document.write( "Question 1091936: A random sample of 22 people employed by the Florida state authority established they earned an average wage (including benefits) of $61.00 per hour. The sample standard deviation was $5.66 per hour. (Use t Distribution Table.)
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document.write( "a. What is the best estimate of the population mean?
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document.write( " Estimated population mean $
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document.write( "b.
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document.write( "Develop a 98% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.)
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document.write( " Confidence interval for the population mean wage is between
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document.write( "c.
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document.write( "How large a sample is needed to assess the population mean with an allowable error of $1.00 at 95% confidence? (Round up your answer to the next whole number.)
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document.write( " Sample size
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Algebra.Com's Answer #706997 by Boreal(15235) You can put this solution on YOUR website! The best estimate of the population mean is $61, the sample mean. \n" ); document.write( "98% CI is 61+/-t df=21,0.99*5.66/sqrt(22); the t is 2.52 \n" ); document.write( "the half interval is 3.04, 2.52*5.66/sqrt(22) \n" ); document.write( "the 98% CI is ($57.96, $63.04) \n" ); document.write( "use a z since we don't know the t for the sample size. Then check to see if this works. \n" ); document.write( "half-interval is 1 \n" ); document.write( "z*5.66/sqrt(n)=1, and z is 1.96, so \n" ); document.write( "1.96*5.66=sqrt (n) \n" ); document.write( "sqrt(n)=11.09 \n" ); document.write( "n=123.06 or 124 \n" ); document.write( "check to see if this works for t. It is a little large at 1.006 half interval or (58.994, 61.006) \n" ); document.write( "if one uses 125, it is +/-1.002. \n" ); document.write( "n=126 gives an error of +/-0.998\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |