Question

Find the 99% confidence interval for the variance and standard deviation of the active ingredient in a 500 mg pill if the sample size is 25 and the variance is 4 mg. Assume the variable is normally distributed

Answer 1

n	25
S	2
df	24	n-1
99% CI	alpha	0.01
Critical values
Upper	45.55851	CHIINV(alpha/2,df)
Lower	9.886234	CHIINV(1-alpha/2,df)
CI (Variance)
Lower	2.10718	(n-1)S^2/Upper critical value
Upper	9.710473	(n-1)S^2/Lower critical value
CI (Standard deviation)
Lower	1.451613	SQRT(LCI Var)
Upper	3.116163	SQRT(UCI Var)

CI (Variance) = (2.11, 9.71)

CI (Standard deviation) = (1.45, 3.12)