document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1149929>Question 1149929</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Waiting times (in hours) at a particular hospital are believed to be approximately normally distributed with a variance of 2.25 hr.<BR>\n" );
document.write( "a.. A sample of 20 outpatients revealed a mean waiting time of 1.52 hours. Construct the 95% CI for the estimate of the population mean.<BR>\n" );
document.write( "b. Suppose that the mean of 1.52 hours had resulted from a sample of 32 patients. Find the 95% CI.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #771262 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=771262\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a.  sample mean(u) of 1.52<BR>\n" );
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document.write( "standard error(SE) of the sample = 2.25/square root(20) = 0.5031<BR>\n" );
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document.write( "alpha(a) = 1 -(95/100) = 0.05<BR>\n" );
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document.write( "critical probability(p*) = 1 -a/2 = 0.975<BR>\n" );
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document.write( "degrees of freedom(DF) = 20 -1 = 19<BR>\n" );
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document.write( "because the sample size(20) is less than 30, we use the student t-distribution tables to find the critical value(CV) which is the t-score for DF=19 and p* = 0.975.  CV = 2.093<BR>\n" );
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document.write( "Margin of Error(ME) = CV * SE = 2.093 * 0.5031 = 1.053<BR>\n" );
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document.write( "For a normal distribution the population mean is also the mean of the sample<BR>\n" );
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document.write( "95% confidence interval(CI) = 1.52 + or minus 1.053 = (0.467, 2.573)<BR>\n" );
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document.write( "b.  SE = 2.25/square root(32) = 0.3977<BR>\n" );
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document.write( "p* = 0.975<BR>\n" );
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document.write( "because the sample size(32) is greater than 30, we can use the Normal distribution table of z-values to find the CV which is the z-score associated with the p*.  CV = 1.96<BR>\n" );
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document.write( "ME = 1.96 * 0.3977 = 0.7795<BR>\n" );
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document.write( "95% confidence interval(CI) = 1.52 + or minus 0.7795 = (0.7405, 2.2995)<BR>\n" );
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