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      <text>41.3.1.1 Two independent populations each with a known variance</text>
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      <text><![CDATA[<h3>Tests concerning two samples<br></h3>
<h5>Two independent populations each with a known variance</h5>
<p>We assume that the populations are normally distributed. This may not always be true and you should note this basic assumption while studying this Section of the Workbook.
  A standard notation often used to describe the populations and samples is:
</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>If you are not familiar with the double suffix notation used to represent the samples, simply remember that a random sample of size \(n_1\) is drawn from \(X_1\sim N(\mu_1,\sigma^2_1)\) and a random sample of size \(n_2\) is drawn from \(X_2\sim N(\mu_2,\sigma^2_2)\). In diagrammatic form the populations may be represented as follows: &nbsp;</p><p><img src="@@PLUGINFILE@@/41.3-figure6-populations.svg" alt="2 populations graphed, population 1 with lower mean and variance, and population 2 with higher mean and variance" width="569" height="158" class="img-fluid atto_image_button_text-bottom"><br></p>

<p>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
\(\qquad\displaystyle H_0:\ \ \mu_1-\mu_2=\) Value&nbsp;</p>

<p>As you might expect, Value will often be zero and we will be trying to detect whether there is any statistically significant evidence of a difference between the means.</p>
<p>We know, from our previous work on continuous distributions 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>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> since \(\bar X_1\) and \(\bar X_2\) are independent. Given the assumptions made we can assert that the quantity \(Z\) defined by</p>
 <p>\(\qquad\displaystyle Z=\dfrac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt {\dfrac{\sigma^2_1}{n_1}+\dfrac{\sigma^2_2}{n_2}}}\) </p>
<p> follows the standard normal distribution \(N(0,1)\).</p>
<p> We are now ready to apply this formula to practical problems in which random samples of different sizes are drawn from normal populations. The conditions for the rejection of \(H_0\) at the 5% and the 1% levels of significance are exactly the same as those previously used for single sample problems.</p>
 
<hr>
<h4 class="HELM_example">Example</h4>
<p>A motor manufacturer wishes to replace steel suspension components by aluminium components to save weight and thereby improve performance and fuel consumption. Tensile strength tests are carried out on randomly chosen samples of two possible components before a final choice is made. The results are:</p>

<table class="HELM_table">
  <thead>
    <tr><th><strong>Component Number</strong></th>
<th><strong>Sample Size</strong></th>
<th><strong>Mean Tensile Strength (kg mm\(^{-2})\)</strong></th>
<th><strong>Standard Deviation(kg mm\(^{-2})\)</strong></th> </tr> </thead>
 <tbody>
    <tr>
      <td>1</td> <td>15</td> <td>90</td> <td>2.3</td></tr>
<tr><td>2</td> <td>10</td> <td>88</td> <td>2.2</td></tr></tbody>
</table>

<p>Is there any difference between the measured tensile strengths at the 5% level of significance?</p>
<h4 class="HELM_solution">Solution</h4>
<p>The null and alternative hypotheses are:</p>
<p>\(\qquad\displaystyle H_0:\ \ \mu_1-\mu_2=0\qquad H_1:\ \ \mu_1-\mu_2\ne 0\)</p>
<p>The null hypothesis represents the statement  'there is no difference in the tensile strengths of the two components.' The test statistic \(Z\) is calculated as:</p>
<p>\(\begin{align}
Z &=\dfrac{(\bar{X_1} - \bar{X_2}) - (\mu_1-\mu_2)}{\sqrt{\dfrac{\sigma^2_1}{n_1}+\dfrac{\sigma^2_2}{n_2}}} \\\\
&=\dfrac{(90-88)-(0)} {\sqrt{\dfrac{2.3^2}{15}+\dfrac{2.2^2}{10}}}\\\\
&=\dfrac{2}{\sqrt{0.3527+0.484}}\\\\
&=2.186
\end{align}\)</p><br>

<p> Since \(2.186 &gt; 1.96\) we conclude that, on the basis of the (limited) evidence available, there is a difference in tensile strength between the components tested.  The manufacturer should carry out more comprehensive tests before making a final decision as to which component to use.  The decision is a serious one with safety implications as well as economic implications.  As well as carrying out more tests the manufacturer should consider the level of rejection of the null hypothesis, perhaps using 1% instead of 5%.  Component 1 appears to be stronger but this may not be the case after more tests are carried out.</p>]]></text>
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