document.write( "Question 1197536: The mayor is interested in finding a 90% confidence interval for the mean number of pounds of trash per person per week that is generated in city. The study included 117 residents whose mean number of pounds of trash generated per person per week was 37.8 pounds and the standard deviation was 6.4 pounds.\r
\n" ); document.write( "\n" ); document.write( "a. To compute the confidence interval use a
\n" ); document.write( "?
\n" ); document.write( " distribution.\r
\n" ); document.write( "\n" ); document.write( "b. With 90% confidence the population mean number of pounds per person per week is between
\n" ); document.write( " and
\n" ); document.write( " pounds.\r
\n" ); document.write( "\n" ); document.write( "c. If many groups of 117 randomly selected members are studied, then a different confidence interval would be produced from each group. About
\n" ); document.write( " percent of these confidence intervals will contain the true population mean number of pounds of trash generated per person per week and about
\n" ); document.write( " percent will not contain the true population mean number of pounds of trash generated per person per week. \n" ); document.write( "

Algebra.Com's Answer #830924 by ewatrrr(24785) $\"\"$ $\"About$

You can put this solution on YOUR website!

\r\n" );
document.write( "Hi  \r\n" );
document.write( "\r\n" );
document.write( "'Standard Distribution': µ = 37.8 and σ = 6.4 lbs \r\n" );
document.write( "n = 117\r\n" );
document.write( " \r\n" );
document.write( " 90% confidence interval:  z = 1.645   |  Invnorm(.90 + .1/2) = Invnorm(.95) \r\n" );
document.write( "   = .97\r\n" );
document.write( "37.8 - .97  < 37.8 + .97\r\n" );
document.write( "36.83 <  < 38.77\r\n" );
document.write( "\r\n" );
document.write( "90% confident,  interval will contain true population mean\r\n" );
document.write( "\r\n" );
document.write( "Wish You the Best in your Studies.\r\n" );
document.write( "

\n" ); document.write( "

\n" );