document.write( "Question 1197535: A researcher is interested in finding a 90% confidence interval for the mean number minutes students are concentrating on their professor during a one hour statistics lecture. The study included 107 students who averaged 32.4 minutes concentrating on their professor during the hour lecture. The standard deviation was 13.7 minutes. Round answers to 3 decimal places where possible.\r
\n" ); document.write( "\n" ); document.write( "a. To compute the confidence interval use a
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\n" ); document.write( "\n" ); document.write( "b. With 90% confidence the population mean minutes of concentration is between
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\n" ); document.write( "\n" ); document.write( "c. If many groups of 107 randomly selected members are studied, then a different confidence interval would be produced from each group. About
\n" ); document.write( " percent of these confidence intervals will contain the true population mean minutes of concentration and about
\n" ); document.write( " percent will not contain the true population mean minutes of concentration. \n" ); document.write( "

Algebra.Com's Answer #830927 by ewatrrr(24785) $\"\"$ $\"About$

You can put this solution on YOUR website!

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document.write( "Hi  \r\n" );
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document.write( "'Standard Distribution': µ = 32.4 and σ =13.7\r\n" );
document.write( "n 107\r\n" );
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document.write( " 90% confidence interval:  z = 1.645  |  Invnorm(.9 + .1/2) = Invnorm(.95) \r\n" );
document.write( "   = 2.179\r\n" );
document.write( "32.5 - 2.179  < 32.5+ 2.179\r\n" );
document.write( "30.321  < 34.679\r\n" );
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document.write( "90% confident,  interval will contain true population mean\r\n" );
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document.write( "Wish You the Best in your Studies.\r\n" );
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