<?xml version="1.0" encoding="utf-8"?>
<document xmlns="http://cnx.rice.edu/cnxml" xmlns:m="http://www.w3.org/1998/Math/MathML" xmlns:md="http://cnx.rice.edu/mdml/0.4" xmlns:bib="http://bibtexml.sf.net/" xmlns:q="http://cnx.rice.edu/qml/1.0" id="new" module-id="" cnxml-version="0.6">
  <title>F Distribution and ANOVA: Test of Two Variances</title>
  <metadata xmlns:md="http://cnx.rice.edu/mdml/0.4">
  <!-- WARNING! The 'metadata' section is read only. Do not edit below.
       Changes to the metadata section in the source will not be saved. -->
  <md:content-id>m17075</md:content-id>
  <md:title>F Distribution and ANOVA: Test of Two Variances</md:title>
  <md:version>1.6</md:version>
  <md:created>2008/06/24 12:34:21 GMT-5</md:created>
  <md:revised>2009/02/05 21:31:47.599 US/Central</md:revised>
  <md:authorlist>
    <md:author id="sdean">
        <md:firstname>Susan</md:firstname>
        <md:surname>Dean</md:surname>
        <md:fullname>Susan Dean</md:fullname>
        <md:email>deansusan@deanza.edu</md:email>
    </md:author>
    <md:author id="billowsky">
        <md:firstname>Barbara</md:firstname>
        <md:surname>Illowsky</md:surname>
        <md:fullname>Dr. Barbara Illowsky</md:fullname>
        <md:email>illowskybarbara@deanza.edu</md:email>
    </md:author>
  </md:authorlist>
  <md:maintainerlist>
    <md:maintainer id="sdean">
        <md:firstname>Susan</md:firstname>
        <md:surname>Dean</md:surname>
        <md:fullname>Susan Dean</md:fullname>
        <md:email>deansusan@deanza.edu</md:email>
    </md:maintainer>
    <md:maintainer id="billowsky">
        <md:firstname>Barbara</md:firstname>
        <md:surname>Illowsky</md:surname>
        <md:fullname>Dr. Barbara Illowsky</md:fullname>
        <md:email>illowskybarbara@deanza.edu</md:email>
    </md:maintainer>
    <md:maintainer id="cnxorg">
        <md:firstname/>
        <md:surname>Connexions</md:surname>
        <md:fullname>Connexions</md:fullname>
        <md:email>cnx@cnx.org</md:email>
    </md:maintainer>
  </md:maintainerlist>
  <md:license href="http://creativecommons.org/licenses/by/2.0/"/>
  <md:licensorlist>
    <md:licensor id="MaxfieldFoundation">
        <md:firstname/>
        <md:surname>Maxfield Foundation</md:surname>
        <md:fullname>Maxfield Foundation</md:fullname>
        <md:email>cnx@cnx.org</md:email>
    </md:licensor>
  </md:licensorlist>
  <md:keywordlist>
    <md:keyword>ANOVA</md:keyword>
    <md:keyword>degrees of freedom</md:keyword>
    <md:keyword>F Distribution</md:keyword>
    <md:keyword>normal distribution</md:keyword>
    <md:keyword>populations</md:keyword>
    <md:keyword>sample</md:keyword>
    <md:keyword>statistics</md:keyword>
    <md:keyword>Test of Two Variances</md:keyword>
    <md:keyword>variance</md:keyword>
  </md:keywordlist>
  <md:subjectlist>
    <md:subject>Mathematics and Statistics</md:subject>
  </md:subjectlist>
  <md:abstract>This module provides the assumptions to be considered in order to calculate a Test of Two Variances and how to execute the Test of Two Variances. An example is provided to help clarify the concept.</md:abstract>
  <md:language>en</md:language>
  <!-- WARNING! The 'metadata' section is read only. Do not edit above.
       Changes to the metadata section in the source will not be saved. -->
</metadata>
  <content>
    <para id="delete_me">Another of the uses of the F distribution is testing two variances. It is often desirable
to compare two variances rather than two averages. For instance, college
administrators would like two college professors grading exams to have the same
variation in their grading. In order for a lid to fit a container, the variation in the lid
and the container should be the same. A supermarket might be interested in the
variability of check-out times for two checkers.</para><para id="element-114">In order to perform a F test of two variances, it is important that the following are true:
<list id="list1" list-type="enumerated">
<item>The populations from which the two samples are drawn are normally distributed.</item>
<item>The two populations are independent of each other.</item>
</list></para><para id="element-503">Suppose we sample randomly from two independent normal populations. Let <m:math><m:msubsup><m:mi>σ</m:mi><m:mn>1</m:mn><m:mn>2</m:mn></m:msubsup></m:math> and
<m:math><m:msubsup><m:mi>σ</m:mi><m:mn>2</m:mn><m:mn>2</m:mn></m:msubsup></m:math>
be the population variances and <m:math><m:msubsup><m:mi>s</m:mi><m:mn>1</m:mn><m:mn>2</m:mn></m:msubsup></m:math> and <m:math><m:msubsup><m:mi>s</m:mi><m:mn>2</m:mn><m:mn>2</m:mn></m:msubsup></m:math> be the sample variances. Let the
sample sizes be <m:math><m:msub><m:mi>n</m:mi><m:mn>1</m:mn></m:msub></m:math> and <m:math><m:msub><m:mi>n</m:mi><m:mn>2</m:mn></m:msub></m:math>. Since we are interested in comparing the two sample
variances, we use the F ratio</para><para id="element-321"><m:math>
	<m:mi>F</m:mi><m:mo>=</m:mo>
	<m:mfrac>
		<m:mrow>
			<m:mo>[</m:mo>
			<m:mfrac>
				<m:mrow>
					<m:mo>(</m:mo>
   					<m:msub><m:mi>s</m:mi><m:mn>1</m:mn></m:msub>
					<m:msup><m:mo>)</m:mo><m:mn>2</m:mn></m:msup>
				</m:mrow>
				<m:mrow>
					<m:msup>
						<m:mrow>
							<m:mo>(</m:mo>
							<m:msub><m:mi>σ</m:mi><m:mn>1</m:mn></m:msub>
							<m:mo>)</m:mo>
						</m:mrow>
						<m:mn>2</m:mn>
					</m:msup>
				</m:mrow>
			</m:mfrac>
			<m:mo>]</m:mo>
		</m:mrow>
		<m:mrow>
			<m:mo>[</m:mo>
			<m:mfrac>
				<m:mrow><m:mo>(</m:mo>
					<m:msub><m:mi>s</m:mi><m:mn>2</m:mn></m:msub>
					<m:msup><m:mo>)</m:mo><m:mn>2</m:mn></m:msup>
				</m:mrow>
				<m:mrow>
					<m:msup>
						<m:mrow>
							<m:mo>(</m:mo>
							<m:msub><m:mi>σ</m:mi><m:mn>2</m:mn></m:msub>
							<m:mo>)</m:mo>
						</m:mrow>
						<m:mn>2</m:mn>
					</m:msup>
				</m:mrow>
			</m:mfrac>
			<m:mo>]</m:mo>
		</m:mrow>
	</m:mfrac>
</m:math></para><para id="element-648"><m:math><m:mi>F</m:mi></m:math> has the distribution <m:math><m:mi>F</m:mi></m:math> ~ <m:math><m:mi>F</m:mi><m:mo>(</m:mo><m:msub><m:mi>n</m:mi><m:mn>1</m:mn></m:msub><m:mo>-</m:mo><m:mn>1</m:mn><m:mo>,</m:mo><m:msub><m:mi>n</m:mi><m:mn>2</m:mn></m:msub><m:mo>-</m:mo><m:mn>1</m:mn><m:mo>)</m:mo></m:math> </para><para id="element-266">where <m:math><m:msub><m:mi>n</m:mi><m:mn>1</m:mn></m:msub><m:mo>-</m:mo><m:mn>1</m:mn></m:math> are the degrees of freedom for the numerator and <m:math><m:msub><m:mi>n</m:mi><m:mn>2</m:mn></m:msub><m:mo>-</m:mo><m:mn>1</m:mn></m:math> are the
degrees of freedom for the denominator.</para><para id="eip-373">If the null hypothesis is <m:math><m:msubsup><m:mi>σ</m:mi><m:mn>1</m:mn><m:mn>2</m:mn></m:msubsup><m:mo>=</m:mo><m:msubsup><m:mi>σ</m:mi><m:mn>2</m:mn><m:mn>2</m:mn></m:msubsup></m:math>, then the F-Ratio becomes  
<m:math>
	<m:mi>F</m:mi><m:mo>=</m:mo>
	<m:mfrac>
		<m:mrow>
			<m:mo>[</m:mo>
			<m:mfrac>
				<m:mrow>
					<m:mo>(</m:mo>
   					<m:msub><m:mi>s</m:mi><m:mn>1</m:mn></m:msub>
					<m:msup><m:mo>)</m:mo><m:mn>2</m:mn></m:msup>
				</m:mrow>
				<m:mrow>
					<m:msup>
						<m:mrow>
							<m:mo>(</m:mo>
							<m:msub><m:mi>σ</m:mi><m:mn>1</m:mn></m:msub>
							<m:mo>)</m:mo>
						</m:mrow>
						<m:mn>2</m:mn>
					</m:msup>
				</m:mrow>
			</m:mfrac>
			<m:mo>]</m:mo>
		</m:mrow>
		<m:mrow>
			<m:mo>[</m:mo>
			<m:mfrac>
				<m:mrow><m:mo>(</m:mo>
					<m:msub><m:mi>s</m:mi><m:mn>2</m:mn></m:msub>
					<m:msup><m:mo>)</m:mo><m:mn>2</m:mn></m:msup>
				</m:mrow>
				<m:mrow>
					<m:msup>
						<m:mrow>
							<m:mo>(</m:mo>
							<m:msub><m:mi>σ</m:mi><m:mn>2</m:mn></m:msub>
							<m:mo>)</m:mo>
						</m:mrow>
						<m:mn>2</m:mn>
					</m:msup>
				</m:mrow>
			</m:mfrac>
			<m:mo>]</m:mo>
		</m:mrow>
	</m:mfrac>
<m:mo>=</m:mo><m:mfrac>
	<m:mrow>
		<m:mo>(</m:mo>
		<m:msub>
			<m:mi>s</m:mi><m:mn>1</m:mn>
		</m:msub>
		<m:msup>
			<m:mo>)</m:mo><m:mn>2</m:mn>
		</m:msup>
	</m:mrow>

	<m:mrow>
		<m:msup>
			
                        <m:mrow>
				<m:mo>(</m:mo>
				<m:msub>
					<m:mi>s</m:mi><m:mn>2</m:mn>
				</m:msub>
                                <m:mo>)</m:mo>
			</m:mrow>
			<m:mn>2</m:mn>
		</m:msup>
	</m:mrow>
</m:mfrac>
</m:math>.  

</para><para id="eip-666">If the two populations have equal variances, then 

<m:math><m:msubsup><m:mi>s</m:mi><m:mn>1</m:mn><m:mn>2</m:mn></m:msubsup></m:math> and <m:math><m:msubsup><m:mi>s</m:mi><m:mn>2</m:mn><m:mn>2</m:mn></m:msubsup></m:math> are close in value and 

<m:math><m:mi>F</m:mi><m:mo>=</m:mo></m:math> 
<m:math><m:mfrac>
	<m:mrow>
		<m:mo>(</m:mo>
		<m:msub>
			<m:mi>s</m:mi><m:mn>1</m:mn>
		</m:msub>
		<m:msup>
			<m:mo>)</m:mo><m:mn>2</m:mn>
		</m:msup>
	</m:mrow>

	<m:mrow>
		<m:msup>
			
                        <m:mrow>
				<m:mo>(</m:mo>
				<m:msub>
					<m:mi>s</m:mi><m:mn>2</m:mn>
				</m:msub>
                                <m:mo>)</m:mo>
			</m:mrow>
			<m:mn>2</m:mn>
		</m:msup>
	</m:mrow>
</m:mfrac>
</m:math> is close to <m:math><m:mn>1</m:mn></m:math>.  
But if the two population variances are very different, <m:math><m:msubsup><m:mi>s</m:mi><m:mn>1</m:mn><m:mn>2</m:mn></m:msubsup></m:math> and <m:math><m:msubsup><m:mi>s</m:mi><m:mn>2</m:mn><m:mn>2</m:mn></m:msubsup></m:math> 
tend to be very different, too.  

Choosing  
<m:math><m:msubsup><m:mi>s</m:mi><m:mn>1</m:mn><m:mn>2</m:mn></m:msubsup></m:math> as the larger sample variance causes the ratio 
 
<m:math><m:mfrac>
	<m:mrow>
		<m:mo>(</m:mo>
		<m:msub>
			<m:mi>s</m:mi><m:mn>1</m:mn>
		</m:msub>
		<m:msup>
			<m:mo>)</m:mo><m:mn>2</m:mn>
		</m:msup>
	</m:mrow>

	<m:mrow>
		<m:msup>
			
                        <m:mrow>
				<m:mo>(</m:mo>
				<m:msub>
					<m:mi>s</m:mi><m:mn>2</m:mn>
				</m:msub>
                                <m:mo>)</m:mo>
			</m:mrow>
			<m:mn>2</m:mn>
		</m:msup>
	</m:mrow>
</m:mfrac>
</m:math> to be greater than <m:math><m:mn>1</m:mn></m:math>. 

If <m:math><m:msubsup><m:mi>s</m:mi><m:mn>1</m:mn><m:mn>2</m:mn></m:msubsup></m:math> and <m:math><m:msubsup><m:mi>s</m:mi><m:mn>2</m:mn><m:mn>2</m:mn></m:msubsup></m:math> are far apart, then 

<m:math><m:mi>F</m:mi><m:mo>=</m:mo></m:math>
<m:math><m:mfrac>
	<m:mrow>
		<m:mo>(</m:mo>
		<m:msub>
			<m:mi>s</m:mi><m:mn>1</m:mn>
		</m:msub>
		<m:msup>
			<m:mo>)</m:mo><m:mn>2</m:mn>
		</m:msup>
	</m:mrow>

	<m:mrow>
		<m:msup>
			
                        <m:mrow>
				<m:mo>(</m:mo>
				<m:msub>
					<m:mi>s</m:mi><m:mn>2</m:mn>
				</m:msub>
                                <m:mo>)</m:mo>
			</m:mrow>
			<m:mn>2</m:mn>
		</m:msup>
	</m:mrow>
</m:mfrac>
</m:math> is a large number.  
   
  

</para><para id="eip-657">Therefore, if <m:math><m:mn>F</m:mn></m:math> is close to <m:math><m:mn>1</m:mn></m:math>, the evidence favors the null hypothesis (the two population variances are equal). But if <m:math><m:mn>F</m:mn></m:math> is much larger than  <m:math><m:mn>1</m:mn></m:math>, then the evidence is against the null hypothesis.</para><para id="element-848"><emphasis>A test of two variances may be left, right, or two-tailed.</emphasis></para><example id="element-315"><para id="element-33">Two college instructors are interested in whether or not there is any
variation in the way they grade math exams. They each grade the same set of 30
exams. The first instructor's grades have a variance of 52.3. The second instructor's
grades have a variance of 89.9. 
</para><exercise id="element-888"><problem id="id17653959">
  <para id="element-185">Test the claim that the first instructor's variance is
smaller. (In most colleges, it is desirable for the variances of exam grades to be nearly
the same among instructors.) The level of significance is 10%.</para>
</problem>

<solution id="id17653979">
  <para id="element-497">Let 1 and 2 be the subscripts that indicate the first and second instructor, respectively.</para><para id="element-6"><m:math><m:msub><m:mi>n</m:mi><m:mn>1</m:mn></m:msub><m:mo>=</m:mo><m:msub><m:mi>n</m:mi><m:mn>2</m:mn></m:msub><m:mo>=</m:mo><m:mn>30</m:mn></m:math>.</para><para id="element-310"><m:math><m:msub><m:mi>H</m:mi><m:mi>o</m:mi></m:msub></m:math>: <m:math><m:msubsup><m:mi>σ</m:mi><m:mn>1</m:mn><m:mn>2</m:mn></m:msubsup><m:mo>=</m:mo><m:msubsup><m:mi>σ</m:mi><m:mn>2</m:mn><m:mn>2</m:mn></m:msubsup></m:math> 
and 

<m:math><m:msub><m:mi>H</m:mi><m:mi>a</m:mi></m:msub></m:math>: 
<m:math><m:msubsup><m:mi>σ</m:mi><m:mn>1</m:mn><m:mn>2</m:mn></m:msubsup></m:math><m:math><m:reln><m:lt/></m:reln></m:math><m:math><m:msubsup><m:mi>σ</m:mi><m:mn>2</m:mn><m:mn>2</m:mn></m:msubsup></m:math></para><para id="element-147"><emphasis>Calculate the test statistic:</emphasis> By the null hypothesis <m:math><m:mo>(</m:mo><m:msubsup><m:mi>σ</m:mi><m:mn>1</m:mn><m:mn>2</m:mn></m:msubsup><m:mo>=</m:mo><m:msubsup><m:mi>σ</m:mi><m:mn>2</m:mn><m:mn>2</m:mn></m:msubsup><m:mo>)</m:mo></m:math>, the F statistic is</para><para id="element-932"><m:math>
	<m:mi>F</m:mi><m:mo>=</m:mo>
	<m:mfrac>
		<m:mrow>
			<m:mo>[</m:mo>
			<m:mfrac>
				<m:mrow>
					<m:mo>(</m:mo>
   					<m:msub><m:mi>s</m:mi><m:mn>1</m:mn></m:msub>
					<m:msup><m:mo>)</m:mo><m:mn>2</m:mn></m:msup>
				</m:mrow>
				<m:mrow>
					<m:msup>
						<m:mrow>
							<m:mo>(</m:mo>
							<m:msub><m:mi>σ</m:mi><m:mn>1</m:mn></m:msub>
							<m:mo>)</m:mo>
						</m:mrow>
						<m:mn>2</m:mn>
					</m:msup>
				</m:mrow>
			</m:mfrac>
			<m:mo>]</m:mo>
		</m:mrow>
		<m:mrow>
			<m:mo>[</m:mo>
			<m:mfrac>
				<m:mrow><m:mo>(</m:mo>
					<m:msub><m:mi>s</m:mi><m:mn>2</m:mn></m:msub>
					<m:msup><m:mo>)</m:mo><m:mn>2</m:mn></m:msup>
				</m:mrow>
				<m:mrow>
					<m:msup>
						<m:mrow>
							<m:mo>(</m:mo>
							<m:msub><m:mi>σ</m:mi><m:mn>2</m:mn></m:msub>
							<m:mo>)</m:mo>
						</m:mrow>
						<m:mn>2</m:mn>
					</m:msup>
				</m:mrow>
			</m:mfrac>
			<m:mo>]</m:mo>
		</m:mrow>
	</m:mfrac>
<m:mo>=</m:mo><m:mfrac>
	<m:mrow>
		<m:mo>(</m:mo>
		<m:msub>
			<m:mi>s</m:mi><m:mn>1</m:mn>
		</m:msub>
		<m:msup>
			<m:mo>)</m:mo><m:mn>2</m:mn>
		</m:msup>
	</m:mrow>

	<m:mrow>
		<m:msup>
			
                        <m:mrow>
				<m:mo>(</m:mo>
				<m:msub>
					<m:mi>s</m:mi><m:mn>2</m:mn>
				</m:msub>
                                <m:mo>)</m:mo>
			</m:mrow>
			<m:mn>2</m:mn>
		</m:msup>
	</m:mrow>
</m:mfrac><m:mo>=</m:mo><m:mfrac><m:mrow><m:mn>52.3</m:mn></m:mrow><m:mrow><m:mn>89.9</m:mn></m:mrow></m:mfrac><m:mo>=</m:mo><m:mn>0.6</m:mn>
</m:math></para><para id="element-569"><emphasis>Distribution for the test:</emphasis> <m:math><m:msub><m:mi>F</m:mi><m:mrow><m:mn>29</m:mn><m:mo>,</m:mo><m:mn>29</m:mn></m:mrow></m:msub><m:mspace width="20pt"/></m:math>where <m:math><m:msub><m:mi>n</m:mi><m:mn>1</m:mn></m:msub><m:mo>-</m:mo><m:mn>1</m:mn><m:mo>=</m:mo><m:mn>29</m:mn></m:math> and <m:math><m:msub><m:mi>n</m:mi><m:mn>2</m:mn></m:msub><m:mo>-</m:mo><m:mn>1</m:mn><m:mo>=</m:mo><m:mn>29</m:mn></m:math>. </para><para id="element-184"><emphasis>Graph:</emphasis><m:math><m:mspace width="20pt"/></m:math><emphasis>This test is left tailed.</emphasis></para><para id="element-975">Draw the graph labeling and shading appropriately.</para><para id="element-825"><figure id="anova_twotests"><media id="id18591379" alt=""><image src="anova_twotests.png" mime-type="image/png" print-width="4in"/></media></figure></para><para id="element-561"><emphasis>Probability statement:</emphasis> <m:math><m:mtext>p-value</m:mtext><m:mo>=</m:mo><m:mi>P</m:mi></m:math>

(<m:math><m:mi>F</m:mi></m:math>

<m:math><m:reln><m:lt/></m:reln></m:math>

<m:math><m:mn>0.582</m:mn></m:math>)

<m:math><m:mo>=</m:mo><m:mn>0.0755</m:mn></m:math></para><para id="element-186"><emphasis>Compare <m:math><m:mi>α</m:mi></m:math> and the p-value:</emphasis>
<m:math><m:mi>α</m:mi><m:mo>=</m:mo><m:mn>0.10</m:mn></m:math><m:math><m:mspace width="20pt"/><m:mi>α</m:mi><m:mo>&gt;</m:mo><m:mtext>p-value</m:mtext></m:math>.</para><para id="element-153"><emphasis>Make a decision:</emphasis> Since <m:math><m:mi>α</m:mi><m:mo>&gt;</m:mo><m:mtext>p-value</m:mtext></m:math>, reject <m:math><m:msub><m:mi>H</m:mi><m:mi>o</m:mi></m:msub></m:math>.</para><para id="element-760"><emphasis>Conclusion:</emphasis> With a 10% level of significance, from the data, there is sufficient
evidence to conclude that the variance in grades for the first instructor is smaller.</para><para id="element-44"><emphasis>TI-83+ and TI-84:</emphasis> Press <code>STAT</code> and arrow over to <code>TESTS</code>. Arrow down to
<code>D:2-SampFTest</code>. Press <code>ENTER</code>. Arrow to <code>Stats</code> and press <code>ENTER</code>. For
<code>Sx1</code>, <code>n1</code>, <code>Sx2</code>, and <code>n2</code>, enter <code><m:math><m:msqrt><m:mo>(</m:mo><m:mn>52.3</m:mn><m:mo>)</m:mo></m:msqrt></m:math></code>, <code>30</code>, <code><m:math><m:msqrt><m:mo>(</m:mo><m:mn>89.9</m:mn><m:mo>)</m:mo></m:msqrt></m:math></code>, and <code>30</code>. Press <code>ENTER</code> after
each. Arrow to <code>σ1:</code> and <code><m:math><m:reln><m:lt/></m:reln></m:math>σ2</code>. Press <code>ENTER</code>. Arrow down to <code>Calculate</code> and
press <code>ENTER</code>. <m:math><m:mi>F</m:mi><m:mo>=</m:mo><m:mn>0.5818</m:mn></m:math> and <m:math><m:mtext>p-value</m:mtext><m:mo>=</m:mo><m:mn>0.0753</m:mn></m:math>. Do the procedure again
and try <code>Draw</code> instead of <code>Calculate</code>.</para>
</solution>
</exercise>
</example>   
  </content>
  
</document>
