document.write( "Question 868431: Although male polar bears weigh only about one pound at birth, the mean weight of a random sample of 10 adult male polar bears is 1200 pounds with a standard deviation of 100 pounds. Construct and interpret a 98% confidence interval for the mean weight of all adult male polar bears. Assume that the population is evenly distributed. \n" ); document.write( "
Algebra.Com's Answer #523872 by jim_thompson5910(35256)![]() ![]() ![]() You can put this solution on YOUR website! Use a table to find that the critical value is \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So we're given\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "xbar = 1200 (sample mean) \n" ); document.write( "t = 2.821 (see above) \n" ); document.write( "s = 100 (sample standard deviation) \n" ); document.write( "n = 10 (sample size)\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Now compute the lower bound (L) and the upper bound (U) of the confidence interval.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Lower Bound:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "L = xbar - t*s/sqrt(n) \n" ); document.write( "L = 1200 - 2.821*100/sqrt(10) \n" ); document.write( "L = 1,110.79214720666\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Upper Bound:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "U = xbar + t*s/sqrt(n) \n" ); document.write( "U = 1200 + 2.821*100/sqrt(10) \n" ); document.write( "U = 1,289.20785279334\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The 98% confidence interval is approximately (1110.79214720666, 1289.20785279334)\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So if we construct 100 confidence intervals, then about 98 of them will contain the true population mean. \n" ); document.write( " |