Question

A sample (n = 10) tuna sushi was taken to determine the amounts of mercury found in them. The sample standard deviation was .3128. Construct a 95% confidence interval estimate of the variance of the amounts of mercury in the population.

Answer 1

Solution :

Given that,

s = 0.3128

Point estimate = s² = 0.097844384

n = 10

Degrees of freedom = df = n - 1 = 10 - 1 = 9

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

²_L = ²_/2,df = 19.023

²_R = ²_{1 - /2,df} = 2.700

The 95% confidence interval for ² is,

(n - 1)s² / ²_/2 < ² < (n - 1)s² / ²_{1 - /2}

9 * 0.097844384 / 19.023 < ² < 9 * 0.097844384 / 2.700

0.05 < ² < 0.33

( 0.05 , 0.33)