>
<th><strong>Circuit 2 - Current (mA)</strong></th></tr></thead>
<tbody>
    <tr>
      <td>1</td> <td>80.1</td> <td>80.7</td></tr>
<tr><td>2</td> <td>82.3</td> <td>81.3</td></tr>
<tr><td>3</td> <td>84.1</td> <td>84.6</td></tr>
<tr><td>4</td> <td>82.6</td> <td>81.7</td></tr>
<tr><td>5</td> <td>85.3</td> <td>86.3</td></tr>
<tr><td>6</td> <td>81.3</td> <td>84.3</td></tr>
<tr><td>7</td> <td>83.2</td> <td>83.7</td></tr>
<tr><td>8</td> <td>81.7</td> <td>84.7</td></tr>
<tr><td>9</td> <td>82.2</td> <td>82.8</td></tr>
<tr><td>10</td> <td>81.4</td> <td>84.4</td></tr>
<tr><td>11</td> <td> </td>    <td>85.2</td></tr>
<tr><td>12</td> <td> </td>    <td>84.9</td></tr></tbody>
</table>
<p>In a previous example we worked on the assumption that the populations from which the samples are drawn have equal variances. Is this assumption valid at the 5% level of significance? Note that the manufacturer may also be interested in knowing whether the variances are equal as well as the means. We shall not address that problem here but it can be argued that equality of variances will facilitate consistent performance from the components.</p>
<h4 class="HELM_solution">Solution</h4>
<p>We form the hypotheses</p>
<p>\(\qquad\displaystyle H_0:\ \ \sigma_1^2=\sigma_2^2 \qquad\displaystyle H_1:\ \ \sigma_1^2\ne\sigma_2^2\)</p>
<p>and perform a two-tailed test.</p>

<p>The sample variances are  \(S_1^2=2.00\)  and  \(S_2^2=2.72\).</p>
 
<p>The test statistic is </p>
<p>\(\qquad\displaystyle F=\frac{S_1^2}{S_2^2}=\frac{2.00}{2.72}=0.735\)</p>
<p>which has an \(F\)-distribution with 9 degrees of freedom in the numerator and 11 degrees of freedom in the denominator.</p>
 
<p>We require two 2.5% tails, that is we require right-tail \(f_{0.025,9,11}=3.59\) and left-tail  \(f_{0.975,9,11}\).  The latter may be approximated as follows:</p>
<p>\(\begin{align}
f_{0.975,9,11} = \frac{1}{f_{0.025,11,9}} & \approx \frac{ \left(\frac{1}{11} - \frac{1}{20} \right) 
\frac{1}{f_{0.025,10,9}} + \left( \frac{1}{10} - \frac{1}{11} \right) \frac{1}{f_{0.025,20,9}} }{\left( \frac{1}{10} - \frac{1}{20} \right) } \\
  & \approx  \frac{ \frac{0.040909}{3.96} + \frac{0.009091}{3.67} }{ 0.05} \\
  & \approx \frac{0.81818}{3.96} + \frac{0.18182}{3.67} = 0.256
\end{align}\)</p>
Since \(0.256&lt;0.735&lt;3.59\)  we conclude that we cannot reject the null hypothesis in favour of the alternative at the 5% level of significance.  The evidence supports the conclusion that the samples have equal variability</p>
<hr>

<p>Note that we can adopt the rule (many statisticians do this) of always dividing the larger \(S^2\) value by the smaller \(S^2\) value so that you only need to look up right tail values.</p>

<p><img src="@@PLUGINFILE@@/41.3-table1-normal.svg" alt="The normal probability distribution table" width="577" height="924" class="img-fluid atto_image_button_text-bottom"><br></p><p><img src="@@PLUGINFILE@@/41.3-table2-tdistribution.svg" alt="Table of t-distribution values" width="594" height="883" class="img-fluid atto_image_button_text-bottom"><br></p><p><img src="@@PLUGINFILE@@/41.3-table3-fdistribution.svg" alt="Table of degrees of freedom for the F-distribution with a 5% tail" width="644" height="762" class="img-fluid atto_image_button_text-bottom"><br></p><p><img src="@@PLUGINFILE@@/41.3-table4-fdistribution2.svg" alt="Table of degrees of freedom for the F-distribution with a 2.5% tail" width="684" height="806" class="img-fluid atto_image_button_text-bottom"><br></p>]]></text>
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