document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1067747>Question 1067747</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 3)	Students from school A and B are compared on the basis of their scores on an aptitude test. Two random samples of 90 and 100 students are selected from school A and B respectively. The sample means are 76.4 and 81.2, where as the sample standard deviation is 8.2 and 7.6 respectively. Establish a 98% confidence interval for the difference in population mean scores between students A and B. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #682910 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=682910\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> t=(mean1-mean2)/ sqrt {(s1^2/n1)+(s2^2/n2)}<BR>\n" );
document.write( "=(76.4-81.2)/sqrt{8.2^2/90 + 7.6^2/100}<BR>\n" );
document.write( "=-4.8/1.15=-4.17\r<BR>\n" );
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document.write( "the SE is sqrt {(s1^2/n1)+(s2^2/n2)}<BR>\n" );
document.write( "8.2^2/90 + 7.6^2/100=1.325<BR>\n" );
document.write( "sqrt of that is 1.15<BR>\n" );
document.write( "98% CI for t df=188 is 2.34(1.15)=+/-2.69<BR>\n" );
document.write( "This is added or subtracted to the mean<BR>\n" );
document.write( "(-7.49, -2.11)<BR>\n" );
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