document.write( "Question 930112: help please!! \r
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\n" ); document.write( "\n" ); document.write( "A random sample of 31 people employed by the Florida state authority established they earned an average wage (including benefits) of $56 per hour. The sample standard deviation was $6.35 per hour.
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\n" ); document.write( "\n" ); document.write( "(a) What is your best estimate of the population mean? (Omit the \"$\" sign in your response.) \r
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\n" ); document.write( "\n" ); document.write( " Estimated population mean $ \r
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\n" ); document.write( "\n" ); document.write( "(b)
\n" ); document.write( "Develop a 99 percent confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.)
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\n" ); document.write( "\n" ); document.write( " Confidence interval for the population mean wage is between and . \r
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\n" ); document.write( "\n" ); document.write( "(c)
\n" ); document.write( "How large a sample is needed to assess the population mean with an allowable error of $2 at 95 percent confidence? (Round up your answer to the next whole number.)
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\n" ); document.write( "\n" ); document.write( " Sample size
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Algebra.Com's Answer #564704 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
Sample: n = 31, mean = $56, s = $6.35
\n" ); document.write( "a) Estimated population mean = $56
\n" ); document.write( "b) ME =2.576(6.35/sqrt(31) = 2.94
\n" ); document.write( "CI: $56±2.94, (53.06, 58.94)
\n" ); document.write( "c) n = (1.96(6.35)/2)^2 = 19.7580, 20 sample size (Always go to next Integer)
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