On a standardized test, 21 students had an average of 82 and a varia.pdf
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On a standardized test, 21 students had an average of 82 and a varia.pdf
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On a standardized test, 21 students had an average of 82 and a variance of 15. Find the 99% confidence interval for the standard deviation of these students\' scores. Solution Given a=0.01, chisquare with 0.005, df=n-1=20 is 7.43 (from chisquare table) chisquare with 0.995, df=20 is 39.997 (from chisquare table) So 99% confidece interval is (sqrt(n-1)*s/sqrt(39.997) , sqrt(n-1)*s/sqrt(7.43)) --> (sqrt(20)/sqrt(39.997), sqrt(20)/sqrt(7.43) ) --> ( 0.7071333, 1.640668).
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On a standardized test, 21 students had an average of 82 and a varia.pdf
- On a standardized test, 21 students had an average of 82 and a variance of 15. Find the 99% confidence interval for the standard deviation of these students' scores. Solution Given a=0.01, chisquare with 0.005, df=n-1=20 is 7.43 (from chisquare table) chisquare with 0.995, df=20 is 39.997 (from chisquare table) So 99% confidece interval is (sqrt(n-1)*s/sqrt(39.997) , sqrt(n-1)*s/sqrt(7.43)) --> (sqrt(20)/sqrt(39.997), sqrt(20)/sqrt(7.43) ) --> ( 0.7071333, 1.640668)