a) The 95% confidence interval for the population standard deviation is (1527.5 lbs, 2072.5 lbs).
b) The 99% confidence interval for the population standard deviation is (1424.8 lbs, 2145.2 lbs).
c) The 99.73% confidence interval for the population standard deviation is (1361.2 lbs, 2218.8 lbs).
a) For a 95% confidence interval, we use the chi-square distribution with degrees of freedom n-1 = 100-1 = 99. The chi-square values for the lower and upper endpoints of the interval are 1.642 and 2.397, respectively. The corresponding standard deviation values are:
lower_95 = sqrt((1.642 * 1800^2) / 100) = 1527.5 lbs
upper_95 = sqrt((2.397 * 1800^2) / 100) = 2072.5 lbs
Therefore, the 95% confidence interval for the population standard deviation is (1527.5 lbs, 2072.5 lbs).
b) For a 99% confidence interval, we use the chi-square distribution with degrees of freedom n-1 = 100-1 = 99. The chi-square values for the lower and upper endpoints of the interval are 0.985 and 2.584, respectively. The corresponding standard deviation values are:
lower_99 = sqrt((0.985 * 1800^2) / 100) = 1424.8 lbs
upper_99 = sqrt((2.584 * 1800^2) / 100) = 2145.2 lbs
Therefore, the 99% confidence interval for the population standard deviation is (1424.8 lbs, 2145.2 lbs).
c) For a 99.73% confidence interval, we use the chi-square distribution with degrees of freedom n-1 = 100-1 = 99. The chi-square values for the lower and upper endpoints of the interval are 0.564 and 2.941, respectively. The corresponding standard deviation values are:
lower_9973 = sqrt((0.564 * 1800^2) / 100) = 1361.2 lbs
upper_9973 = sqrt((2.941 * 1800^2) / 100) = 2218.8 lbs
Therefore, the 99.73% confidence interval for the population standard deviation is (1361.2 lbs, 2218.8 lbs).
Question
the standard deviation of the breaking strengths of 10o cables tested by a company was 1800 lbs. find the a) 95% , b) 99%, c) 99.73% confidence interval for the standard deviation in the breaking strength of all cables produced by the company.