document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1138901>Question 1138901</A>: <I><SPAN STYLE=\"word-break: break-all;\"> It is said that happy and healthy workers are efficient and productive. A company that manufactures exercising machines wanted to know the percentage of large companies that provide on-site health club facilities. A sample of 200 such companies showed that 150 of them provide such facilities on site. \r<BR>\n" );
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document.write( "What is the point estimate of the percentage of all such companies that provide such facilities on site and Construct a 98% confidence interval for the percentage of all such companies that provide such facilities on site as well as find the margin of error for this estimate?\r<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 #756939 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=756939\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The point estimate is 150/200 or 0.75 as the proportion.<BR>\n" );
document.write( "a 98% CI is z* sqrt(0.75*0.25/200) and z 0.99 is 2.33<BR>\n" );
document.write( "half-interval is 2.33*sqrt(0.0009375) or 0.071, 0.07 here<BR>\n" );
document.write( "98%CI is (0.68, 0.82) with margin of error of 7%<BR>\n" );
document.write( "(0.679, 0,821) is to 3 decimal places<BR>\n" );
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