document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1114655>Question 1114655</A>: <I><SPAN STYLE=\"word-break: break-all;\"> You randomly select and weigh 30 samples of an allergy medicine and the sample standard deviation is 1.20 milligrams. Assuming the weights are normally distributed, construct 99% confidence intervals for the population variance\r<BR>\n" );
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document.write( "(a) Complete the table:<BR>\n" );
document.write( "Population Parameter Being Estimated<BR>\n" );
document.write( "Point Estimate<BR>\n" );
document.write( "Sample Size n =<BR>\n" );
document.write( "Margin of Error<BR>\n" );
document.write( "Distribution Used<BR>\n" );
document.write( "Reason for Using that Distribution<BR>\n" );
document.write( "Confidence Level<BR>\n" );
document.write( "Confidence Interval<BR>\n" );
document.write( "Units for the Confidence Interval (if applicable)\r<BR>\n" );
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document.write( "(b) Interpret the confidence interval (don’t forget the units): </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #729600 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=729600\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Problem asks to construct a 99% confidence interval for the population variance<BR>\n" );
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document.write( "The variance is a non-negative number. This means the domain of the probability distribution is not (&#8722;&#8734;,&#8734;), therefore the normal distribution cannot be the distribution of a variance. The correct PDF must have a domain of [0,&#8734;), it can be shown that if the original population of data is normally distributed, then the expression (n&#8722;1)s^2/&#963;^2 has a chi-square distribution with n&#8722;1 degrees of freedom.<BR>\n" );
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document.write( "alpha(a) = 1 -(99/100) = 0.01<BR>\n" );
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document.write( "a/2 = 0.005<BR>\n" );
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document.write( "critical probability(p*) = 1 - 0.005 = 0.995<BR>\n" );
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document.write( "degrees of freedom(df) = 30 -1 = 29<BR>\n" );
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document.write( "chi^2 (0.995) distribution with df = 29 is 13.121<BR>\n" );
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document.write( "chi^2 (0.005) distribution with df = 29 is 52.336<BR>\n" );
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document.write( "now we evaluate (n-1)s^2/chi^2 for 13.121 and 52.336<BR>\n" );
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document.write( "for 13.121<BR>\n" );
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document.write( "(30-1)(1.20)^2/13.121 = 3.1827<BR>\n" );
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document.write( "for 52.336<BR>\n" );
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document.write( "(30-1)(1.20)^2/52.336 = 0.7979<BR>\n" );
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document.write( "99% confidence interval for variance is [0.7979, 3.1827] milligrams<BR>\n" );
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document.write( " </SPAN>\n" );
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