document.write( "Question 1123006: In a random sample of 25 people, the mean commute time to work was 33.8 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a t-distribution to construct a 95% confidence interval for the population mean. What is the margin or error? \n" ); document.write( "
Algebra.Com's Answer #739215 by jim_thompson5910(35256)\"\" \"About 
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Use a T-table such as this one to look up the critical T value that corresponds to a 95% confidence interval.
\n" ); document.write( "Since n = 25 is the sample size, this means the degrees of freedom (df) is
\n" ); document.write( "df = n-1
\n" ); document.write( "df = 25-1
\n" ); document.write( "df = 24
\n" ); document.write( "Use a highlighter to mark the entire row that has df = 24.
\n" ); document.write( "Mark the column that is over top the 95% confidence interval.
\n" ); document.write( "At the intersection of this row and column is the value 2.064
\n" ); document.write( "This value is approximate. To get more accuracy, you'll need to use the Tinv function on your calculator.\r
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\n" ); document.write( "\n" ); document.write( "Margin of Error = (critical T value)*(standard deviation)/sqrt(sample size)
\n" ); document.write( "Margin of Error = (2.064)*(7.2)/sqrt(25)
\n" ); document.write( "Margin of Error = (2.064)*(7.2)/5
\n" ); document.write( "Margin of Error = 14.8608/5
\n" ); document.write( "Margin of Error = 2.97216 (which is approximate)\r
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\n" ); document.write( "\n" ); document.write( "Side Note: the mean is not used at all to compute the margin of error.
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