document.write( "Question 1165343: A fitness center is interested in finding a 98% confidence interval for the mean number of days per week that Americans who are members of a
\n" ); document.write( "fitness club go to their fitness center. Records of 244 members were looked at and their mean number of visits per week was 3.5 and the
\n" ); document.write( "standard deviation was 1.8.\r
\n" ); document.write( "\n" ); document.write( "I understand a and b, but need help with c.\r
\n" ); document.write( "\n" ); document.write( "a. To compute the confidence interval use a t distribution.\r
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\n" ); document.write( "\n" ); document.write( "b. With 98% confidence the population mean number of visits per week is between 3.23 and 3.77 visits.\r
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\n" ); document.write( "\n" ); document.write( "c. If many groups of 244 randomly selected members are studied, then a different confidence interval would be produced from each group.
\n" ); document.write( "About ______ percent of these confidence intervals will contain the true population mean number of visits per week and about_________
\n" ); document.write( " percent will not contain the true population mean number of visits per week.
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Algebra.Com's Answer #789847 by Boreal(15235)\"\" \"About 
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About 98% of the confidence intervals of samples of 244 will contain the true population mean number (the parameter) and about 2% will not contain the true population mean number. One doesn't know which 98, however, or which 2. \n" ); document.write( "
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