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)\"\" \"About 
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sample size(n) is 106 body temperatures
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\n" ); document.write( "degrees of freedom = 106 - 1 = 105
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\n" ); document.write( "the mean of the sample is 98.90
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\n" ); document.write( "standard deviation of the sample is 0.61
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\n" ); document.write( "We use the chi-square distribution to construct the 95% confidence interval(CI) for the population standard deviation
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\n" ); document.write( "alpha(a) = 1 -(95/100) = 0.05
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\n" ); document.write( "critical probability(p*) = 1 -(a/2) = 0.975
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\n" ); document.write( "chi-square(0.975) = 78.536
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\n" ); document.write( "chi-square(0.025) = 135.247
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\n" ); document.write( "population variance(0.975) = ((106-1)*(0.61)^2)/78.536 = 0.4975
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\n" ); document.write( "population variance(0.025) = ((106-1)*(0.61)^2)/135.247 = 0.2889
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\n" ); document.write( "take the square roots to get standard deviation
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\n" ); document.write( "the 95% confidence interval is (0.5375, 0.7053)
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