document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=927121>Question 927121</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In a random sample of 150 households with an Internet connection,33 said that they had changed their Internet service provider within the past six months.\r<BR>\n" );
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document.write( "a)Finda99%confidence interval for the proportion of customers who changed their Internet service provider within the past six months. <BR>\n" );
document.write( "b)Find the sample size needed for a99% confidence interval to specify the proportion to within ±0.03.<BR>\n" );
document.write( "c)If no estimate of the proportion is available,how large should the sample be? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #562739 by </B><A HREF=/tutors/aboutme.mpl?userid=ewatrrr><B>ewatrrr(24785)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ewatrrr><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=ewatrrr><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=562739\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a) p = 33/150 = .22<BR>\n" );
document.write( "ME = 2.576sqrt((.22*.78)/150) = .034<BR>\n" );
document.write( "CI: .22 ± .034<BR>\n" );
document.write( "..........<BR>\n" );
document.write( "b) n = .22*.78(2.576/.03)^2 = 1265.2  0r 1266 (next whole number)<BR>\n" );
document.write( "......<BR>\n" );
document.write( "c) If no estimate of the proportion is available, how large should the sample be. <BR>\n" );
document.write( "Larger the sample, smaller the ME is. Use Your own parameters. </SPAN>\n" );
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