document.write( "Question 1060969: An IQ test is designed so that the mean is 100 and the standard deviation is 17 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 99​% confidence that the sample mean is within 4 IQ points of the true mean. Assume that sigmaσequals=17 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.
\n" ); document.write( "The required sample size is?
\n" ); document.write( "Would it be reasonable to sample this number of​ students?\r
\n" ); document.write( "\n" ); document.write( "No. This number of IQ test scores is a fairly large number.\r
\n" ); document.write( "\n" ); document.write( "No. This number of IQ test scores is a fairly small number.\r
\n" ); document.write( "\n" ); document.write( "Yes. This number of IQ test scores is a fairly small number.\r
\n" ); document.write( "\n" ); document.write( "Yes. This number of IQ test scores is a fairly large number.
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Algebra.Com's Answer #675796 by rothauserc(4718)\"\" \"About 
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margin of error(M.E.) = 4
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\n" ); document.write( "alpha(a) = 1 - (99/100) = 0.01
\n" ); document.write( "critical probability(p*) = 1 -(a/2) = 1 - 0.005 = 0.995
\n" ); document.write( "critical value(cv) is the value associated with p* = 2.58
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\n" ); document.write( "let n be the sample size, then
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\n" ); document.write( "n = (cv^2 * standard deviation^2) / M.E.^2 = (2.58^2 * 17^2) / 4^2 = 120.23 is approximately 120
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\n" ); document.write( "The required sample size is 120
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\n" ); document.write( "Yes. This number of IQ test scores is a fairly large number
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\n" ); document.write( "Note from the Central Limit Theorem when we have the population
\n" ); document.write( "standard deviation and a sample size > 30, we can assume a normal
\n" ); document.write( "distribution
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