document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1058357>Question 1058357</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Question<BR>\n" );
document.write( "A.	A survey on the Cypriot market on the mean amount spent per customer for lunch meals at restaurant, data were collected for 25 customers in Nicosia. Assume a sample standard deviation of 3 and a sample of 20.<BR>\n" );
document.write( "1.	Find the 90% confidence interval.<BR>\n" );
document.write( "2.	Find the 95% confidence interval.<BR>\n" );
document.write( "3.	Test that the mean amount spent is larger than 18euros (value from tables=2.131).<BR>\n" );
document.write( "B.	For the same project data were also collected for 49 customers in Larnaca. They found that they spent on average 18 Euros with standard deviation of 2 Euros.<BR>\n" );
document.write( "4.	Find the 90% confidence interval.<BR>\n" );
document.write( "5.	Find the 99% confidence interval.<BR>\n" );
document.write( "6.	 Test that the mean amount spent is indifferent to 15 Euros (value from tables=1.96).<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 #673421 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=673421\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I am assuming the mean is 20 and the standard deviation is 3<BR>\n" );
document.write( "the interval is +/-t(0.95,df=24)s/sqrt(n)<BR>\n" );
document.write( "for 90% interval, it is +/- 1.711*3/5=1.03 so the interval is (18.97,21.03)<BR>\n" );
document.write( "for 95%, t changes to 2.064 and the interval width 1.24, so (18.76,21.24)<BR>\n" );
document.write( "t=(18-21)*5/3=-5. Virtually the entire t-distribution would be greater than 18, so that one could say with extremely high confidence that the value is >18 euros.<BR>\n" );
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document.write( "t(0.95, df=48) is 1.677<BR>\n" );
document.write( "s/sqrt (n)=2/7<BR>\n" );
document.write( "interval width is 0.48.  Interval itself is (17.52,18.48)<BR>\n" );
document.write( "t(0.99 df=48) is 2.682, width 0.77, Interval (17.23,18.77)<BR>\n" );
document.write( "t for 15 euros is (15-18)/(2/7)=-10.5 and the whole distribution would be greater than 15 euros.<BR>\n" );
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