The lower bound is ?

Question

A simple random sample of size n is drawn from a population that is known to be normally distributed. The sample​ variance S^2 is determined to be 12.1.

a- Construct a​ 90% confidence interval for σ2 if the sample​ size, n, is 20.

The lower bound is ___?

​(Round to two decimal places as​ needed.)

Answer 1

To construct confidence interval for variance you need critical values for Chi-Square distribution with df = n-1.

You can find the table of critical values for Chi-Square in many places, like

https://www.mathsisfun.com/data/chi-square-table.html

For 90% CI with df=19 the critical values are: c₁ = 30.144 (look it in the column 0.05) and c₂ = 10.117 (look it in the column 0.95).

Confidence interval for σ² is calculated as:

The lower bound is (n-1)s²/c₁ = 19×12.1²/30.144 ≈7.63

The upper bound is (n-1)s²/c₂ = 19×12.1²/10.117 ≈ 22.72

However, you can avoid the table and calculations by hand if you use the site www.wolframalpha.com

Just enter in the Search field: confidence interval variance

Then enter in Computational Inputs your confidence level, sample size, and sample variance.