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) ![]() You can put this solution on YOUR website! 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 \n" ); document.write( " \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 \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Side Note: the mean is not used at all to compute the margin of error. \n" ); document.write( " |