document.write( "Question 1076761: Suppose a random sample of 36 is selected from a population with a standard deviation of 12. If the sample mean is 98, the 99% confidence interval to estimate the population mean is \n" ); document.write( "
Algebra.Com's Answer #691369 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! CI= mean +/- t (df=35, 0.995)*s/sqrt (n) \n" ); document.write( "t value is 2.71 \n" ); document.write( "98+/- 2.71*12/6= \n" ); document.write( "98+/- 5.42 \n" ); document.write( "(92.58, 103.42) \n" ); document.write( " |