document.write( "Question 1082720: The following data represent the asking price of a simple random sample of homes for sale. Construct a 99​% confidence interval with and without the outlier included. Comment on the effect the outlier has on the confidence interval.\r
\n" ); document.write( "\n" ); document.write( "​$223,000
\n" ); document.write( "$279,900
\n" ); document.write( "$19,900
\n" ); document.write( "​$143,000
\n" ); document.write( "​$205,800
\n" ); document.write( "​$181,500
\n" ); document.write( "​$459,900
\n" ); document.write( "​$212,000
\n" ); document.write( "​$187,500
\n" ); document.write( "​$201,500
\n" ); document.write( "​$147,800
\n" ); document.write( "​$264,900
\n" ); document.write( "

Algebra.Com's Answer #696784 by Boreal(15235)\"\" \"About 
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mean with is $210,558 with s=$102,878
\n" ); document.write( "99%CI is ($118,320, $302,800)
\n" ); document.write( "without the outlier mean is $227,891 with s=$87,618
\n" ); document.write( "99%CI is ($144,170, $311,620)
\n" ); document.write( "This is done using t.995, df=11 * s/sqrt(12) for the first and 1 less in n and df for the second. That is the interval, and it is added to and subtracted from the sample mean.
\n" ); document.write( "The interval is narrower, because the variability is less. The mean is larger
\n" ); document.write( " because a small number outlier was removed.
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