document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1129146>Question 1129146</A>: <I><SPAN STYLE=\"word-break: break-all;\"> On a survey on entertainment activities, 322 of 2,023 adult Americans surveyed report reported that they have watched a movie in cinema at least once in the past year. Assume that this sample is representative of the population of adult Americans.<BR>\n" );
document.write( "A. Find the sample proportion of adult Americans who have seen a movie in cinema in the last year at least once. <BR>\n" );
document.write( "C.calculate the margin of error. <BR>\n" );
document.write( "D. Construct 90% confidence level <BR>\n" );
document.write( "E. Construct 95% confidence level </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #745699 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=745699\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A.  sample proportion(P) = 322/2023 = 0.1592<BR>\n" );
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document.write( "Standard Error(SE) of P = square root(0.1592 * (1-0.1592) / 2023) = 0.0001<BR>\n" );
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document.write( "Margin of error(ME) = Critical value(CV) x SE of P<BR>\n" );
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document.write( "Since the sample size is sufficiently large, that is, sample size is greater than 30, we can use the Normal Distribution to determine CV<BR>\n" );
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document.write( "To determine the CV for a confidence level use the following steps<BR>\n" );
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document.write( "alpha = 1 - (confidence level/100)<BR>\n" );
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document.write( "critical probability(p*) = 1 - (a/2)<BR>\n" );
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document.write( "CV is the z-score associated with p*<BR>\n" );
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document.write( "CV for a 90% confidence level is 1.645<BR>\n" );
document.write( "CV for a 95% confidence level is 1.960<BR>\n" );
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document.write( "C.<BR>\n" );
document.write( "ME for 90% = 1.645 * 0.0001 = 0.0001645 is approximately 0.0002<BR>\n" );
document.write( "ME for 95% = 1.960 * 0.0001 = 0.0001960 is approximately 0.0002<BR>\n" );
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document.write( "D. 0.1592 + or - 0.0002 is (0.1590, 0.1594)<BR>\n" );
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document.write( "E. 0.1592 + or - 0.0002 is (0.1590, 0.1594)<BR>\n" );
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document.write( "Note the problem does not state the accuracy level so rounding to 4 significant digits yields the same confidence interval for D and E<BR>\n" );
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document.write( " </SPAN>\n" );
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