document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1193340>Question 1193340</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In a sample of credit card holders the mean monthly value of credit card purchases was $ 400 and the sample variance was 90 ($ squared). Assume that the population distribution is normal. Answer the following, rounding your answers to two decimal places where appropriate.\r<BR>\n" );
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document.write( "(a) Suppose the sample results were obtained from a random sample of 11 credit card holders. Find a 95% confidence interval for the mean monthly value of credit card purchases of all card holders.\r<BR>\n" );
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document.write( "(b) Suppose the sample results were obtained from a random sample of 19 credit card holders. Find a 95% confidence interval for the mean monthly value of credit card purchases of all card holders.\r<BR>\n" );
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document.write( "Before you answer (b) consider whether the confidence interval will be wider than or narrower than the confidence interval found for (a). Then check that your answer verifies this. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #825351 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=825351\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Because b has more information, its interval will be narrower. Ci also is inversely proportional to the square root of the sample size and directly proportional to the t-value which is also less for a larger n, everything else's being the same.<BR>\n" );
document.write( "95% half-interval of a is t(0.975,df=10)*9.49/sqrt(11); the 9.49 is the sd, the sqrt(90), and the units are $.<BR>\n" );
document.write( "=2.228*9.49/sqrt(11)=6.375 or 6.38<BR>\n" );
document.write( "the interval is ($393.62, $406.38)<BR>\n" );
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document.write( "For the second, t is 2.101, and half-interval is 2.101*9.49/sqrt(19)=4.57<BR>\n" );
document.write( "interval is ($395.43, $404.57) It is narrower  </SPAN>\n" );
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