document.write( "Question 1197539: You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately
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Algebra.Com's Answer #830871 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Hi  \r\n" );
document.write( " 90% confident  ⇒  z = 1.645 (See below)\r\n" );
document.write( "σ = 50.2,   ME = 1.5\r\n" );
document.write( "How large of a sample size is required? \r\n" );
document.write( "\"ME+=+z%2Asigma%2Fsqrt%28n%29\"\r\n" );
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document.write( "n = \"%28+z%2Asigma%2FME%29%5E2\"\r\n" );
document.write( "n = \"%28%281.64%2A50.2%29%2F%281.5%29%5E2+%29%29\" = 36.7  0r sample Size = 37 (always round Up)\r\n" );
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document.write( " = CI	z = value\r\n" );
document.write( "90%	z =1.645   Invnorm(.95)  2-sided (.90 + .10/2 = .95 ) \r\n" );
document.write( "92%	z = 1.751\r\n" );
document.write( "95%	z = 1.96   \r\n" );
document.write( "98%	z = 2.326  \r\n" );
document.write( "99%	z = 2.576\r\n" );
document.write( "Wish You the Best in your Studies.\r\n" );
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