document.write( "Question 844304: Suppose that the amount of time teenagers spend on the internet per week is normally distributed with a standard deviation of 1.5 hours. A sample of 100 teenagers is selected at random, and the sample mean computed as 6.5 hours. Determine and interpret the 99% confidence interval estimate of the population mean. \n" ); document.write( "

Algebra.Com's Answer #508664 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "Hi,
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document.write( "ME = 
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document.write( " 6.5 - ME<  <  6.5 + ME
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document.write( "Note:
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document.write( "80%	z = 1.28155
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document.write( "90%	z =1.645
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document.write( "92%	z = 1.751
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document.write( "95%	z = 1.96
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document.write( "98%	z = 2.326
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document.write( "99%	z = 2.576
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