document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=367001>Question 367001</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1. A 99% confidence interval estimate for a population mean was computed to be (29.2, 46.6). Determine the mean of the sample, which was used to determine the interval estimate (show all work).\r<BR>\n" );
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document.write( "2. A study was conducted to estimate the mean amount spent on birthday gifts for a typical family having two children. A sample of 170 was taken, and the mean amount spent was $248.72. Assuming a standard deviation equal to $48.31, find the 90% confidence interval for m, the mean for all such families (show all work). </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #261637 by </B><A HREF=/tutors/aboutme.mpl?userid=robertb><B>robertb(5830)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=robertb><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=robertb><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=261637\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1. (29.2 + 46.6)/2 = 37.9, the sample mean.<BR>\n" );
document.write( "2.  248.72 +-(1.645)*(48.31)/<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28170%29 ALIGN=MIDDLE  ALT=\"sqrt%28170%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >= 248.72+-6.095, so the 90% C.I. is <BR>\n" );
document.write( "(242.625, 254.815). </SPAN>\n" );
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