document.write( "Question 1166620: A researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 30 bacteria reveals a sample mean of
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document.write( "64
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document.write( " hours with a standard deviation of
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document.write( "5.8
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document.write( " hours. He would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.8 hours at a 98% level of confidence.\r
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document.write( "What sample size should you gather to achieve a 0.8 hour margin of error?\r
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document.write( "He would need to sample
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document.write( " bacteria. \n" );
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Algebra.Com's Answer #791287 by Boreal(15235) You can put this solution on YOUR website! start with z0.99 since we aren't sure yet of df for t \n" ); document.write( "that is 2.326 \n" ); document.write( "the half-interval is 0.8 and is equal to z*s/sqrt(n) for the first approximation \n" ); document.write( "0.8=2.326*5.8/sqrt(n) \n" ); document.write( "0.8sqrt*n=2.326*5.8 \n" ); document.write( "square both sides \n" ); document.write( "0.64n=182.00 \n" ); document.write( "n=284.375 or 285 \n" ); document.write( "At this point the t-value is a few more df than 285. From the calculator, n=288 or df=287 gives the desired margin of error, and that is the answer.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |