document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1181382>Question 1181382</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Refer to the technology output given to the right that results from measured hemoglobin levels (g/dL) in 100 randomly selected adult females. <BR>\n" );
document.write( "The confidence level of 95% was used.<BR>\n" );
document.write( "a. Express the confidence interval in the format that uses the \"less than\" symbol. Assume that the original listed data use two decimal places, <BR>\n" );
document.write( "and round the confidence interval limits accordingly.<BR>\n" );
document.write( "b. Identify the best point estimate of μ and the margin of error.<BR>\n" );
document.write( "c. In constructing the confidence interval estimate of μ, <BR>\n" );
document.write( "why is it not necessary to confirm that the sample data appear to be from a population with a normal distribution?<BR>\n" );
document.write( "TInterval<BR>\n" );
document.write( "(12.800,13.318)<BR>\n" );
document.write( "x=13.059<BR>\n" );
document.write( "Sx=1.307<BR>\n" );
document.write( "n=100<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 #850039 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=850039\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's the breakdown of how to interpret the technology output and answer the questions:\r<BR>\n" );
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document.write( "**(a) Confidence Interval Format:**\r<BR>\n" );
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document.write( "The technology output gives the 95% confidence interval as (12.800, 13.318).  In the \"less than\" symbol format, this is expressed as:\r<BR>\n" );
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document.write( "12.80 g/dL < μ < 13.32 g/dL\r<BR>\n" );
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document.write( "**(b) Point Estimate and Margin of Error:**\r<BR>\n" );
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document.write( "* **Best Point Estimate of μ:** The best point estimate of the population mean (μ) is the sample mean (x̄), which is given as x̄ = 13.059 g/dL.\r<BR>\n" );
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document.write( "* **Margin of Error:** The margin of error is half the width of the confidence interval.  You can calculate it as follows:\r<BR>\n" );
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document.write( "   Margin of Error = (Upper Limit - Lower Limit) / 2<BR>\n" );
document.write( "   Margin of Error = (13.318 - 12.800) / 2<BR>\n" );
document.write( "   Margin of Error = 0.518 / 2<BR>\n" );
document.write( "   Margin of Error = 0.259 g/dL\r<BR>\n" );
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document.write( "**(c) Normality Assumption:**\r<BR>\n" );
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document.write( "Because the sample size is large (n = 100), the Central Limit Theorem applies.  The Central Limit Theorem states that the distribution of sample means will be approximately normal, *regardless* of the shape of the population distribution, as long as the sample size is sufficiently large (generally considered to be n ≥ 30).  Therefore, it is *not* necessary to confirm that the sample data appear to be from a normally distributed population when constructing this confidence interval.<BR>\n" );
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