document.write( "Question 1197514: tate cooperative department is performing a survey to determine the annual salary earned by the managers numbering 3000 in the cooperative sector within the state. how large a sample size it should take in order to estimate the mean annual earnings within plus and minus 1000 and at 95 % confidence level the standard deviation of annual earning of the entire population is known to be 3000 \n" ); document.write( "

Algebra.Com's Answer #830808 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "Hi  \r\n" );
document.write( " 95% confident  ⇒  z = 1.96 (See below)\r\n" );
document.write( "σ  = 3000,   ME = 1000\r\n" );
document.write( "How large of a sample size is required? ME large ⇒ expect small sample size\r\n" );
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document.write( "0r\r\n" );
document.write( "n = \r\n" );
document.write( "n =  = 34.5744  0r sample Size = 35 (always round Up)\r\n" );
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document.write( " = CI	z = value\r\n" );
document.write( "90%	z =1.645    \r\n" );
document.write( "92%	z = 1.751\r\n" );
document.write( "95%	z = 1.96   Invnorm(.975)  2-sided (.95 + .05/2 = .975 )\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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