Question

I'm needing assistance with calculating the Chi-Squared upper and lower bounds. I'm working on an example in which I am trying to find a 95% confidence interval estimate for a population variance and this is what the example states:

Find the chi-squared upper and lower bounds, X²_{6, .025} =12.832 (this is the upper bound) and X²_{6, .975} = .0831 (this is the lower bound) with 5 (=6-1) degrees of freedom (because remember, the degrees of freedom is “n-1”). Also, we want an area of .025 ABOVE our upper value and an area of .975 ABOVE our lower value because we want a 95% confidence interval…so we want to cover 95% of the area.

I'm not understanding where the .025 and the .975 came from. Can I please have assistance in how to calculate these bounds?

Thanks!

Answer 1

Solution :

Given that,

n = 6

Degrees of freedom = df = n - 1 = 6 - 1 = 5

At 95% confidence level the ² value is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

1 - / 2 = 1 - 0.025 = 0.975

Upper bound = ²_L = ²_/2,df = 12.832

Upper bound = ²_R = ²_{1 - /2,df} = 0.831