document.write( "Question 1120471: A publisher wants to estimate the mean length of time (in minutes) all adults spend reading newspapers. To determine this estimate, the publisher takes a random sample of 15 people and obtains the results below. From past studies, the publisher assumes sigma is 2.1 minutes and that the population of times is normally distributed.
\n" ); document.write( "7
\n" ); document.write( "12
\n" ); document.write( "7
\n" ); document.write( "12
\n" ); document.write( "11
\n" ); document.write( "9
\n" ); document.write( "12
\n" ); document.write( "10
\n" ); document.write( "8
\n" ); document.write( "10
\n" ); document.write( "6
\n" ); document.write( "8
\n" ); document.write( "7
\n" ); document.write( "9
\n" ); document.write( "7
\n" ); document.write( "
\n" ); document.write( "Construct the 90% and 99% confidence intervals for the population mean. Which interval is wider? If convenient, use technology to construct the confidence intervals.
\n" ); document.write( "The 90% confidence interval is ?
\n" ); document.write( "(Round to one decimal place as needed.)
\n" ); document.write( "The 99% confidence interval is?
\n" ); document.write( "(Round to one decimal place as needed.)
\n" ); document.write( "Which interval is wider?
\n" ); document.write( "The 90% confidence interval?
\n" ); document.write( "The 99% confidence interval? \n" ); document.write( "

Algebra.Com's Answer #736136 by Boreal(15235) $\"\"$ $\"About$

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The mean is 9
\n" ); document.write( "Assuming the sd is known is consistent with using a z-test
\n" ); document.write( "the 90% interval will be mean+/-1.645, the z value for 90%*2.1/sqrt(15). The 99% interval will be using 2.576 in place of 1.645
\n" ); document.write( "The intervals are (8.1, 9.9) for a 90% CI. To be more confident, one needs a wider interval and the 99%CI is (7.6, 10.4). Units are minutes. \n" ); document.write( "

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