document.write( "Question 1131798: A simple random sample from a population with a normal distribution of 106 body temperatures has x =98.90 degrees and s =0.61 degrees. Construct a 95% confidence interval estimate of the standard deviation of body temperature of all healthy humans.
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Algebra.Com's Answer #748527 by rothauserc(4718)![]() ![]() You can put this solution on YOUR website! sample size(n) is 106 body temperatures \n" ); document.write( ": \n" ); document.write( "degrees of freedom = 106 - 1 = 105 \n" ); document.write( ": \n" ); document.write( "the mean of the sample is 98.90 \n" ); document.write( ": \n" ); document.write( "standard deviation of the sample is 0.61 \n" ); document.write( ": \n" ); document.write( "We use the chi-square distribution to construct the 95% confidence interval(CI) for the population standard deviation \n" ); document.write( ": \n" ); document.write( "alpha(a) = 1 -(95/100) = 0.05 \n" ); document.write( ": \n" ); document.write( "critical probability(p*) = 1 -(a/2) = 0.975 \n" ); document.write( ": \n" ); document.write( "chi-square(0.975) = 78.536 \n" ); document.write( ": \n" ); document.write( "chi-square(0.025) = 135.247 \n" ); document.write( ": \n" ); document.write( "population variance(0.975) = ((106-1)*(0.61)^2)/78.536 = 0.4975 \n" ); document.write( ": \n" ); document.write( "population variance(0.025) = ((106-1)*(0.61)^2)/135.247 = 0.2889 \n" ); document.write( ": \n" ); document.write( "take the square roots to get standard deviation \n" ); document.write( ": \n" ); document.write( "************************************************ \n" ); document.write( "the 95% confidence interval is (0.5375, 0.7053) \n" ); document.write( "************************************************ \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( " |