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      <text>41.3.1.3 Two independent populations each with unknown variance continued</text>
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      <text><![CDATA[<p><strong>Case (ii) - Unknown and unequal variances</strong></p>
<p>In this case we will take \(\sigma_1^2\ne\sigma_2^2\) so that in diagrammatic form the populations may be represented as shown below. </p><img src="@@PLUGINFILE@@/41.3-figure8-unequalvariances.svg" alt="2 populations with unequal means and unequal variances graphed" width="569" height="158" class="img-fluid atto_image_button_text-bottom"><br></p>

<p>Again, when we look at hypothesis testing using two means, we will be considering the difference \(\mu_1-\mu_2\) of the means and writing null hypotheses of the form</p>
<p>\(\qquad\displaystyle H_0:\ \ \mu_1-\mu_2=\) Value</p>
<p>and again Value will often be zero and we will be trying to detect whether there is any statistically significant difference between the means.</p>
,p>In the case where we assume unequal variances, there is no exact statistic which we can use to test the validity or otherwise of the null hypothesis \(H_0:\ \ \mu_1-\mu_2=\) Value. However, the following approximation in the Key Point allows us to overcome this problem.</p>

<div class="HELM_keypoint">
<h4>Key Point</h4>
Provided that the null hypothesis is true, the statistic 

\(\displaystyle T=\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{S_c\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\)
will it <i>approximately</i> follow Student's \(t\)-distribution with the number of degrees of freedom given by the expression:
\[
\nu=\dfrac{\left(\dfrac{S_1^2}{n_1}+\dfrac{S_2^2}{n_2}\right)^2}{\dfrac{\left(\dfrac{S_1^2}{n_1}
\right)^2}{n_1-1}+
\dfrac{\left(\dfrac{S_2^2}{n_2}\right)^2}{n_2-1}}\]
</div>

<p>Essentially, this means that the actual test procedure is similar to that used previously but with \(T\) and the number of degrees of freedom \(\nu\) calculated using the above formulae.
We are now ready to apply these formulae to practical problems in which random samples of different sizes with unknown and unequal variances are drawn from independent normal populations. We will illustrate the test procedure by reworking an Example and Task done previously but we will assume unequal rather than equal variances.
This next Example is a repeat of the previous example but here assuming unequal variances.</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 are 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>On the assumption that the populations from which the samples are drawn <strong>do not have equal variances</strong>, should the manufacturer replace the old circuit design by the new one? Use the 5% level of significance. </p>
<h4 class="HELM_solution">Solution</h4>
<p>If the average current flows are represented by \(\mu_1\)  and \(\mu_2\)   we form the hypotheses</p>
<p>\(\qquad\displaystyle H_0:\ \ \mu_1-\mu_2=0\qquad H_1:\ \ \mu_1-\mu_2\ne 0\)</p>
<p>The sample means are \(\bar X_1=82.42\)  and  \(\bar X_2=83.72\).</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>\(\qquad\displaystyle T=\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}}
=\frac{82.42-83.72}{\sqrt{\frac{2.00}{10}+\frac{2.72}{12}}}=-\frac{1.3}{\sqrt{0.427}}=-1.990\)</p>
<p>The number of degrees of freedom is given by</p>
<p>\(\begin{align}
\nu&=\dfrac{\left(\dfrac{S_1^2}{n_1}+\dfrac{S_2^2}{n_2}\right)^2}
{\dfrac{\left(\dfrac{S_1^2}{n_1}\right)^2}{n_1+1}+
\dfrac{\left(\dfrac{S_2^2}{n_2}\right)^2}{n_2+1}} - 2 \\ &=
\dfrac{\left(\dfrac{2.00}{10}+\dfrac{2.72}{12}\right)^2}{\dfrac{(2.00/10)^2}{11}
+\dfrac{(2.72/12)^2}{13}} - 2 =\dfrac{0.182}{0.004+0.004}\approx
21\end{align}\)</p>

<p>From \(t\)-tables, the critical values (two-tailed test, 5% level of significance) are  \(\pm 2.080\).  Since   \(-2.080&lt;-1.990&lt;2.080\) we conclude that there is insufficient evidence to reject the null hypothesis in favour of the alternative at the 5% level of significance.</p>]]></text>
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<file name="41.3-figure8-unequalvariances.svg" path="/" encoding="base64">[encoded data, 23736 characters]</file>
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