document.write( "Question 1152625: A sample of 91 drivers who were ticketed in a 55 mi/hr zone was taken in the month of January. The average sample speed was 66.2 mi/hr with a sample standard deviation of 3.4 mi/hr. Find the 99% confidence interval of the population standard deviation. Assume that the sample has been selected from a population that has a normal distribution. \n" ); document.write( "

Algebra.Com's Answer #775038 by VFBundy(438) $\"\"$ $\"About$

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SD of the sample = $\"SD%2Fsqrt%28n%29\"$ = $\"3.4%2Fsqrt%2891%29\"$ = 0.36
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\n" ); document.write( "Because we are looking at a 99% confidence interval, z* will be 2.576.
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\n" ); document.write( "The confidence interval will be: (z*)(0.36) = (2.576)(0.36) = 0.9
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\n" ); document.write( "Subtract/add 0.9 from/to the sample mean of 66.2. We get 65.3 and 67.1.
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\n" ); document.write( "The 99% confidence interval is 65.3 to 67.1.
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