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      <text>41.3.1.2 Two independent populations each with unknown variance</text>
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      <text><![CDATA[<p><h5>Two independent populations each with an unknown variance</h5></p>
<p>Again we assume that the populations are normally distributed and use the same standard notation used previously to describe the populations and samples, namely:</p>
<table class="HELM_table">
  <thead>
    <tr><th><strong>Population</strong></th>
<th><strong>Sample</strong></th></tr></thead>
 <tbody>
    <tr>
      <td>\[X_1\sim N(\mu_1,\sigma^2_1)\]</td> <td>\[x_{11},x_{12},x_{13},\cdots ,x_{1n_1}\] with \(n_1\) members.</td></tr>
<tr><td>\[X_2\sim N(\mu_2,\sigma^2_2)\]</td> <td>\[x_{21},x_{22},x_{23},\cdots ,x_{2n_2}\] with \(n_2\) members.</td></tr></tbody>
</table>

<p>There are two distinct cases to consider. Firstly, we will assume that although the variances are unknown, they are in fact equal. Secondly, we will assume that the unknown variances are not necessarily equal.</p>

<p><strong>Case (i) - Unknown but equal variances</strong></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>We will take \(\sigma^2_1=\sigma^2_2=\sigma^2\) so that in diagrammatic form the populations are:&nbsp;</p><p><img src="@@PLUGINFILE@@/41.3-figure7-equalvariance.svg" alt="2 normally distributed populations with different means but the same variance graphed" width="569" height="159" class="img-fluid atto_image_button_text-bottom"><br></p>

<p>The results from our work on continuous distributions tell us that:</p>
<p>\(\qquad\displaystyle \text{E}(\bar X_1-\bar X_2)=\text{E}(\bar X_1)-\text{E}(\bar X_2)=\mu_1-\mu_2\) </p>
<p>as before, and that</p>
<p>\(\qquad\displaystyle \text{V}(\bar X_1-\bar X_2)=\text{V}(\bar X_1)-\text{V}(\bar X_2)=\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}\)</p>
<p>Given that we do not know the value of \(\sigma\), we must estimate it. This is done by combining (or pooling) the sample variances say \(S_1^2\) and \(S_2^2\) for samples 1 and 2 respectively according to the formula:</p>
<p>\(\qquad\displaystyle S_c^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}\) </p>
<p>Notice that</p>
<p>\(\qquad\displaystyle S_c^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}=\frac{(n_1-1)S_1^2}{n_1+n_2-2}+ \frac{(n_2-1)S_2^2}{n_1+n_2-2}\)</p>
<p>so that you can see that \(S_c^2\) is a weighted average of \(S_1^2\) and \(S_2^2\). In fact, each sample variance is weighted according to the number of degrees of freedom available. </p> 
<p> Notice also that the first sample contributes \(n_1-1\) degrees of freedom and the second sample contributes \(n_2-1\) degrees of freedom so that \(S_c^2\) has \(n_1+n_2-2\) degrees of freedom.</p>
<p>Since we are estimating unknown variances, the quantity \(T\) defined by</p>
<p>\(\qquad\displaystyle T=\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{S_c\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\) </p>
<p>will follow Student’s \(t\)-distribution with \(n_1+n_2-2\) degrees of freedom.</p>
<p>We are now ready to apply this formula to practical problems in which random samples of different sizes with unknown but equal variances are drawn from independent normal populations. The conditions for the rejection of \(H_0\) at the 5% and the 1% levels of significance are found from tables of the \(t\)-distribution (Table 2), a copy of which is included to the end of this Section.</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 following results are obtained.</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>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 pooled estimate of the variance is </p>
<p>\(\qquad S_c^2=\dfrac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}=\dfrac{9\times 2.00+11\times 2.72}{20}=2.396\) </p>
<p> The test statistic is</p>
<p>\(\qquad\displaystyle T=\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{S_c\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} = \frac{82.42-83.72}{\sqrt{2.396}\sqrt{\frac{1}{10}+\frac{1}{12}}}=-1.267.\)</p>
<p> From \(t\)-tables, the critical values with 20 degrees of freedom and a two-tailed test are \(\pm 2.086\). Since \(-2.086 &lt; -1.267 &lt; 2.086\) we conclude that we cannot reject the null hypothesis in favour of the alternative. A 95% confidence interval for the difference between the mean currents is given by \(\displaystyle \hat{x_1} − \hat{x}_2 \pm 2.086 \times S_c\sqrt{ \frac{1}{n_1}+ \frac{1}{n_2}}\). The confidence interval is \(-2.683 &lt; \mu_1 - \mu_2 &lt; 0.083\).</p>]]></text>
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