Question

use the given information to find the number of degrees of freedom the critical values x2l and x2r and the confidence interval estimate of the standard deviation it is reasonable to assume that a simple random sample has been selected from a population with a normal distribution platelet counts of women 98% confidence in equals 28 s equals 65.5 what is the degrees of freedom...X2L...X2R..CI estimate of the standard deviation?

Answer 1

df = n -1 = 28 -1 = 27

Chi square critical values at 0.02 level with 27 df = L = 12.879 , R = 46.963

98% confidence interval for is

sqrt [(n-1) S² /  R ] < < sqrt [(n-1) S² /  L ]

sqrt [ 27 * 65.5² / 46.963 ] < < sqrt [ 27 * 65.5² / 12.879 ]

49.66 < < 94.84

98% CI Is ( 49.66 , 94.84)