document.write( "Question 866629: The population of the scores of all high school seniors that took the SAT-M test (mathematics component of the SAT test) last year followed a Normal distribution, with mean mu and standard deviation = 100. You read a report that says, “On the basis of a simple random sample of 500 high school seniors that took the SAT-M test this year, a confidence interval for the population (mu) is 512.00 ± 11.52.” \r
\n" ); document.write( "\n" ); document.write( "A 95% confidence interval for the population (mu) would be\r
\n" ); document.write( "\n" ); document.write( "Question 4 options:
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\n" ); document.write( "503.24 to 520.77\r
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\n" ); document.write( "500.48 to 523.52\r
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\n" ); document.write( "316 to 708\r
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\n" ); document.write( "None of the above
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Algebra.Com's Answer #522387 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
The lower bound L and the upper bound U are equal to \r
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\n" ); document.write( "\n" ); document.write( "L = xbar - ME
\n" ); document.write( "U = xbar + ME\r
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\n" ); document.write( "\n" ); document.write( "where xbar is the sample mean and ME is the margin of error. In this case, \r
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\n" ); document.write( "\n" ); document.write( "xbar = 512
\n" ); document.write( "ME = 11.52\r
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\n" ); document.write( "\n" ); document.write( "These are both given when they report the confidence interval (CI) of 512.00 ± 11.52\r
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\n" ); document.write( "\n" ); document.write( "So,\r
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\n" ); document.write( "\n" ); document.write( "L = xbar - ME
\n" ); document.write( "L = 512 - 11.52
\n" ); document.write( "L = 500.48\r
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\n" ); document.write( "\n" ); document.write( "and\r
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\n" ); document.write( "\n" ); document.write( "U = xbar + ME
\n" ); document.write( "U = 512 + 11.52
\n" ); document.write( "U = 523.52\r
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\n" ); document.write( "\n" ); document.write( "Making the final answer to be 500.48 to 523.52
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