document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1112685>Question 1112685</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Another Question , Help Please \r<BR>\n" );
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document.write( "A survey of 20 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of $57.00 per hour. The sample standard deviation was $5.60 per hour. (Use  t Distribution Table.)\r<BR>\n" );
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document.write( "Q1) What is the best estimate of the population mean?\r<BR>\n" );
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document.write( "Q2) Develop a 90% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.)\r<BR>\n" );
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document.write( "Confidence interval for the population mean wage is between ...?... and ...?...\r<BR>\n" );
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document.write( "Q3)How large a sample is needed to assess the population mean with an allowable error of $1.00 at 90% confidence? (Round up your answer to the next whole number.) Sample size ?<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 #727722 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=727722\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Q1.  Use the sample mean of $57.00<BR>\n" );
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document.write( "Q2.  Standard Error(SE) = $5.60/square root(2) = 1.2522<BR>\n" );
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document.write( "alpha(a) = 1 - (90/100) = 0.10<BR>\n" );
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document.write( "critical probability(p*) = 1 - (a/2) = 0.95<BR>\n" );
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document.write( "degrees of freedom(DF) = 20 - 1 = 19<BR>\n" );
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document.write( "the critical value(CV) is the t-statistic having 19 DF and a cumulative probability of 0.95.  From the t-distribution tables, the CV is 1.729<BR>\n" );
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document.write( "Margin of Error(ME) is CV * SE = 1.729 * 1.2522 = 2.1650 approximately 2.17<BR>\n" );
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document.write( "90% confidence interval is $57.00 + or - 2.17 = ($54.83, $59.17)<BR>\n" );
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document.write( "Q3.  We have no standard deviation for the population, so we must estimate it.<BR>\n" );
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document.write( "Since our sample size is 20 which is < 30, we must use the t-statistic and the standard deviation of the sample($5.60)<BR>\n" );
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document.write( "The problem states that the allowable error is $1.00, which I assume describes the margin of error(ME)<BR>\n" );
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document.write( "ME = 1<BR>\n" );
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document.write( "estimated population standard deviation = 5.60<BR>\n" );
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document.write( "CV is 1.729<BR>\n" );
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document.write( "ME = CV * standard deviation/square root(n)<BR>\n" );
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document.write( "n = ((1.729 * 5.60)/1)^2 = 93.7488<BR>\n" );
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document.write( "sample size(n) is 94<BR>\n" );
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