document.write( "Question 972972: The average weight of 40 randomly selected minivans was 4150 pounds with a standard deviation of 480. Find a point estimate of the population mean. Find the 99% confidence interval of the true mean weight of the minivans. \n" ); document.write( "
Algebra.Com's Answer #595238 by Boreal(15235)\"\" \"About 
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Point estimate is the sample mean, which is 4150 pounds. It is the best \"guess\" one has.
\n" ); document.write( "z.99=2.58
\n" ); document.write( "99% CI =+/-2.58 *SE of the mean.\r
\n" ); document.write( "\n" ); document.write( "SE= 480/sqrt (40) ;;It is the standard deviation/sqrt (sample size)\r
\n" ); document.write( "\n" ); document.write( "75.894 is SE
\n" ); document.write( "To the nearest pound, z*SE=+/-196\r
\n" ); document.write( "\n" ); document.write( "CI = (3954,4346) units are pounds\r
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