document.write( "Question 1122465: A recent study of 25 Orlando, FL residents revealed that they had lived in their current house for a mean of 11.5 years, with a sample standard deviation of 2.7 years.\r
\n" ); document.write( "\n" ); document.write( "1. What is the population mean? unknown or 11.5\r
\n" ); document.write( "\n" ); document.write( "2. What is the best estimate of the population mean? 11.5\r
\n" ); document.write( "\n" ); document.write( "3. n: 25
\n" ); document.write( "4. x-bar: 11.5
\n" ); document.write( "5. 𝜎: 2.7
\n" ); document.write( "6. Standard Error: 0.5400\r
\n" ); document.write( "\n" ); document.write( "Then I got lost here:\r
\n" ); document.write( "\n" ); document.write( "7. Develop a 90% confidence interval for the population mean?
\n" ); document.write( "distribution:
\n" ); document.write( "critical value:
\n" ); document.write( "Lower bound:
\n" ); document.write( "Upper bound: \r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

Algebra.Com's Answer #739686 by Boreal(15235) $\"\"$ $\"About$

You can put this solution on YOUR website!
This is a sample. The population itself is not defined, but it if is anything larger than this sample, which it likely is, then the population mean is unknown.\r
\n" ); document.write( "\n" ); document.write( "The best estimate is 11.5, x-bar is 11.5, s (not sigma) is 2.7, and SE is 0.54
\n" ); document.write( "90% CI is +/-t df=24 *SE
\n" ); document.write( "=+/-1.711*2.7=+/-4.62
\n" ); document.write( "mean +/-4.62
\n" ); document.write( "(6.88, 16.12) units are years\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

\n" );