document.write( "Question 1197458: You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you believe the population standard deviation is approximately σ=30.4 dollars. You would like to be 90% confident that your estimate is within 5 dollar(s) of average spending on the birthday parties. How many moms do you have to sample? Do not round mid-calculation.
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Algebra.Com's Answer #830738 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\r\n" );
document.write( "σ = 30.4,   ME = 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" );
document.write( "0r\r\n" );
document.write( "n = \"%28+z%2Asigma%2FME%29%5E2\"\r\n" );
document.write( "n = \"%28%281.645%2A30.4%29%2F5%29%5E2+%29%29\" = 100.0320  0r sample Size =101 (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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