document.write( "Question 1036654: 63, 66, 66, 66, 66, 67, 67, 67, 67, 68, 68, 69, 69, 69, 69, 69, 69, 70, 70, 70, 70, 70, 71, 71, 71, 72, 72, 72, 72, 74, 75, 76, 76
\n" ); document.write( "(As any data value that is more than two standard deviations from the mean of the data set)
\n" ); document.write( "I am not understanding when it asks me to use the sample mean and the standard deviation to calculate the upper and lower \"limits\" for being considered an outlier?\r
\n" ); document.write( "\n" ); document.write( "I know my sample mean is 69.606 and my standard deviation is 3.01. \n" ); document.write( "
Algebra.Com's Answer #651367 by Boreal(15235)\"\" \"About 
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Typically, we use 95% confidence intervals. A CI is a per cent, not a probability. We don't know the parameter of interest and never will, but we can create an interval where we are highly confident (95%) that the parameter will be there. In reality, the parameter is or isn't, which is a 0/100% probability issue and not useful. If we created 100 such CIs, 95 of them would contain the parameter. We don't know which 95.
\n" ); document.write( "For a sample, the CI is a +/- t (degrees of freedom=n-1,.975)*s/sqrt(n), where n is the sample size and the t-value may be looked up. There isn't enough information here but the 95% CI is 69.606 +/- t*(3.01)/sqrt(n). Anything outside of that would be considered an outlier, although realistically it can happen, just not at the level we consider significant.\r
\n" ); document.write( "\n" ); document.write( "Given the definition of an outlier that is >2 SDs from the mean, any data element that is more than 6.02 units (2*3.01) from the mean would be an outlier. That would be <63.586 or >75.626 \n" ); document.write( "
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