document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1187723>Question 1187723</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A recent survey revealed that an American's Christmas spending averaged $830. Use this as the population mean American's Christmas spending. Suppose Americans' Christmas spending is normally distributed with a standard deviation of $220. A random sample of size 100 is selected from the population of American consumers. The middle 90% of the sample mean spending will be bounded by what two values? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #819043 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=819043\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This is a 90% confidence interval.<BR>\n" );
document.write( "the half-interval is z(0.95)*sigma/sqrt(n)<BR>\n" );
document.write( "=1.645*220/sqrt(100)<BR>\n" );
document.write( "=36.19<BR>\n" );
document.write( "The mean of this sampling distribution Is 830, same as population mean<BR>\n" );
document.write( "the interval is ($793.81, $866.19) </SPAN>\n" );
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