Question

The sugar content of the syrup in canned peaches is normally distributed. A random sample of n = 12 cans yields a sample standard deviation of s = 5.2 milligrams. Calculate the 99% two-sided CI for σ.

Answer 1

Answer:

3.30185$\leq$ σ $\leq$ 8.7636

Step-by-step explanation:

the fromula for 100(1-α)% confidence (two sided) for σ is

√(η-1)*s^2/x^2_α/2,V $\leq$ σ $\leq$ √(η-1)*s^2/x^2_(1-α/2,V)               ∴ (v=n+1)

now data given is n = 12 and s=5.2 mg

for 99% CI ,α=0.05  and (n-1)=9 degree of freedom

from chi- square table x^2_(0.025,9)=19.02 ; x_(0.975,9)=2.7

substitute them in above expression we get

3.30185$\leq$ σ $\leq$ 8.7636