Question

The amount of lateral expansion (mils) was determined for a sample of n = 5 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.84 mils. Assuming normality, derive a 95% CI for σ2 and for σ. (Round your answers to two decimal places.)

Answer 1

Answer: A 95 % confidence interval for $\sigma^2$ :-

$2.90$

A 95 % confidence interval for $\sigma$ :-

$1.70$

Step-by-step explanation:

Given : Sample size : $n=5$

Standard deviation : $\sigma : 2.84$

Significance level : $\alpha=1-0.95=0.05$

We assume this is normal distribution

By using chi-square distribution, the critical value will be :-

$\chi^2_{n-1,\alpha/2}=\chi^2_{4,0.025}=11.14$

$\chi^2_{n-1,1-\alpha/2}=\chi^2_{4,0.975}=0.48$

The confidence interval for population variance:-

$\dfrac{(n-1)s^2}{\chi^2_{n-1\alpha/2}}$

$=\dfrac{(2.84)^2(4)}{11.14}$

For standard deviation just take square root in the above inequality,

$\sqrt{2.90}$