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  <question type="description">
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      <text>41.3.1.4 The F-test</text>
    </name>
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      <text><![CDATA[<h5>The \(F\)-test</h5>
<p>In the tests above, we distinguished between the cases of equal and unequal variances of samples chosen from independent normal populations. As you have seen, the analysis changes according to the assumptions made, conclusions reached and recommendations made - accepting or rejecting a null hypothesis for example - may also change. In view of this, we may wish to test in order to decide whether the assumption that the variances \(\sigma_1^2\) and \(\sigma_2^2\) of the independent normal populations shown in the diagram below, may be regarded as equal.</p><img src="@@PLUGINFILE@@/41.3-figure9-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>Essentially, we will test the null hypothesis</p>
<p>\(\qquad\displaystyle H_0:\ \ \sigma_1^2=\sigma_2^2\)</p>
<p>against one of the alternatives</p>
<p>\(\qquad\displaystyle H_1:\ \ \sigma_1^2\ne\sigma_2^2\qquad H_1:\ \ \sigma_1^2&gt;\sigma_2^2\qquad H_1:\ \ \sigma_1^2&lt;\sigma_2^2\)</p>
<p>In order to do this, we use the \(F\)-distribution. The hypothesis test for the equality of two variances \(\sigma_1^2\) and \(\sigma_2^2\) is encapsulated in the following Key Point.</p>

<div class="HELM_keypoint">
<h4>Key Point</h4>
<p>Consider a random sample of size \(n_1\)  taken from a normal population with mean \(\mu_1\)  and variance \(\sigma_1^2\)  and a random sample of size \(n_1\)  taken from a second normal population with mean \(\mu_2\)  and variance  \(\sigma_2^2\).  Denote the respective sample variances by \(S_1^2\)  and \(S_2^2\)  and assume that the populations are independent.  </p>
<p>The ratio</p>
<p>\[\displaystyle F=\frac{S_1^2}{\sigma_1^2}/\frac{S_2^2}{\sigma_2^2}\]</p>
<p>follows an \(F\) distribution in which the numerator has \(n_1-1\)  degrees of freedom and the denominator has \(n_2-1\)  degrees of freedom.</p>
<p> Note that if the null hypothesis \(H_0:\ \sigma_1^2=\sigma_2^2\)  is true, then the value of \(F\) reduces to the ratio of the sample variances and that in this case</p>
<p>\[\displaystyle F=\frac{S_1^2}{S_2^2}\]</p>
</div>

<p><strong>Note</strong></p>
<p>Recall that if a random sample of size \(n_1\) is taken from a normal population with mean \(\mu_1\) and variance \(\sigma_1^2\) and if the sample variance is denoted by \(S_1^2\), the random variable</p>
<p>\(\qquad\displaystyle X_1^2=\frac{(n_1-1)S_1^2}{\sigma_1^2}\)</p>
<p>has a \(\chi^2\) distribution with \(n_1-1\) degrees of freedom. Similarly, if a random sample of size \(n_2\) is taken from a normal population with mean \(\mu_2\) and variance \(\sigma_2^2\) and if the sample variance is denoted by \(S_2^2\), the random variable</p>
<p>\(\qquad\displaystyle X_2^2=\frac{(n_2-1)S_2^2}{\sigma_2^2}\)</p>
<p>has a \(\chi^2\) distribution with \(n_2-1\) degrees of freedom. This means that the ratio</p>
<p>\(\qquad\displaystyle F=\frac{S_1^2}{\sigma_1^2}/\frac{S_2^2}{\sigma_2^2}\)</p>
<p>is a ratio of \(\chi^2\) random variables with \(n_1-1\) degrees of freedom in the numerator and \(n_2-1\) degrees of freedom in the denominator. Under the null hypothesis</p>
<p>\(\qquad\displaystyle H_0:\ \ \sigma_1^2=\sigma_2^2\)</p>
<p>we know that the expression for \(F\) reduces to</p>
<p>\(\qquad\displaystyle F=\frac{S_1^2}{S_2^2}\)</p>
<p>and we say that \(F\) has an \(F\)-distribution with \(n_1-1\) degrees of freedom in the numerator and \(n_2-1\) degrees of freedom in the denominator. This distribution is denoted by</p>
<p>\(\qquad\displaystyle F_{n_1-1, n_2-1}\)</p>
<p>and some tabulated values are given in Tables 3 and 4 at the end of this section.</p>
<p>If you check Tables 3 and 4, you will find that only right-tail values are given. The left-tail values are calculated by using the following formula:</p>
<p>\(\qquad\displaystyle f_{{1-\alpha},\ n_1-1,\ n_2-1}=\frac{1}{ f_{\alpha,\ n_2-1,\ n_1-1}}\)</p>
<p>Note the reversal in the order in which the expressions for the number of degrees of freedom occur. </p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>The following is an extract from the \(F\)-distribution tables (5% tail) given at the end of this Workbook. </p><img src="@@PLUGINFILE@@/431.3-figure10-fdistribution.svg" alt="The f-distribution with 5% tail and the corresponding table of degrees of freedom" width="575" height="298" class="img-fluid atto_image_button_text-bottom"><br>
<p style="text-align: center">Write down or calculate as appropriate, the following values of \(F\) from the table:</p>
<table class="HELM_table">
  <thead>
    <tr><th><b>Right-tail Values</b></th>  <th><b>Left-tail Values</b></th></tr></thead>
<tbody>
    <tr>
      <td>\[f_{0.05,4,3}\]</td> <td>\[f_{0.95,4,3}\]</td></tr>
<tr><td>\[f_{0.05,8,2}\]</td> <td>\[f_{0.95,8,2}\] </td></tr>
<tr><td>\[f_{0.05,7,8}\]</td> <td>\[f_{0.95,7,8}\]</td></tr></tbody>
</table>

<h4 class="HELM_solution">Solution</h4>
<p>The right-tail values are read directly from the tables.  The left-tail values are calculated
using the formula given above.


</p><table class="HELM_table">
  <thead>
    <tr><th><b>Right-tail Values</b></th>  <th><b>Left-tail Values</b></th></tr></thead>
<tbody>
    <tr>
      <td>\[f_{0.05,4,3} = 9.12\]</td> <td>\[f_{0.95,4,3} = \frac{1}{f_{0.05,3,4}} = \frac{1}{6.59} = 0.152\]</td></tr>
<tr><td>\[f_{0.05,8,2} = 19.37\]</td> <td>\[f_{0.95,8,2} = \frac{1}{f_{0.05,2,8}} = \frac{1}{4.46} = 0.224 \]</td></tr>
<tr><td>\[f_{0.05,7,8} = 3.50 \]</td> <td>\[f_{0.95,7,8} = \frac{1}{f_{0.05,8,7}} = \frac{1}{3.73} = 0.268 \]</td></tr></tbody>
</table>
<p></p>]]></text>
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<file name="431.3-figure10-fdistribution.svg" path="/" encoding="base64">[encoded data, 179328 characters]</file>
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