document.write( "Question 478987: The Human Relations Department of Smallville Co. would like to include a dental plan as part of its benefits package. The question is, how much does a typical employee and his/her family spend on annual dental expenses? A sample of 45 employees reveals the mean amount spent last year was $1,820 with a standard deviation of $660. Assume a normal distribution of annual dental expenses.\r
\n" ); document.write( "\n" ); document.write( " A) Construct a 95% or .95 confidence interval estimate for the population mean.
\n" ); document.write( " B) The information from part A, above, was given to the President who indicated the Company could afford $1,700 per year. Is it possible the population mean could be $1,700? Justify your answer.
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Algebra.Com's Answer #328214 by stanbon(75887)\"\" \"About 
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The Human Relations Department of Smallville Co. would like to include a dental plan as part of its benefits package. The question is, how much does a typical employee and his/her family spend on annual dental expenses?
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\n" ); document.write( "A sample of 45 employees reveals the mean amount spent last year was $1,820 with a standard deviation of $660. Assume a normal distribution of annual dental expenses.
\n" ); document.write( "A) Construct a 95% or .95 confidence interval estimate for the population mean.
\n" ); document.write( "ME = 1.96*660/sqrt(45) = 192.84
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\n" ); document.write( "95%CI: 1820-192.84 < u < 1820+192.84
\n" ); document.write( "1627.16 < u < 2012.84
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\n" ); document.write( "\n" ); document.write( "B) The information from part A, above, was given to the President who indicated the Company could afford $1,700 per year. Is it possible the population mean could be $1,700? Justify your answer.
\n" ); document.write( "We have 95% confidence the population mean is beween 1627.16 and 2012.84
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\n" ); document.write( "We might be wrong.
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.\r
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