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<document xmlns="http://cnx.rice.edu/cnxml" xmlns:md="http://cnx.rice.edu/mdml/0.4" xmlns:bib="http://bibtexml.sf.net/" xmlns:m="http://www.w3.org/1998/Math/MathML" id="new">
  <name>Confidence Intervals: Confidence Interval, Single Population Mean, Population Standard Deviation Known, Normal</name>
  <metadata>
  <md:version>1.6</md:version>
  <md:created>2008/06/06 16:34:42 GMT-5</md:created>
  <md:revised>2008/07/18 15:51:41.833 GMT-5</md:revised>
  <md:authorlist>
      <md:author id="billowsky">
      <md:firstname>Barbara</md:firstname>
      
      <md:surname>Illowsky</md:surname>
      <md:email>illowskybarbara@deanza.edu</md:email>
    </md:author>
      <md:author id="sdean">
      <md:firstname>Susan</md:firstname>
      
      <md:surname>Dean</md:surname>
      <md:email>deansusan@deanza.edu</md:email>
    </md:author>
  </md:authorlist>

  <md:maintainerlist>
    <md:maintainer id="cnxorg">
      <md:firstname/>
      
      <md:surname>Connexions</md:surname>
      <md:email>cnx@cnx.org</md:email>
    </md:maintainer>
  </md:maintainerlist>
  
  <md:keywordlist>
    <md:keyword>elementary</md:keyword>
    <md:keyword>statistics</md:keyword>
  </md:keywordlist>

  <md:abstract/>
</metadata>
  <content>
    <para id="delete_me">To construct a confidence interval for a single unknown population mean <m:math><m:mi>μ</m:mi></m:math> <emphasis>where
the population standard deviation is known,</emphasis> we need <m:math>
<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>
</m:math>

as an estimate for <m:math><m:mi>μ</m:mi></m:math> and a
margin of error. Here, the margin of error is called the <term src="#ebpbound">error bound for a
population mean</term> (abbreviated <emphasis>EBM</emphasis>). The margin of error depends on the
<term src="#conflevel">confidence level</term> (abbreviated <emphasis>CL</emphasis>). The confidence level is the probability that the
confidence interval produced contains the true population parameter. Most often, it
is the choice of the person constructing the confidence interval to choose a
confidence level of 90% or higher because he wants to be reasonably certain of his
conclusions.</para><example id="element-258"><para id="element-914">Suppose the sample mean is 7 and the error bound for the mean is 2.5.</para><exercise id="element-566"><?solution_in_back?><problem>
  <para id="element-588">
<m:math>
<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>
<m:mo>=</m:mo>
</m:math> 
_______ and 
<m:math>
<m:mtext>EBM</m:mtext>
<m:mo>=</m:mo>
</m:math>
_______.
  </para>
</problem>

<solution>
  <para id="element-689">
<m:math>
<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>
<m:mo>=</m:mo>
</m:math> 
7 and 
<m:math>
<m:mtext>EBM</m:mtext>
<m:mo>=</m:mo>
</m:math>
2.5.
  </para>
</solution>
</exercise><para id="element-146">The confidence interval is
<m:math>
<m:mo>(</m:mo>
<m:mn>7</m:mn>
<m:mo>-</m:mo>
<m:mn>2.5</m:mn>
<m:mo>,</m:mo>
<m:mn>7</m:mn>
<m:mo>+</m:mo>
<m:mn>2.5</m:mn>
<m:mo>)</m:mo>
</m:math>.</para><para id="element-722">If the confidence level (CL) is 95%, then we say we are 95% confident that the true
population mean is between 4.5 and 9.5.</para><para id="element-944">A confidence interval for a population mean with a known standard deviation is
based on the fact that the sample means follow an approximately normal
distribution. Suppose we have constructed the 90% confidence interval (5, 15)
where <m:math>
<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>
<m:mo>=</m:mo>
<m:mn>10</m:mn>
</m:math>

and 
<m:math>
<m:mtext>EBM</m:mtext>
<m:mo>=</m:mo> 
<m:mn>5</m:mn>
</m:math>. To get a 90% confidence interval, we must include
the central 90% of the sample means. If we include the central 90%, we leave out
a total of 10 % or 5% in each tail of the normal distribution. To capture the central
90% of the sample means, we must go out 1.645 standard deviations on either side
of the calculated sample mean. The 1.645 is the z-score from a standard normal
table that has area to the right equal to 0.05
(5% area in the right tail). The graph shows the general situation.</para><para id="element-903"><media type="image/png" src="conf_intvls_cfpm1.png">
  <param name="alt" value="Normal distribution curve with values of 5 and 15 on the x-axis. Vertical upward lines from points 5 and 15 extend to the curve. The confidence interval area between these two points is equal to 0.90."/>
  <param name="width" value="500"/>
  <param name="print-width" value="5in"/>
</media></para><para id="element-17">To summarize, resulting from the Central Limit Theorem,</para><para id="element-206"><m:math>
<m:apply>
  <m:conjugate/>
  <m:ci>X</m:ci>
</m:apply>
</m:math>
is normally distributed, that is, 






<m:math>
<m:apply>
  <m:conjugate/>
  <m:ci>X</m:ci>
</m:apply></m:math> ~
<m:math>
<m:mn>N</m:mn>
<m:mo>(</m:mo>
<m:msub>
<m:mi>μ</m:mi>
<m:mi>X</m:mi>
</m:msub>
<m:mo>,</m:mo>
<m:mfrac>
<m:mrow>
<m:msub>
<m:mi>σ</m:mi>
<m:mi>X</m:mi>
</m:msub>
</m:mrow>
<m:mrow>
<m:msqrt>
<m:mi>n</m:mi>
</m:msqrt>
</m:mrow>
</m:mfrac>
<m:mo>)</m:mo>
<m:mspace width="35pt"/>
</m:math></para><para id="element-870"><emphasis>Since the population standard deviation, σ, is known, we use a normal curve.</emphasis></para><para id="element-436">The confidence level, <m:math><m:mi>CL</m:mi></m:math>, is 
<m:math>
<m:mi>CL</m:mi>
<m:mo>=</m:mo>
<m:mn>1</m:mn>
<m:mo>-</m:mo>
<m:mi>α</m:mi>
</m:math>. Each of the tails contains an area
equal to
<m:math>
<m:mfrac>
<m:mi>α</m:mi>
<m:mn>2</m:mn>
</m:mfrac>
</m:math>
.</para><para id="element-901">The z-score that has area to the right of 
<m:math>
<m:mfrac><m:mn>α</m:mn><m:mn>2</m:mn></m:mfrac>
</m:math> is 
<m:math>
<m:msub>
<m:mi>z</m:mi>
<m:mrow>
<m:mfrac>
<m:mi>α</m:mi>
<m:mn>2</m:mn>
</m:mfrac>
</m:mrow>
</m:msub>
</m:math>.</para><para id="element-599">For example, if
<m:math>
<m:mfrac><m:mi>α</m:mi>
<m:mn>2</m:mn></m:mfrac>
<m:mo>=</m:mo>
<m:mn>0.025</m:mn>
</m:math>,
then area to the right 
<m:math>
<m:mo>=</m:mo>
<m:mn>0.025</m:mn>
</m:math>
and area to the left 
<m:math>
<m:mo>=</m:mo>
<m:mn>1</m:mn>
<m:mo>-</m:mo>
<m:mn>0.025</m:mn>
<m:mo>=</m:mo>
<m:mn>0.975</m:mn>
</m:math>
and
<m:math>
<m:msub>
<m:mi>z</m:mi>
<m:mrow>
<m:mfrac>
<m:mi>α</m:mi>
<m:mn>2</m:mn>
</m:mfrac>
</m:mrow>
</m:msub>
<m:mo>=</m:mo>
<m:msub>
<m:mi>z</m:mi>
<m:mn>0.025</m:mn>
</m:msub>
<m:mo>=</m:mo>
<m:mn>1.96</m:mn>
</m:math>
using a calculator, computer or table. Using the TI83+ or 84 calculator function, <code>invNorm</code>, you can verify this result.
<code>invNorm</code><m:math><m:mo>(</m:mo><m:mn>.975</m:mn><m:mo>,</m:mo><m:mn>0</m:mn><m:mo>,</m:mo><m:mn>1</m:mn><m:mo>)</m:mo><m:mo>=</m:mo><m:mn>1.96</m:mn></m:math>.</para><para id="element-786">The error bound formula for a single population mean when the variance is known is
</para><para id="element-855"><m:math>
<m:mi>EBM</m:mi>
<m:mo>=</m:mo>
<m:msub>
<m:mi>z</m:mi>
<m:mrow>
<m:mfrac>
<m:mi>α</m:mi>
<m:mn>2</m:mn>
</m:mfrac>
</m:mrow>
</m:msub>
<m:mo>⋅</m:mo>
<m:mfrac>
<m:mi>σ</m:mi>
<m:mrow>
<m:msqrt>
<m:mi>n</m:mi>
</m:msqrt>
</m:mrow>
</m:mfrac>
</m:math>
</para><para id="element-498">The confidence interval has the format 
<m:math>
<m:mo>(</m:mo>
<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>
<m:mo>−</m:mo>
<m:mi>EBM</m:mi>
<m:mo>,</m:mo> 
<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>
<m:mo>+</m:mo>
<m:mi>EBM</m:mi>
<m:mo>)</m:mo>
</m:math>.</para><para id="element-538">The graph gives a picture of the entire situation.</para><para id="element-43"><m:math>
<m:mi>CL</m:mi>
<m:mo>+</m:mo>
<m:mfrac>
<m:mi>α</m:mi>
<m:mn>2</m:mn>
</m:mfrac>
<m:mo>+</m:mo>
<m:mfrac>
<m:mi>α</m:mi>
<m:mn>2</m:mn>
</m:mfrac>
<m:mo>=</m:mo>
<m:mi>CL</m:mi>
<m:mo>+</m:mo>
<m:mi>α</m:mi>
<m:mo>=</m:mo>
<m:mn>1</m:mn>
</m:math>.</para><para id="element-848"><media type="image/png" src="conf_intvls_cfpm2.png">
  <param name="alt" value="Normal distribution curve displaying the confidence interval formulas and corresponding area formulas."/>
  <param name="width" value="300"/>
  <param name="print-width" value="3in"/>
</media></para>
</example><example id="element-106"><exercise id="element-899"><problem>
  <para id="element-230">
    Suppose scores on exams in statistics are normally distributed with an
unknown population mean and a population standard deviation of 3 points. A sample
of 36 scores is taken and gives a sample mean (sample average score) of 68. Find a
90% confidence interval for the true (population) mean of statistics exam scores.

  </para><list id="element-220" type="bulleted"><item>The first solution is step-by-step.</item>
<item>The second solution uses the TI-83+ and TI-84 calculators.</item></list>
</problem>

<solution>
  <para id="element-429">To find the confidence interval, you need the sample mean, 
<m:math>
<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>
</m:math>, and the EBM.</para><para id="element-253"><list id="list1223426346354" type="named-item"><?mark .?><item><name>a</name><m:math>

<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>

<m:mo>=</m:mo>
<m:mn>68</m:mn>
</m:math>
</item>
<item><name>b</name>
<m:math>
<m:mi>EBM</m:mi>
<m:mo>=</m:mo>
<m:msub>
<m:mi>z</m:mi>
<m:mrow>
<m:mfrac>
<m:mi>α</m:mi>
<m:mn>2</m:mn>
</m:mfrac>
</m:mrow>
</m:msub>
<m:mo>⋅</m:mo>
<m:mo>(</m:mo>
<m:mfrac>
<m:mi>σ</m:mi>
<m:mrow>
<m:msqrt>
<m:mi>n</m:mi>
</m:msqrt>
</m:mrow>
</m:mfrac>
<m:mo>)</m:mo>
</m:math>
</item>
<item><name>c</name>
<m:math>
<m:mi>σ</m:mi>
<m:mo>=</m:mo>
<m:mn>3</m:mn>
</m:math>
</item>
<item><name>d</name>
<m:math>
<m:mi>n</m:mi>
<m:mo>=</m:mo>
<m:mn>36</m:mn>
</m:math>
</item>
</list></para><para id="element-403"><m:math>
<m:mi>CL = 0.90</m:mi>
</m:math>
 so 
<m:math>
<m:mi>a</m:mi>
<m:mo>=</m:mo> 
<m:mn>1</m:mn>
<m:mo>-</m:mo>
<m:mi>CL</m:mi>
<m:mo>=</m:mo>
<m:mn>1</m:mn>
<m:mo>-</m:mo>
<m:mn>0.90</m:mn>
<m:mo>=</m:mo>
<m:mn>0.10</m:mn>
</m:math></para><para id="element-997">Since
<m:math>
<m:mfrac>
<m:mi>α</m:mi>
<m:mn>2</m:mn>
</m:mfrac>
<m:mo>=</m:mo>
<m:mn>0.05</m:mn>

</m:math>, then
<m:math>
<m:msub>
<m:mi>z</m:mi>
<m:mrow>
<m:mfrac>
<m:mi>α</m:mi>
<m:mn>2</m:mn>
</m:mfrac>
</m:mrow>
</m:msub>
<m:mo>=</m:mo>
<m:msub>
<m:mi>z</m:mi>
<m:mn>.05</m:mn>
</m:msub>
<m:mo>=</m:mo>
<m:mn>1.645</m:mn>
</m:math></para><para id="element-65">from a calculator, computer or standard normal table</para><para id="element-475">Therefore,
<m:math>
<m:mi>EBM</m:mi>
<m:mo>=</m:mo>
<m:mn>1.645</m:mn>
<m:mo>⋅</m:mo>
<m:mo>(</m:mo>
<m:mfrac>
<m:mn>3</m:mn>
<m:mrow>
<m:msqrt>
<m:mn>36</m:mn>
</m:msqrt>
</m:mrow>
</m:mfrac>
<m:mo>)</m:mo>
<m:mo>=</m:mo>
<m:mn>0.8225</m:mn>
</m:math>
</para><para id="element-808">This gives 
<m:math>
<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>
<m:mo>-</m:mo>
<m:mi>EBM</m:mi>
<m:mo>=</m:mo>
<m:mn>68</m:mn>
<m:mo>-</m:mo>
<m:mn>0.8225</m:mn>
<m:mo>=</m:mo>
<m:mn>67.18</m:mn>
</m:math></para><para id="element-739">and 
<m:math>
<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>
<m:mo>+</m:mo>
<m:mi>EBM</m:mi>
<m:mo>=</m:mo>
<m:mn>68</m:mn>
<m:mo>+</m:mo>
<m:mn>0.8225</m:mn>
<m:mo>=</m:mo>
<m:mn>68.82</m:mn>
</m:math></para><para id="element-290">The 90% confidence interval is <emphasis>(67.18, 68.82).</emphasis></para>
</solution>
<solution>
<para id="element-430">The TI-83+ and TI-84 caculators simplify this whole procedure. Press
<code>STAT</code> and arrow over to <code>TESTS</code>. Arrow down to <code>7:ZInterval</code>. Press
<code>ENTER</code>. Arrow to <code>Stats</code> and press <code>ENTER</code>. Arrow down and enter 3 for
<m:math>
<m:mi>σ</m:mi></m:math>, 68 for <m:math>
<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>
</m:math>

, 36 for 
<m:math>
<m:mi>n</m:mi></m:math>, and .90 for <code>C-level</code>. Arrow down to <code>Calculate</code> and
press <code>ENTER</code>. The confidence interval is (to 3 decimal places) (67.178,
68.822).</para>
</solution></exercise><para id="element-21"><emphasis> We can find the error bound from the confidence interval.</emphasis> From the upper
value, subtract the sample mean <emphasis>or</emphasis> subtract the lower value from the upper value
and divide by two. The result is the error bound for the mean (EBM).
</para><para id="element-638"><m:math>
<m:mi>EBM</m:mi>
<m:mo>=</m:mo> 
<m:mn>68.822</m:mn>
<m:mo>-</m:mo>
<m:mn>68</m:mn>
<m:mo>=</m:mo>
<m:mn>0.822</m:mn>
<m:mspace width="5pt"/>
</m:math>
 or 
<m:math>
<m:mspace width="5pt"/>
<m:mi>EBM</m:mi>
<m:mo>=</m:mo>
<m:mfrac>
<m:mrow>
<m:mo>(</m:mo>
<m:mn>68.822</m:mn>
<m:mo>−</m:mo> 
<m:mn>67.178</m:mn>
<m:mo>)</m:mo>
</m:mrow>
<m:mrow>
<m:mn>2</m:mn>
</m:mrow>
</m:mfrac>
<m:mo>=</m:mo>
<m:mn>0.822</m:mn>
</m:math></para><para id="element-500"><emphasis>We can interpret the confidence interval in two ways:</emphasis>
<list id="list-1" type="enumerated">
<item>We are 90% confident that the true population mean for statistics exam scores is
between 67.178 and 68.822.</item>
<item>Ninety percent of all confidence intervals constructed in this way contain the true
average statistics exam score. For example, if we constructed 100 of these
confidence intervals, we would expect 90 of them to contain the true population mean
exam score.</item>
</list></para><para id="element-394">Now for the same problem, find a 95% confidence interval for the true (population)
mean of scores. Draw the graph. The sample mean, standard deviation, and sample
size are:</para><exercise id="element-971"><?solution_in_back?>

<problem>

<list id="list1634453" type="named-item"><?mark .?><item>
<name>a</name>
<m:math>

<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>
<m:mo>=</m:mo>

</m:math>
</item>
<item>
<name>b</name>
<m:math>
<m:mi>σ</m:mi>
<m:mo>=</m:mo>

</m:math>
</item>
<item>
<name>c</name>
<m:math>
<m:mi>n</m:mi>
<m:mo>=</m:mo>

</m:math>
</item>

</list>


</problem>

<solution>
<list id="list123353453" type="named-item"><?mark .?><item>
<name>a</name>
<m:math>

<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>
<m:mo> = </m:mo>
<m:mn> 68</m:mn>
</m:math>
</item>
<item>
<name>b</name>
<m:math>
<m:mi>σ</m:mi>
<m:mo> = </m:mo>
<m:mn>3</m:mn>
</m:math>
</item>
<item>
<name>c</name>
<m:math>
<m:mi>n</m:mi>
<m:mo> = </m:mo>
<m:mn>36</m:mn>
</m:math>
</item>

</list>

</solution>
</exercise><para id="element-879">The confidence level is <m:math><m:mi>CL</m:mi> <m:mo>= </m:mo><m:mn>0.95</m:mn></m:math>.<m:math><m:mspace width="20pt"/></m:math> Graph:</para><para id="element-116">The confidence interval is (use technology)</para><exercise id="element-454"><problem>
  <para id="element-630">
    <m:math>
<m:mo>(</m:mo>

<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>
<m:mo>-</m:mo>
<m:mi>EBM</m:mi>
<m:mo>,</m:mo>
<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>
<m:mo>+</m:mo>
<m:mi>EBM</m:mi>
<m:mo>)</m:mo>
<m:mo>=</m:mo>
<m:mtext>(_______, _______)</m:mtext>
</m:math>.
The error bound <m:math><m:mi>EBM </m:mi><m:mo>=</m:mo></m:math> _______.
  </para>
</problem>

<solution>
  <para id="element-878">
    <m:math>
<m:mo>(</m:mo>

<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>
<m:mo>-</m:mo>
<m:mi>EBM</m:mi>
<m:mo>,</m:mo>
<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>
<m:mo>+</m:mo>
<m:mi>EBM</m:mi>
<m:mo>)</m:mo>
<m:mo>=</m:mo>
<m:mtext>(67.02 , 68.98)</m:mtext>
</m:math>.
The error bound <m:math><m:mi>EBM </m:mi><m:mo>=</m:mo></m:math> 0.98.
  </para>
</solution>
</exercise><para id="element-552">We can say that we are 95 % confident that the true population mean for
statistics exam scores is between 67.02 and 68.98 and that 95% of all
confidence intervals constructed in this way contain the true average
statistics exam score.</para>
</example><example id="element-542"><para id="element-338">Suppose we change the previous problem.</para><exercise id="element-442"><problem>
  <para id="element-98">
Leave everything the same except the sample size.
  </para><para id="element-521"><list id="yep" type="named-item"><?mark .?><item><name>a</name><m:math>

<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>

<m:mo>=</m:mo>
<m:mn>68</m:mn>
</m:math>
</item>
<item><name>b</name>
<m:math>
<m:mi>σ</m:mi>
<m:mo>=</m:mo>
<m:mn>3</m:mn>
</m:math>
</item>
<item><name>c</name>
<m:math>
<m:msub>
<m:mi>z</m:mi>
<m:mfrac>
<m:mrow>
<m:mi>α</m:mi>
</m:mrow>
<m:mrow>
<m:mn>2</m:mn>
</m:mrow>
</m:mfrac>
</m:msub>
<m:mo>=</m:mo>
<m:mn>1.645</m:mn>
</m:math>
</item>
</list></para>
</problem>

<solution>
  <para id="element-124">If we <emphasis>increase</emphasis> the sample size <m:math> <m:mi>n</m:mi> </m:math> to 100, we <emphasis>decrease</emphasis> the error bound.</para><para id="element-412"><m:math>
<m:mi>EBM</m:mi>
<m:mo>=</m:mo>
<m:msub>
<m:mi>z</m:mi>
<m:mrow>
<m:mfrac>
<m:mi>α</m:mi>
<m:mn>2</m:mn>
</m:mfrac>
</m:mrow>
</m:msub>
<m:mo>⋅</m:mo>
<m:mo>(</m:mo>
<m:mfrac>
<m:mi>σ</m:mi>
<m:mrow>
<m:msqrt>
<m:mi>n</m:mi>
</m:msqrt>
</m:mrow>
</m:mfrac>
<m:mo>)</m:mo>
<m:mo>=</m:mo>
<m:mn>1.645</m:mn>
<m:mo>⋅</m:mo>
<m:mo>(</m:mo>
<m:mfrac>
<m:mi>3</m:mi>
<m:mrow>
<m:msqrt>
<m:mn>100</m:mn>
</m:msqrt>
</m:mrow>
</m:mfrac>
<m:mo>)</m:mo>
<m:mo>=</m:mo>
<m:mn>0.4935</m:mn>
</m:math></para>
</solution>

<solution>
  <para id="element-125">If we <emphasis>decrease</emphasis> the sample size <m:math> <m:mi>n</m:mi> </m:math> to 25, we <emphasis>increase</emphasis> the error bound.
</para><para id="element-9"><m:math>
<m:mi>EBM</m:mi>
<m:mo>=</m:mo>
<m:msub>
<m:mi>z</m:mi>
<m:mrow>
<m:mfrac>
<m:mi>α</m:mi>
<m:mn>2</m:mn>
</m:mfrac>
</m:mrow>
</m:msub>
<m:mo>⋅</m:mo>
<m:mo>(</m:mo>
<m:mfrac>
<m:mi>σ</m:mi>
<m:mrow>
<m:msqrt>
<m:mi>n</m:mi>
</m:msqrt>
</m:mrow>
</m:mfrac>
<m:mo>)</m:mo>
<m:mo>=</m:mo>
<m:mn>1.645</m:mn>
<m:mo>⋅</m:mo>
<m:mo>(</m:mo>
<m:mfrac>
<m:mi>3</m:mi>
<m:mrow>
<m:msqrt>
<m:mn>25</m:mn>
</m:msqrt>
</m:mrow>
</m:mfrac>
<m:mo>)</m:mo>
<m:mo>=</m:mo>
<m:mn>0.987</m:mn>
</m:math></para>
</solution></exercise><exercise id="element-821"><problem>
  <para id="element-320">Leave everything the same except for the confidence level. We increase the confidence level from 0.90 to 0.95.</para><para id="element-319"><list id="list234u23523" type="named-item"><?mark .?><item><name>a</name><m:math>

<m:apply>
  <m:conjugate/>
  <m:ci>x</m:ci>
</m:apply>

<m:mo>=</m:mo>
<m:mn>68</m:mn>
</m:math>
</item>
<item><name>b</name>
<m:math>
<m:mi>σ</m:mi>
<m:mo>=</m:mo>
<m:mn>3</m:mn>
</m:math>
</item>
<item><name>c</name>
<m:math>
<m:msub>
<m:mi>z</m:mi>
<m:mfrac>
<m:mrow>
<m:mi>α</m:mi>
</m:mrow>
<m:mrow>
<m:mn>2</m:mn>
</m:mrow>
</m:mfrac>
</m:msub>
<m:mo>=</m:mo>
<m:mn>1.645</m:mn>
</m:math>
</item></list></para>
</problem>

<solution>
  <para id="element-567"><figure id="conf_intvls_cfpm3"><subfigure>
     <media type="image/png" src="conf_intvls_cfpm3-1.png">
<param name="alt" value="Normal distribution curve with 0.90 confidence interval area blocked off and corresponding residual areas."/>
<param name="width" value="250"/>
<param name="print-width" value="3in"/>
</media>
  </subfigure>
<subfigure>
     <media type="image/png" src="conf_intvls_cfpm3-2.png">
<param name="alt" value="Normal distribution curve with 0.95 confidence interval area blocked off and corresponding residual areas."/>
<param name="width" value="250"/>
<param name="print-width" value="3in"/>
</media>
  </subfigure></figure></para><para id="element-674">The 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98). The 95% confidence interval is wider. <emphasis>If you look at the graphs, because the area 0.95
is larger than the area 0.90, it makes sense that the 95% confidence interval is wider.</emphasis></para>
</solution>
</exercise>
</example>   
  </content>
  <glossary>

  <definition id="conflevel">
    <term>Confidence Level</term>
    <meaning>
The percent expression for the probability that the confidence interval contains the true population parameter. That is, for ex., if CL=90%, then in 90 out of 100 samples the interval estimate will enclose the true population parameter.
    </meaning>
  </definition>

  <definition id="ebmbound">
    <term>Error Bound for a Population Mean (EBM)</term>
    <meaning>
      The margin of error. Depends on the confidence level, sample size, and known or estimated population standard deviation.
    </meaning>
  </definition>


</glossary>

</document>
